Practice Problems

1. Classify each column

   Column	Level	Why
   Membership number	Nominal	Numeric, but it is a name for a person, not a quantity
   Age in years	Ratio	Equal intervals and a real zero, so 40 is twice 20
   Postcode district	Nominal	A label with digits in it
   Satisfaction 1–5	Ordinal	Ordered, but the gap 1→2 need not equal 4→5
   Books borrowed	Ratio	A count; zero books means zero borrowing
   Temperature °C	Interval	Equal intervals, arbitrary zero: 20 °C is not twice as warm as 10 °C
   Branch	Nominal	Unordered categories
   Queue position	Ordinal	Ranks; the gap between 1st and 2nd is not a fixed quantity

2. Which can be averaged Age, books borrowed and temperature. Interval and ratio data support addition and division, which is all a mean is.
3. The postcode mean The digits in M14 encode a place, not an amount. Averaging them produces a number that refers to nothing: there is no location "17.4", and moving from M14 to M20 is not "six units" of anything.
4. Mean satisfaction of 3.8 Strictly, a mean assumes equal spacing, and Likert points are only ordered. In practice psychology and market research report Likert means constantly, and with five or more points and roughly symmetric responses the distortion is small. Report the median and the full response distribution alongside it, and the reader can judge for themselves.

Three of the eight columns are nominal, two are ranks, one is interval and two are ratio. Only age, books borrowed and temperature are safely averageable, and satisfaction is the one worth a footnote.

From §1.2 Types of Data.

1. Mean x̄ = Σx / n Σx = 12+14+15+16+17+19+21+25+41 = 180 x̄ = 180 / 9 = 20.0 loaves
2. Median The values are already in order and n = 9 is odd, so the median is the 5th value. Median = 17 loaves
3. Deviations and their squares (x − x̄): −8, −6, −5, −4, −3, −1, 1, 5, 21 (x − x̄)²: 64, 36, 25, 16, 9, 1, 1, 25, 441 Σ(x − x̄)² = 618
4. Variance, then standard deviation s² = Σ(x − x̄)² / (n − 1) = 618 / 8 = 77.25 s = √77.25 = 8.789 ≈ 8.79 loaves Divide by n − 1, not n: these nine days are a sample of the bakery's trading, not the whole of it.
5. Which center to plan with The mean of 20 is dragged up by the single festival day. Drop that day and the mean of the remaining eight is 139 / 8 = 17.4, which lands right next to the median. For an ordinary Tuesday the median of 17 is the honest planning figure, and the festival day belongs in the plan as its own event.

Mean = 20.0, median = 17, s = 8.79 loaves. Order for about 17, and order extra for festival days separately.

Rounding: keep the full 8.7892 if the figure feeds another calculation, and round to 8.79 only when reporting.

From §1.3 Describing Data · check it in the Descriptives calculator.

1. Formula z = (x − μ) / σ
2. Statistics z = (68 − 60) / 10 = 8 / 10 = 0.80
3. Research Methods z = (74 − 65) / 12 = 9 / 12 = 0.75
4. Compare The raw marks say Methods (74 > 68). The z-scores say Statistics, by a whisker: 0.80 SDs above her cohort against 0.75. The bigger raw mark came on the paper where everyone scored higher and the spread was wider.
5. T-scores T = 50 + 10z Statistics: 50 + 10(0.80) = 58.0 Methods: 50 + 10(0.75) = 57.5

Statistics z = 0.80 (T = 58.0), Methods z = 0.75 (T = 57.5). Her Statistics paper was marginally the stronger performance, and the raw marks pointed the other way.

From §1.5 Z-Scores & The Normal Distribution.

1. Below 60 bpm: standardize z = (60 − 72) / 8 = −12 / 8 = −1.50 P(Z < −1.50) = .0668 About 6.7% of adults, or roughly 1 in 15.
2. Between 64 and 84 bpm: two z-scores z₁ = (64 − 72) / 8 = −1.00 z₂ = (84 − 72) / 8 = +1.50
3. Subtract the two cumulative areas P(−1.00 < Z < 1.50) = P(Z < 1.50) − P(Z < −1.00) = .9332 − .1587 = .7745 About 77.5%.
4. The top 5%: go backwards from the area The cutoff leaves .95 below it, so z = 1.645 x = μ + zσ = 72 + 1.645(8) = 72 + 13.16 = 85.16 Round to 85 bpm for the protocol.

6.7% below 60 bpm, 77.5% between 64 and 84 bpm, and the top 5% sit above roughly 85 bpm.

Rounding: 1.645 is the conventional two-decimal z for the 95th percentile (1.6449 exactly). Using 1.64 gives 85.12 and using 1.65 gives 85.20, so at whole-bpm precision the choice does not matter here. It would matter if the protocol were written to one decimal.

From §1.5 Z-Scores & The Normal Distribution · check the areas in Statistical Tables.

1. Marginal probability P(cycles) = 70 / 200 = .35
2. Conditional probability: the denominator shrinks P(cycles | under 30) = 48 / 100 = .48 Conditioning on "under 30" throws away the other row, so the denominator is that row's total, not 200.
3. Joint probability P(under 30 and cycles) = 48 / 200 = .24
4. Independence check If independent: P(A and B) = P(A) × P(B) P(under 30) × P(cycles) = .50 × .35 = .175 Observed joint = .24 ≠ .175
5. Read it the other way too P(cycles | under 30) = .48 against P(cycles) = .35. Knowing someone's age band changes the probability, which is exactly what dependence means.

.35, .48, .24, and no: the two variables are dependent. Under-30s cycle at nearly twice the rate of over-30s (48% against 22%).

From §1.7 Probability Basics · test the dependence formally with a chi-square test (Problem 19 below).

1. Formula SE = σ / √n
2. n = 25 SE = 60 / √25 = 60 / 5 = 12 ms
3. n = 100 SE = 60 / √100 = 60 / 10 = 6 ms Four times the sample halves the standard error, because √4 = 2. Precision is bought with the square root, which is why the second hundred participants buy far less than the first hundred.
4. Solve for n 4 = 60 / √n → √n = 60 / 4 = 15 n = 15² = 225 drivers
5. SD against SE The SD describes how much individual drivers differ from each other, and it does not shrink when you collect more of them. The SE describes how much a sample mean would bounce around from study to study, and it shrinks as √n grows.

SE = 12 ms at n = 25, 6 ms at n = 100, and 225 drivers are needed for an SE of 4 ms.

From §1.9 Sampling Distributions.

1. Why normality applies here The central limit theorem is about the distribution of the sample mean, not about the raw deliveries. With n = 36 the sampling distribution of x̄ is close to normal even though individual delivery times are not. The skew has to be extreme, or n small, for this to fail.
2. Standard error SE = σ / √n = 9 / √36 = 9 / 6 = 1.5 min
3. P(x̄ > 30) — standardize the mean, not a delivery z = (30 − 28) / 1.5 = 2 / 1.5 = 1.333 P(Z > 1.333) = 1 − .9088 = .0912
4. P(27 < x̄ < 29) z₁ = (27 − 28) / 1.5 = −0.667 z₂ = (29 − 28) / 1.5 = +0.667 P(−0.667 < Z < 0.667) = .7475 − .2525 = .4950
5. Sanity check A single delivery taking over 30 minutes is common. A mean of 36 deliveries doing so is not, at about 9%. Averaging is what kills the variability.

The CLT covers the sample mean regardless of the skew. P(x̄ > 30) = .091, and P(27 < x̄ < 29) = .495, so almost exactly half of such samples land within a minute of the true mean.

Rounding: carry z = 1.333, not 1.3. Rounding to 1.3 gives .0968 instead of .0912, an error of more than half a percentage point.

From §1.10 Central Limit Theorem.

1. Formula: t, not z, because σ is unknown CI = x̄ ± t* × (s / √n), df = n − 1
2. Standard error SE = 1.5 / √25 = 1.5 / 5 = 0.30 h
3. Critical value df = 25 − 1 = 24, t*(.025, 24) = 2.064
4. Margin of error 2.064 × 0.30 = 0.619 h
5. The interval 6.8 ± 0.619 = [6.18, 7.42] hours
6. Against the 8-hour guidance 8 sits well outside the interval, so 8 hours is not a plausible value for the population mean given this sample. Equivalently, a one-sample t-test against 8 would reject at α = .05. What the interval does not say is that 95% of first-years sleep between 6.18 and 7.42 hours: the interval is about the mean, not about individuals.
7. With n = 100 SE = 1.5 / 10 = 0.15, df = 99, t* = 1.984 margin = 1.984 × 0.15 = 0.298 CI = [6.50, 7.10] hours Four times the sample gives an interval about half as wide, and t* also drifts down towards 1.96 as df grows.

95% CI [6.18, 7.42] hours. First-years sleep less than the recommended 8 hours, and quadrupling the sample narrows the interval to [6.50, 7.10].

From §1.11 Confidence Intervals · look up t* in Statistical Tables.

1. What p actually is If the feed truly made no difference, 3% of trials run this way would produce a difference of at least 0.42 kg in one direction or the other. It is the probability of data this extreme given the null hypothesis, written P(data | H₀).
2. Marek's error He has flipped the conditional into P(H₀ | data). Those two are not interchangeable, and turning one into the other needs a prior probability that the test never supplies.
3. The neighbor's version Same inversion, wearing a hat. 1 − p is not the probability that the alternative is true, and it is not the probability the result replicates either.
4. The decision p = .030 < α = .05 → reject H₀ If the feed genuinely does nothing, this rejection is a Type I error. Setting α = .05 caps that error rate at 5% over the long run; it says nothing about whether this particular result is the mistake.
5. The 8-plot repeat No. A non-significant result means the data are compatible with no effect, not that the effect is zero. With 8 plots the test has very little power, so a real effect of this size would be missed most of the time. Failing to detect it is a Type II error. Report the confidence interval: if it stretches from −0.3 to +1.1 kg, the study is simply too small to settle the question.
6. Twenty null trials Expected false positives = 20 × .05 = 1 One society in twenty would find "significant" evidence for a feed that does nothing. If only the exciting result gets into the newsletter, the newsletter reports an effect that is not there.

p is P(data at least this extreme | H₀ true), so both Marek and his neighbor have inverted the conditional. The decision is to reject at α = .05, with Type I error as the risk. The 8-plot follow-up shows nothing either way, and one false positive per twenty null trials is the expected background rate.

From §1.12 Hypothesis Testing Logic and §3.1 Effect Size & Power.

1. Hypotheses H₀: μ = 65 H₁: μ ≠ 65 Two-tailed, because the consumer group would report the course being better than claimed as well as worse.
2. Standard error SE = s / √n = 6.4 / √16 = 6.4 / 4 = 1.60
3. Test statistic t = (x̄ − μ₀) / SE = (61.5 − 65) / 1.60 t = −3.5 / 1.60 = −2.1875 df = n − 1 = 15
4. Critical value t*(.025, 15) = 2.131 |−2.1875| = 2.188 > 2.131 → reject H₀
5. p-value p = 2 × P(T₁₅ > 2.1875) = .045
6. Effect size d = (x̄ − μ₀) / s = −3.5 / 6.4 = −0.55 A medium shortfall of just over half a standard deviation.
7. Write-up Graduates typed more slowly than the advertised 65 wpm, M = 61.5, SD = 6.4, t(15) = −2.19, p = .045, d = −0.55, 95% CI [58.09, 64.91].

Reject H₀. The 65 wpm claim is not supported, though at p = .045 with n = 16 the evidence is thin, and the confidence interval reaches almost up to 65.

Rounding matters here. Carry t = −2.1875 rather than rounding the standard error first; this test clears its critical value by 0.06, so an early round-off can flip the decision. Report t to two decimals and p to three.

From §1.13 One-Sample & Paired t-Tests · format the sentence with the APA formatter.

1. Why paired Each after-score is tied to a specific person's before-score. Subtracting within a person removes that person's baseline memory ability from the comparison, which is where most of the noise lives. An independent test would treat the twenty scores as twenty unrelated people and throw that advantage away.
2. Mean difference Σd = 3+5−1+4+6+2+0+7+3+1 = 30 d̄ = 30 / 10 = 3.00 items
3. SD of the differences (d − d̄): 0, 2, −4, 1, 3, −1, −3, 4, 0, −2 (d − d̄)²: 0, 4, 16, 1, 9, 1, 9, 16, 0, 4 → Σ = 60 s_d = √(60 / 9) = √6.667 = 2.582
4. Standard error and t SE = s_d / √n = 2.582 / √10 = 2.582 / 3.162 = 0.8165 t = d̄ / SE = 3.00 / 0.8165 = 3.674 df = n − 1 = 9
5. Decision t*(.025, 9) = 2.262 3.674 > 2.262 → reject H₀ p = 2 × P(T₉ > 3.674) = .005
6. Effect size for a paired design d_z = d̄ / s_d = 3.00 / 2.582 = 1.16 Large. Note that d_z is scaled by the SD of the differences, so it is not comparable with a between-groups d computed from raw score SDs. Say which one you used.

Recall improved by 3.00 items on average, t(9) = 3.67, p = .005, d_z = 1.16. With ten students and no control group, this shows the scores went up; it does not rule out practice or test-order effects.

From §1.13 One-Sample & Paired t-Tests · effect-size conversions in the converter.

1. Pooled variance s²_p = [(n₁−1)s₁² + (n₂−1)s₂²] / (n₁ + n₂ − 2) = [11(64) + 11(81)] / 22 = (704 + 891) / 22 = 1595 / 22 = 72.50 → s_p = √72.50 = 8.515
2. Standard error of the difference SE = s_p × √(1/n₁ + 1/n₂) = 8.515 × √(1/12 + 1/12) = 8.515 × √0.1667 = 8.515 × 0.4082 = 3.476
3. Test statistic t = (78.0 − 71.5) / 3.476 = 6.5 / 3.476 = 1.870 df = 12 + 12 − 2 = 22
4. Decision t*(.025, 22) = 2.074 1.870 < 2.074 → fail to reject H₀, p = .075
5. Effect size d = (M₁ − M₂) / s_p = 6.5 / 8.515 = 0.76 By Cohen's benchmarks that is medium-to-large, just short of the 0.80 mark, sitting inside a non-significant test.
6. Assess the conclusion "Equally well" is not supported. A non-significant result means the data do not rule out zero difference, and here the best estimate of the difference is 6.5 marks with d = 0.76. With 12 per group the study had about 43% power to detect an effect that size, so it was more likely to miss it than to find it. The honest sentence is that the trial was too small to decide.
7. Sample size for a replication Two-sample t, d = 0.76, α = .05, power = .80 n = 29 per group (58 children total) 28 per group falls just short at 79.8% power, which is why the answer rounds up rather than down. Planning for a medium d = 0.50 instead would need 64 per group. Powering a study on the effect size your own underpowered pilot produced is optimistic, because a pilot that reached significance has almost certainly overestimated the effect.

t(22) = 1.87, p = .075, d = 0.76. Not significant, not "no difference": a large point estimate measured with far too little precision. A replication needs at least 29 per group.

From §1.14 Independent Samples t-Test and §3.1 Effect Size & Power · redo the sample size in the Power calculator.

1. The three probabilities you are handed P(S) = 240 / 400 = .60 P(M) = 180 / 400 = .45 P(S and M) = 120 / 400 = .30
2. General addition rule P(S or M) = P(S) + P(M) − P(S and M) = .60 + .45 − .30 = .75 Adding .60 and .45 gives 1.05, which is already impossible. The 120 students in the overlap were counted inside each total, so exactly one copy of them has to come back out.
3. Check it by counting people 240 + 180 − 120 = 300 students → 300 / 400 = .75
4. Neither, by the complement rule P(neither) = 1 − P(S or M) = 1 − .75 = .25 → 100 students
5. Statistics but not methods 240 − 120 = 120 students → .30 Methods but not statistics is 180 − 120 = 60, and the four groups 120 + 120 + 60 + 100 rebuild the 400.
6. Independence check If independent: P(S and M) = P(S) × P(M) = .60 × .45 = .27 Observed: .30 ≠ .27 P(M | S) = 120 / 240 = .50 against P(M) = .45 Not independent, though the dependence is mild. Taking statistics lifts the chance of taking methods from .45 to .50, which is a real association and a small one.

.75, .25, 120 students, and no: .30 against the .27 independence would require. The gap is small, so a formal test would be worth running before making anything of it.

From §1.7 Probability Basics · the same table shape is tested for independence in §1.16 Chi-Square Tests.

1. Split the cohort by the condition Ill: 10,000 × .02 = 200 Healthy: 10,000 − 200 = 9,800
2. Split each branch by the test result Ill and positive: 200 × .95 = 190 Ill and negative: 200 − 190 = 10 Healthy and positive: 9,800 × .10 = 980 Healthy and negative: 9,800 − 980 = 8,820 The four leaves add back to 10,000, which is the arithmetic check worth doing every time.
3. P(ill | positive): condition on the positives Positives in total: 190 + 980 = 1,170 P(ill | positive) = 190 / 1,170 = .162 → 16.2% The same answer from Bayes' rule, with the law of total probability underneath the line: (.02 × .95) / (.02 × .95 + .98 × .10) = .019 / .117 = .162.
4. P(healthy | negative): condition on the negatives Negatives in total: 10 + 8,820 = 8,830 P(healthy | negative) = 8,820 / 8,830 = .9989 → 99.9% A negative result here is close to conclusive, while a positive one is mostly a false alarm. The same test does two very different jobs depending on which way it points.
5. What "accurate" means, and what it hides Correct results: 190 + 8,820 = 9,010 → 9,010 / 10,000 = 90.1% The 95% in the leaflet is the sensitivity, P(positive | ill), which describes the test's behavior on people already known to be ill. The reader of the letter is in the opposite position and needs P(ill | positive), which is 16.2%. Overall accuracy, 90.1% here, is a third number again, and it is flattered by the 8,820 healthy people the test clears without effort.

190 / 10 / 980 / 8,820. P(ill | positive) = 16.2%, P(healthy | negative) = 99.9%, overall accuracy 90.1%. Roughly five out of six positive letters go to someone who does not have the condition, which is why a positive screen is normally followed by a second, more specific test rather than a diagnosis.

From §1.7 Probability Basics · rebuild the tree with the sliders in that lesson's Base-Rate Tree.

