Question

Math SAT: The math SAT test was originally designed to have a mean of 500 and a standard deviation of 100. The mean math SAT score last year was 515 but the standard deviation was not reported. You read in an article for an SAT prep course that states in a sample of 75 students, the mean math score was 546, but they did not disclose the standard deviation. Assume the population standard deviation (σ) for all prep course students is 100 and test the claim that the mean score for prep course students is above the national average of 515. Use a 0.01 significance level.

(a) What type of test is this?

-This is a right-tailed test.

-This is a two-tailed test.    

-This is a left-tailed test.

(b) What is the test statistic? Round your answer to 2 decimal places.
z-_x =

(c) What is the P-value of the test statistic? Use the answer found in the z-table or round to 4 decimal places.
P-value =

(d) What is the critical value of z? Use the answer found in the z-table or round to 3 decimal places.
z_α =

(e) What is the conclusion regarding the null hypothesis?

-reject H₀

-fail to reject H₀    

(f) Choose the appropriate concluding statement.

-The data supports the claim that the mean score for all students taking the prep course is above the national average.

-There is not enough data to support the claim that the mean score for all students taking the prep course is above the national average.     

-We reject the claim that the mean score for all students taking the prep course is above the national average.

-We have proven that the mean score for all students taking the prep course is above the national average.

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis: μ = 515
Alternative Hypothesis: μ > 515

a)
Rejection Region
This is right tailed test, for α = 0.01

b)
Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (546 - 515)/(100/sqrt(75))
z = 2.68

c)
P-value Approach
P-value = 0.0037
As P-value < 0.01, reject the null hypothesis.

d)
Critical value of z is 2.33.

e)
reject H0

f)
-The data supports the claim that the mean score for all students taking the prep course is above the national average.