Paired t test practice: 8 problems with solutions

Ask the same question of every design: is each observation in one group tied to one specific observation in the other? If yes, the data are paired and you work with differences. If no, the groups are independent and you work with two samples.

(a) Paired. Each runner supplies both times, so runner 7's two times belong together and their difference is meaningful. Randomizing the order controls for fatigue and learning; it does not make the two times independent.

(b) Two-sample. Nine plants got A and nine different plants got B. Equal group sizes are a coincidence here, not evidence of pairing, because no plant in group A is linked to a particular plant in group B.

(c) Paired. This is a matched-pairs design: the twins are matched on genetics and upbringing, and the treatment is randomly assigned within each pair. There are 12 differences, one per twin pair, so $df = 11$ rather than 23.

(d) Two-sample. Different students, different grades, and unequal sizes 40 and 35. Unequal sizes alone settle it, since a paired procedure needs one difference per pair and therefore equal counts.

Build the differences. Using $d = \text{after} - \text{before}$, the six differences are $5, 3, 8, 2, 6, 0$ records per hour. From here the before and after columns are never used again.

State. Let $\mu_d$ be the true mean change in data-entry speed (after minus before) for workers who take this workshop. $H_0: \mu_d = 0$ versus $H_a: \mu_d > 0$, with $\alpha = 0.05$.

Plan. Paired $t$ test, which is a one-sample $t$ test on the differences. Random: the 6 workers were randomly selected. 10%: 6 pairs is less than 10% of all workers who could take the workshop. Normal/large sample: $n = 6 < 30$, so assume the differences come from a roughly normal population; the six values show no strong skew and no outlier.

Find the mean difference. $\bar{d} = \frac{5 + 3 + 8 + 2 + 6 + 0}{6} = \frac{24}{6} = 4$ records per hour.

Find the standard deviation of the differences. The squared deviations from 4 are $1, 1, 16, 4, 4, 16$, summing to $42$, so $s_d = \sqrt{\frac{42}{6 - 1}} = \sqrt{8.4} = 2.8983$.

Find the standard error and the test statistic. $SE = \frac{s_d}{\sqrt{n}} = \frac{2.8983}{\sqrt{6}} = \frac{2.8983}{2.4495} = 1.1832$, so $t = \frac{4}{1.1832} = 3.38$ with $df = 6 - 1 = 5$.

Find the p-value. Right-tailed. In the $df = 5$ row, $3.38$ falls between $3.365$ (tail $0.01$) and $4.032$ (tail $0.005$), so the p-value is between $0.005$ and $0.01$. Technology gives $0.0098$.

Conclude. The p-value is below $0.05$, so reject $H_0$. There is convincing evidence that mean data-entry speed is higher after the workshop than before it. Because there is no control group, the increase cannot be separated from practice effects or from anything else that changed over the same period.

State. Let $\mu_d$ be the true mean difference in peak impact force (usual insole minus new insole), in newtons, for runners like these. Estimate $\mu_d$ with a 95% confidence interval.

Plan. Paired $t$ interval, which is a one-sample $t$ interval on the differences. Random: the 10 runners are a random sample, and the order of the two insoles was randomized within each runner. 10%: 10 pairs is under 10% of the runners the company could test. Normal/large sample: $n = 10 < 30$, so assume the differences are roughly normal with no strong skew or outliers.

Find the standard error. $SE = \frac{s_d}{\sqrt{n}} = \frac{21}{\sqrt{10}} = \frac{21}{3.1623} = 6.6408$ N.

Find the critical value and margin of error. $df = 10 - 1 = 9$, and the 95% column of the $t$-table gives $t^* = 2.262$. Then $ME = 2.262 \times 6.6408 = 15.02$ N.

Build the interval. $\bar{d} \pm ME = 18 \pm 15.02 = (2.98,\ 33.02)$ N.

Interpret. We are 95% confident that the interval from $2.98$ N to $33.02$ N captures the true mean reduction in peak impact force when runners switch from their usual insole to the new one.

Answer the implied question. The whole interval lies above 0, so a true mean difference of zero is not plausible at 95% confidence: the new insole does reduce mean peak impact force. The interval is wide because $n$ is only 10, so the size of the reduction is poorly pinned down, anywhere from a token 3 N to a substantial 33 N.

