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AP Statistics 7.6 Confidence Intervals for the Difference of Two Means - MCQs - Exam Style Questions

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Question

A film student was interested in whether there was a difference in the mean file sizes of all movie downloads from a popular streaming service for two genres—comedy and drama. The film student selected a random sample of comedy movies and a random sample of drama movies from the streaming service. The summary statistics are shown in the table.
Summary Statistics
 Number in SampleMean (in megabytes)Standard Deviation (in megabytes)
Comedy32468113
Drama41508141
In a 95 percent confidence interval, which of the following is the standard error of the difference of the two sample means between all comedy movie downloads and all drama movie downloads from the streaming service?
(A) \(\displaystyle \sqrt{\frac{113^{2}}{32}+\frac{141^{2}}{41}}\)
(B) \(\displaystyle \sqrt{\frac{113^{2}}{32}-\frac{141^{2}}{41}}\)
(C) \(\displaystyle \sqrt{\frac{113^{2}+141^{2}}{32+41}}\)
(D) \(\displaystyle \sqrt{\frac{113^{2}-141^{2}}{32+41}}\)
(E) \(\displaystyle \frac{113}{\sqrt{32}}+\frac{141}{\sqrt{41}}\)
▶️ Answer/Explanation
For a two-sample mean difference (Welch), the standard error is
\[ SE_{\;\bar X_C-\bar X_D} =\sqrt{\frac{s_C^{2}}{n_C}+\frac{s_D^{2}}{n_D}}. \]
Substitute \(s_C=113,\; n_C=32,\; s_D=141,\; n_D=41\):
\[ SE=\sqrt{\frac{113^{2}}{32}+\frac{141^{2}}{41}}. \]
(Numerically, \(SE\approx \sqrt{399.8+484.4}\approx \sqrt{884.2}\approx 29.7\) megabytes.)
✅ Answer: (A)

Question

A biologist wants to estimate the difference between the mean body lengths of green and brown stinkbugs. A random sample of 20 green stinkbugs has a mean body length of 16.22 millimeters (mm) and a standard deviation of 1.34 mm. A random sample of 20 brown stinkbugs has a mean body length of 13.41 mm and a standard deviation of 0.73 mm. What is the standard error of the difference (green – brown) between the sample means?
(A) \(\sqrt{\frac{(1.34)^{2}+(0.73)^{2}}{40}}\)
(B) \(\sqrt{\frac{(1.34)^{2}-(0.73)^{2}}{40}}\)
(C) \(\sqrt{\frac{(1.34)^{2}+(0.73)^{2}}{20}}\)
(D) \(\sqrt{\frac{(1.34)^{2}-(0.73)^{2}}{20}}\)
(E) \(\sqrt{\frac{(1.34)-(0.73)}{20}}\)
▶️ Answer/Explanation
Detailed solution

The formula for the standard error of the difference between two independent sample means (\(\bar{x}_1 – \bar{x}_2\)) is:
\(SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\)

Given:
– Green bugs: \(s_1 = 1.34\), \(n_1 = 20\)
– Brown bugs: \(s_2 = 0.73\), \(n_2 = 20\)

Substituting the values into the formula gives:
\(SE = \sqrt{\frac{(1.34)^2}{20} + \frac{(0.73)^2}{20}} = \sqrt{\frac{(1.34)^2 + (0.73)^2}{20}}\)
✅ Answer: (C)

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