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