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High school students can be categorized into two groups by the amount of activities they are involved in. Let group 1 consist of all high school students who are very involved in sports and other activities and group 2 consist of all high school students who aren't. The distributions of GPAs in both groups are approximately normal. The mean and standard deviation for group 1 are 2.9 and 0.4, respectively. The mean and standard deviation for group 2 are 2.7 and 0.5, respectively. Independent random samples of 50 high school students are to be selected from both groups (for a total of 100 students).
-If the sample mean GPA is to be calculated for both groups and we calculate the difference as involved in activities - not so involved in activities, what is the standard deviation of the sampling distribution of the difference in sample means?
A) 0.0082
B) 0.0905
C) 0.18
D) 0.45

Question

Use the following information for questions:
 High school students can be categorized into two groups by the amount of activities they are involved in. Let group 1 consist of all high school students who are very involved in sports and other activities and group 2 consist of all high school students who aren't. The distributions of GPAs in both groups are approximately normal. The mean and standard deviation for group 1 are 2.9 and 0.4, respectively. The mean and standard deviation for group 2 are 2.7 and 0.5, respectively. Independent random samples of 50 high school students are to be selected from both groups (for a total of 100 students).
-If the sample mean GPA is to be calculated for both groups and we calculate the difference as involved in activities - not so involved in activities, what is the standard deviation of the sampling distribution of the difference in sample means?
A) 0.0082
B) 0.0905
C) 0.18
D) 0.45

Quizplus · Accepted Answer

The standard deviation of the sampling distribution of the difference in sample means can be found using the formula $\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$, where $\sigma_1$ and $\sigma_2$ are the standard deviations of the two groups, and $n_1$ and $n_2$ are the sample sizes of the two groups. Plugging in the given values, we get $\sqrt{\frac{0.4^2}{50} + \frac{0.5^2}{50}} = \sqrt{\frac{0.16}{50} + \frac{0.25}{50}} = \sqrt{0.0082} = 0.0905$.

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