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 2 of 3: compute the value of the test statistic. round your answer to three decimal places.

Explanation:

Step1: Calculate pooled - variance

The formula for the pooled - variance $s_p^2$ is $s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}$, where $n_1 = 18$, $s_1 = 3.6$, $n_2=11$, $s_2 = 6.6$.
First, calculate $(n_1 - 1)s_1^2=(18 - 1)\times3.6^2=17\times12.96 = 220.32$.
Second, calculate $(n_2 - 1)s_2^2=(11 - 1)\times6.6^2=10\times43.56 = 435.6$.
Then, $s_p^2=\frac{220.32 + 435.6}{18+11 - 2}=\frac{655.92}{27}\approx24.293$.

Step2: Calculate the test - statistic

The formula for the two - sample t - test statistic (equal variances) is $t=\frac{\bar{x}_1-\bar{x}_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$, where $\bar{x}_1 = 77.7$, $\bar{x}_2 = 81.9$, $s_p=\sqrt{24.293}\approx4.929$, $n_1 = 18$, $n_2 = 11$.
First, calculate $\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}=\sqrt{\frac{1}{18}+\frac{1}{11}}=\sqrt{\frac{11 + 18}{18\times11}}=\sqrt{\frac{29}{198}}\approx\sqrt{0.1465}\approx0.383$.
Then, $t=\frac{77.7-81.9}{4.929\times0.383}=\frac{- 4.2}{1.888}\approx - 2.224$.

Answer:

$-2.224$