(a) Build the differences. With $d = \text{after} - \text{before}$, the eight differences are $3, 3, 5, 2, 3, 4, 1, 5$ cm. Every one is positive, which is the first sign that the pairing is carrying real information.

(a) State and plan. Let $\mu_d$ be the true mean increase in vertical jump. $H_0: \mu_d = 0$ versus $H_a: \mu_d > 0$ at $\alpha = 0.05$. Paired $t$ test on the 8 differences, with $df = 7$. The athletes are a convenience group rather than a random sample, so treat the conclusion as applying to these athletes; $n = 8 < 30$, and the differences show no strong skew and no outliers.

(a) Compute. $\bar{d} = \frac{3 + 3 + 5 + 2 + 3 + 4 + 1 + 5}{8} = \frac{26}{8} = 3.25$ cm. The squared deviations from 3.25 sum to $13.5$, so $s_d = \sqrt{\frac{13.5}{7}} = \sqrt{1.9286} = 1.3887$ cm. Then $SE = \frac{1.3887}{\sqrt{8}} = \frac{1.3887}{2.8284} = 0.4910$ and $t = \frac{3.25}{0.4910} = 6.62$.

(a) Conclude. With $df = 7$, a $t$ of $6.62$ lies far beyond every critical value the $t$-table prints in that row, so the p-value is smaller than the smallest tail probability the table carries; technology gives $0.00015$. Reject $H_0$: there is very strong evidence that mean vertical jump increased.

(b) The incorrect analysis, computed anyway. Treating the columns as independent samples: the after column has $\bar{x}_1 = 56.75$ cm with $s_1 = 4.0267$, and the before column has $\bar{x}_2 = 53.5$ cm with $s_2 = 5.0990$. Then $\frac{s_1^2}{n_1} = \frac{16.2143}{8} = 2.0268$ and $\frac{s_2^2}{n_2} = \frac{26}{8} = 3.25$, so $SE = \sqrt{2.0268 + 3.25} = \sqrt{5.2768} = 2.2971$ cm.

(b) Its test statistic and p-value. $t = \frac{56.75 - 53.5}{2.2971} = \frac{3.25}{2.2971} = 1.41$, with conservative $df = \min(8, 8) - 1 = 7$, giving a one-sided p-value of $0.100$. That is greater than $0.05$, so the incorrect analysis fails to reject $H_0$ and finds no convincing evidence of an increase. Even the more generous Welch value $df = 13.29$ only lowers it to $0.090$, so the decision does not change.

(c) Why they differ. Both analyses have the same numerator: the estimated increase is $3.25$ cm either way. The denominators are what separate them. The paired standard error, $0.4910$, is built from how much the individual changes vary, and the changes range only from 1 to 5 cm. The two-sample standard error, $2.2971$, is built from how much athletes differ from each other, and jump heights range from 46 to 63 cm. Ignoring the pairing throws the between-athlete variation into the error term, inflating the standard error by a factor of about $\frac{2.2971}{0.4910} = 4.7$ and shrinking $t$ by the same factor.

(c) Which is right. The paired analysis. The two columns are not independent, since athlete 3's before and after values are both driven by athlete 3, so the two-sample procedure's independence condition is violated and its p-value is not trustworthy. The practical cost is a Type II error: a real effect that the paired test finds overwhelming evidence for, the two-sample test misses entirely.

(a) State. Let $\mu_d$ be the true mean drop in cholesterol (before minus after), in mg/dL. A diet that lowers cholesterol makes the before value larger, so $H_0: \mu_d = 0$ versus $H_a: \mu_d > 0$, with $\alpha = 0.05$.

(a) Plan. Paired $t$ test on the 12 differences. Random: the 12 patients are a random sample. 10%: 12 pairs is under 10% of the patients this diet could be given to. Normal/large sample: $n = 12 < 30$, so assume the differences are roughly normal with no strong skew or outliers.

(a) Compute. $SE = \frac{s_d}{\sqrt{n}} = \frac{22.0}{\sqrt{12}} = \frac{22.0}{3.4641} = 6.3509$, so $t = \frac{14.5}{6.3509} = 2.28$ with $df = 12 - 1 = 11$.

