Question

a professor believes that, for the introductory art history classes at his university, the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. he collects data from a random sample of 200 students in evening classes and finds that they have a mean test - score of 87.8. he knows the population standard deviation for the evening classes to be 6.7 points. a random sample of 250 students from morning classes results in a mean test score of 88.6. he knows the population standard deviation for the morning classes to be 1.8 points. test his claim with a 98% level of confidence. let students in the evening classes be population 1 and let students in the morning classes 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 0.02 level of significance to support the professors claim that the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. we fail to reject the null hypothesis and conclude that there is sufficient evidence at a 0.02 level of significance to support the professors claim that the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. we fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.02 level of significance to support the professors claim that the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. we reject the null hypothesis and conclude that there is insufficient evidence at a 0.02 level of significance to support the professors claim that the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes.

Explanation:

Step1: Identify the null and alternative hypotheses

The null hypothesis $H_0:\mu_1\geq\mu_2$ and the alternative hypothesis $H_1:\mu_1 < \mu_2$, where $\mu_1$ is the mean of population 1 (evening - classes) and $\mu_2$ is the mean of population 2 (morning - classes). The significance level $\alpha=1 - 0.98 = 0.02$.

Step2: Calculate the test - statistic for two - sample z - test

The formula for the two - sample z - test statistic when the population standard deviations $\sigma_1$ and $\sigma_2$ are known is $z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}}$. Here, $\bar{x}_1 = 87.8$, $\bar{x}_2 = 88.6$, $\sigma_1 = 6.7$, $\sigma_2 = 1.8$, $n_1 = 200$, $n_2 = 250$, and under $H_0$, $\mu_1-\mu_2 = 0$.
\[

$$\begin{align*} z&=\frac{(87.8 - 88.6)-0}{\sqrt{\frac{6.7^{2}}{200}+\frac{1.8^{2}}{250}}}\\ &=\frac{- 0.8}{\sqrt{\frac{44.89}{200}+\frac{3.24}{250}}}\\ &=\frac{-0.8}{\sqrt{0.22445 + 0.01296}}\\ &=\frac{-0.8}{\sqrt{0.23741}}\\ &=\frac{-0.8}{0.48725}\\ &\approx - 1.64 \end{align*}$$

\]
The critical value for a one - tailed test with $\alpha = 0.02$ is $z_{\alpha}=- 2.05$.

Step3: Draw a conclusion

Since $z=-1.64>-2.05$ (the critical value), we fail to reject the null hypothesis. This means that there is insufficient evidence at a 0.02 level of significance to support the professor's claim that the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes.

Answer:

We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.02 level of significance to support the professor's claim that the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes.