A sociologist develops a test to measure attitudes about public transportation, and 27 randomly selected subjects are given the test.Their mean score is 76.2 and their standarddeviation is 21.4.Construct the 95% confidence interval for the mean score of all such subjects.
A)(64.2, 83.2)
B)(64.2, 88.2)
C)(74.6, 77.8)
D)(69.2, 83.2)
E)(67.7, 84.7)

Question

Quizplus · Accepted Answer

To construct a 95% confidence interval for the mean, we use the formula: $\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, $n$ is the sample size, and $t_{\alpha/2}$ is the t-score for a 95% confidence level with $n-1$ degrees of freedom. Given $\bar{x} = 76.2$, $s = 21.4$, and $n = 27$, and assuming the t-score for 26 degrees of freedom at a 95% confidence level is approximately 2.06 (this value can vary slightly depending on the source, but it's around 2.06 for most t-distribution tables), the margin of error is $2.06 \cdot \frac{21.4}{\sqrt{27}} \approx 8.5$. Therefore, the 95% confidence interval is $76.2 \pm 8.5$, which gives us (67.7, 84.7).

A Sociologist Develops a Test to Measure Attitudes About Public