Free-response practice · Question 3 style · 10 points

Warm-up routines and vertical jump

A strength coach for a college club volleyball team wants to know whether a dynamic warm-up (leg swings, skips, and lunges done while moving) leads to higher vertical jumps than static stretching (stretches held in place for 30 seconds). Fourteen players on the team volunteered for a study. Each player completed both routines, one week apart, and a coin flip determined which routine each player did first. Right after each routine, the player's vertical jump height, in centimeters (cm), was measured with a jump mat.

The table summarizes the results, and the dotplot shows the 14 differences in jump height (dynamic minus static).

Summary of vertical jump heights for the 14 players
Jump height (cm)	Mean	Standard deviation
After static stretching	47.88	5.65
After dynamic warm-up	49.20	5.91
Difference (dynamic minus static)	1.32	1.93

Differences in jump height for the 14 players (dynamic minus static)

At the 0.05 level of significance, complete the appropriate inference procedure to determine whether there is convincing statistical evidence that, for players similar to those in the study, the mean vertical jump height after the dynamic warm-up is greater than the mean vertical jump height after static stretching.

Use the given information to respond to parts A, B, C, and D. Label any subparts (e.g., i and ii) that may be present.

A.

i. Identify the appropriate inference procedure.

ii. Explain why the procedure you identified in (i) is appropriate for the way these data were collected.
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Score part A
- 1 point Identifies the procedure as a one-sample t-test for a population mean difference (also called a matched pairs or paired t-test for the mean difference). A correct formula, $t = \frac{{\overline{x}}_{d} - 0}{s_{d} / \sqrt{n}}$ , also earns the point.Earns it: "one-sample t-test for a population mean difference"; "matched pairs t-test for the mean difference"Does not: "two-sample t-test for a difference between two population means" (the two sets of jumps come from the same players, so they are not independent samples); "one-sample z-test for a mean" (the population standard deviation is unknown, so the procedure uses t); "paired t-interval" (the question asks for a test, not an estimate)
- 1 point Explains that each player was measured after both routines, so a player's two jump heights are paired rather than independent, and the analysis uses the one difference per player.Earns it: "Every player did both warm-ups, so the two measurements for a player are paired. We analyze the 14 differences, one per player."Does not: "Because the sample size is small." (does not address how the data were collected); "Because there are two groups of jumps." (that reasoning would point to a two-sample procedure); "Because the data are quantitative." (true, but it does not explain why a paired procedure is needed)
Model response
i. A one-sample t-test for a population mean difference (a matched pairs t-test).

ii. Each player did both routines, so the two jump heights for a player are paired, not independent. The test is carried out on the 14 differences, one for each player.
B.

Complete the inference procedure, including calculating the appropriate statistics.
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Score part B
- 1 point Hypotheses: states $H_{0} : μ_{d} = 0$ and $H_{a} : μ_{d} > 0$ . Hypotheses written with the opposite order of subtraction ( $H_{a} : μ_{d} < 0$ for static minus dynamic) earn the point if they match the parameter definition.Earns it: " $H_{0} : μ_{d} = 0$ , $H_{a} : μ_{d} > 0$ "Does not: " $H_{0} : μ_{d} = 0$ , $H_{a} : μ_{d} \neq 0$ " (the question asks whether jumps are greater after the dynamic warm-up, which is a one-sided alternative); " $H_{0} : {\overline{x}}_{d} = 0$ , $H_{a} : {\overline{x}}_{d} > 0$ " (hypotheses are about the population mean difference, not the sample mean difference)
- 1 point Parameter: defines $μ_{d}$ as the true (population) mean difference in vertical jump height, dynamic minus static, for players similar to those in the study. The order of subtraction must be stated.Earns it: " $μ_{d}$ is the true mean difference in vertical jump height (dynamic minus static) for players like those in the study."Does not: " $μ_{d}$ is the mean difference in jump height for the 14 players." (that is the sample mean difference, 1.32 cm); " $μ_{d}$ is the true mean difference in jump height." (the order of subtraction is not stated)
- 1 point Randomization condition: states that the condition is met because the order of the two routines was randomly assigned (by a coin flip) for each player. The 10% condition is not needed because the players are volunteers in a randomized experiment; a response that checks it anyway is not penalized.Earns it: "The order of the warm-ups was randomly assigned to each player with a coin flip."Does not: "The 14 players were randomly selected." (the players volunteered); "Random condition: met." (no justification)
- 1 point Sample data condition: states that although there are fewer than 30 differences ( $n = 14$ ), the dotplot of the differences shows no strong skewness and no outliers, so the condition is met. The check must be about the differences.Earns it: " $n = 14 < 30$ , but the dotplot of differences has no strong skew and no outliers."Does not: " $n \geq 30$ " (there are only 14 differences); "By the central limit theorem, the sampling distribution is approximately normal." (the sample is too small to rely on that without checking the data); "The jump heights for each routine look roughly symmetric." (checks the two separate sets of jumps instead of the differences)
- 1 point Test statistic and p-value: reports $t \approx 2.56$ with 13 degrees of freedom and a p-value of about 0.012. A p-value from a t table stated as between 0.01 and 0.025 also earns the point. An arithmetic or transcription error can be ignored if correct work is shown.Earns it: " $t = \frac{1.32 - 0}{1.93 / \sqrt{14}} = 2.56$ , df = 13, p-value = 0.012"Does not: " $t = 0.60$ , p-value = 0.28" (uses the two separate standard deviations, a two-sample calculation); " $t = 2.56$ , p-value = 0.024" (a two-sided p-value for a one-sided alternative)
Model response
Let $μ_{d}$ be the true mean difference in vertical jump height (dynamic minus static), in cm, for college club volleyball players similar to those in the study.

