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.

Answer 1

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 step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 25 - 1 = 24

95% confidence interval

Now, 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 $1 - \frac{1 - 0.95}{2} = 0.975$. 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 end of the interval is the sample mean subtracted by M. So it is 76.2 - 44.167 = 32.033

The upper end of the interval is the sample mean added to M. So it is 76.2 + 44.167 = 120.367

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