Question

question 2 of 6, step 2 of 3
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 2 of 3: compute the value of the test statistic. round your answer to two decimal places.
answer

Explanation:

Step1: Identify the formula for two - sample z - test statistic

The formula for the two - sample z - test statistic when population standard deviations $\sigma_1$ and $\sigma_2$ are known is $z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}}$. Since the claim is $\mu_1<\mu_2$, under the null hypothesis $H_0:\mu_1-\mu_2 = 0$. So the formula simplifies to $z=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}}$.

Step2: Assign the given values

We have $\bar{x}_1 = 87.8$ (mean of evening - class students), $\bar{x}_2=88.6$ (mean of morning - class students), $\sigma_1 = 6.7$, $n_1 = 200$, $\sigma_2 = 1.8$, and $n_2=250$.
Substitute these values into the formula:
\[

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

\]

Answer:

$-1.64$