1. Two conditions to check Every P(X = x) lies between 0 and 1 ✓ .10 + .25 + .35 + .20 + .10 = 1.00 ✓
2. Mean: μ = Σ x · p 0(.10) + 1(.25) + 2(.35) + 3(.20) + 4(.10) = 0 + .25 + .70 + .60 + .40 = 1.95 bikes The mean of a random variable need not be a value the variable can take. No hour ever sees 1.95 bikes returned.
3. Variance by the definition: σ² = Σ (x − μ)² · p (0−1.95)²(.10) = 3.8025 × .10 = .380250 (1−1.95)²(.25) = 0.9025 × .25 = .225625 (2−1.95)²(.35) = 0.0025 × .35 = .000875 (3−1.95)²(.20) = 1.1025 × .20 = .220500 (4−1.95)²(.10) = 4.2025 × .10 = .420250 σ² = 1.2475
4. Variance by the shortcut: σ² = E(X²) − μ² E(X²) = 0(.10) + 1(.25) + 4(.35) + 9(.20) + 16(.10) = 5.05 σ² = 5.05 − 1.95² = 5.05 − 3.8025 = 1.2475 ✓ σ = √1.2475 = 1.117 bikes The two routes must agree exactly, and the second is far less error-prone on a calculator, because it never subtracts the mean five separate times.
5. A tail probability is a sum over the table P(X ≥ 2) = .35 + .20 + .10 = .65
6. Why the binomial shortcut does not apply μ = np and σ = √(np(1 − p)) belong to a binomial count, which needs a fixed number of independent trials each carrying the same success probability. Nothing here supplies an n or a p: the table was measured, not derived from trials. So the general formulas are the only route. Read the other way, the binomial shortcut is what you get when you run these same sums on a distribution that does have that structure.

Valid distribution; μ = 1.95, σ² = 1.2475, σ = 1.117, P(X ≥ 2) = .65. The binomial shortcut needs a fixed n and a constant p, and this distribution has neither.

From §1.8 The Binomial Distribution · the same two sums, run on a binomial, give np and √(np(1 − p)).

1. Median first, because the quartiles are built from it n = 15 is odd, so the median is the 8th value. Median = 14 min
2. Quartiles: median of each half, the median itself left out Lower half (7 values): 3, 6, 7, 9, 10, 12, 12 → Q1 = 4th = 9 Upper half (7 values): 15, 16, 18, 19, 21, 24, 41 → Q3 = 4th = 19 IQR = 19 − 9 = 10 min With an odd n the middle value belongs to neither half. Include it in both and you get different quartiles, which is one of the two or three most common ways to lose a mark here.
3. Five-number summary Min = 3 · Q1 = 9 · Median = 14 · Q3 = 19 · Max = 41
4. Fences, then flags Lower: Q1 − 1.5 × IQR = 9 − 15 = −6 Upper: Q3 + 1.5 × IQR = 19 + 15 = 34 Only 41 lies outside [−6, 34] A negative lower fence is fine and quite common. It says nothing in this batch could have been low enough to flag, since the times cannot go below zero.
5. The boxplot Box from 9 to 19, with the median line at 14, sitting left of center inside the box. The left whisker runs to the minimum, 3. The right whisker stops at 24, the largest value still inside the upper fence, not at the maximum. The 41 is drawn as its own point beyond the whisker.
6. The software answer R: Q1 = 9.5, Q3 = 18.5, IQR = 9 Fences: 9.5 − 13.5 = −4 and 18.5 + 13.5 = 32 The quartiles differ because R interpolates between neighboring values instead of taking the median of a half. The box would be drawn very slightly narrower, and 41 is still the only flagged time. Different convention, same conclusion here, which is the usual outcome and exactly why it is worth knowing that both exist.

Five-number summary 3, 9, 14, 19, 41. Fences at −6 and 34, so the 41 is the single flagged time. Box 9 to 19 with the median at 14, whiskers to 3 and 24, and 41 as a separate point. R's quartiles (9.5 and 18.5) move the fences to −4 and 32 and flag the same value.

Rounding: none needed. Every quartile here lands on a data value or halfway between two, so the arithmetic is exact.

From §1.3 Describing Data and §1.4 Visualizing Data · check it in the Descriptives calculator, prefilled, where the Quartiles switch turns the textbook answer into R's.

1. Standard error, then the margin σ is known, so this is a z-interval and nothing is estimated in the denominator. SE = σ/√n = 4/√9 = 4/3 = 1.3333 mmHg m = z* × SE = 1.960 × 1.3333 = 2.6133 mmHg
2. The 95% interval 138 ± 2.6133 = [135.39, 140.61] mmHg
3. The same data at 99% m = 2.576 × 1.3333 = 3.4347 mmHg 138 ± 3.4347 = [134.57, 141.43] mmHg The width goes from 5.23 to 6.87 mmHg, about 31% wider, which is exactly the ratio 2.576/1.960. Nothing about the data changed. The extra width is the price of a procedure that misses one time in a hundred instead of one time in twenty.
4. How many readings for m ≤ 1.5 Rearrange m = z*σ/√n for n. n = (z*σ/m)² = (1.960 × 4 / 1.5)² = 5.2267² = 27.32 Round up: n = 28 readings The rounding direction is worth a mark on its own. With 27 readings the margin comes out at 1.5088, which is over the limit; with 28 it is 1.4816. Rounding down breaks the promise the clinic just made.
5. What the interval does not say The true mean is a fixed number, not a random one. This particular interval either contains it or it does not, and no probability attaches to that fact once the data are in. The 95% belongs to the procedure: build intervals this way forever and 95% of them capture the truth. Read the number as a statement about the method, and the sentence becomes true.

SE = 1.3333 mmHg. The 95% interval is 138 ± 2.61 = [135.39, 140.61] and the 99% interval is 138 ± 3.43 = [134.57, 141.43], about 31% wider. A margin of 1.5 mmHg needs n = 28 readings. The "95% probability" reading is wrong because confidence describes the long-run behavior of the procedure, not any one interval.

Rounding: z* values taken as the printed 1.960 and 2.576. Intervals given to 2 dp; the sample size is always rounded up.

From §1.11 Confidence Intervals · watch it assemble in the Interval Builder, prefilled, and read z* off the printed Table D.

1. Hypotheses, from the wording H₀: μ = 1,200    H₁: μ < 1,200 "Do not last as long as claimed" names a direction, so this is a left-tailed test. Had the group asked whether the advertised figure was simply wrong, H₁ would be μ ≠ 1,200.
2. Standard error and statistic SE = σ/√n = 100/√25 = 100/5 = 20 h z = (x̄ − μ₀)/SE = (1,165 − 1,200)/20 = −35/20 = −1.75
3. The p-value, left tail One tail, so read the area below −1.75 straight from Table A. p = P(Z < −1.75) = .0401
4. Two decisions, two thresholds At α = .05: .0401 < .05, so reject H₀ At α = .01: .0401 > .01, so fail to reject H₀ The evidence did not change between those two lines. The same z = −1.75 sits past the .05 cutoff of −1.645 and short of the .01 cutoff of −2.326. This is why α belongs in the write-up: "significant" means nothing without the threshold it was judged against.
5. The two-sided version p = 2 × P(Z < −1.75) = 2 × .0401 = .0801 Doubling it pushes the result past .05, so a two-sided test keeps H₀ at both thresholds. That is the whole reason the tail has to be fixed in advance. Choosing it once the sample mean is on the table means always taking the half that helps, which turns a nominal 5% false-alarm rate into a real 10% one.
6. The sentence to write At α = .05, one-tailed: the 25 bulbs give evidence that mean lifetime falls short of the advertised 1,200 hours, z = −1.75, p = .040. The shortfall estimated here is 35 hours, which is worth reporting beside the p-value, since a 3% shortfall may or may not matter to a buyer.

H₀: μ = 1,200 against H₁: μ < 1,200, one-tailed because the group suspected a shortfall. SE = 20, z = −1.75, one-tailed p = .0401: reject at α = .05, fail to reject at α = .01. Two-sided the p-value is .0801, which rejects at neither.

Rounding: p-values to 4 dp from Table A; a printed table gives .0401 for z = −1.75 directly.

From §1.12 Hypothesis Testing Logic · load it into the explorer, prefilled, or look the area up on the printed Table A.

1. Standard error, then the multiplier σ is unknown and s has been estimated from the same sixteen shots, so this is a t procedure with df = n − 1 = 15. SE = s/√n = 3.2/√16 = 3.2/4 = 0.800 s Table D, row 15, 95% column: t* = 2.131
2. The 95% interval m = t* × SE = 2.131 × 0.800 = 1.705 s 27.4 ± 1.705 = [25.70, 29.11] s
3. The verdict, read off the interval 25 lies below 25.70, so it is outside the interval. A 95% interval holds exactly the values a two-sided test at α = .05 would not reject, so the test rejects H₀: μ = 25. t = (x̄ − μ₀)/SE = (27.4 − 25)/0.800 = 3.00 On row 15 of Table D, 3.00 falls between the .005 column (2.947) and the .0025 column (3.286), so the upper-tail area is between .0025 and .005 and the two-tailed p is between .005 and .01. Exactly, p = .0090. The interval and the test agree, as they must.
4. The same data at 99% t*(15, 99%) = 2.947 → m = 2.947 × 0.800 = 2.357 s 27.4 ± 2.357 = [25.04, 29.76] s Still above 25, though only just. The two-tailed p of .0090 is under .01, which says the same thing: this result also clears the stricter threshold. Had p landed at, say, .03, the 99% interval would have swallowed 25 while the 95% one did not.
5. Is t allowed here Sixteen observations put this in the middle band of the rule of thumb, 15 to 40, where mild skewness is acceptable and outliers are not. The description gives mild skew and no unusual values, so the procedure is defensible. Under 15 observations the same data would have needed a shape much closer to normal, and a single wild shot would have been a problem at any sample size, because it moves the mean and inflates s at once.
6. The sentence to write Sixteen shots averaged 27.4 s (SD = 3.2), significantly longer than the 25 s the recipe calls for, t(15) = 3.00, p = .009, 95% CI [25.70, 29.11]. The interval is the useful half of that sentence: the machine is running long by somewhere between 0.7 and 4.1 seconds, and whether that matters is a question about coffee rather than about statistics.

SE = 0.800 s. The 95% interval is 27.4 ± 1.71 = [25.70, 29.11], which excludes 25, so a two-sided test rejects at α = .05; t(15) = 3.00 with p between .005 and .01 on Table D (.0090 exactly). At 99% the interval is [25.04, 29.76] and still excludes 25. With n = 16, mild skew and no outliers, the middle band of the rule of thumb allows the t procedure.

Rounding: t* values as Table D prints them (2.131 and 2.947); interval limits to 2 dp.

From §1.13 One-Sample & Paired t-Tests · mark t = 3.00 on the t-distribution viz, read t* off the printed Table D, or paste a column into the descriptives calculator.

1. Standard error of the difference Each group contributes its own variance over its own n. The two contributions add before the square root; the two standard errors are never added. s₁²/n₁ = 29.16/12 = 2.4300    s₂²/n₂ = 16.81/15 = 1.1207 SE = √(2.4300 + 1.1207) = √3.5507 = 1.884 cm t = (46.8 − 41.9)/1.884 = 4.90/1.884 = 2.600
2. The conservative degrees of freedom df = min(n₁ − 1, n₂ − 1) = min(11, 14) = 11 Table D, row 11, 95% column: t* = 2.201 2.600 is beyond 2.201, so H₀ of equal means is rejected at α = .05. On the same row, 2.600 sits between the .02 column (2.328) and the .01 column (2.718). Those headings are one-tail areas, so the one-tail figure lands between .01 and .02, and doubling puts the two-tailed p between .02 and .04. Exactly, p = .0247.
3. The same statistic on the software's curve Nothing about the data changed: the standard error is still 1.884 and t is still 2.600. Only the curve those numbers are read against moves, from 11 degrees of freedom to 20.12. df = 20.12 → p = .0171 Both reject at α = .05, so the two answers agree here. They would not always. A statistic sitting between the two critical values, anywhere from 2.085 to 2.201 on this data, clears the software's threshold and misses the hand rule's, which is exactly the direction the conservative rule is designed to err in.
4. The interval for the difference 4.90 ± 2.201 × 1.884 = 4.90 ± 4.147 95% CI = [0.75, 9.05] cm Zero is outside it, which is the rejection restated. The width is the more honest finding: twenty-seven athletes locate the advantage of the dynamic routine somewhere between three quarters of a centimeter and nine, and only the top of that range would interest a coach. At the Welch df the interval tightens slightly, to [0.97, 8.83].
5. Which one to write down In an exam, the conservative one. It needs nothing beyond a square root, and every error it makes runs the same way: holding the fewest degrees of freedom of the three, it draws the fattest tails, so its critical value is the highest and its p the biggest. A result that clears that bar would have cleared either of the others too, which is why a marker can take it on trust. Away from the exam hall, quote whichever df your software printed and name which one it was.

SE = 1.884 cm and t = 2.600. At the conservative df = 11, t* = 2.201, so H₀ is rejected at α = .05, with Table D bracketing the two-tailed p between .02 and .04 (.0247 exactly). At the Welch df of 20.12 the p is .0171; both reject, though they need not always. The conservative 95% CI for the difference is [0.75, 9.05] cm. Write the conservative answer in an exam, because it errs only toward caution.

Rounding: SE and t to three decimal places as a calculator gives them; t* as Table D prints it; interval limits to 2 dp. Pooling instead, which these unequal spreads do not obviously justify, would give s_p = 4.716, t = 2.683 on 25 df and p = .0128.

From §1.14 Independent Samples t-Test · compare the three rules live in the Tug-of-War with the conservative df selected, or look t* up on the printed Table D.

1. Margins first, then the expected counts Every expected count is its own row total times its own column total, divided by the grand total. Doing the margins first means each one is a single multiplication and a single division. Rows: 70 + 32 + 18 = 120   30 + 28 + 22 = 80    Columns: 100, 60, 40    N = 200 Members: 120 × 100/200 = 60   120 × 60/200 = 36   120 × 40/200 = 24 Casual: 80 × 100/200 = 40   80 × 60/200 = 24   80 × 40/200 = 16 Check the margins of the expected table before going on: 60 + 36 + 24 = 120 and 40 + 24 + 16 = 80, so the expected counts reproduce the observed totals, which they always must. The smallest expected count is 16, so every one clears 1, none is below 5, and the conditions hold comfortably.
2. One component per cell Subtract, square, divide by the expected count. Keep four decimal places until the final addition. Members: (70 − 60)²/60 = 1.6667   (32 − 36)²/36 = 0.4444   (18 − 24)²/24 = 1.5000 Casual: (30 − 40)²/40 = 2.5000   (28 − 24)²/24 = 0.6667   (22 − 16)²/16 = 2.2500 χ² = 1.6667 + 0.4444 + 1.5000 + 2.5000 + 0.6667 + 2.2500 = 9.0278
3. Degrees of freedom, then Table F df = (2 − 1) × (3 − 1) = 2 On the df 2 row, 9.0278 sits between the .02 column (7.824) and the .01 column (9.210), so the p-value is between .01 and .02. Since that whole range is below .05, reject the hypothesis that rider type and verdict are independent. Exactly, p = .0110.
4. Where the 9.03 came from The largest component is 2.50, casual riders who said better. Independence expects 40 of them and only 30 said so. The second largest, 2.25, is its mirror image: 22 casual riders called the stations worse where 16 were expected. Members run the other way in both cells. So the finding is not that riders are split on the docks in general, it is that casual users are markedly less enthusiastic than members, and both ends of their answer distribution show it. The smallest component, 0.44, says the middle option is answered almost identically by the two groups.
5. The two conditional distributions A significant χ² is a claim that these two rows differ. Divide each row by its own total and the claim becomes readable. Members: 70/120 = 58.3%   32/120 = 26.7%   18/120 = 15.0% Casual: 30/80 = 37.5%   28/80 = 35.0%   22/80 = 27.5% Everybody: 100/200 = 50%   60/200 = 30%   40/200 = 20% Members split 58 / 27 / 15 and casual users 38 / 35 / 28, against the 50 / 30 / 20 of the survey as a whole. Independence would have both rows sitting on that last line. The sentence to write: members are the ones happy with the new docks, and approval falls off steeply among casual users, a quarter of whom call the stations worse.

Expected counts 60, 36, 24 for members and 40, 24, 16 for casual riders; all conditions met. Components 1.6667, 0.4444, 1.5000, 2.5000, 0.6667 and 2.2500 give χ² = 9.03 on 2 df. Table F brackets p between .01 and .02 (p = .0110 exactly), so rider type and verdict are associated at α = .05. The biggest contributor is casual riders who chose "better", 30 against an expected 40. Row percentages: members 58.3 / 26.7 / 15.0, casual 37.5 / 35.0 / 27.5.

Rounding: components to 4 dp so the sum is right in the second decimal; χ² to 2 dp and percentages to 1 dp in the report. Reporting it: χ²(2, N = 200) = 9.03, p = .011. Problem 19 takes a different 2 × 3 table on to Cramér's V.

From §1.16 Chi-Square Tests · type the table into the Build a Table explorer to see the same components, the same row percentages and the tint, or look the critical values up on the printed Table F.

1. Conditions, then the proportions Each group needs at least 10 successes and 10 failures. Here it is 205 and 45 in one group, 148 and 52 in the other, so there is nothing to worry about on that front. Both samples are also tiny next to the population of patients a clinic sees. p̂₁ = 205/250 = 0.820    p̂₂ = 148/200 = 0.740    p̂₁ − p̂₂ = 0.080
2. The test, which believes H₀ H₀ says the two population proportions are equal. If that is true there is only one proportion to estimate, and every one of the 450 patients helps estimate it. That single pooled value goes into both halves of the standard error. p̂ = (205 + 148)/(250 + 200) = 353/450 = 0.7844 SE = √(0.7844 × 0.2156 × (1/250 + 1/200)) = √(0.16909 × 0.00900) = √0.0015218 = 0.03901 z = 0.080 / 0.03901 = 2.051    two-sided p = 2 × P(Z > 2.05) = .0403 Table A gives the area left of 2.05 as .9798, so the upper tail is .0202 and doubling it gives .0404. Reject H₀ at α = .05, though not at .01.
3. The interval, which believes nothing An interval has no hypothesized value to lean on, so each group estimates its own spread from its own data and the two variances add. SE = √(0.82 × 0.18/250 + 0.74 × 0.26/200) = √(0.0005904 + 0.0009620) = √0.0015524 = 0.03940 0.080 ± 1.96 × 0.03940 = 0.080 ± 0.077 95% CI = [0.003, 0.157]
4. Why the two standard errors differ A proportion's spread is decided by the proportion itself, so choosing a value for p also chooses a standard error. The test assumes H₀ is true and H₀ hands it one shared proportion, so it would be inconsistent to test that hypothesis using two separate estimates. The interval assumes nothing at all and has only the two observed proportions to work with. The two standard errors come out at 0.03901 and 0.03940 here, close enough that the conclusion is safe either way, but the two formulas answer different questions and swapping them costs marks.
5. The sentence, and the same test in its other costume Text reminders were followed by attendance more often than email reminders, 82.0% against 74.0%, z = 2.05, p = .040, 95% CI of the difference [0.3%, 15.7%]. Laid out as a 2 × 2 table of attended against missed, the same 450 patients give χ²(1, N = 450) = 4.21, and 4.2055 is 2.0507 squared. The p-value is .040 either way. The z version is the one to report, because the interval it comes with is the part the practice manager can act on: the advantage could be as small as a third of a percentage point.

p̂₁ = .820, p̂₂ = .740, difference .080. Pooled p̂ = .7844, test SE = .03901, z = 2.05, two-sided p = .040, so reject H₀ at α = .05. The 95% interval uses the unpooled SE = .03940 and runs [0.003, 0.157]. χ² on the same table is 4.21 on 1 df, which is z² exactly, with the same p.

