Question

a professor is concerned that the two sections of college algebra that he teaches are not performing at the same level. to test his claim, he looks at the mean exam score for a random sample of students from each of his classes. in class 1, the mean exam score for 18 students is 77.7 with a standard deviation of 3.6. in class 2, the mean exam score for 11 students is 81.9 with a standard deviation of 6.6. test the professors claim at the 0.01 level of significance. assume that both populations are approximately normal and that the population variances are equal. let class 1 be population 1 and let class 2 be population 2. step 3 of 3: draw a conclusion and interpret the decision. answer we reject the null hypothesis and conclude that there is sufficient evidence at a 0.01 level of significance to say that the mean exam scores for the two classes are different. we fail to reject the null hypothesis and conclude that there is sufficient evidence at a 0.01 level of significance to say that the mean exam scores for the two classes are different. we fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.01 level of significance to say that the mean exam scores for the two classes are different. we reject the null hypothesis and conclude that there is insufficient evidence at a 0.01 level of significance to say that the mean exam scores for the two classes are different.

Explanation:

Step1: Determine degrees of freedom

The degrees of freedom $df=n_1 + n_2-2=18 + 11-2=27$.

Step2: Find critical - value

For a two - tailed test at $\alpha = 0.01$ with $df = 27$, the critical values are $t_{\alpha/2,df}=t_{0.005,27}\approx\pm 2.771$.

Step3: Calculate the test - statistic

First, calculate the pooled variance $s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}=\frac{(18 - 1)\times3.6^2+(11 - 1)\times6.6^2}{18 + 11-2}=\frac{17\times12.96+10\times43.56}{27}=\frac{220.32 + 435.6}{27}=\frac{655.92}{27}\approx24.293$.
The test - statistic $t=\frac{\bar{x}_1-\bar{x}_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$, where $\bar{x}_1 = 77.7$, $\bar{x}_2 = 81.9$, $s_p=\sqrt{24.293}\approx4.929$, $n_1 = 18$, $n_2 = 11$.
$t=\frac{77.7 - 81.9}{4.929\sqrt{\frac{1}{18}+\frac{1}{11}}}=\frac{- 4.2}{4.929\sqrt{\frac{11 + 18}{18\times11}}}=\frac{-4.2}{4.929\sqrt{\frac{29}{198}}}\approx\frac{-4.2}{4.929\times0.382}\approx\frac{-4.2}{1.883}\approx - 2.23$.

Step4: Make a decision

Since $|t|=2.23<2.771$, we fail to reject the null hypothesis.

Answer:

We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.01 level of significance to say that the mean exam scores for the two classes are different.