Answer :
Answer:
The 98% confidence interval for the mean age of students at the time they take the comprehensive exam for all students enrolled in graduate programs that require students to take comprehensive exams is between 26.2 and 28.8 years. This means that we are 98% sure that the mean age of all students taking the exam is between 26.2 and 28.8 years.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used 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 = 31 - 1 = 30
98% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 30 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.98}{2} = 0.99[/tex]. So we have T = 2.457
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}} = 2.457\frac{2.9}{\sqrt{31}} = 1.3[/tex]
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 27.5 - 1.3 = 26.2 years
The upper end of the interval is the sample mean added to M. So it is 27.5 + 1.3 = 28.8 years
The 98% confidence interval for the mean age of students at the time they take the comprehensive exam for all students enrolled in graduate programs that require students to take comprehensive exams is between 26.2 and 28.8 years. This means that we are 98% sure that the mean age of all students taking the exam is between 26.2 and 28.8 years.