Rounding: proportions to 3 dp, standard errors to 5 dp so the z survives to two decimals, interval limits to 3 dp. Reading z = 2.05 off Table A gives p = .0404 rather than .0403; both round to .040.

From §1.15 Inference for Proportions · set the counts in the One Proportion, or Two explorer, or check the areas on the printed Table A.

1. Means and standard deviations x̄ = 30 / 5 = 6.0 nurses    ȳ = 180 / 5 = 36.0 minutes Σ(x − x̄)² = 16 + 4 + 0 + 4 + 16 = 40    s_x = √(40/4) = √10 = 3.162 Σ(y − ȳ)² = 81 + 4 + 16 + 36 + 25 = 162    s_y = √(162/4) = √40.5 = 6.364
2. The table the formula asks for

   x	y	x − x̄	y − ȳ	(x − x̄)/sx	(y − ȳ)/sy	product
   2	45	−4	+9	−1.265	+1.414	−1.789
   4	34	−2	−2	−0.632	−0.314	+0.199
   6	40	0	+4	0.000	+0.629	0.000
   8	30	+2	−6	+0.632	−0.943	−0.596
   10	31	+4	−5	+1.265	−0.786	−0.994

   Σ products = −3.180    r = −3.180 / (5 − 1) = −0.795 The shortcut route agrees, and is faster if you trust it: Σ(x − x̄)(y − ȳ) = −64, so r = −64 / √(40 × 162) = −64 / 96.00 = −0.795. Use whichever the question asks for; the standardized table is the one that shows your working.
3. Interpretation The sign is negative, so more nurses goes with shorter waits. The size, |r| = .80, is a strong linear association by any of the usual benchmarks, though five wards is a very thin basis for it. r² = (−0.795)² = 0.632, so about 63% of the variation in waiting times across these five wards lines up with the variation in staffing, leaving 37% to everything else: how sick the arriving patients were, how the ward is laid out, and plain chance.
4. The ward that contributes nothing The six-nurse ward. Its standardized x is exactly zero, because it sits on the mean staffing level, and zero times its standardized y of +0.629 is zero. It waited four minutes longer than average and the coefficient never hears about it. Influence in a correlation comes from distance from BOTH means, which is the same arithmetic that makes r so easy for a single far-out point to move.

x̄ = 6.0, ȳ = 36.0, s_x = 3.162, s_y = 6.364. The five standardized products are −1.789, +0.199, 0.000, −0.596 and −0.994, totalling −3.180, so r = −3.180/4 = −0.795. r² = 0.632. The six-nurse ward contributes nothing, because its standardized x is zero.

Rounding: standardized values and products to 3 dp, r to 3 dp. Carrying only 2 dp through the table gives −0.794, which still rounds to −0.79.

From §1.17 Correlation · step the same shape through the 📏 Correlation, Step by Step viz, or load these five wards straight into the explorer: ?data=2:45,4:34,6:40,8:30,10:31.

1. Slope, then intercept b₁ = r · s_y / s_x = 0.42 × 11.4 / 1.2 = 0.42 × 9.5 = 3.99 b₀ = ȳ − b₁ x̄ = 63.5 − 3.99 × 6.8 = 63.5 − 27.132 = 36.368 ŷ = 36.37 + 3.99x
2. Prediction and slope in words ŷ = 36.368 + 3.99 × 8 = 36.368 + 31.92 = 68.29 A student sleeping 8 hours is predicted to score about 68 marks. The slope says that one extra hour of sleep goes with about 4 more marks, on average, across these 60 students. It is an association and not a promise: nobody was assigned a bedtime, so a student who sleeps well may differ in other ways that also help.
3. The intercept b₀ = 36.4 is the mark the line predicts for a student who slept zero hours. Nobody in this sample slept less than 4.5, so that value sits well outside the data and the line has no evidence behind it there. Read it as the arithmetic that pins the line to (x̄, ȳ) rather than as a claim about sleepless students. The honest range for prediction is 4.5 to 9.0 hours, where the line runs from 54.3 to 72.3 marks.
4. Swapping x and y b₁′ = r · s_x / s_y = 0.42 × 1.2 / 11.4 = 0.0442 hours per mark The inversion would have given 1/3.99 = 0.251, nearly six times too big. Correlation is symmetric and regression is not: the two slopes multiply to r² = 0.1764, not to 1, and they agree only when r = 1. Least squares minimizes vertical distances, and swapping the axes changes which direction "vertical" points, so it is genuinely a different line.

ŷ = 36.37 + 3.99x. At 8 hours the prediction is 68.3 marks, and each extra hour of sleep is worth about 4 marks. The intercept 36.4 is the predicted mark at zero hours, far outside the observed 4.5 to 9.0 range and not interpretable. Reversing the roles gives b₁′ = 0.0442 hours per mark, not 1/3.99 = 0.251, because b₁ · b₁′ = r² = .1764.

Rounding: the slope is exact at 3.99 because 11.4/1.2 = 9.5 exactly; carry the intercept to 3 dp (36.368) before rounding the printed line to 2 dp.

From §1.18 Simple Linear Regression · the same identity is in the fourth block of the formula sheet, and the Correlation & Regression Calculator fits a line to raw columns when you have them.

1. Predictions and residuals

   x	1	2	3	4	5	6	7	8
   ŷ = 17 + 2x	19	21	23	25	27	29	31	33
   y	12	20	26	30	32	32	30	26
   residual y − ŷ	−7	−1	+3	+5	+5	+3	−1	−7

   Σ residuals = (−7 − 1 + 3 + 5) + (5 + 3 − 1 − 7) = 0 The total is zero, and it always is for a least-squares line, so it confirms the arithmetic without saying anything about the fit.
2. The pattern Plotted against dose, the residuals rise from −7 to +5 and fall back to −7: a clean arch, negative at both ends and positive through the middle. That is the signature of a curved relationship fitted with a straight line. Yield climbs with dose, flattens, then falls off as the dose becomes too much, and no straight line can bend to follow it. The line is too high at both ends and too low in the middle, by up to seven tomatoes.
3. Why R² does not notice R² measures how much of the variation the line removes, not whether what is left has a shape. Here Σ(y − ȳ)² = 336 and the residuals leave 49 + 1 + 9 + 25 + 25 + 9 + 1 + 49 = 168, exactly half, so R² = 1 − 168/336 = .500 and r = .707. A rising line does capture the general upward drift over most of the range, which is enough for half the variation. The half it leaves behind is not noise; it is the curve. This is exactly why a residual plot is worth drawing even when the summary number looks acceptable, and it is Anscombe's second dataset in a different costume.
4. The other grower's wedge A wedge is non-constant spread, which regression calls heteroscedasticity. The line's shape may be perfectly right; what fails is the assumption that the scatter around it is the same size everywhere. Predictions at the wide end are much less reliable than the narrow end, and the standard errors, confidence intervals and p-values that Stats 2 computes all rest on one shared spread, so they come out wrong for both ends at once. The usual repairs are a transformation of the outcome or a standard error that does not assume constant spread.

Predicted values 19, 21, 23, 25, 27, 29, 31, 33; residuals −7, −1, +3, +5, +5, +3, −1, −7, summing to zero. The residual plot is an arch, so the true relationship is curved and a straight line is the wrong shape. R² = .500 anyway, because R² measures variation removed rather than pattern left behind. The wedge is heteroscedasticity, which invalidates the standard errors rather than the shape of the line.

Rounding: none needed. Every predicted value and every residual is a whole number here, and R² is exactly .500 because Σ residual² = 168 is exactly half of Σ(y − ȳ)² = 336.

From §1.18 Simple Linear Regression · see the arch for yourself by loading these eight pots with the residual plot open: ?data=…&resid=1, then press "Snap to best fit". The full treatment is §2.4 Regression Diagnostics.

1. The four words Population: all 18,400 enrolled undergraduates. Sample: the 175 who returned a usable form. Parameter: the mean weeknight sleep of the whole population, one fixed number nobody has measured. Statistic: the 6.8 hours, computed from the sample, and a different number in every sample the registry could have drawn.
2. Notation μ = mean sleep of all 18,400 (parameter, unknown) x̄ = 6.8 hours (statistic, computed) p = true share sleeping under 7 hours (parameter, unknown) p̂ = observed share (statistic, computed) The population gets a Greek letter or a plain p; the sample gets a bar or a hat. Writing μ where you mean x̄ is the single most expensive slip on an early exam, because every sentence after it is then about the wrong quantity.
3. The observed share p̂ = 63 / 175 = 0.36 36% of the students who replied sleep under 7 hours. Note the denominator: 175, the people who answered, not the 250 who were asked.
4. The 75 who did not reply The selection was random, but the participation was not. The 75 silent students chose themselves out, and whatever kept them from answering may well be related to sleep, so the 175 are no longer a random slice of the university. Raising the sample to 2,000 buys a more precise estimate of the same biased quantity: a bigger sample shrinks the standard error and leaves the bias exactly where it was. The repair is chasing the non-respondents, not adding more names.
5. Sample size against population size 18,400 / 250 = 73.6, so the sample is about 1.4% of the university The usual caution is that the population should be at least ten or twenty times the sample. At 73.6 times, this one clears it comfortably, which is why the standard error depends on n alone and the size of the university never enters the formula.

Population: all 18,400 undergraduates. Sample: the 175 usable replies. Parameter: μ, their true mean sleep. Statistic: x̄ = 6.8 hours. The observed share is p̂ = 63/175 = 0.36. The 30% non-response is a bias that a larger sample cannot fix, and at 1.4% of the population the sampling fraction is small enough to ignore.

From §1.1 What Is Statistics? and §1.6 Producing Data & Sampling Design.

1. Sort first; nothing else works until you have 6, 7, 8, 9, 10, 11, 12, 14, 18, 22, 26, 31, 47 n = 13 is odd, so the median is the 7th value. Median = 12 min
2. Quartiles: median of each half, the middle value in neither Lower half (6 values): 6, 7, 8, 9, 10, 11 → Q1 = (8 + 9)/2 = 8.5 Upper half (6 values): 14, 18, 22, 26, 31, 47 → Q3 = (22 + 26)/2 = 24 IQR = 24 − 8.5 = 15.5 min Five-number summary: 6 · 8.5 · 12 · 24 · 47
3. Fences under the textbook quartiles 1.5 × IQR = 1.5 × 15.5 = 23.25 Lower: 8.5 − 23.25 = −14.75 Upper: 24 + 23.25 = 47.25 47 sits inside the upper fence by a quarter of a minute, so this rule flags nothing at all. The longest wait of the morning is, by this convention, an ordinary long wait.
4. The same rule on R's quartiles IQR = 22 − 9 = 13, so 1.5 × IQR = 19.5 Fences: 9 − 19.5 = −10.5 and 22 + 19.5 = 41.5 Now 47 is outside, and the boxplot draws it as a separate point. Two defensible conventions, one dataset, opposite verdicts on the one call the manager cares about. It happens when a value sits near a fence, which is exactly where the argument always is. SPSS, incidentally, returns 8.5 and 24 here and agrees with the textbook rule, so the disagreement is not a simple hand-versus-software split.
5. One number for a typical wait Mean = 221 / 13 = 17.0 min · Median = 12 min Five minutes apart on thirteen calls. The mean is carrying the 47 and the 31, and no caller that morning waited 17 minutes; the median is a wait a caller actually had. Report the median for a typical wait, and the maximum alongside it, since a 47-minute hold is the thing a manager needs to hear about and a median is designed not to mention it.

Five-number summary 6, 8.5, 12, 24, 47. Textbook fences −14.75 and 47.25 flag nothing; R's quartiles put the fences at −10.5 and 41.5 and flag the 47. Mean 17.0 against a median of 12: report the median, with the longest wait beside it.

Rounding: none. Every quartile lands halfway between two whole values, and 221/13 is exactly 17.

From §1.3 Describing Data and §1.4 Visualizing Data · load it into the Descriptives calculator, prefilled and flip the Quartiles switch to watch the flag appear.

1. Under the label z = (500 − 502) / 4 = −0.50 Table A at z = −0.50 gives .3085 Table A always reports the area to the LEFT, which is what this question wants, so the entry is the answer with no arithmetic after it. About 30.85% of bags are under the label.
2. Above 508 g z = (508 − 502) / 4 = 1.50 Table A at 1.50 gives .9332, the area BELOW P(X > 508) = 1 − .9332 = .0668
3. Between 499 g and 506 g z₁ = (499 − 502)/4 = −0.75 → .2266 z₂ = (506 − 502)/4 = 1.00 → .8413 P(499 < X < 506) = .8413 − .2266 = .6147 A "between" question is two lookups and one subtraction, larger area first. Subtracting the z-scores instead of the areas is the standard way to lose this mark.
4. How many of the 240 240 × .3085 = 74.04, so about 74 bags A proportion becomes a count by multiplying, and the answer is rounded to a whole bag because bags are whole.
5. The lightest 1%: read the table backwards Hunt inside the body of Table A for the area nearest .0100. It carries .0099 at z = −2.33 and .0102 at z = −2.32, so −2.33 is the nearer entry. x = μ + zσ = 502 + (−2.33)(4) = 502 − 9.32 = 492.68 g The lightest 1% of bags weigh under about 492.7 g, roughly 7 g short of the label. Software solving for the exact quantile returns z = −2.3263 and x = 492.69 g; on paper the table's entry is the answer, and the gap between the two is a hundredth of a gram.

.3085 under the label, .0668 above 508 g, .6147 between 499 g and 506 g, about 74 of the 240 bags under the label, and the lightest 1% below 492.68 g.

Rounding: Table A carries four decimal places, so the three proportions are exact as read. The 1% cutoff inherits the table's two-decimal z and lands 0.01 g from the exact value.

From §1.5 Z-Scores & The Normal Distribution · every area here is on the printed Table A, and the look-it-up drill below it sets the same three moves at random.

1. Count outcomes, then apply the general addition rule A: (1,6) (2,5) (3,4) (4,3) (5,2) (6,1) → P(A) = 6/36 B: six rolls with a 6 first, six with a 6 second, the double 6 counted twice → 11/36 A and B: (1,6) and (6,1) → 2/36 P(A or B) = 6/36 + 11/36 − 2/36 = 15/36 = 5/12 = .4167 The overlap has to come back out once, because the two outcomes in it were counted inside both totals. Notice that B itself needed the same correction while it was being counted.
2. Why 4 × 1/6 is not a probability Multiplying a probability by the number of trials is not one of the rules. Follow it to six rolls and it returns 6 × 1/6 = 1, certainty, though six rolls plainly can miss; follow it to seven and it returns more than 1. The reasoning is the addition rule applied to events that are not disjoint, since two of the four rolls can both show a 6. P(no 6 in four rolls) = (5/6)⁴ = 625/1296 = .4823 P(at least one 6) = 1 − 625/1296 = 671/1296 = .5177 "At least one" is the signal to work with the complement: one multiplication and one subtraction, instead of adding up four overlapping cases.
3. The second bet P(no double 6 on one roll of two dice) = 35/36 P(no double 6 in 24 rolls) = (35/36)²⁴ = .5086 P(at least one double 6) = 1 − .5086 = .4914 Just under a half, where the first bet was just over. De Méré lost slowly on one and won slowly on the other, and the two probabilities differ by less than three points, which is why it took him a long run of evenings to notice.
4. How many rolls he needed Want (35/36)ⁿ < .5, so n > ln(.5) / ln(35/36) = 24.6 n = 25: 1 − (35/36)²⁵ = .5055 One more roll turns the bet from a slow loss into a slow win. His mistake was a proportional one: four rolls of one die out of six faces felt like it should scale to 24 rolls of two dice out of 36 pairs, and that scaling is not what the complement rule does.

P(A) = 6/36, P(B) = 11/36, P(A and B) = 2/36, P(A or B) = 15/36 = .4167. The first bet pays .5177, not 2/3. The second pays .4914, and would have needed 25 rolls to clear a half.

Rounding: probabilities to 4 dp. The two exact fractions are 671/1296 and 1 − (35/36)²⁴.

From §1.7 Probability Basics · all 36 outcomes are laid out in the dice lab.

1. The tree, with the second layer conditioned on the first Flat then flat: (4/12)(3/11) = 12/132 = 1/11 = .0909 Flat then good: (4/12)(8/11) = 32/132 Good then flat: (8/12)(4/11) = 32/132 Good then good: (8/12)(7/11) = 56/132 = 14/33 = .4242 The four leaves come to 12 + 32 + 32 + 56 = 132 out of 132, which is the check worth doing before going any further. P(at least one flat) = 1 − 14/33 = 19/33 = .5758
2. Two probabilities that look alike and are not P(2nd flat | 1st flat) = 3/11 = .2727 P(2nd flat) = 12/132 + 32/132 = 44/132 = 1/3 = .3333 The unconditional answer is 1/3, exactly what the first draw was. Before you look at the first battery, the second one is just as likely to be flat as the first; once you have seen the first come out flat, the drawer holds one fewer flat battery and the odds shorten.
3. Independence, by inspection Not independent. Taking a battery out changes what is left, so the second draw runs on a different drawer depending on how the first went. You can say that without arithmetic: sampling without replacement from a finite pile is dependent by construction. The numbers confirm it, since .2727 is not .3333.
4. With replacement P(both flat) = (1/3)² = 1/9 = .1111 P(neither flat) = (2/3)² = 4/9 = .4444 P(at least one flat) = 1 − 4/9 = 5/9 = .5556 Both matching outcomes get more likely and the mixed outcome gets less likely, because replacing removes the pull back toward balance: without replacement, drawing a flat one leaves the drawer richer in good ones. Every second-layer branch becomes 4/12 or 8/12 again, which is what independence means here.

Without replacement: P(both flat) = 1/11 = .0909, P(neither) = 14/33 = .4242, P(at least one flat) = 19/33 = .5758. P(2nd flat | 1st flat) = 3/11 against an unconditional 1/3, so the draws are dependent. With replacement the three become 1/9, 4/9 and 5/9.

Rounding: probabilities to 4 dp; every value here is an exact fraction.

From §1.7 Probability Basics.

