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 [tex]1 - \frac{1 - 0.95}{2} = 0.975[/tex]. 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).
