Don’t know how to solve **Hypothesis Testing: Comparing Writing Requirements in Political Science and History Classes****Context:**The goal is to determine if political science classes require less writing compared to history classes. The test involves comparing the average number of pages assigned for essay writing in both types of classes.- **Political Science Classes:** - Sample Size: 44 classes - Average Pages: 15.39 - Standard Deviation: 2.47- **History Classes:** - Sample Size: 52 classes - Average Pages: 19.87 - Standard Deviation: 3.16**Statistical Significance:**The test is conducted at a \(\alpha = 0.01\) level of significance. **Step-by-Step Guide:****Step 1: Determine the Null and Alternative Hypotheses.**- Null Hypothesis (\(H_0\)): \(\mu_1 = \mu_2\) - This implies that the average number of pages assigned in political science and history classes is the same.- Alternative Hypothesis (\(H_a\)): \(\mu_1 < \mu_2\) - This suggests that political science classes require fewer pages than history classes.**Step 2: Determine the Test Statistic and p-value.**- Test Statistic: \(t\)- \(p\)-value: (to be calculated based on the test statistic)**Step 3: Find the Critical Value for the Rejection Region (Use the \(t\) value).**- This step involves calculating the critical \(t\) value to determine the rejection region for the null hypothesis.**Decision:**Based on the calculated values, you can decide to:- Reject the null hypothesis if the \(p\)-value is less than the significance level (\(\alpha = 0.01\)).- Fail to reject the null hypothesis if the \(p\)-value is greater than the significance level.Upon calculation, you will determine whether the writing requirements significantly differ between the two types of classes.

Solution: Given information: Sample statisticfor political sciences n1=44x1=15.39s1=2.47Sample statistic for history n2=52x2=19.87 s2=3.16Assumed that the population variances are equal. To find pooled variance s2p=(n1-1)×s12+(n2-1)×s22n1+n2-2s2p=43×2.472+51×3.16244+52-2s2p=771.604394=8.208556 step 1) The null and alternative hypothesis are Ho:μ1=μ2 VS Ha:μ1<μ2 step 2) Test statistic t=x1-x2s2p×1n1+1n2t=15.39-19.878.208556×144+152t=15.39-19.878.208556×0.041958t=-4.480.344415=-7.633733t=-7.6337 Test statistic t= -7.6337 To find p value p value =P(t< -7.6337) p value = 0.0000 From Excel step 3) The critical value α=0.01 is -tα,n1+n2-2=-t0.01,94=-2.3667 From Excel The critical value for the rejection region is -2.3667 Rejection region: Reject Ho if t< -2.3667 Here Reject the null hypothesis because t=-7.6337< -2.3667 Option A is Correct