1. The four conditions A fixed number of trials, n = 8. Two outcomes per trial, right or wrong. The guesses are independent, since one answer tells the student nothing about the next. And the success probability is the same each time, p = 1/4, because every question has four options. All four hold. X ~ B(n = 8, p = 0.25)
2. C(8, 3), then the probability C(8,3) = 8! / (3! 5!) = (8 × 7 × 6) / (3 × 2 × 1) = 336 / 6 = 56 P(X = 3) = 56 × (0.25)³ × (0.75)⁵ = 56 × 0.015625 × 0.2373047 = .2076 Cancel before multiplying: 8 × 7 × 6 over 3 × 2 × 1 is three factors on top and three underneath, never the full 40,320. The three pieces are the number of ways to choose which questions come out right, the probability of getting those three right, and the probability of getting the other five wrong.
3. P(X ≤ 1) is two terms P(X = 0) = (0.75)⁸ = .1001 P(X = 1) = 8 × (0.25) × (0.75)⁷ = 8 × 0.25 × 0.1334839 = .2670 P(X ≤ 1) = .1001 + .2670 = .3671 C(8,0) = 1 and C(8,1) = 8, so neither term needs the formula written out.
4. Center and spread μ = np = 8 × 0.25 = 2 σ = √(np(1 − p)) = √(8 × 0.25 × 0.75) = √1.5 = 1.2247
5. Passing by guessing P(X ≥ 5) = .0230713 + .0038452 + .0003662 + .0000153 = .0273 About 1 in 37. The pass mark of 5 sits (5 − 2)/1.2247 = 2.45 standard deviations above what guessing delivers on average, which is what a test designer is buying with four options rather than two: at two options the mean guess score would be 4 and half the room would pass on luck.

X ~ B(8, 0.25). C(8,3) = 56 and P(X = 3) = .2076. P(X ≤ 1) = .3671. μ = 2 correct, σ = 1.2247. The chance of passing by guessing is .0273, about 1 in 37.

Rounding: probabilities to 4 dp. The exact values are 13608/65536 for P(X = 3) and 6561/65536 for P(X = 0), and σ = √1.5 exactly.

From §1.8 The Binomial Distribution · see the whole distribution at n = 8, p = .25, and its shape beside the others in the distribution playground.

1. One cup: standardize the cup z = (244 − 250) / 15 = −0.40 P = .3446 The denominator is σ itself, because a single pour is what varies.
2. Nine cups: standardize the mean SE = σ/√n = 15/√9 = 5 ml z = (244 − 250) / 5 = −1.20 P = .1151 Same 6 ml shortfall, divided by a smaller denominator, so it counts as a bigger departure. Using σ = 15 here instead of SE = 5 is the classic error, and it returns the answer to part 1.
3. Thirty-six cups SE = 15/√36 = 2.5 ml z = −6 / 2.5 = −2.40 P = .0082
4. Why .3446, .1151 and .0082 Averaging cancels: one stingy cup is unremarkable, nine cups averaging 6 ml low means nine pours leaned the same way at once, and thirty-six doing it is close to evidence that the machine itself has drifted rather than that the day was unlucky. Quadrupling n halves the standard error, which is why the step from 9 to 36 costs more than the step from 1 to 9 bought.
5. Which answer leans on normality Part 1 is about a single observation, so it needs the pours themselves to be normal; there is no averaging to rescue it and no theorem that helps. Parts 2 and 3 are about a sample mean, where the central limit theorem does the work, but n = 9 is small enough to still lean on the population being roughly normal. At n = 36 the sampling distribution of the mean would be near normal even from a moderately skewed machine.

.3446 for one cup, .1151 for the mean of 9, .0082 for the mean of 36. The shortfall is the same 6 ml every time; what shrinks is the standard error, from 15 to 5 to 2.5. Only the first answer depends on the individual pours being normal.

Rounding: areas to 4 dp as Table A carries them. All three z-scores land on exact two-decimal table rows.

From §1.9 Sampling Distributions and §1.10 Central Limit Theorem · the three areas are on the printed Table A.

1. The 95% interval SE = σ/√n = 6/√36 = 1.0 Ω z* = 1.96, from the ∞ row of Table D margin = 1.96 × 1.0 = 1.96 Ω 102.2 ± 1.96 → [100.24, 104.16] 100 is not in that interval, which is already the answer to part 3 in disguise.
2. The test z = (x̄ − μ₀)/SE = (102.2 − 100)/1.0 = 2.20 Table A at 2.20 gives .9861 below, so the upper tail is .0139 Two-sided p = 2 × .0139 = .0278 .0278 < .05, so reject H₀ Two-sided, because the contract lets the buyer complain about resistors that are too high or too low, and that decision was made by the contract rather than by looking at the 102.2.
3. Why they cannot disagree The 95% interval is precisely the set of claimed means that a two-sided test at α = .05 would not reject. Both are asking whether μ₀ sits more than 1.96 standard errors from x̄: the interval measures that distance outward from the sample mean, the test measures the same distance inward from the claim. Here 100 is 2.20 standard errors away, which is past 1.96, so 100 falls outside the interval and the test rejects it. One piece of arithmetic, read two ways.
4. The same delivery at the 1% level z* = 2.576 → 102.2 ± 2.576 → [99.62, 104.78] 100 IS inside this interval p = .0278 > .01, so fail to reject H₀ They agree again, in the other direction. Demanding 99% confidence widens the interval until it swallows the claim, and demanding a 1% significance level raises the bar until the same evidence no longer clears it. Nothing about the resistors changed between those two lines.
5. Which error is live after each decision At α = .05 the buyer rejected, so the only error available is Type I: sending back a delivery that was on spec all along, and paying for a return and a delay over a batch that was fine. At α = .01 the buyer did not reject, so the live error is Type II: accepting a delivery whose mean really is about 2 Ω high, and building it into a product. Tightening α from .05 to .01 does not reduce the chance of being wrong; it trades one way of being wrong for the other, and which trade is right depends on which mistake costs more here.

SE = 1.0 Ω. The 95% interval is [100.24, 104.16] and excludes 100; z = 2.20 with a two-sided p of .0278 rejects at α = .05. At the 1% level the interval [99.62, 104.78] contains 100 and the test fails to reject, because .0278 > .01. Rejecting risks a Type I error, failing to reject risks a Type II.

Rounding: z* taken as 1.96 and 2.576, the values printed under Table D. The exact quantiles are 1.95996 and 2.57583, which shift each endpoint by under a hundredth of an ohm. The tail area .0139 is Table A's entry at z = 2.20.

From §1.11 Confidence Intervals and §1.12 Hypothesis Testing Logic · build the interval in the Interval Builder, prefilled, then run the same numbers through the test explorer, prefilled.

1. Group means and the grand mean Silence: Σ = 20, M = 4 Music: Σ = 30, M = 6 Café: Σ = 40, M = 8 Grand: Σ = 90, N = 15, GM = 6
2. Between-groups sum of squares SS_B = Σ n_j (M_j − GM)² = 5(4−6)² + 5(6−6)² + 5(8−6)² = 5(4) + 5(0) + 5(4) = 40
3. Within-groups sum of squares Silence: (−1)²+(1)²+0²+(−2)²+(2)² = 10 Music: (−2)²+(1)²+0²+(−1)²+(2)² = 10 Café: (1)²+(−2)²+0²+(2)²+(−1)² = 10 SS_W = 10 + 10 + 10 = 30 The three groups happen to have identical spread, which is the homogeneity of variance an ANOVA wants.
4. The table

   Source	SS	df	MS	F
   Between (condition)	40	2	20.00	8.00
   Within (error)	30	12	2.50
   Total	70	14

   df_B = k − 1 = 2, df_W = N − k = 12, and SS_B + SS_W = SS_Total confirms the arithmetic.
5. Decision F(2, 12) = 8.00, F*(.05, 2, 12) = 3.89 8.00 > 3.89 → reject H₀, p = .006
6. Effect size η² = SS_B / SS_Total = 40 / 70 = .571 57% of the variance in errors missed is associated with the sound condition. η² is biased upward in small samples; ω² (omega squared) = .483 is the less flattering estimate.
7. Why not three t-tests Three pairwise tests at α = .05 push the chance of at least one false positive to about 14% rather than 5% (1 − .95³ = .143 if the three were independent; sharing the same data pulls it a little below that, but nowhere near back to .05). The ANOVA asks one question at one α, and the follow-up comparisons then get their own correction (Problem 14).

Sound condition affected proofreading, F(2, 12) = 8.00, p = .006, η² = .57. The omnibus test says the three means are not all equal; it does not say which pairs differ.

From §2.2 One-Way ANOVA.

1. Pairwise standard error: use MS_W, not just the two groups SE = √(MS_W × (1/n₁ + 1/n₂)) = √(2.50 × (1/5 + 1/5)) = √(2.50 × 0.40) = √1.00 = 1.00 Pooling all three groups' error gives df = 12 rather than the 8 a standalone pair would have.
2. The three comparisons

   Comparison	Difference	t	p
   Silence vs music	2	2.00	.069
   Silence vs café	4	4.00	.002
   Music vs café	2	2.00	.069

   Each t = difference / 1.00, evaluated on df = 12
3. Bonferroni α_adjusted = .05 / 3 = .0167 Silence vs café: .002 < .0167 → significant The other two: .069 > .0167 → not significant
4. Tukey's HSD as a cross-check HSD = q*(.05, k = 3, df = 12) × √(MS_W / n) = 3.77 × √(2.50 / 5) = 3.77 × 0.707 = 2.67 Any difference above 2.67 is significant. The 4-point silence-café gap clears it; the two 2-point gaps do not. Both methods reach the same verdict, with Tukey slightly less conservative, as it is built for exactly this set of all-pairs comparisons.
5. What the correction cost Uncorrected, silence vs music at p = .069 would still not have reached .05, so nothing was lost here. In general Bonferroni trades power for protection, and with many comparisons that trade gets expensive. Tukey and Holm both keep the family-wise error rate at .05 while giving some of that power back.

Only silence against café chatter survives correction (difference = 4 errors, t(12) = 4.00, p = .002). Music sits between the two and is not distinguishable from either at α = .05.

From §2.3 Post-Hoc Tests.

1. Degrees of freedom df_caffeine = 2 − 1 = 1 df_sleep = 2 − 1 = 1 df_interaction = 1 × 1 = 1 df_error = N − (number of cells) = 40 − 4 = 36
2. Error mean square MS_error = 36,000 / 36 = 1,000
3. The table

   Source	SS	df	MS	F	p	partial η²
   Caffeine	16,000	1	16,000	16.00	<.001	.308
   Sleep	49,000	1	49,000	49.00	<.001	.576
   Caffeine × Sleep	9,000	1	9,000	9.00	.005	.200
   Error	36,000	36	1,000

   partial η² = SS_effect / (SS_effect + SS_error)
4. Decisions F*(.05, 1, 36) = 4.11 All three Fs exceed 4.11, so both main effects and the interaction are significant.
5. Simple effects of caffeine Rested: 420 − 410 = 10 ms Deprived: 520 − 450 = 70 ms
6. Interpret the interaction Caffeine does something quite different depending on sleep. It buys a rested person 10 ms, which is close to nothing, and a sleep-deprived person 70 ms. Reporting the caffeine main effect on its own ("caffeine improved reaction time by 40 ms") would describe an average that applies to neither group well. With a significant interaction, lead with the simple effects and treat the main effects as background.

Both main effects and the interaction are significant. Sleep deprivation is the larger influence (partial η² = .58), and caffeine helps mainly when you are short of sleep: 70 ms recovered when deprived against 10 ms when rested.

From §2.4 Factorial ANOVA (Two-Way).

1. Degrees of freedom df_conditions = k − 1 = 2 df_error = (n − 1)(k − 1) = 11 × 2 = 22 The error term is smaller than a between-subjects design's would be, because each typist serves as their own control.
2. Mean squares and F MS_conditions = 84.5 / 2 = 42.25 MS_error = 159.2 / 22 = 7.236 F = 42.25 / 7.236 = 5.84 F*(.05, 2, 22) = 3.44 → significant, p = .009
3. What Mauchly's test found Sphericity means the variances of the differences between every pair of conditions are equal. Mauchly's p = .03 rejects that assumption. When sphericity fails, the F ratio is positively biased: the real Type I error rate runs above the nominal 5%, so the uncorrected p of .009 is too generous.
4. Greenhouse–Geisser correction Multiply both df by ε = .71 df_conditions = 2 × .71 = 1.42 df_error = 22 × .71 = 15.62 F itself is unchanged at 5.84; only the reference distribution moves, and it moves against you.
5. Re-decide F(1.42, 15.62) = 5.84, p = .020 Still below .05, so the conclusion survives. The p-value roughly doubled, which is what the correction costs. An uncorrected p near .04 would have flipped.
6. Effect size partial η² = 84.5 / (84.5 + 159.2) = 84.5 / 243.7 = .347

Layout affected typing speed, F(1.42, 15.62) = 5.84, p = .020, partial η² = .35, with Greenhouse–Geisser applied because Mauchly's test rejected sphericity. Report the corrected df with decimals; they are meant to look like that.

From §2.5 Repeated Measures ANOVA.

1. The failed assumption s²_B / s²_A = 47.61 / 9.61 = 4.95 Clinic B's variance is roughly five times Clinic A's. Levene's p = .004 says a variance gap this large would be rare if the population variances were equal, so homogeneity of variance is rejected. Note that Levene's test is itself sensitive to sample size; with unequal n and a visible five-fold gap, the plot and the ratio agree with it here.
2. Normality of what Of the sampling distribution of each group mean, not of the raw session counts. With n = 45 the central limit theorem largely covers Clinic B despite the skew. Clinic A's n = 15 is the more fragile half, and it is the one with the tidier distribution.
3. Welch's t — no pooling SE = √(s₁²/n₁ + s₂²/n₂) s₁²/n₁ = 9.61 / 15 = 0.6407 s₂²/n₂ = 47.61 / 45 = 1.0580 SE = √1.6987 = 1.3033 t = (8.2 − 6.4) / 1.3033 = 1.8 / 1.3033 = 1.381
4. Welch–Satterthwaite degrees of freedom df = (s₁²/n₁ + s₂²/n₂)² / [ (s₁²/n₁)²/(n₁−1) + (s₂²/n₂)²/(n₂−1) ] = (1.6987)² / [ 0.6407²/14 + 1.0580²/44 ] = 2.8855 / [ 0.029318 + 0.025442 ] = 2.8855 / 0.054760 df = 52.69 t*(.025, 52.69) = 2.006 → 1.381 < 2.006, p = .173
5. What pooling would have done The pooled SD is 6.20, dominated by the larger, wider Clinic B. Student's t would give 0.97 on df = 58, p = .334. Both tests fail to reject, but they disagree substantially about how far apart the clinics are, which is why the choice has to be made before looking at which answer you prefer.
6. Welch or Mann–Whitney For Mann–Whitney: session counts are bounded below at zero and badly skewed, and the median is arguably the more meaningful summary for a service report. Against: Mann–Whitney tests stochastic dominance rather than means, and with unequal spreads its interpretation is no cleaner than Welch's. Welch handles unequal variances directly and keeps the outcome on its natural scale. Running one, reporting it, and saying why is better practice than running both and reporting the smaller p.

Homogeneity of variance fails. Welch's t(52.69) = 1.38, p = .173: the clinics do not differ detectably in sessions attended, and Welch is the defensible default whenever the variances are unequal.

From §2.6 ANOVA & t-Test Assumptions · the assumption-checks poster lists the fix for each violation.

1. Pool and rank, smallest first

   Value	9	11	12	14	15	17	18	20	22	31
   Shelter	B	B	A	B	A	B	A	B	A	A
   Rank	1	2	3	4	5	6	7	8	9	10

   No ties, so no shared ranks to average.
2. Rank sums R_A = 3 + 5 + 7 + 9 + 10 = 34 R_B = 1 + 2 + 4 + 6 + 8 = 21 Check: 34 + 21 = 55 = N(N+1)/2 = 10(11)/2 ✓
3. Convert rank sums to U U = R − n(n+1)/2 U_A = 34 − 5(6)/2 = 34 − 15 = 19 U_B = 21 − 15 = 6 Check: U_A + U_B = 25 = n_A × n_B ✓
4. Decision U = min(19, 6) = 6 U*(.05, n₁ = n₂ = 5, two-tailed) = 2 6 > 2 → fail to reject H₀ Mann–Whitney is one of the few tests where you reject when the statistic is small. The exact two-tailed p here is .222.
5. The floor on p Five against five gives C(10,5) = 252 possible arrangements. Exactly one of them puts every Shelter B dog ahead of every Shelter A dog, and one puts the reverse, so the smallest attainable two-tailed p is 2/252 = .0079. The rejection region at α = .05 is U ≤ 2, which is four arrangements in each tail out of 126, so a design this small can only detect something close to complete separation.

U = 6, p = .222: no detectable difference in days to adoption. Shelter B looks faster in every summary you could compute, and with five dogs each the test cannot tell that apart from chance.

From §2.1 Non-Parametric Alternatives.

1. Expected counts E = (row total × column total) / N First year: 120 × 60 / 180 = 40 (all three columns) Final year: 60 × 60 / 180 = 20 (all three columns) All expected counts are 20 or above, comfortably clear of the "at least 5" rule, so no correction or exact test is needed.
2. Cell contributions χ² = Σ (O − E)² / E (30−40)²/40 = 100/40 = 2.500 (45−40)²/40 = 25/40 = 0.625 (twice) (30−20)²/20 = 100/20 = 5.000 (15−20)²/20 = 25/20 = 1.250 (twice)
3. Total and df χ² = 2.500 + 0.625 + 0.625 + 5.000 + 1.250 + 1.250 = 11.25 df = (rows − 1)(columns − 1) = (2 − 1)(3 − 1) = 2
4. Decision χ²*(.05, 2) = 5.99 11.25 > 5.99 → reject H₀, p = .004
5. Effect size Cramér's V = √( χ² / (N × min(r−1, c−1)) ) = √( 11.25 / (180 × 1) ) = √0.0625 = .25 A small-to-moderate association.
6. Where the association lives The single largest contribution is the final-year "alone" cell at 5.00, where 30 students chose solitary revision against 20 expected. Only a quarter of first-years revise alone, and three-quarters work with someone; among finalists it is half and half. Reading the biggest cell contributions is how you turn one significant χ² into a sentence about people.

Revision preference depends on year group, χ²(2, N = 180) = 11.25, p = .004, Cramér's V = .25. Final-years revise alone far more than first-years do.

From §1.16 Chi-Square Tests · check the critical value in Statistical Tables.

