Question

a data set includes data from student evaluations of courses. the summary statistics are ( n = 91,overline{x}=3.34,s = 0.65 ). 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. ( h_{0}:mu = 3.50 )
( h_{1}:mu>3.50 )
b. ( h_{0}:mu
eq3.50 )
( h_{1}:mu = 3.50 )
c. ( h_{0}:mu = 3.50 )
( h_{1}:mu
eq3.50 )
d. ( h_{0}:mu = 3.50 )
( h_{1}:mu<3.50 )

Explanation:

Step1: Determine null and alternative hypotheses

The null hypothesis \(H_{0}\) is the claim we assume to be true. The alternative hypothesis \(H_{1}\) is the claim we test against \(H_{0}\). Here, the claim is about the population mean \(\mu\). The null hypothesis is \(H_{0}:\mu = 3.50\) (the assumed value). Since we are testing if the mean is different (not just greater or less) from \(3.50\), the alternative hypothesis is \(H_{1}:\mu
eq3.50\)

Step2: Calculate the test - statistic

The formula for the \(t\) - test statistic (since population standard deviation \(\sigma\) is unknown) is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\)
Given \(\bar{x} = 3.34\), \(\mu = 3.50\), \(s = 0.65\), \(n=91\)

$$t=\frac{3.34 - 3.50}{0.65/\sqrt{91}}$$

$$t=\frac{- 0.16}{0.65/9.539}$$

$$t=\frac{-0.16}{0.0681}\approx - 2.35$$

Step3: Find the P - value

The degrees of freedom \(df=n - 1=91-1 = 90\). Since it is a two - tailed test (\(H_{1}:\mu
eq3.50\)), we find the probability that \(|t|>2.35\) with \(df = 90\). Using a \(t\) - distribution table or technology, the P - value is approximately \(0.021\)

Step4: Make a decision

Since the significance level \(\alpha = 0.05\) and \(P - value=0.021<0.05\), we reject the null hypothesis.

Answer:

The null and alternative hypotheses are \(H_{0}:\mu = 3.50\) and \(H_{1}:\mu
eq3.50\) (Option C). The test statistic \(t\approx - 2.35\), the P - value is approximately \(0.021\). Since \(P - value<\alpha\), we reject \(H_{0}\). There is sufficient evidence to reject the claim that the population mean of student course evaluations is equal to \(3.50\)