Question

A data set includes data from student evaluations of courses. The summary statistics are n=89​, x=3.44​, s=0.67. Use a 0.05 significance level to test the claim that the population of student course evaluations has a mean equal to 3.50. Assume that a simple random sample has been selected. Identify the null and alternative​ hypotheses, test​ statistic, P-value, and state the final conclusion that addresses the original claim.

What are the null and alternative​ hypotheses?
A.
H0​: μ=3.50
H1​: μ>3.50
B.
H0​: μ=3.50
H1​: μ<3.50
C.
H0​: μ≠3.50
H1​: μ=3.50
D.
H0​: μ=3.50
H1​: μ≠
(I also need the test statistic and p-value) thank you so much in advance :)

Answer 1

We're told that "the claim that the population of student course evaluations has a mean equal to 3.50". So this means μ=3.50 makes up the null H0

The alternative would be H1: μ ≠ 3.50 since it's the opposite of the claim made in the null.

We go with answer choice D to form the null and alternative hypotheses.

The sign ≠ in the alternative hypothesis tell us that we have a two tail test.

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Let's compute the test statistic

z = (xbar - mu)/(s/sqrt(n))

z = (3.44 - 3.50)/(0.67/sqrt(89))

z = -0.84483413122896

z = -0.84

The test statistic is roughly -0.84

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Despite not knowing what sigma is (aka the population standard deviation), we can see that n > 30 is the case. So we can use the Z distribution. This is the standard normal distribution. When n > 30, the T distribution is fairly approximately the same as the Z distribution.

Use a calculator or a Z table to determine that

P(Z < -0.84) = 0.2005

which is approximate

Because we're doing a two-tail test, this means we double that result to get 2*0.2005 = 0.401

The p-value is roughly 0.401

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Since the p-value is larger than alpha = 0.05, we don't have enough evidence to reject the null. So you can say that we fail to reject the null, or we accept the null.

The conclusion based on that means that μ=3.50 must be true (unless other evidence comes along to disprove this). In other words, the mean evaluation score from students appears to be 3.50