1. Means x̄ = 42 / 6 = 7.0 ȳ = 30 / 6 = 5.0
2. Deviations and cross-products

   x	y	x−x̄	y−ȳ	(x−x̄)²	(y−ȳ)²	product
   4	3	−3	−2	9	4	6
   5	5	−2	0	4	0	0
   6	4	−1	−1	1	1	1
   8	5	1	0	1	0	0
   9	7	2	2	4	4	4
   10	6	3	1	9	1	3
   Sums				28	10	14

3. Pearson's r r = Σ(x−x̄)(y−ȳ) / √( Σ(x−x̄)² × Σ(y−ȳ)² ) = 14 / √(28 × 10) = 14 / √280 = 14 / 16.733 r = .837
4. Significance test t = r√(n−2) / √(1 − r²), df = n − 2 = 4 = .837 × √4 / √(1 − .70) = 1.673 / √0.30 = 1.673 / 0.548 = 3.055 t*(.025, 4) = 2.776 → 3.055 > 2.776, reject H₀, p = .038 Significant, but only just, and on six people. A correlation of .84 needs to be enormous to clear the bar at n = 6.
5. r² r² = .837² = .70 70% of the variance in mood is shared with sleep in this sample. "Shared", not "caused by", and in a sample this small the estimate is very unstable.
6. Two non-causal readings Reverse direction: low mood disrupts sleep, which is well documented, so mood could be driving sleep rather than the other way round. Third variable: workload, pain, a new baby, or alcohol would each push sleep down and mood down at the same time, producing the correlation with no arrow between the two measured variables at all. Only random assignment to a sleep condition would separate these.

r(4) = .84, p = .038, r² = .70. Strong and significant in this sample, and entirely consistent with mood driving sleep or with something else driving both.

From §1.17 Correlation · try the correlation calculator.

1. Slope b = Σ(x−x̄)(y−ȳ) / Σ(x−x̄)² = 14 / 28 = 0.50
2. Intercept: the line goes through (x̄, ȳ) a = ȳ − b·x̄ = 5.0 − 0.50(7.0) = 5.0 − 3.5 = 1.50 ŷ = 1.50 + 0.50x
3. Interpret the slope Each extra hour of sleep is associated with half a point more mood on the 1–10 scale. The intercept of 1.50 is the predicted mood at zero hours of sleep, which is outside the data and not meaningful on its own.
4. Prediction and residual at x = 6 ŷ = 1.50 + 0.50(6) = 4.50 observed y = 4 residual = y − ŷ = 4 − 4.50 = −0.50 This person's mood was half a point below what their sleep predicted.
5. All six residuals, as a check −0.5, +1.0, −0.5, −0.5, +1.0, −0.5 Σ residuals = 0 ✓ (always true for a least-squares fit with an intercept) SS_residual = 0.25 + 1 + 0.25 + 0.25 + 1 + 0.25 = 3.00
6. Tie it back to r² SS_total = 10, SS_residual = 3.00, SS_regression = 7.00 r² = 7.00 / 10 = .70 ✓ matches Problem 20 s_e = √(SS_residual / (n − 2)) = √(3/4) = 0.87
7. Why not predict at 2 hours The data run from 4 to 10 hours. Predicting at 2 assumes the straight line keeps going where nobody was measured, and there is no evidence in these six people that it does. Mood is bounded at 1, and the true relationship almost certainly bends at the extremes. Extrapolation is the most common way a correct regression produces a wrong number.

ŷ = 1.50 + 0.50x. At 6 hours the model predicts 4.50, the person scored 4, so the residual is −0.50. The fit explains 70% of the variance in mood over the observed range of 4 to 10 hours, and says nothing about 2.

From §1.18 Simple Linear Regression · fit it in the correlation & regression calculator.

1. The fan Heteroscedasticity: residual variance grows with the fitted value, which is typical for prices. The coefficients stay unbiased, so the slope estimates are still fine on average. What breaks is everything built on the standard errors: the t-tests, the p-values and the confidence intervals are all wrong, usually too narrow at the top end.
2. Leverage Cutoff = 2(k + 1)/n = 2(2 + 1)/40 = 6/40 = 0.15 h = 0.31, about twice the cutoff Case 31 sits far out on the predictors: a very large house, or an odd area-to-bedrooms combination. High leverage alone is not a crime, it only means the case has the opportunity to pull the line.
3. Cook's D Common cutoffs: D > 1, or the stricter 4/n = 4/40 = 0.10 D = 0.68 → below 1, but nearly seven times 4/n High leverage plus a large Cook's D means this case is not merely unusual, it is actively moving the fitted coefficients. Refit without it and see how much the slopes shift.
4. Multicollinearity VIF = 1 / (1 − r²) = 1 / (1 − .92²) = 1 / (1 − .8464) = 1 / .1536 = 6.51 Above the usual warning level of 5. Floor area and bedroom count carry nearly the same information, so the model cannot tell their contributions apart. The symptom to expect is a significant overall F with two unstable, possibly non-significant individual coefficients, one of which may even carry the wrong sign.
5. What to do, in order Investigate case 31 as a data point before deleting anything, since a genuine mansion is information rather than error, and report the fit with and without it if it matters. For the fan, model log(price) instead of price, or keep the model and use heteroscedasticity-robust standard errors. For the VIF, drop one predictor or combine the two into a single size measure, and be explicit that the model now answers a narrower question. Report the diagnostics you ran, including the ones that came back clean.

Three problems, three different fixes: heteroscedasticity invalidates the standard errors, case 31 has both the leverage and the influence to bend the line, and a VIF of 6.51 means the two predictors are nearly the same variable. None of them is a reason to abandon the model, and all three change what the coefficients mean.

From §2.8 Regression Diagnostics and §2.13 Assumptions of Regression.

1. Grand mean Equal group sizes, so it is the mean of the three means x̄ = (12.0 + 15.0 + 18.0) / 3 = 15.0 cm With unequal group sizes this shortcut fails and the grand mean has to be the weighted average Σnᵢx̄ᵢ / N.
2. Between groups SS_G = Σ nᵢ(x̄ᵢ − x̄)² = 8(12 − 15)² + 8(15 − 15)² + 8(18 − 15)² = 8(9) + 8(0) + 8(9) = 144.00
3. Within groups, from the SDs SS_E = Σ (nᵢ − 1)sᵢ² = 7(3.2² + 2.9² + 3.5²) = 7(10.24 + 8.41 + 12.25) = 7(30.90) = 216.30 This is the form to reach for whenever a question hands you standard deviations instead of raw scores. It works because each group's SD is already built on nᵢ − 1, so multiplying by that df undoes the division and returns the group's own sum of squares.
4. Total, and the check SS_T = 144.00 + 216.30 = 360.30 Nothing else was available to compute SS_T from here, since the raw scores are gone. When you do have them, computing all three separately and seeing them add up is the cheapest error check there is.
5. Degrees of freedom and mean squares df_G = I − 1 = 2 df_E = N − I = 24 − 3 = 21 df_T = 23 MS_G = 144.00 / 2 = 72.00 MS_E = 216.30 / 21 = 10.30
6. The F ratio F = 72.00 / 10.30 = 6.99 on (2, 21) 6.99 > 3.47, so reject H₀ at α = .05 (exact p = .005)
7. Effect size and pooled SD η² = 144.00 / 360.30 = .40 s_p = √10.30 = 3.21 cm Two of these numbers do different jobs. η² says the routine accounts for about 40% of the variation in reach. s_p is the error term any follow-up comparison between two routines will use, and it is close to the individual group SDs, which is what you would hope given the ANOVA assumes they are estimating the same thing.

SS_G = 144.00 on 2 df, SS_E = 216.30 on 21, SS_T = 360.30 on 23. MS_G = 72.00, MS_E = 10.30, F(2, 21) = 6.99, p = .005. Warm-up routine affected reach, η² = .40, s_p = 3.21 cm.

From §2.2 One-Way ANOVA. Check it in Build the ANOVA Table or in the APA formatter.

1. The two sums of squares SS_T = SS_G + SS_E, so SS_G = 396.00 − 216.00 = 180.00 The partition identity is doing all the work here. Two of the three sums of squares always determine the third.
2. The missing df df_T = df_G + df_E, so df_E = 39 − 3 = 36 Read the design back out of the df while you are here: df_G = I − 1 = 3 means four groups, and df_T = N − 1 = 39 means forty participants. Both match the scenario, which is a check worth making every time.
3. Mean squares MS_G = 180.00 / 3 = 60.00 MS_E = 216.00 / 36 = 6.00
4. The F ratio and the decision F = 60.00 / 6.00 = 10.00 on (3, 36) 10.00 > 2.87, so reject H₀ (exact p = .00006) A common slip is to divide the sums of squares instead: 180.00 / 216.00 = 0.83, which would have said the opposite. F is a ratio of mean squares, so each sum of squares has to meet its own df first.
5. Effect size and pooled SD η² = 180.00 / 396.00 = .45 s_p = √6.00 = 2.45 The leaflet version accounts for about 45% of the variation in comprehension. What the table cannot say is which versions differ, and the pooled SD is exactly what the follow-up comparisons will need to find out.

SS_G = 180.00, df_E = 36, MS_G = 60.00, MS_E = 6.00, F(3, 36) = 10.00, p < .001. Reject H₀: the four leaflets are not equally well understood, η² = .45, s_p = 2.45.

From §2.2 One-Way ANOVA. The exam mode of Build the ANOVA Table generates more of these.

1. The error term, from the SDs SS_E = Σ (nᵢ − 1)sᵢ² = 9(6.0² + 5.0² + 7.0² + 6.0²) = 9(36 + 25 + 49 + 36) = 9(146) = 1314.00 df_E = N − I = 40 − 4 = 36,  MS_E = 1314.00 / 36 = 36.50 Everything after this borrows MS_E. It is the one number the contrast needs from the ANOVA table, which is why the table is worth building even when the question was never about all four groups.
2. The omnibus F, for the record x̄ = (52 + 58 + 61 + 65) / 4 = 59.00 SS_G = 10[(−7)² + (−1)² + 2² + 6²] = 10(49 + 1 + 4 + 36) = 900.00 MS_G = 900.00 / 3 = 300.00,  F = 300.00 / 36.50 = 8.22 on (3, 36) 8.22 > 2.87, so reject H₀ (exact p = .00027), η² = 900 / 2214 = .41 The equal group sizes are what let the grand mean be the plain average of the four means. A planned contrast does not actually need this test to have been significant, since it was not chosen by looking at the means, but reporting it costs one line and tells a reader the design worked.
3. The coefficients a = [1, −⅓, −⅓, −⅓] Σaᵢ = 1 − ⅓ − ⅓ − ⅓ = 0 ✓ One group on one side, the other three averaged on the other. Coefficients of [1, −1, −1, −1] would sum to −2 and would not be a contrast at all: the three treatment weights have to share the −1 between them, not each take it.
4. Estimate and standard error ψ̂ = 52.00 − (58.00 + 61.00 + 65.00)/3 = 52.00 − 61.33 = −9.33 Σ aᵢ²/nᵢ = (1 + ⅑ + ⅑ + ⅑) / 10 = (4/3) / 10 = .1333 SE = √(36.50 × .1333) = √4.8667 = 2.21
5. The test t = −9.33 / 2.21 = −4.23 on df = N − I = 36 |−4.23| > 2.028, so reject H₀ (exact two-tailed p = .00015) The direction was in the protocol, so a one-tailed test at .05 would have been legitimate here and would halve that p to .000077. It changes nothing about this decision, and the write-up has to say the direction was planned.
6. The interval −9.33 ± 2.028 × 2.21 = −9.33 ± 4.47 95% CI [−13.81, −4.86] percentage points Reading it out loud: workers given no break caught somewhere between about 5 and 14 percentage points fewer of the planted errors than the average of the three break conditions. That sentence is what the study was for, and the t alone cannot produce it.
7. The same contrast, in whole numbers a = [3, −1, −1, −1]:  ψ̂ = 156.00 − 184.00 = −28.00 Σ aᵢ²/nᵢ = (9 + 1 + 1 + 1)/10 = 1.20,  SE = √(36.50 × 1.20) = 6.62 t = −28.00 / 6.62 = −4.23, unchanged. CI [−41.42, −14.58] Everything scaled by three: the estimate, the standard error and both interval limits. The t is a ratio of two of them, so it cannot move. Type the whole numbers into SPSS and report the fractional version, or divide the output by three yourself.

MS_E = 36.50 on 36 df; F(3, 36) = 8.22, p < .001, η² = .41. The planned contrast of no break against the average of the three break conditions gives ψ̂ = −9.33, SE = 2.21, t(36) = −4.23, p < .001, 95% CI [−13.81, −4.86]. Any break beats none, by roughly 5 to 14 percentage points.

From §2.3 Post-Hoc Tests. Type the coefficients into Contrast Builder to check every line of it.

1. The omnibus, and the shared standard error F = 433.33 / 100.00 = 4.33 on (2, 72); 4.33 > 3.12, so reject H₀ (exact p = .017) s_p = √MS_E = √100.00 = 10.00 SE = 10.00 × √(1/25 + 1/25) = 10.00 × √.08 = 2.83 Equal group sizes mean one standard error covers all three pairs. With unequal n each pair would get its own, and only s_p would be shared. η² = 866.67 / 8066.67 = .11, which is a modest effect for a significant omnibus and worth carrying into the write-up.
2. Three t statistics A vs. B: t = (64.0 − 70.0) / 2.83 = −2.12 A vs. C: t = (64.0 − 72.0) / 2.83 = −2.83 B vs. C: t = (70.0 − 72.0) / 2.83 = −0.71 All three on df = N − I = 75 − 3 = 72 Every one of these uses 72 degrees of freedom, not the 48 that two groups of 25 would give on their own. That is the pooled error term earning its keep.
3. Two verdicts each Uncorrected, against t* = 1.993: A vs. B reject, A vs. C reject, B vs. C keep Bonferroni, against t** = 2.451: A vs. B keep, A vs. C reject, B vs. C keep Exact p-values: .037, .006 and .482 The correction costs the A-versus-B comparison, whose t of 2.12 sits between the two critical values. The same decision in the other convention: multiply each p by 3 to get .112, .018 and 1.000, then read them against .05. Two roads, same three answers, and SPSS drives the second.
4. The write-up Revision format affected exam scores, F(2, 72) = 4.33, p = .017, η² = .11. With Bonferroni-corrected pairwise comparisons on the pooled error term, self-testing outperformed rereading, M_diff = 8.00, t(72) = 2.83, p = .018 corrected. Neither the rereading-versus-summarizing difference of 6.00 points, t(72) = 2.12, p = .112 corrected, nor the summarizing-versus-self-testing difference of 2.00 points, t(72) = 0.71, p = 1.000 corrected, reached significance. Two habits that reviewers notice. The word Bonferroni appears, because a bare p-value from a family of comparisons cannot be interpreted without it. And the corrected p-values are labeled as corrected, so nobody has to guess whether .018 was multiplied by three already.

s_p = 10.00, shared SE = 2.83 on 72 df. t = −2.12, −2.83 and −0.71 for A vs. B, A vs. C and B vs. C. Uncorrected, the first two are significant; under Bonferroni only A vs. C survives, since 2.12 falls between t* = 1.993 and t** = 2.451. Corrected p-values .112, .018 and 1.000.

From §2.3 Post-Hoc Tests. The All pairs panel of Contrast Builder lays out both verdicts side by side, and the APA formatter handles the omnibus sentence.

1. Pool, rank, and total Sorted: 12, 15, 17, 18, 20, 22, 23, 25, 26, 28, 30, 31, 33, 36, 40 Workbook ranks: 1, 2, 3, 5, 7, 9, 12 → R₁ = 39 App ranks: 4, 6, 8, 10, 11, 13, 14, 15 → R₂ = 81 Check: 39 + 81 = 120 = N(N + 1)/2 = 15 × 16 / 2 Rank the pooled set, not each group separately. Nothing ties here, so every rank is a whole number; a tie would take the average of the ranks the tied values were competing for. The check on the two sums is worth thirty seconds, because a misranked value is the single commonest way this question goes wrong.
2. Both U values U₁ = n₁n₂ + n₁(n₁ + 1)/2 − R₁ = 56 + 28 − 39 = 45 U₂ = n₁n₂ + n₂(n₂ + 1)/2 − R₂ = 56 + 36 − 81 = 11 Check: 45 + 11 = 56 = n₁n₂ The statistic is U = 11, the smaller of the two Working the second one out from its own rank sum rather than by subtracting from 56 costs one line and checks the ranking again. The smaller U belongs to the app, which is the group sitting higher in the pooled order, and that is what makes it the interesting one.
3. The table, which runs the other way U = 11 against U*(7, 8) = 10 at two-tailed .05 11 > 10, so keep H₀ A U table prints a ceiling. The statistic counts how much the two groups still overlap, so it shrinks as they separate, and only a value at or below the cell counts as significant. Eleven misses by one. The exact two-tailed p is .054, which says the same thing from the other direction and shows how close this was.
4. One tail, and who is allowed to use it One-tailed at .05: U*(7, 8) = 13, and 11 ≤ 13, so reject H₀ Exact one-tailed p = .027 The arithmetic is not in question; the timing is. A direction chosen before the data arrived is a legitimate one-tailed test and this result rejects. A direction chosen after seeing that the app came out ahead is not a test at all, it is the same two-tailed .054 relabeled, and a marker will treat it as such. If your table prints only two-tailed columns, double the α before you look: a directional test at .05 takes its cell from the .10 column, because half of a two-tailed α is reserved for the tail nobody is watching.

R₁ = 39 and R₂ = 81, summing to 120. U₁ = 45, U₂ = 11, so U = 11. Against U* = 10 at two-tailed .05 the null survives, exact p = .054. A one-tailed test predicted in advance would reject against U* = 13, exact p = .027, but only if the direction was stated first.

From §2.1 Non-Parametric Alternatives. The Mann–Whitney tab of the rank ledger prints the same arithmetic for its own data, and Table U is the table this problem reads.

1. Differences, and one that is not a difference after − before: +5, +8, 0, +11, +3, −1, +7, +14, +2 The third participant scored 19 both times, so that pair is dropped n = 8 A pair with a difference of exactly zero has no direction, and the whole test is built on attaching a sign to a rank. There is no honest sign to give it, so it goes, and n becomes the number of pairs left. This is the step most often skipped, and it changes both the ranking and the row of the table you read.
2. Rank the sizes, then sort by sign |d| sorted: 1, 2, 3, 5, 7, 8, 11, 14 → ranks 1 to 8 +5 → 4, +8 → 6, +11 → 7, +3 → 3, −1 → 1, +7 → 5, +14 → 8, +2 → 2 T₊ = 4 + 6 + 7 + 3 + 5 + 8 + 2 = 35 T₋ = 1 Check: 35 + 1 = 36 = n(n + 1)/2 = 8 × 9 / 2 Rank the absolute sizes, ignoring sign, and only then look back at which sign each rank came from. The one person who got worse moved least, so the negative side collects a single rank of 1. Had two differences tied in size they would share the average of their two ranks, and the sums would still have to reach 36.
3. Both decisions W = min(35, 1) = 1 At .05: 1 ≤ 3, so reject H₀ At .01: 1 > 0, so keep H₀ Exact two-tailed p = .016 Like U, W is a ceiling, and for the same reason: it counts the ranks piled on the losing side, so agreement drives it down. The .01 cell of 0 means that with only eight pairs the evidence has to be perfect, every single person moving the same way, before the test can reject at that level. The exact p of .016 sits between the two, which is exactly what the two decisions say.
4. Keeping the zero Two things break at once. The unchanged participant has to be given a signed rank, and there is no sign to give, so whichever side receives it is an invention. And n becomes 9, which sends you to a different row of the table, where the cell is 5 rather than 3. Here the verdict happens to survive both errors, which is the worst kind of luck: the procedure was wrong and the answer came out right anyway. The honest reading of a dropped pair is not that data was discarded but that this participant contributes nothing to a question about direction, and the smaller n reports that loss of information rather than hiding it.

