## A sociologist develops a test to measure attitudes towards public transportation, and 25 randomly selected subjects are given the test. The

Question

A sociologist develops a test to measure attitudes towards public transportation, and 25 randomly selected subjects are given the test. The sample mean score is 76.2 and the sample standard deviation is 21.4. The population is normally distributed. What is the 95% confidence interval for the mean score of all such subjects? Round to 3 decimal places.

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2021-10-08T13:52:56+00:00
2021-10-08T13:52:56+00:00 1 Answer
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## Answers ( )

Answer:The 95% confidence interval for the mean score of all such subjects is (32.033, 120.367).

Step-by-step explanation:We have the sample standard deviation, so we use the t-distribution to solve this question.

The

first stepto solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. Sodf = 25 – 1 = 24

95% confidence intervalNow, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.039

The margin of error is:M = T*s = 2.0639*21.4 = 44.167

In which s is the standard deviation of the sample.

The

lower endof the interval is the sample mean subtracted by M. So it is 76.2 – 44.167 = 32.033The

upper endof the interval is the sample mean added to M. So it is 76.2 + 44.167 = 120.367The 95% confidence interval for the mean score of all such subjects is (32.033, 120.367).