(a) p-value and conclusion. Right-tailed. In the $df = 11$ row, $2.28$ falls between $2.201$ (tail $0.025$) and $2.718$ (tail $0.01$), so the p-value is between $0.01$ and $0.025$; technology gives $0.0216$. That is below $0.05$, so reject $H_0$. There is convincing evidence that mean total cholesterol dropped over the eight weeks. With no control group, the drop cannot be attributed to the diet rather than to the passage of eight weeks.

(b) What changes. Every difference flips sign, so $\bar{d}$ becomes $-14.5$ mg/dL and $t$ becomes $-2.28$. The alternative must flip with it: a drop in cholesterol now means a negative difference, so the analyst writes $H_a: \mu_d < 0$ and takes a left tail.

(b) What does not change. $s_d$ is unchanged at $22.0$, since flipping every sign shifts the mean but not the spread. So $SE = 6.3509$, $df = 11$, and $|t| = 2.28$ are all identical, and the left-tail p-value is the same $0.0216$ as the right-tail p-value was. The conclusion is word for word the same.

(b) The interval. Using $t^* = 2.201$ at $df = 11$, $ME = 2.201 \times 6.3509 = 13.98$ mg/dL. The first analyst reports $14.5 \pm 13.98 = (0.52,\ 28.48)$; the second reports $(-28.48,\ -0.52)$. Same width, same distance from zero, mirrored across it. Both exclude 0 and both say the same thing about cholesterol; only the direction the sign encodes has changed, which is why a response must always state which way it subtracted.

(a) State. Let $\mu_d$ be the true mean difference in final score (simulator minus road) for students like these. $H_0: \mu_d = 0$ versus $H_a: \mu_d \ne 0$, with $\alpha = 0.05$. The alternative is two-sided because the school is asking whether the types differ, not whether one is better.

(a) Plan. Paired $t$ test on the 15 differences, one per matched pair. Random: the lesson type was randomly assigned within each pair, which is the form the random condition takes in a matched-pairs experiment. The 10% condition does not apply, because the pairs are not a sample drawn from a larger population of pairs. Normal/large sample: $n = 15 < 30$, so assume the differences are roughly normal with no strong skew or outliers.

(a) Compute. $SE = \frac{5.1}{\sqrt{15}} = \frac{5.1}{3.8730} = 1.3168$ points, so $t = \frac{2.4}{1.3168} = 1.82$ with $df = 15 - 1 = 14$.

(a) p-value. Two-sided, so use $|t| = 1.82$. In the $df = 14$ row it falls between $1.761$ (tail $0.05$) and $2.145$ (tail $0.025$), so one tail holds between $0.025$ and $0.05$ and the two-sided p-value is between $0.05$ and $0.10$. Technology gives $0.0898$.

(a) Conclude. The p-value $0.0898$ exceeds $\alpha = 0.05$, so fail to reject $H_0$. There is not convincing evidence that the two lesson types differ in true mean final score. Note the wording: this is a failure to find a difference, not evidence that none exists, and with only 15 pairs the test has limited power to detect a difference of a couple of points.

(b) Scope of inference. Lesson type was randomly assigned within each pair, so if a difference had been found it could have been attributed to the lesson type rather than to student ability, which the matching already controlled. But the 15 pairs were not randomly selected from any larger population, so no conclusion generalizes beyond students like these. Randomization supports causation; random selection supports generalization, and only the first is present here.

(a) Hypotheses. Let $\mu_d$ be the true mean change in score (post minus pre). The output is one-sided with a positive $t$, so it tests $H_0: \mu_d = 0$ against $H_a: \mu_d > 0$, with $df = 20 - 1 = 19$. Confirm the printout internally: $SE = \frac{8.4}{\sqrt{20}} = \frac{8.4}{4.4721} = 1.878$ and $t = \frac{3.6}{1.878} = 1.917$, both matching.

(b) Interpret the p-value. If the course truly had no effect on mean score, so that $\mu_d = 0$, then about 3.5% of random samples of 20 participants would produce a mean improvement of $3.6$ points or more. The probability is computed assuming $H_0$ is true, and it describes results at least as extreme as the one observed.