One zero difference is dropped, so n = 8. T₊ = 35 and T₋ = 1, summing to 36. W = 1, which rejects at two-tailed .05 against W* = 3 but not at .01 against W* = 0. Exact two-tailed p = .016.

From §2.1 Non-Parametric Alternatives. The Wilcoxon tab of the rank ledger shows the same ledger for its own seven pairs, and Table W holds both cells.

1. Two columns of ranks Exercise is already in order, so rank x = 1, 2, 3, 4, 5, 6, 7, 8 Wellbeing sorted: 41, 44, 47, 52, 55, 58, 63, 70 rank y in the original order = 4, 1, 2, 3, 7, 5, 8, 6 d = −3, 1, 1, 1, −2, 1, −1, 2 d² = 9, 1, 1, 1, 4, 1, 1, 4 → Σd² = 22 Rank the two columns separately, then subtract row by row. The signs of d do not matter once you square, but they are worth writing down anyway: their sum has to be zero, which catches a misranked row before it reaches the formula.
2. The shortcut ρ_s = 1 − 6Σd² / (n(n² − 1)) = 1 − 6(22) / (8 × 63) = 1 − 132 / 504 = 1 − .2619 = .738 Pearson's r on the raw numbers would give .656. The two heavy exercisers stretch the x axis and flatten the linear fit, while the ranks treat them simply as the sixth, seventh and eighth largest. That gap is the reason for choosing ρ_s here, and it is worth a sentence in any write-up.
3. Two decisions, and one boundary At two-tailed .05: .738 against a cell of .738 The rule is reject when |ρ_s| REACHES the critical value, so reject H₀ At two-tailed .01: .738 < .881, so keep H₀ Exact two-tailed p = .046 Landing exactly on the cell is not a coincidence to be embarrassed about; it is what a discrete statistic does. Σd² = 22 is the very value that defines the .05 cell for n = 8, and the exact two-tailed probability of a result this extreme is .046, comfortably under .05. Had the rule been a strict inequality, this genuinely significant result would have been thrown away. Spearman's table is a floor rather than a ceiling, because ρ_s is a correlation and grows with the strength of the relationship, unlike U and W.
4. What a tie costs The two tied scores share the average of the two ranks they were competing for, and the y column stops being the whole numbers 1 to 8. The shortcut was derived on the assumption that both columns are exactly those whole numbers, so it is no longer algebraically exact and will drift a little from the truth. The fix is to go back to the definition and run Pearson's formula on the two rank columns, which is what Spearman's ρ_s has always been and what SPSS and JASP compute whether or not anything ties. With one tied pair in eight the drift is small, but "small" is not a number you can quote.

Σd² = 22, so ρ_s = 1 − 132/504 = .738. That reaches the two-tailed .05 cell of .738 exactly, so the null is rejected (exact p = .046), but it falls short of the .01 cell of .881. Ties would break the shortcut and force Pearson's formula on the ranks.

From §2.1 Non-Parametric Alternatives and §1.17 Correlation. The Spearman tab of the rank ledger builds the same d² column live, and Table ρ holds both cells.

1. The square, with its margins Row A₁: 84 and 64  →  margin (84 + 64)/2 = 74 Row A₂: 62 and 58  →  margin (62 + 58)/2 = 60 Column B₁: 84 and 62  →  margin (84 + 62)/2 = 73 Column B₂: 64 and 58  →  margin (64 + 58)/2 = 61 Grand mean = (84 + 64 + 62 + 58)/4 = 268/4 = 67 Two arithmetic checks come free. The two row margins average to the grand mean, (74 + 60)/2 = 67, and so do the two column margins, (73 + 61)/2 = 67. Both hold only because every cell has the same ten students in it; with unequal cells a margin is a weighted average and these tidy identities go away.
2. Two main effects Main effect of A = 60 − 74 = −14 minutes (the café costs 14) Main effect of B = 73 − 61 = +12 minutes (the phone away buys 12) Each is a statement about an average over the other factor. The café figure averages over the two phone conditions; the phone figure averages over the two places.
3. Four simple main effects, and their difference A at B₁: 62 − 84 = −22    A at B₂: 58 − 64 = −6 Mean of the two: (−22 + −6)/2 = −14, which is the main effect of A B at A₁: 84 − 64 = +20    B at A₂: 62 − 58 = +4 Mean of the two: (20 + 4)/2 = +12, which is the main effect of B Difference of differences = −6 − (−22) = +16 Difference the other factor first and you get 4 − 20 = −16, the same size with the other sign. There is only one interaction here, and which sign it wears depends on which factor you subtracted first. Both simple main effects of A point the same way and both simple main effects of B point the same way, so this is a spreading interaction rather than a crossover: the phone matters in both places, five times as much in the quiet room.
4. The picture Two points on the left at A₁, 84 for B₁ and 64 for B₂; two on the right at A₂, 62 and 58. The B₁ line falls steeply from 84 to 62; the B₂ line drops gently from 64 to 58. The B₁ line stays above the B₂ line across the whole plot, so the lines do not cross inside the range, and their different slopes are the interaction. A 20-minute gap on the left has narrowed to a 4-minute gap on the right.
5. What the summary gets wrong Twelve minutes is the correct main effect of B, so the number is not wrong. What is misleading is the implied "wherever you work". The phone bought 20 minutes in the quiet room and 4 in the café, and 12 is the average of two numbers, neither of which is 12. With an interaction this size the honest summary reports it first and then gives the two simple main effects: leaving the phone in another room helped in both settings, substantially in the quiet room and barely in the café.

Margins 74 and 60 down the rows, 73 and 61 across the columns, grand mean 67. Main effects −14 for place and +12 for the phone. Simple main effects −22 and −6 for place, +20 and +4 for the phone, each pair averaging to its own main effect; the difference of differences is 16, a spreading interaction. The twelve-minute claim is the right average and the wrong story.

From §2.4 Factorial ANOVA. Type the four cell means into the Interaction Plot Explorer and every number above is printed for you.

1. Reading the design off the df column df_A = I − 1 = 1  →  I = 2 levels of medium df_B = J − 1 = 2  →  J = 3 delays, and (I − 1)(J − 1) = 2 checks out df_T = N − 1 = 47  →  N = 48 students IJ = 6 cells, so n = 48/6 = 8 per cell A two-way table carries its own design in the df column, which is why an exam can hand you nothing else and still expect the sample size back. The same trick works in reverse: quote F(2, 42) in a paper and a reader can reconstruct three levels and 48 people.
2. The error row, then the ratios SS_E = SS_T − SS_A − SS_B − SS_AB = 838 − 96 − 150 − 88 = 504.00 df_E = N − IJ = 48 − 6 = 42, and 1 + 2 + 2 + 42 = 47 as it must MS_E = 504/42 = 12.00 MS_B = 150/2 = 75.00 F_A = 96/12 = 8.00    F_B = 75/12 = 6.25    F_AB = 44/12 = 3.67 All three F ratios divide by the same MS_E, which is the only reason the missing error row had to be recovered first. Its square root, s_p = √12 = 3.46 points, is the residual spread worth quoting beside the means.
3. Three decisions and an effect size Medium: 8.00 > 4.07, so reject (exact p = .007) Delay: 6.25 > 3.22, so reject (exact p = .004) A × B: 3.67 > 3.22, so reject (exact p = .034) η_p² for the interaction = 88 / (88 + 504) = 88/592 = .149 Partial η² puts each effect over itself plus the error rather than over the total, so the three values here (.160, .229 and .149) are not shares of one pie and do not add to anything. The interaction is the smallest of the three and still comfortably past its critical value, because 42 error degrees of freedom buy a good deal of precision.
4. The reading order The interaction line goes first, because it is significant and it changes what the two lines above it are allowed to say. The medium effect of 8.00 and the delay effect of 6.25 are both real, but each is an average over the other factor, and a significant A × B says the medium's advantage is not the same at every delay. The write-up leads with the interaction, then reports each main effect as qualified by it, and then tests the simple main effects that say what actually happened: the medium's effect at each of the three delays, or the delay's effect within each medium. Reporting the two main effects first and mentioning the interaction as an afterthought inverts the logic of the table.

Two levels of medium, three delays, 48 students, 8 per cell. SS_E = 504.00 on 42 df, MS_E = 12.00; MS_B = 75.00; F(1, 42) = 8.00, F(2, 42) = 6.25 and F(2, 42) = 3.67, all three significant at .05, with η_p² = .149 for the interaction. Report the interaction first and treat both main effects as qualified by it.

From §2.4 Factorial ANOVA and §2.2 One-Way ANOVA. The printed F tables hold both criticals, and Where the Variance Goes shows what the error row would have swallowed had the delay factor been left out.

1. Reading a raw coefficient Among students who sleep the same amount, each extra hour of weekly study is associated with 1.30 more exam points. The phrase doing the work is "who sleep the same amount": every coefficient in a multiple regression is the effect of its predictor with the others held fixed. Nothing here licenses the word causes, since nobody was assigned their study hours.
2. Standardizing β = b × (SD of the predictor) / (SD of the outcome) β_study = 1.30 × 4.0 / 12.4 = 5.20 / 12.4 = .419 β_sleep = 3.20 × 1.2 / 12.4 = 3.84 / 12.4 = .310 The journalist has it backwards. Study hours vary far more between students (SD = 4.0 hours) than sleep does (SD = 1.2 hours), so an hour of study is a much smaller step through the actual data than an hour of sleep. Once both are expressed in standard deviations, studying is the stronger predictor. Raw coefficients are only comparable when the predictors share a scale, and hours-per-week and hours-per-night do not.
3. A prediction ŷ = 29.57 + 1.30(10) + 3.20(7) = 29.57 + 13.00 + 22.40 = 64.97 ≈ 65.0 points Both values sit near the middle of their observed ranges, which is the only region where a prediction is safe.
4. Adjusted R² and the intercept adj R² = 1 − (1 − R²)(n − 1)/(n − k − 1) = 1 − (.66)(119/117) = 1 − .671 = .329 ≈ .33 Barely below R², because two predictors is a cheap model at n = 120. The intercept of 29.57 is the predicted score for a student who studies zero hours a week and sleeps zero hours a night. No such student exists in the data, and the model has no information about that corner of the space, so the number is an artifact of where the plane happens to cross the axis.

β_study = .42 against β_sleep = .31, so studying is the stronger predictor and the headline is wrong. Predicted score at 10 study hours and 7 sleep hours is 65.0, adjusted R² = .33.

From §2.9 Multiple Regression · the same distinction shows up in the effect-size converter.

1. The covariate's grand mean X̄ = (52.0 + 45.0) / 2 = 48.5 Equal group sizes, so the simple average is the grand mean. ANCOVA compares the groups at this pretest value.
2. Adjusted means adjusted Ȳ_j = Ȳ_j − b(X̄_j − X̄) program: 61.4 − 0.62(52.0 − 48.5) = 61.4 − 2.17 = 59.23 comparison: 55.0 − 0.62(45.0 − 48.5) = 55.0 + 2.17 = 57.17 The program group is slid down because it started 3.5 points above average, and the comparison group is slid up for the mirror-image reason.
3. The adjusted effect 59.23 − 57.17 = 2.06 points shortcut: adjusted = raw − b × (covariate gap) = 6.4 − 0.62(7.0) = 6.4 − 4.34 = 2.06 Of the 6.4-point advantage, 4.34 points were already there in September. What the term added, at least as far as this analysis can see, is 2.06 points, and F(1, 57) = 1.48, p = .229 says even that is within what chance produces at this sample size.
4. Rewriting the claim "Children in the program finished 6.4 points ahead, but they also began the year 7.0 points ahead. Adjusting for the starting gap leaves a 2.06-point difference that is not statistically distinguishable from zero, F(1, 57) = 1.48, p = .229." Note also that families chose the program, so even the adjusted figure is an observational comparison. Pretest score is the one baseline difference that was measured; motivation, home support and reading habits were not.
5. Homogeneity of regression slopes ANCOVA fits one slope to both groups and reports a single adjusted gap. That gap only exists if the pretest predicts the posttest equally in each group. Test it by adding the group × pretest interaction: if the interaction is significant, the lines are not parallel and there is no single number to report. You would then fit the interaction model and describe how the program's effect changes across the pretest range, which is often the more interesting finding anyway.

Adjusted means 59.23 and 57.17, an adjusted difference of 2.06 points against a raw difference of 6.4. Two thirds of the headline gap was baseline imbalance, and what remains is not significant.

From §2.15 ANCOVA · the parallel-lines assumption belongs to §2.11 Interactions.

1. Odds ratios OR (1 lesson) = e^b₁ = e^0.45 = 1.57 OR (5 lessons) = e^5b₁ = e^2.25 = 9.49 Each lesson multiplies the odds of passing by about 1.57. Five lessons multiply them by 1.57⁵, which is 9.49, not by 5 × 1.57. Odds ratios compound.
2. Probabilities at 6 and 8 lessons P = 1 / (1 + e^{−(b₀ + b₁x)}) x = 6: η = −2.80 + 2.70 = −0.10 → P = 1/(1 + 1.105) = .475 x = 8: η = −2.80 + 3.60 = 0.80 → P = 1/(1 + 0.449) = .690 increase = .690 − .475 = .215
3. Probabilities at 14 and 16 lessons x = 14: η = −2.80 + 6.30 = 3.50 → P = .971 x = 16: η = −2.80 + 7.20 = 4.40 → P = .988 increase = .988 − .971 = .017 The same two-lesson step carries the same odds ratio in both places, e^0.90 = 2.46, and yet the probability moves 21.5 points in the middle of the curve and 1.7 points near the top. This is why a logistic model has no single "effect on probability". The odds ratio is constant everywhere; the probability change depends entirely on where you start.
4. The 50% point P = .50 when the log-odds = 0, so b₀ + b₁x = 0 x = −b₀ / b₁ = 2.80 / 0.45 = 6.22 lessons Just over six lessons is where a learner becomes as likely to pass as to fail. The intercept alone answers a related question: at zero lessons the log-odds are −2.80, so P = .057, a 6% chance of passing untaught.

OR = 1.57 per lesson and 9.49 per five lessons. The same two extra lessons raise the pass probability by .215 around the middle of the curve and by .017 near the top, and the coin-flip point is 6.22 lessons.

From §2.16 Logistic Regression · the same curve reappears as a psychometric function in §3.16.

1. Marginal means immediate = (82 + 68)/2 = 75 delayed = (78 + 52)/2 = 65 easy = (82 + 78)/2 = 80 hard = (68 + 52)/2 = 60 grand mean = (82 + 68 + 78 + 52)/4 = 70 Main effect of feedback = 75 − 65 = 10 points. Main effect of difficulty = 80 − 60 = 20 points.
2. Simple effects feedback effect on easy puzzles: 82 − 78 = 4 points feedback effect on hard puzzles: 68 − 52 = 16 points interaction contrast = 4 − 16 = −12 points The 10-point main effect is the average of a 4 and a 16. It describes neither condition.
3. The F-ratios F = MS_effect / MS_error, and each effect here has df = 1 feedback: F(1, 56) = 1500 / 90 = 16.67, p < .001 difficulty: F(1, 56) = 6000 / 90 = 66.67, p < .001 interaction: F(1, 56) = 540 / 90 = 6.00, p = .017 partial η² = SS_effect / (SS_effect + SS_error), with SS_error = 90 × 56 = 5040 = .23 (feedback), .54 (difficulty), .10 (interaction)
4. Fixing the abstract The sentence is arithmetically true and practically misleading. A significant interaction means the feedback effect is not one number, so quoting the average as though it applied everywhere hides the finding. Something closer to the data: "Feedback timing interacted with difficulty, F(1, 56) = 6.00, p = .017, partial η² = .10. Immediate feedback helped very little on easy puzzles (4 points) and substantially on hard ones (16 points)." When an interaction is significant, describe the simple effects first and let the main effects sit in the table.

Both main effects and the interaction are significant, Fs(1, 56) = 16.67, 66.67 and 6.00. Because the feedback benefit is 4 points on easy puzzles and 16 on hard ones, the abstract's single 10-point claim describes no condition in the study.

From §2.4 Factorial ANOVA and §2.11 Interactions in Regression.

1. The recommendation Within-subjects, analyzed with a one-way repeated-measures ANOVA. Typing speed varies a lot between people and has nothing to do with the interfaces, so in a between-subjects design it lands in the error term and buries the effect. Having each person serve as their own control removes it before the test ever runs.
2. The trade What you gain is power: the same 90 sessions now compare designs within a person rather than across people, and the error term shrinks by whatever the individual differences were worth. What you pay for is order: someone on their third interface is faster because they have solved two puzzles already, and possibly more bored. Practice, fatigue and carryover are all real, and none of them exist in a between-subjects design.
3. Counting the orders 3! = 3 × 2 × 1 = 6 possible orders With 30 participants you can run complete counterbalancing, five people per order, which balances practice effects across designs exactly. A Latin square using three sequences is the cheaper alternative when the number of conditions makes 6 impractical, and it balances first-order carryover rather than everything. Also vary the specific puzzles across designs so nobody sees the same problem twice.
4. Sphericity and ε Sphericity requires the variances of the differences between every pair of conditions to be equal. When it fails, the F-test's degrees of freedom are too generous and the p-value comes out too small. Greenhouse–Geisser multiplies both df by ε to compensate. uncorrected: F(2, 58) corrected: F(2 × 0.72, 58 × 0.72) = F(1.44, 41.76) Report the fractional degrees of freedom and name the correction. An ε of 0.72 is a meaningful departure, so with three conditions this is not optional.
5. When a mixed model wins Repeated-measures ANOVA needs a complete row for every participant, so one missing session throws that person out entirely. A mixed model keeps partial data, handles unequal numbers of trials per person, allows the covariance structure to be modeled rather than assumed, and extends to designs where the repeated factor is continuous. If sessions get canceled or the design grows a second random factor such as stimulus items, switch.

Run it within-subjects with a repeated-measures ANOVA, counterbalance across all 6 orders with 5 participants each, and report the Greenhouse–Geisser corrected test as F(1.44, 41.76).

From §1.5 Between vs Within Designs, §2.5 Repeated Measures ANOVA and §3.10 Mixed Models.

