Short Answer

In the calorie-estimation study, the standard deviation (SDSD) of the difference scores is 24.2724.27 with a mean (MM) of 8.508.50. If the standard deviation had been much smaller (e.g., 5.005.00) while the mean difference (M=8.50M = 8.50) and sample size (N=10N = 10) remained the same, analyze how this change would affect the standard error, the computed tt score, and the final hypothesis decision.

Question: In the calorie-estimation study, the standard deviation (SDSD) of the difference scores is 24.2724.27 with a mean (MM) of 8.508.50. If the standard deviation had been much smaller (e.g., 5.005.00) while the mean difference (M=8.50M = 8.50) and sample size (N=10N = 10) remained the same, analyze how this change would affect the standard error, the computed tt score, and the final hypothesis decision.

Sample answer: A smaller standard deviation (SDSD) decreases the standard error of the difference scores. Because the standard error is in the denominator of the tt-test formula, a smaller standard error increases the calculated tt score. This higher tt score would likely exceed the critical value of 1.8331.833, leading the researcher to reject the null hypothesis and conclude that the training program significantly changed calorie estimates.

Key points:

  • A smaller standard deviation decreases the standard error.
  • A smaller standard error increases the calculated tt score.
  • An increased tt score makes it more likely to exceed the critical value and reject the null hypothesis.

Rubric: The answer should analyze three connected components: 1. A smaller standard deviation reduces the standard error. 2. A smaller standard error increases the calculated tt score. 3. A higher calculated tt score increases the likelihood of rejecting the null hypothesis.

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Updated 2026-05-27

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Research Methods in Psychology - 4th American Edition @ KPU

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