) The GPA of accounting students in a university is known to be normally distributed. A random sample of 20 accounting students results in a mean of 2.92 and a standard deviation of 0.16. Construct the 95% confidence interval for the mean GPA of all accounting students at this university.

The 95% confidence interval for the mean GPA of all accounting students at this university is between 2.5851 and 3.2549

Step-by-step explanation:

We are in posessions of the sample's standard deviation. So we use the student's 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 = 20 - 1 = 19

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of ). So we have T = 2.0930

The margin of error is:

M = T*s = 2.0930*0.16 = 0.3349.

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 2.92 - 0.3349 = 2.5851

The upper end of the interval is the sample mean added to M. So it is 2.92 + 0.3349 = 3.2549

The 95% confidence interval for the mean GPA of all accounting students at this university is between 2.5851 and 3.2549

the answer is somewhere there guys it really . there im just not gonna answer

step-by-step explanation:

but i'm guessing ur first option tho