1. The right effect size Three groups compared with a one-way ANOVA means Cohen's f, not d. Cohen's benchmarks for f are .10 small, .25 medium, .40 large. Writing "medium effect" without naming the metric is the first thing to fix, because d = 0.5 and f = 0.25 are different quantities that happen to share the label.
2. Power at the planned n noncentrality λ = f²N = 0.25² × 60 = 3.75 df₁ = k − 1 = 2, df₂ = N − k = 57 power ≈ .37 If the effect she expects is really there, this design finds it about 37% of the time. She is more likely to end the year with an ambiguous null than with an answer, and a null from a study this size tells nobody anything.
3. The sample that works λ = 0.25² × 159 = 9.94, df₁ = 2, df₂ = 156 → power = .805 159 in total, so 53 per group. Set the power calculator to one-way ANOVA, f = 0.25, k = 3, α = .05, power = .80 and it returns the same figure. Two things to say in the write-up: "medium" is the weakest possible justification for an effect size, and the smallest effect that would actually change teaching practice is the better anchor. If prior studies are the source, discount them, because published effects run inflated.
4. Observed power Observed power is computed from the effect size she happens to obtain, which makes it a rearrangement of her own p-value rather than new information. A p just above .05 always yields observed power just under 50%, whatever the design. The question worth answering after a null result is different: what is the smallest effect this study could have detected with 80% power? Answer that from the design, before the data.
5. One more thing to warn her about She mentioned a possible follow-up with a second factor. Interactions typically need considerably more data than main effects of the same nominal size, because the interaction contrast is a difference of differences and its standard error is correspondingly larger. Power the study for the effect she most wants to detect, and if that is the interaction, plan the sample around it.

She needs Cohen's f = 0.25, not d. At 20 per group her power is about .37; reaching .80 takes 159 participants, 53 per group. Observed power after the fact adds nothing.

From §3.2 Power Analysis · run the numbers in Power & Sample Size or sketch the whole design in Plan My Analysis.

1. Tolerance and VIF tolerance = 1 − R²ⱼ VIF = 1 / tolerance

   Predictor	R²ⱼ	Tolerance	VIF
   Age	.923	.077	12.99
   Years of experience	.887	.113	8.85
   Years since qualifying	.861	.139	7.19

2. Reading the thresholds The common alarm level is VIF > 5, with VIF > 10 as the lenient version. All three predictors clear the strict threshold and age clears the lenient one. A tolerance of .077 says that only 7.7% of the variance in age is unique to age; the other 92.3% is already carried by the other two predictors.
3. Why the F and the t-tests disagree They are asking different questions. The F-test asks whether the three predictors together explain performance, and they clearly do: R² = .31. Each t-test asks whether one predictor adds anything once the other two are in the model, and since the three are nearly the same variable, none of them adds much on top of the others. This pattern is the textbook fingerprint of multicollinearity. The negative coefficient on experience is the same problem in a different costume: with predictors this entangled, the coefficient estimates become unstable, their standard errors inflate, and the fitted values can be reached by combinations that make no substantive sense. Do not interpret that minus sign as evidence that experience harms performance.
4. Three fixes and their costs Drop predictors and keep one: simplest, and it changes the question. The surviving coefficient is now the effect of that variable plus everything it stands in for, so name it honestly (say "career stage", not "age"). Combine them into a composite or a principal component: keeps all the information and buys interpretability problems, since a component score is no longer a thing anyone can act on. Regularize with ridge regression: stabilizes the estimates by accepting a little bias, which is the right call when prediction matters more than individual coefficients. There is also a fourth option worth naming, which is to collect data where the predictors come apart, for instance including career changers whose age and experience diverge.
5. Predictions are fine Multicollinearity inflates the variance of the coefficient estimates, not the fitted values. Within the range of predictor combinations actually observed, this model predicts as well as its R² suggests. What it cannot do is tell you which of the three variables is responsible, and that is usually the question that prompted the study.

VIFs of 12.99, 8.85 and 7.19, all above the usual alarm level of 5. The predictors are nearly the same variable, so the model as a whole is significant while no individual coefficient can be trusted, including its sign. Prediction survives; interpretation does not.

From §2.12 Multicollinearity · compare with the diagnostics in §2.8 and Problem 22 above.

1. Dropout by arm intervention: 18 / 90 = 20.0% control: 6 / 90 = 6.7% overall: 24 / 180 = 13.3% Differential dropout is a finding, not a nuisance. Three times as many people left the intervention arm, and the week-4 side-effect reports suggest why. A trial where the treatment drives people out has learned something important about the treatment.
2. Naming the mechanism Not MCAR: missingness is clearly related to arm, to baseline weight and to week-4 side effects, all of which are observed. That is the MAR pattern, and it is the workable case, because the observed variables carry the information needed to correct for it. What you cannot rule out is MNAR: if people quit precisely because the scale was not moving, missingness depends on the unobserved week-12 weight itself, and no amount of cleverness with the observed data settles that. The distinction is untestable from the data. It is an assumption you argue for and then probe.
3. Direction of the bias Complete-case analysis keeps the people who stayed, and in the intervention arm those are disproportionately the ones for whom it was working and tolerable. The 20% who left were heavier at baseline and struggling with side effects. So the −5.8 kg is an estimate of what the treatment does for people who can stick with it, which flatters it relative to what it does for everyone offered it. The control arm loses far fewer people, so the comparison is unbalanced on top of being optimistic.
4. Three fixes, graded Mean imputation is the worst of the three. It inserts values with zero variability, shrinks the standard deviation, weakens every correlation, and leaves the software believing it has a full 180 cases, so the standard errors and p-values come out overconfident. Last observation carried forward is better than it looks in this specific trial, since carrying a week-4 weight forward in a study where weight is falling understates the intervention's benefit and is therefore conservative. It is still an assumption dressed up as data, it does not propagate any uncertainty, and regulators have moved away from it. Multiple imputation is the right tool: build the imputation model from baseline weight, arm, week-4 weight and side-effect reports, generate several completed datasets, analyze each, and pool. The extra uncertainty from the holes then flows honestly into the standard errors. A mixed model fitted to all timepoints reaches the same place under MAR and needs no imputation step at all.
5. What the paper should say How much was missing and from which arm; a comparison of dropouts and completers on every baseline variable available; the mechanism assumed and the reasoning behind it; the primary analysis under MAR, by multiple imputation or a mixed model; and a sensitivity analysis that asks what happens if the dropouts did badly, for instance assuming they returned to baseline weight. If the conclusion survives that, say so. If it does not, that is the result.

20% dropout in the intervention arm against 6.7% in the control arm, related to observed variables, so MAR at best and possibly MNAR. Complete-case analysis keeps the people it worked for and overstates the benefit. Use multiple imputation or a mixed model, and report a sensitivity analysis alongside it.

From §3.14 Missing Data · the reporting standard belongs to §3.4 From Output to Results.

1. Natural frequencies, 100,000 people

   	Test positive	Test negative	Total
   Has the condition	760	40	800
   Does not have it	7,936	91,264	99,200
   Total	8,696	91,304	100,000

   800 × .95 = 760 true positives 99,200 × .08 = 7,936 false positives Counting people instead of manipulating probabilities is the single most reliable way to get this right, and it is how the answer becomes obvious rather than surprising.
2. The answer she wants P(condition | positive) = 760 / 8,696 = .0874 About 9%. Of every 100 positive results the program issues, roughly 91 are false alarms. The test has not failed; it is doing exactly what its specificity promises. The trouble is that 8% of a very large healthy group outnumbers 95% of a very small sick one.
3. A second positive posterior odds = prior odds × likelihood ratio LR⁺ = sensitivity / (1 − specificity) = .95 / .08 = 11.875 prior odds = .008 / .992 = .00806 after one positive: .00806 × 11.875 = .0958 → P = .0958/1.0958 = .0874 ✓ after two positives: .0958 × 11.875 = 1.1372 → P = 1.1372/2.1372 = .532 The first test moved her from 0.8% to 8.7%, and the second from 8.7% to 53%. Yesterday's posterior is today's prior, and the odds form makes the update a single multiplication. The word "independent" is carrying real weight here: if both tests share a systematic quirk, for instance both cross-reacting with the same protein, the second result adds far less than this calculation assumes.
4. A negative result P(clear | negative) = 91,264 / 91,304 = .9996 99.96%. When the condition is rare, a negative result is extremely informative and a positive one barely is. Screening tests are usually built this way on purpose, because the positives get followed up and the negatives get sent home.
5. Why the intuition fails Sensitivity answers "if you are sick, will the test say so?" She is asking the reverse: "the test said so, am I sick?" Those two conditional probabilities are only equal when the base rate is 50%, and here it is 0.8%. Confusing them is base-rate neglect, and it is not a beginner's mistake. Doctors asked this exact question in studies of medical decision-making get it wrong at high rates too, until the numbers are presented as counts of people.

A single positive leaves her at about 9%, two independent positives at 53%, and a negative result clears her with 99.96% probability. Nothing is wrong with the test; the base rate is doing the work.

From §3.7 Bayesian Thinking · the same 2 × 2 is the confusion matrix in §5.4 of the ML course.

1. The update Beta(a, b) prior + k successes in n trials → Beta(a + k, b + n − k) = Beta(8 + 33, 12 + 27) = Beta(41, 39) The Beta is conjugate to the binomial, which is why the update is addition rather than integration. A Beta(a, b) prior is worth a + b − 2 pretend observations, so this one carried the weight of 18 previous users.
2. Summarizing the posterior mean = 41 / 80 = .5125 mode = 40 / 78 = .5128 SD = √[(41 × 39) / (80² × 81)] = √(1599 / 518400) = √.003085 = .0555 Mean and mode agree to three decimals because the posterior is nearly symmetric at these values. They part company when a or b is small.
3. Why .5125 and not .55 The prior mean was 8/20 = .40, and the prior brought 18 pseudo-observations to a dataset of 60. The posterior lands between the two, pulled toward the prior in proportion to how much evidence each side carries. With a flat Beta(1, 1) prior instead, the posterior would be Beta(34, 28) with mean .548, essentially the raw .55. Which of those two answers is right depends on a question the arithmetic cannot settle: if last quarter's users really are informative about this quarter's, the shrinkage is information rather than distortion, and if they are not, that prior should not have been used.
4. P(θ > .50) and the interval z = (.50 − .5125) / .0555 = −.225 P(θ > .50) ≈ 1 − Φ(−.225) = .589 The exact posterior probability is .5889, so the normal approximation is safe here. About a 59% chance the new flow beats a coin flip, which is a much more honest summary than "33 out of 60 is over half". The interval in words: given the prior and these 60 users, there is a 95% probability that the true completion rate lies between .404 and .621. That sentence is legal for a credible interval and illegal for a confidence interval, which is most of the reason people reach for Bayesian methods.
5. What to tell the team The interval spans .404 to .621, so the data do not yet distinguish a mild improvement from a mild regression. 60 users is not enough. Rerun the numbers after a few hundred and the interval will tighten around whatever is true.

Posterior Beta(41, 39): mean .5125, mode .5128, SD .0555. The prior's 18 pseudo-observations pull the estimate below the observed .55, and P(θ > .50) is about .59.

From §3.8 Bayesian Estimation, building on §3.7.

1. Participant A H = 34/40 = .85 FA = 12/40 = .30 d′ = z(H) − z(FA) = 1.036 − (−0.524) = 1.56 c = −[z(H) + z(FA)] / 2 = −[1.036 − 0.524] / 2 = −0.26 A negative c is a liberal criterion: A says "old" more readily than a neutral observer would, which buys hits and pays for them in false alarms.
2. Bias as a likelihood ratio, and the ROC area β = e^c·d′ = e^{(−0.256)(1.561)} = e^−0.400 = 0.67 AUC = Φ(d′/√2) = Φ(1.561 / 1.414) = Φ(1.104) = .865 β below 1 says the same thing as negative c: at A's criterion the evidence is not yet more likely under "old" than under "new", and A commits anyway. The AUC of .865 is the bias-free summary. It would be identical for an observer with A's sensitivity and any criterion whatsoever.
3. Participant C, and the percent-correct trap H = 38/40 = .95 FA = 20/40 = .50 d′ = 1.645 − 0.000 = 1.64 c = −[1.645 + 0]/2 = −0.82 A's percent correct = (34 + 28)/80 = .775 C's percent correct = (38 + 20)/80 = .725 C scores five points worse on percent correct and yet has slightly higher sensitivity, 1.64 against 1.56. The whole difference is strategy: C is far more liberal (c = −0.82) and pays for a near-perfect hit rate with 20 false alarms. Percent correct blends skill and strategy into one number and then hides which one moved. That is the argument for reporting d′ and c separately, and it is why a memory drug that "improves recognition" tells you nothing until you also know how often it produced false alarms.
4. Participant B and the ceiling H = 40/40 = 1.00, and z(1) is infinite, so d′ cannot be computed. The standard repair is the log-linear correction: add 0.5 to every cell count and 1 to every trial total, applied to every participant rather than only to the offending one. H = (40 + 0.5)/(40 + 1) = 40.5/41 = .988 FA = (8 + 0.5)/41 = 8.5/41 = .207 d′ = 2.251 − (−0.816) = 3.07 c = −[2.251 − 0.816]/2 = −0.72 Applying the correction to everyone keeps the estimates comparable; correcting only the extreme participants would bias that person relative to the rest. Name whichever rule you used in the methods section, and treat a d′ above 3 with some caution, because rates this close to the ceiling make the estimate fragile.

A: d′ = 1.56, c = −0.26, β = 0.67, AUC = .865. C: d′ = 1.64 with a much more liberal c = −0.82, which is why C has the higher sensitivity and the lower percent correct. B needs the log-linear correction, giving d′ = 3.07.

From §3.17 Signal Detection Theory · the ROC half of it is §5.5 in ML & AI.

1. PSE and JND quiet: PSE = 3.60 / 0.0090 = 400.0 ms quiet: JND = 1.0986 / 0.0090 = 122.1 ms loud: PSE = 2.94 / 0.0084 = 350.0 ms loud: JND = 1.0986 / 0.0084 = 130.8 ms The PSE is where the fitted curve crosses 50%, so it is the duration that feels exactly midway between the anchors. The JND is half the distance between the 25% and 75% points, which is the curve's steepness expressed in milliseconds.
2. Weber fractions quiet: 122.1 / 400.0 = .305 loud: 130.8 / 350.0 = .374 Dividing precision by the point it was measured at makes the two conditions comparable despite sitting at different places on the duration axis.
3. Checking the JND from the curve 75% point: b₀ + b₁x = ln 3 → x = (1.0986 + 3.60)/0.0090 = 522.1 ms 25% point: b₀ + b₁x = −ln 3 → x = (−1.0986 + 3.60)/0.0090 = 277.9 ms half the span = (522.1 − 277.9)/2 = 122.1 ms ✓ The ln 3 shortcut comes straight from the logistic: at 75% the odds are 3 to 1, so the linear predictor equals ln 3.
4. Shift or slope Mostly a shift. The PSE drops by 50 ms while the JND rises by only 8.7 ms, so the curve slid left without flattening much. Louder probes are judged longer, which is a bias in perceived duration rather than a loss of timing precision. Had the JND doubled with the PSE unmoved, the finding would have been the opposite kind: the same subjective midpoint, measured with a noisier instrument. In a regression this dissociation is two terms. The condition main effect tests the shift, and the condition × duration interaction tests the slope change.
5. 400 rather than 500 500 ms is the arithmetic midpoint of 200 and 800. 400 ms is their geometric mean, √(200 × 800), and bisection points landing near the geometric mean is one of the standard findings in timing research. It suggests the internal scale for duration is compressive, so the subjective middle sits below the arithmetic one. Getting 400 here is a signature, not an error.

Quiet: PSE 400.0 ms, JND 122.1 ms, Weber .305. Loud: PSE 350.0 ms, JND 130.8 ms, Weber .374. A 50 ms leftward shift with almost no change in slope, so louder tones feel longer without being harder to time.

From §3.16 Psychometric Functions · the fitting machinery is §2.16 Logistic Regression.

1. Families and links

   Outcome	Family · link	eb means
   a. Attacks per month	Poisson · log (negative binomial if overdispersed)	Rate ratio
   b. Readmitted yes/no	Binomial · logit	Odds ratio
   c. Cost in euros	Gamma · log	Multiplicative change in mean cost
   d. Recalled out of 20	Binomial · logit, as successes out of 20	Odds ratio per item
   e. Blood pressure	Gaussian · identity (ordinary regression)	Nothing; read b directly

   Two are worth pausing on. Outcome (d) looks like a count but has a known ceiling of 20, which makes it a binomial proportion rather than a Poisson count; modeling the raw proportion with ordinary regression would happily predict recall of 23 items. Outcome (c) is the classic case for Gamma: costs cannot go negative, and their variance grows with their mean, which is exactly the mean-variance relationship Gamma encodes.
2. Interpreting the drug coefficient rate ratio = e^b₁ = e^−0.51 = 0.60 placebo rate = e^b₀ = e^0.92 = 2.51 attacks per month drug rate = e^{b₀ + b₁} = e^0.41 = 1.51 attacks per month check: 2.51 × 0.60 = 1.51 ✓ The drug multiplies the attack rate by 0.60, a reduction of about 40%. Coefficients are additive on the log scale and multiplicative on the count scale, which is the same pattern as odds ratios in logistic regression.
3. Overdispersion deviance / df = 412.6 / 178 = 2.32 A Poisson model assumes variance equals mean, which puts this ratio near 1. At 2.32 the counts vary more than twice as much as Poisson allows, and the consequence is standard errors that are too small and p-values that are too impressive. Migraine counts cluster within people, so this is the expected outcome rather than a surprise. Refit as a negative binomial, which adds its own dispersion parameter, or use quasi-Poisson to rescale the standard errors. The point estimate of the rate ratio will barely move; the confidence interval around it will widen, correctly.

Poisson-log, binomial-logit, Gamma-log, binomial-logit again, and plain Gaussian. The drug cuts the attack rate from 2.51 to 1.51 per month, a rate ratio of 0.60, but a deviance-to-df ratio of 2.32 means the Poisson standard errors are not trustworthy until the model is refitted as negative binomial.

From §3.9 Generalized Linear Models · the logit branch is §2.16.