(c) What is wrong with the student's sentence. A p-value is not the probability that a hypothesis is true. It reverses the conditioning: the correct statement is the probability of data this extreme given that $H_0$ holds, while the student states the probability that $H_0$ holds given the data. In this framework $\mu_d$ is a fixed unknown number, not a random one, so 'the probability the course has no effect' has no value to report. The sentence also misreads 'no effect' as an event with a probability rather than as an assumption the calculation was made under.

(d) Build the interval. $df = 19$, so the 95% column gives $t^* = 2.093$, and $ME = 2.093 \times 1.878 = 3.931$ points. The interval is $3.6 \pm 3.931 = (-0.331,\ 7.531)$ points.

(d) Reconcile. The interval contains 0, so at 95% confidence a true mean change of zero is still plausible. That looks like a contradiction with a p-value of $0.035$ until you match the tails: a 95% confidence interval corresponds to a two-sided test at $\alpha = 0.05$, and the two-sided p-value here is $2 \times 0.035 = 0.070$, which exceeds $0.05$. So the interval and the two-sided test agree, and the one-sided test rejects only because it spends all $0.05$ in one tail. A one-sided alternative must be chosen from the research question before the data are seen, never after seeing which direction the sample went.

(a) Random and 10%. The 9 routes are the routes the city studied, not a random sample of routes, so any conclusion is about these routes; the 10% condition is not relevant for the same reason.

(a) The shape condition, and the problem. With $n = 9 < 30$, the differences themselves must show no strong skew and no outliers. Sorted, they are $0.6, 1.2, 1.4, 1.9, 2.0, 2.1, 2.5, 3.0, 8.7$. The median is $2.0$, and with the median excluded $Q_1 = \frac{1.2 + 1.4}{2} = 1.3$ and $Q_3 = \frac{2.5 + 3.0}{2} = 2.75$, so $IQR = 1.45$ and the upper fence is $2.75 + 1.5 \times 1.45 = 4.925$. The value $8.7$ sits well beyond it, so it is an outlier and the shape condition fails. A $t$ procedure here is not fully justified, and anything computed from it should be reported with that caveat.

(b) All 9 routes. $\bar{d} = \frac{23.4}{9} = 2.6$ minutes. The squared deviations from $2.6$ sum to $45.88$, so $s_d = \sqrt{\frac{45.88}{8}} = \sqrt{5.735} = 2.3948$ and $SE = \frac{2.3948}{\sqrt{9}} = \frac{2.3948}{3} = 0.7983$ minutes.

(b) The interval. $df = 8$, so $t^* = 2.306$ and $ME = 2.306 \times 0.7983 = 1.841$ minutes. The interval is $2.6 \pm 1.841 = (0.759,\ 4.441)$ minutes.

(c) The other 8 routes. Dropping $8.7$ gives $\bar{d} = \frac{14.7}{8} = 1.8375$ minutes, squared deviations summing to $4.019$, so $s_d = \sqrt{\frac{4.019}{7}} = \sqrt{0.5741} = 0.7577$ and $SE = \frac{0.7577}{\sqrt{8}} = 0.2679$ minutes. With $df = 7$ and $t^* = 2.365$, $ME = 2.365 \times 0.2679 = 0.6336$ and the interval is $(1.204,\ 2.471)$ minutes.

(c) Compare. One route moved the center from $1.84$ to $2.60$ minutes and inflated $s_d$ from $0.758$ to $2.395$, more than tripling it. The interval went from $1.27$ minutes wide to $3.68$ minutes wide. Both intervals lie entirely above 0, so the direction of the finding is stable, but the estimated size of the improvement is not.

(d) What to report. Both intervals exclude 0, so travel time on these routes fell after the new signals went in, and the finding does not hinge on the outlier. What the city cannot report is a precise figure: the estimate is $2.6$ minutes with all routes and $1.84$ minutes without one of them, so the honest statement names both and identifies route 7 as unusual. Dropping a data point to get a tidier answer is not allowed; the outlier should be investigated, since a route that improved by 8.7 minutes probably had something else change too. And because the routes were not randomly selected and no control routes were tracked, the improvement cannot be attributed to the signal system alone.