How do I find the confidence interval?

Question

A research wants to determine whether high school students who attend an SAT prep course score significantly different on the SAt than student who do not attend the prep course. For those who do not attend the course, the population mean is 1050. The 16 students who attend the prep course average 1150 on the SAT, with a sample standard deviation of 100. On the basis of these data, can the researcher conclude that the prep course has significant difference on the SAT scores? Set alpha equal to .01.

Calculate the 95% confidence interval.

The mean you will use for this calculation is

a. 1050

b. 1150

What is the new critical value you will use for this calculation?

What are the two values?

____ ≤ population mean ≤ _____

I'm not asking for the answer - just some help!

Answer 1

Critical value with significance level 0.01 is 2.947 (t-distribution with 15 degrees of freedom).

(The question does not specify a direction (improve etc.,) , so it is a two-tailed test.)

H0: mu = 1050

HA: mu not = 1050

Sample mean 1150

Standard deviation = 100

Standard error of mean = sd / sqrt(n)

SE = 100/4

Standard error of mean 25

t = (xbar-mu) /se

t = (1150-1050) / 25

t = 4

Computed t exceeds critical t of 2.947. Reject H0. The two scores differ significantly at 0.01 level.

95 % confidence interval: The t critical value is 2.131

The mean you will use for this calculation is

b. 1150 (the sample mean)

Sample mean 1150

Standard deviation = 100

Standard error of mean = sd / sqrt(n)

SE = 100/4

Standard error of mean 25

Confidence limits 1150-(25)(2.131)

and 1150+(25)(2.131)

95 % limits are (1096.725 , 1203.275)