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

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

-reject *H*_{0}

-fail to reject
*H*_{0}

(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.

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designed to have a mean of 500 and a standard deviation of 100. The
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