An educator wanted to look at the study habits of university students. As part of the research, data was collected for three variables - the amount of time (in hours per week) spent studying, the amount of time (in hours per week) spent playing video games and the GPA - for a sample of 20 male university students. As part of the research, a 95% confidence interval for the average GPA of all male university students was calculated to be: (2.95, 3.10). Which of the following statements is true? (Can you please do it by hand and explain-Thanks)

A) In construction of the confidence interval, at-value with 20 degrees of freedom was used.
B) In construction of the confidence interval, a z-value with 20 degrees of freedom was used.
C) In construction of the confidence interval, at-value with 19 degrees of freedom was used.
D) In construction of the confidence interval, a z-value was used.

C) In construction of the confidence interval, a t-value with 19 degrees of freedom was used.

Step-by-step explanation:

When to use t or z distribution?

If we know the population standard deviation, we use the z distribution(find a z value related to the confidence level).

Otherwise, if we only have the sample standard deviation, the t-distribution is used.

In this problem, we have no information about the standard deviation of GPA for male university adults, just for the sample, so the t-distribution, and a respective t value is used.

To find a t-value, we also have to find our degrees of freedom. The degrees of freedom(df) is the sample size subtracted by 1.

For this interval, we have a sample of 20 students. So, we have 20-1 = 19 degrees of freedom.

So the correct answer is:

C) In construction of the confidence interval, a t-value with 19 degrees of freedom was used.

i can you with this question sweetie!

step-by-step explanation:

72 is the answer to this problem