To test the effectiveness of new educational materials, students were separated into two groups. The test group used the new materials, but the control group did not. When the two groups were tested, these were the results:
\begin{tabular}{|c|c|}
\hline Test Group & Control Group \\
\hline $\bar{x}_{1}=60$ & $\bar{x}_{2}=55.5$ \\
\hline$s_{1}=9.8$ & $s_{2}=6.4$ \\
\hline$n_{1}=61$ & $n_{2}=41$ \\
\hline
\end{tabular}
ii. Test the claim that the population mean for those using the new materials (test group) actually is higher than the mean of students using the traditional materials (control group), using a level of significance of 0.1 .
a. What type of test will be used in this problem?
A test for the difference between two means
b. Enter the null hypothesis for this test.
c. Enter the alternative hypothesis for this test.
d. Is the original claim located in the null or alternative hypothesis?
\[
\text { Alternative Hypothesis } \checkmark{ }^{\circ}
\]
e. What is the test statistic for the given statistics? (round to two decimals)
f. What is the $p$-value for this test? (round to 4 decimals)

Step 1 ：We are given the following data for the test and control groups: \(\bar{x}_{1}=60\), \(s_{1}=9.8\), \(n_{1}=61\) for the test group and \(\bar{x}_{2}=55.5\), \(s_{2}=6.4\), \(n_{2}=41\) for the control group.

Step 2 ：We are asked to test the claim that the population mean for those using the new materials (test group) is higher than the mean of students using the traditional materials (control group), using a level of significance of 0.1.

Step 3 ：The type of test to be used in this problem is a test for the difference between two means.

Step 4 ：The null hypothesis for this test is that the population means of the test group and the control group are equal.

Step 5 ：The alternative hypothesis for this test is that the population mean of the test group is greater than the population mean of the control group.

Step 6 ：The original claim is located in the alternative hypothesis.

Step 7 ：We calculate the test statistic using the formula: \(t = \frac{\bar{x}_{1} - \bar{x}_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}}\). Substituting the given values, we get \(t \approx 2.81\).

Step 8 ：We calculate the degrees of freedom as \(df = n_{1} + n_{2} - 2 = 100\).

Step 9 ：We calculate the p-value using the t-distribution table or a statistical software. The p-value is approximately 0.0030.

Step 10 ：Since the p-value is less than the level of significance (0.1), we reject the null hypothesis. This suggests that the population mean for those using the new materials is higher than the mean of students using the traditional materials.

Step 11 ：The final answer is: The test statistic is approximately \(\boxed{2.81}\) and the p-value is approximately \(\boxed{0.0030}\).