A study was done on proctored and nonproctored tests. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below.
\begin{tabular}{|c|c|c|}
\hline & Proctored & Nonproctored \\
\hline $\boldsymbol{\mu}$ & $\mu_{1}$ & $\mu_{2}$ \\
\hline $\mathbf{n}$ & 33 & 32 \\
\hline$\overline{\mathbf{x}}$ & 76.65 & 83.57 \\
\hline $\mathbf{s}$ & 10.24 & 19.46 \\
\hline
\end{tabular}
a. Use a 0.05 significance level to test the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests.

What are the null and alternative hypotheses?
A.
\[
\begin{array}{l}
H_{0}: \mu_{1}=\mu_{2} \\
H_{1}: \mu_{1} \neq \mu_{2}
\end{array}
\]
C.
\[
\begin{array}{l}
H_{0}: \mu_{1}=\mu_{2} \\
H_{1}: \mu_{1}> \mu_{2}
\end{array}
\]
B.
\[
\begin{array}{l}
H_{0}: \mu_{1} \neq \mu_{2} \\
H_{1}: \mu_{1}< \mu_{2}
\end{array}
\]
D.
\[
\begin{array}{l}
H_{0}: \mu_{1}=\mu_{2} \\
H_{1}: \mu_{1}< \mu_{2}
\end{array}
\]

The test statistic, $t$, is $\square$. (Round to two decimal places as needed.)

Step 1 ：The null hypothesis (H0) is usually a statement of no effect or no difference. In this case, it would be that the mean scores for proctored and nonproctored tests are equal. The alternative hypothesis (H1) is what we are testing for. In this case, it would be that the mean score for nonproctored tests is higher than for proctored tests. So, the null and alternative hypotheses should be: \(H_{0}: \mu_{1} = \mu_{2}\) and \(H_{1}: \mu_{1} < \mu_{2}\). This corresponds to option D in the question.

Step 2 ：The test statistic, t, can be calculated using the formula for a two-sample t-test. This requires the sample means, sample sizes, and sample standard deviations for the two groups. These values are given in the table. The formula for the t statistic is: \(t = (x̄1 - x̄2) / \sqrt{(s1^2/n1) + (s2^2/n2)}\) where \(x̄1\) and \(x̄2\) are the sample means, \(s1\) and \(s2\) are the sample standard deviations, and \(n1\) and \(n2\) are the sample sizes.

Step 3 ：We can plug in the given values into this formula to calculate the t statistic. Given \(x1 = 76.65\), \(x2 = 83.57\), \(s1 = 10.24\), \(s2 = 19.46\), \(n1 = 33\), and \(n2 = 32\), we find that \(t = -1.7860449222425094\).

Step 4 ：The calculated t statistic is -1.79 when rounded to two decimal places. This value is negative because the mean score for proctored tests is less than the mean score for nonproctored tests, which is consistent with the claim that students taking nonproctored tests get a higher mean score.

Step 5 ：Final Answer: The null and alternative hypotheses are: \(H_{0}: \mu_{1}=\mu_{2}\) and \(H_{1}: \mu_{1}<\mu_{2}\). The test statistic, \(t\), is \(\boxed{-1.79}\).