Stats: can you check my answer?

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.

The appropriate statistcal procedure for this example would be a

a. t-test

b. z-test

(I said B)

Is this is a one-tailed or two-tailed test?

(I said two-tailed)

The most appropriate null hypothesis would be

a. μ SAT prep = 1050

b. μ SAT prep = 1150

c. μ SAT prep ≤ 1050

d. μ SAT prep ≥ 1050

(I said A)

Set up the criteria for making a decision. Find the critical value using an alpha = .01.

I have no idea how to do this one. Maybe you could just give me the formula or tell me how to solve it without giving the answer?

What is the numeric value of your standard error?

No idea.

Based on your results, you would

A. reject the null hypothesis

b. fail to reject the null hypothesis

Any help is appreciated greatly!!!

Answer 1

The appropriate statistical procedure for this example would be a

a. t-test

*because you are using sample s.d. You don't know the population s.d.

Is this is a one-tailed or two-tailed test?

two-tailed, but in reality, the question should really ask if prep>no prep.

The most appropriate null hypothesis would be

a. μ SAT prep = 1050

Set up the criteria for making a decision. Find the critical value using an alpha = .01.

Null Hypothesis μ = 1050. Alternative Hypothesis μ does not = 1050.

t(0.01/2),16-1 = t(0.005),15 = 2.947

Criteria: Reject null if T(observed) > 2.947 OR T(observed) < -2.947

T(observed) = (1150 - 1050)/standard error

What is the numeric value of your standard error?

Standard Error = s/n^0.5 = 100/4 = 25

Based on your results, you would:

T(observed) = 150/25 = 6 > 2.947

Thus, we reject the null hypothesis and conclude that

μ STAT Prep does not = 1050.