1. The Kaplan–Meier table S(t) = Π (1 − dⱼ / nⱼ) over all event times up to t

   t	At risk	Events	1 − d/n	S(t)
   3	10	1	.900	.9000
   5+	9	0	censored	.9000
   7	8	1	.875	.7875
   9	7	2	.714	.5625
   12+	5	0	censored	.5625
   14	4	1	.750	.4219
   18+	3	0	censored	.4219
   20	2	1	.500	.2109
   24+	1	0	censored	.2109

   Two things drive the whole table. Censored patients count in the at-risk column right up to the moment they leave, and then simply drop out of it. The at-risk number falls by both events and censorings, which is why only 4 people are at risk at month 14 even though 6 have not yet relapsed on paper.
2. Two readings off the curve S(12) = .5625 S(15) = .4219 Nothing happens at month 12 except a censoring, so the curve is flat across it and S(12) is still the value set at month 9. The staircase only steps down at event times.
3. Median survival The median is the first time at which S(t) drops to .50 or below. At month 9 the curve is at .5625, still above half. At month 14 it reaches .4219. Median survival = 14 months. Read it off the vertical axis at 0.5 and drop to the horizontal one; do not average the observed times.
4. The two bad averages all ten times: (3+5+7+9+9+12+14+18+20+24)/10 = 121/10 = 12.1 months events only: (3+7+9+9+14+20)/6 = 62/6 = 10.3 months The first treats every censored time as though the patient relapsed at that moment, which understates survival, since we only know they lasted at least that long. The second is worse: throwing out censored cases deletes precisely the people who were doing well, so it drops from 12.1 to 10.3 and moves further from the truth. The Kaplan–Meier median of 14 months is above both, because it is the only one of the three that uses the censored patients' partial information instead of destroying it.
5. Why it stops at .2109 The longest observation, at 24 months, is censored. The curve has no information about what happens after that point, so it simply stops rather than pretending everyone eventually relapses. Estimates from the flat tail of a survival curve rest on very few people, which is why a curve that ends with 1 or 2 patients at risk should be read with the confidence band, not the line.

S steps to .9000, .7875, .5625, .4219 and .2109, giving a median survival of 14 months and S(12) = .5625, S(15) = .4219. Both naive averages (12.1 and 10.3 months) understate survival, and the curve ends at .2109 because the last patient was censored.

From §3.13 Survival Analysis.

1. AIC for each model M1: −2(−455.2) + 2(3) = 910.4 + 6 = 916.4 M2: −2(−449.8) + 2(4) = 899.6 + 8 = 907.6 M3: −2(−449.1) + 2(5) = 898.2 + 10 = 908.2 M4: −2(−447.3) + 2(8) = 894.6 + 16 = 910.6 Ranking, best first: M2, M3, M4, M1. Lower is better, and the numbers themselves mean nothing in isolation. Only differences between models fitted to the same data are interpretable.
2. ΔAIC ΔM2 = 0.0 ΔM3 = 0.6 ΔM4 = 3.0 ΔM1 = 8.8 The common reading: Δ (delta, the gap from the best model) under about 2 means the models are effectively indistinguishable, 4 to 7 means considerably less support, and above 10 means essentially none. So M2 and M3 are tied, M4 is trailing but not dead, and M1 is clearly out. The honest summary is that this dataset cannot separate M2 from M3.
3. Akaike weights and the evidence ratio relative likelihood = e^−Δ/2 M1: e^−4.4 = 0.012 M2: e⁰ = 1.000 M3: e^−0.3 = 0.741 M4: e^−1.5 = 0.223 sum = 1.976 w(M2) = 1.000/1.976 = .506 w(M3) = 0.741/1.976 = .375 evidence ratio = .506/.375 = 1.35 = e^0.3 M2 is 1.35 times better supported than M3. That is close to nothing. For comparison, M2 over M1 is e^4.4 = 81 times better supported, which is what a real difference looks like.
4. The nested test χ² = 2(ln L₃ − ln L₂) = 2(−449.1 + 449.8) = 2(0.7) = 1.4 df = 5 − 4 = 1, p = .24 Motivation does not earn its parameter. The likelihood-ratio test and AIC agree, which is what usually happens with one extra parameter, since AIC charges 2 and the χ² critical value at df = 1 is 3.84. They can disagree in either direction as models grow.
5. Why the best fit loses, and what BIC says BIC: M1 924.8 M2 918.7 M3 922.1 M4 932.9 M4's log-likelihood is the highest because adding parameters can never make in-sample fit worse; setting a new coefficient to zero is always available, so the fit can only improve or stay put. AIC charges 2 per parameter for exactly this reason, and M4's four extra parameters bought only 2.5 units of log-likelihood, which is not enough. BIC picks M2 as well and does so more decisively, because its penalty of ln 120 = 4.79 per parameter is more than twice AIC's. Agreement between the two is reassuring; the interesting case is when they disagree, which usually means a moderately sized effect is on the edge of paying for itself. What would actually settle M2 against M3 is not another criterion but new data, or cross-validation on the data at hand.

AIC = 916.4, 907.6, 908.2 and 910.6, so M2 wins with M3 only ΔAIC = 0.6 behind, an evidence ratio of 1.35 that settles nothing. The likelihood-ratio test agrees that motivation is not worth its parameter (χ²(1) = 1.4, p = .24), and BIC picks M2 more decisively.

From §2.14 Model Comparison · the held-out version is §3.11 Cross-Validation.

1. Self-selection Nobody assigned anyone to anything. Students decided for themselves whether to use the module, so the two groups differ in whatever made them decide, and the module is only one candidate explanation for the nine-point gap. The t-test is valid as a description of two groups that already existed. It is the word "raises" that has no support, because a comparison of self-selected groups cannot distinguish an effect of the treatment from a difference between the people who chose it.
2. Three plausible confounds Prior mathematical preparation: students who already find statistics tractable are more likely to engage with extra material and more likely to score well. Conscientiousness: the trait that produces module use also produces lecture attendance, problem sets and revision. Available time: students working long hours in paid jobs have less of everything, including both module use and study time. Each of these causes both the exposure and the outcome, which is the definition of a confound.
3. Does controlling for GPA fix it? It helps and it does not fix it. Adjusting for prior GPA removes the part of the gap explained by measured prior ability, which is real progress, and the adjusted estimate will almost certainly be smaller than nine points. What remains is everything that predicts both usage and outcome but was never measured: motivation this term, hours available, whether a friend recommended the module. Statistical adjustment can only remove confounding you have data on. This is where the language of causal diagrams earns its keep, because drawing the assumed structure makes visible which variables you would need and which you actually have.
4. The study that would settle it Randomize access. Give the module to a random half of the cohort and withhold it from the other half for the term, or stagger the rollout so every student eventually gets it but the order is random, which is usually easier to get past an ethics committee and a students' union. Pre-register the primary outcome and the analysis. Analyze by assigned group rather than by actual usage, because the moment you analyze by who chose to log in, self-selection is back. If randomization is genuinely impossible, a regression-discontinuity or difference-in-differences design around a rollout date beats a naive comparison.
5. An honest headline "Students who chose to use the new module scored nine points higher, though they may have differed before the term began." Longer, duller, and true. The internal version should add that a randomized rollout next year would answer the question properly.

Self-selection. The groups were formed by student choice, so the nine-point gap confounds the module with prior ability, conscientiousness and available time. Controlling for GPA shrinks the estimate without licensing causal language; only random assignment does that.

From §1.4 Experiments & Random Assignment and §1.6 Quasi-Experiments.

1. The frame and the volunteers Two filters run before anyone answers. Coverage: only people who visit the council website or follow its social accounts ever see the link, which excludes residents without reliable internet, those who do not follow local government, and anyone whose news comes through other channels. Self-selection: of the people who saw it, the ones who answered are those who cared enough to click, and strong feelings on both sides drive response far more than mild ones. Organized campaigns on either side can mobilize supporters through a single post. The realized sample is people who saw the link and chose to respond, which is not the adult population of the city.
2. Why 4,300 does not help Sample size fixes variance, not bias. A convenience sample's estimate is centered in the wrong place, and collecting more of the same kind of response tightens the cloud around that wrong center. 4,300 self-selected responses give a more precise wrong answer than 430 would. The only cure for bias is a better sampling method.
3. The margin of error The ±1.5% figure is the most misleading number in the report. That formula assumes a probability sample in which every resident had a known, non-zero chance of selection, and here nobody was selected at all; people arrived. Quoting it attaches the authority of survey statistics to a number that has none of its properties. It describes sampling error alone, and sampling error is not this survey's problem.
4. One more threat Nothing stops one person responding repeatedly, or a campaign asking supporters to. An open link with no authentication has no way to enforce one response per resident, and no way to confirm that respondents live in the city at all.
5. What to do, and what the survey is worth For an estimate: draw a probability sample from the electoral register, stratify by neighborhood and age band so every part of the city is represented, follow up non-responders more than once, and report the response rate alongside the result. That design supports both the percentage and its margin of error. The open survey is not worthless, though. As a consultation it surfaces arguments, local knowledge and objections the council would not otherwise hear. It should be reported as what it is: 4,300 people chose to tell us this, with no claim about the other 255,700.

A self-selected sample drawn from an incomplete frame. The 78% describes people who saw the link and chose to respond, the sample size makes that no more trustworthy, and the ±1.5% margin of error is meaningless because the formula assumes probability sampling that never happened.

From §1.8 Sampling Methods and §1.9 Designing Surveys.

1. Post-hoc exclusion "After inspecting the data we excluded 7 participants" is an outlier rule chosen after seeing what it would do, and a 2 SD cutoff is lenient enough to remove roughly 5% of a normal distribution by construction. Calling the excluded responses "clearly unrepresentative" is a judgment made with knowledge of the result. Fix: specify the exclusion rule before data collection, apply it blind to condition, and report the analysis both with and without exclusions.
2. Optional stopping "Because the effect was not significant, we collected a further 30 participants" is the textbook definition of data-dependent stopping. Every additional look is another chance for the p-value to wander below .05, and this practice alone can push a nominal 5% false-positive rate toward 20% or more. Fix: fix the sample size in advance from a power analysis, or use a pre-planned sequential design with corrected thresholds, which is a legitimate and sometimes required approach. What is not legitimate is deciding after each look.
3. Selective reporting of outcomes Four measures were collected and one is reported. The reader has no way to know whether reaction time was the planned primary outcome or the one that survived four attempts. Fix: report every measure collected, nominate a primary outcome in advance, and correct for multiplicity across the rest or label them exploratory.
4. HARKing "As predicted" cannot be assessed by anyone, and given the three practices above there is good reason to doubt it. Presenting a finding located by exploration as a prediction converts a hypothesis-generating result into apparent confirmation. Fix: separate confirmatory from exploratory sections explicitly, and let a preregistration carry the claim about what was predicted.
5. The degrees of freedom 60 recruited − 7 excluded + 30 added = 83 participants two groups, so df_error = 83 − 2 = 81 ✓ The arithmetic is fine. Every number in the paragraph is internally consistent, which is exactly why this manuscript would pass a statistical-consistency screen and still be untrustworthy. The problem is the path taken to those numbers, and no recomputation can detect it.
6. Why this outranks fraud Fabrication is rare, with under 1% of surveyed researchers admitting it, and it is unambiguous when caught. These practices are common: in a survey of more than 2,000 psychologists, around 63% admitted failing to report all dependent measures and about 56% admitted deciding whether to collect more data after checking significance. They also feel like reasonable judgment calls at the time, which is why they persist. Rare deliberate fraud damages a few papers; widespread gray-zone flexibility fills the literature with effects that evaporate on replication, and that is the mechanism behind the replication crisis.

Post-hoc outlier exclusion, optional stopping, selective reporting of one outcome out of four, and HARKing. The degrees of freedom check out perfectly, which is the point: consistent numbers are no defense against a flexible analysis path.

From §4.5 Questionable Research Practices, §1.11 The Replication Crisis and §1.12 Preregistration.

1. Step-by-step verdict

   Step	Verdict	Why
   a. Listwise deletion	Defensible, needs justification	Fine if the data are plausibly MCAR and the loss is small, but it must be reported and it contradicts step (b)
   b. Blanks → 0	Error	0 is not a valid response on a 1–7 scale
   c. 999 left in	Error	999 is a missing-value code, not an age
   d. Reverse by ×(−1)	Error	Produces −1 to −7, destroying the scale
   e. Dropped lowest scorers	Serious error	Selection on the outcome variable
   f. Merged on initials	Error	Collides, breaks on name changes, and is identifying
   g. Overwrote the raw file	Serious error	The original data no longer exists

2. Fixing (b) and (d) For (b), leave missing values missing and decide an item-level rule instead: compute the scale when at least 80% of its items were answered, prorating from the items that were, and set it to missing otherwise. Filling with 0 invents the most extreme response available and drags every affected participant's total toward the floor. correct reverse-scoring: x' = (min + max) − x = 8 − x so 1 → 7, 2 → 6, 4 → 4, 7 → 1 Multiplying by −1 gives a range of −1 to −7, which is not the reversed scale, and it corrupts every scale total that includes those three items without producing a single value that looks obviously wrong. Recompute the reversed items from the raw file, which is a problem given step (g).
3. The true mean age total of all 240 ages = 240 × 33.2 = 7,968 the three sentinels contribute 3 × 999 = 2,997 remaining total = 7,968 − 2,997 = 4,971 true mean = 4,971 / 237 = 21.0 years Three cells inflated the reported mean by 12.2 years in a sample of 240. This is what missing-value sentinels do when a pipeline treats them as data, and it is why the first step of any cleaning workflow is to check every variable's minimum and maximum against its plausible range.
4. The most serious step Step (e), with step (g) close behind. Removing the six lowest wellbeing scores is not cleaning; it is deleting cases on the basis of the outcome variable, which changes the mean by construction and does so in a direction the researcher can see while doing it. If non-serious responding is a genuine concern, screen for it with something independent of the outcome: attention-check items, implausibly fast completion times, straight-lining down a column. Step (g) is a different kind of damage. Raw data should be read-only and every transformation should live in a script, so that the cleaned file can always be regenerated and any mistake in this log can be undone. As things stand, none of these errors can be reversed.

Five clear errors: 0 for missing on a 1–7 scale, a 999 sentinel treated as an age (which inflated the mean from 21.0 to 33.2), reverse-scoring by −1 instead of 8 − x, dropping the lowest scorers, and merging on initials. Overwriting the raw file is what makes all of them permanent.

From §2.4 The Cleaning Workflow, §2.5 Outliers and §2.9 Reproducible Workflows · the worked version is the survey-cleaning guide.

1. The catalog of problems The truncated axis is the big one: starting at 7.2 instead of 0 turns a 0.27-point gap into a bar nearly three times the height of its neighbor, since A occupies 25% of the plot area and B occupies 70%. Bars encode value by length, so a bar chart with a truncated axis is misreporting its own data. Beyond that: no uncertainty is shown, so a reader cannot tell a solid difference from noise; the 3-D perspective distorts the apparent heights and adds nothing; two bars summarize 79 people and hide every distribution; and the caption asserts a conclusion ("clearly outperforms") that the figure does not establish and, as question 3 shows, is wrong.
2. Where the 27% came from 7.62 − 7.35 = 0.27 points relative change = 0.27 / 7.35 = 3.7% Someone read the difference of 0.27 as "27%". The actual relative increase in the mean is 3.7%, and even that is a shaky thing to report, because a 1–10 satisfaction scale has no meaningful zero, so percentage changes on it depend on an arbitrary origin. The honest statement is the difference itself: Layout B scored 0.27 points higher on a 10-point scale.
3. Is it distinguishable from zero? s²_pooled = [40(1.42²) + 37(1.38²)] / 77 = (80.66 + 70.46)/77 = 1.963 s_pooled = 1.401 SE = 1.401 × √(1/41 + 1/38) = 1.401 × .2252 = 0.3155 t(77) = 0.27 / 0.3155 = 0.86, p = .39 d = 0.27 / 1.401 = 0.19 No. A difference this size arises about 39% of the time when the two layouts are identical. The figure's entire visual argument rests on a difference the data cannot distinguish from nothing, and the two-million rollout has no evidence behind it. With d = 0.19, detecting an effect this small with 80% power would need over 400 stores per layout rather than 40.
4. The figure to send instead Plot the individual store scores as two columns of points, jittered so they do not overlap, with the group mean marked and a 95% confidence interval drawn around it. Run the y-axis across the full 1 to 10 scale, or across a clearly labeled subrange with the truncation stated in the caption. Drop the third dimension. Make the caption descriptive: "Figure 2. Mean customer satisfaction by store layout, with 95% confidence intervals. Individual stores shown as points; n = 41 and 38." Then let the overlapping intervals say what they say. A figure whose job is to persuade is not a figure, and if the recommendation is to keep testing, the figure should make that obvious at a glance.

Truncated axis, no error bars, gratuitous 3-D, a conclusion in the caption, and a "+27%" that is really a 0.27-point difference (3.7% in relative terms). The difference is not distinguishable from zero, t(77) = 0.86, p = .39, d = 0.19, so the rollout rests on nothing.

From §3.3 Tables & Figures That Don't Lie · check the size of the effect in the effect-size converter.

1. Error 1: a leading zero on p written: p = 0.062 correct: p = .062 Drop the leading zero from any statistic that cannot exceed 1 in magnitude: p, r, R², η², β and Cramér's V. A probability of 0.062 is never going to be 1.062, so the zero carries no information.
2. Error 2: a missing zero on d written: d = .49 correct: d = 0.49 The mirror image of the first rule, and the one people miss because they over-apply the first. Cohen's d can comfortably exceed 1, so it keeps its leading zero. The same goes for t, F, M and SD. The test is not "is it a decimal" but "can this quantity exceed 1".
3. Error 3: the wrong df on the correlation written: r(60) = .34 correct: r(58) = .34 The number in parentheses after r is degrees of freedom, not sample size, and for a correlation that is n − 2. With 60 participants the value is 58. Writing r(60) tells a careful reader you had 62 people, which contradicts the methods section. The t-test in the same sentence has it right, and the two matching by coincidence here is a good reminder to check rather than pattern-match.
4. The corrected sentence "Participants receiving immediate feedback (M = 78.4, SD = 8.6) scored higher on accuracy than those receiving delayed feedback (M = 74.1, SD = 8.9), though the difference was not statistically significant, t(58) = 1.90, p = .062, d = 0.49. Accuracy correlated with practice time, r(58) = .34, p = .008."
5. Checking the numbers SE = √[(8.6² + 8.9²)/30] = √(153.17/30) = 2.26 t = (78.4 − 74.1)/2.26 = 1.90 ✓, two-tailed p = .062 ✓ s_pooled = √[(8.6² + 8.9²)/2] = 8.75, so d = 4.3/8.75 = 0.49 ✓ for r = .34 with df = 58: t = .34√(58/(1 − .34²)) = 2.75, p = .008 ✓ Everything reconciles, which is what the consistency check in the APA formatter does automatically. Formatting errors and arithmetic errors are separate species, and a sentence can have either without the other.
6. What the supervisor circled "Were more accurate" states a difference at p = .062. That is not a formatting error; it is a claim the test does not support. The correction above replaces it with a description of the observed means plus an explicit statement that the difference was not significant, which is the honest form. Resist both temptations here: do not write "approached significance", and do not pretend the 4.3-point gap was not observed. Report the estimate, report that it is compatible with no difference at this sample size, and let the confidence interval carry the uncertainty.

Leading zero on p (should be .062), missing leading zero on d (should be 0.49), and r(60) where the degrees of freedom are n − 2 = 58. The numbers themselves all reconcile; the fourth problem, calling p = .062 a difference, is a writing error rather than a formatting one.

From §3.2 Reporting Statistics in APA Style and §3.5 Non-Significant Results · the rules are on the APA cheat sheet and applied by the APA formatter.