$H_{0} : μ_{d} = 0$ and $H_{a} : μ_{d} > 0$

The randomization condition is met because a coin flip randomly assigned the order of the two routines for each player. There are only 14 differences, fewer than 30, but the dotplot of the differences shows no strong skewness and no outliers, so the sample data condition is met. The 10% condition is not needed because the players are volunteers in an experiment.

$t = \frac{1.32 - 0}{1.93 / \sqrt{14}} \approx 2.56$ with $df = 13$ , and the p-value is $P (t \geq 2.56) \approx 0.012$ .
C.

Justify a conclusion in context.
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Score part C
- 1 point Compares the p-value with the significance level and makes a formal decision about the null hypothesis: because $0.012 < 0.05$ , reject $H_{0}$ . The p-value used in the comparison must be correct.Earns it: "The p-value of 0.012 is less than $α = 0.05$ , so reject $H_{0}$ ."Does not: "p is less than alpha." (the values are not given); "0.012 < 0.05, so we accept $H_{a}$ is true." (not a formal decision about the null hypothesis, and a test never proves a hypothesis); "0.024 < 0.05, reject $H_{0}$ " (uses an incorrect p-value)
- 1 point States that there is convincing statistical evidence that, for players similar to those in the study, the true mean vertical jump height is greater after the dynamic warm-up than after static stretching. The conclusion must be in terms of the alternative hypothesis, use non-definitive language, and refer to the population mean difference (or the population means) in context.Earns it: "There is convincing statistical evidence that, for players like these, the true mean jump height is greater after the dynamic warm-up than after static stretching."Does not: "The dynamic warm-up makes every player jump higher." (definitive, and about individuals rather than a mean); "There is convincing evidence that the 14 players jumped higher after the dynamic warm-up." (a statement about the sample, not the population mean difference)
Model response
Because the p-value of 0.012 is less than $α = 0.05$ , reject $H_{0}$ . There is convincing statistical evidence that, for college club volleyball players similar to those in the study, the true mean vertical jump height after the dynamic warm-up is greater than the true mean vertical jump height after static stretching.
D.

Interpret the p-value from part B in context.
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Score part D
- 1 point Interprets the p-value as the probability, computed assuming the null hypothesis is true (the true mean difference in jump height for players like these is 0 cm), of getting a sample mean difference of 1.32 cm or greater, or equivalently a test statistic of 2.56 or greater. The assumption that the null hypothesis is true and the "or greater" tail must both be present.Earns it: "Assuming the true mean difference (dynamic minus static) is 0 cm, there is a 0.012 probability of getting a sample mean difference of 1.32 cm or more by chance."Does not: "There is a 0.012 probability that the null hypothesis is true." (the p-value is calculated by assuming the null hypothesis is true; it is not the probability that it is true); "There is a 1.2% chance that the dynamic warm-up does not help." (same error, in context); "The probability of getting a mean difference of 1.32 cm is 0.012." (leaves out the assumption that $H_{0}$ is true and the "or greater" tail)
Model response
Assuming that the true mean difference in jump height (dynamic minus static) for players like these is 0 cm, there is about a 0.012 probability of getting a sample mean difference of 1.32 cm or greater just by chance.

Your score: 0 of 10. Each point is earned on its own, the way the exam scores it.