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 88.9
. He knows the population standard deviation for the evening classes to be 2.5
points. A random sample of 250
students from morning classes results in a mean test score of 89.4
. He knows the population standard deviation for the morning classes to be 3.2
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.

To test the professor's claim, we can conduct a hypothesis test for the difference between two population means. Since the population standard deviations are known, we will use the **z-test** for two independent samples.

### Step 1: State the hypotheses
We will set up the null and alternative hypotheses as follows:

- \( H_0 \): \( \mu_1 = \mu_2 \) (The mean test scores of the evening and morning classes are the same)
- \( H_1 \): \( \mu_1 < \mu_2 \) (The mean test score of the evening classes is lower than the mean test score of the morning classes)

This is a **left-tailed test** because the professor believes the mean of the evening class is lower than that of the morning class.

### Step 2: Compute the z-test statistic

The formula for the test statistic is:

\[
z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}
\]

Where:
- \( \bar{x}_1 = 88.9 \) is the sample mean for evening classes (Population 1).
- \( \bar{x}_2 = 89.4 \) is the sample mean for morning classes (Population 2).
- \( \sigma_1 = 2.5 \) is the population standard deviation for evening classes.
- \( \sigma_2 = 3.2 \) is the population standard deviation for morning classes.
- \( n_1 = 200 \) is the sample size for evening classes.
- \( n_2 = 250 \) is the sample size for morning classes.

Now, let's calculate the z-value.

\[
z = \frac{88.9 - 89.4}{\sqrt{\frac{2.5^2}{200} + \frac{3.2^2}{250}}}
\] 

Let me calculate this for you.The value of the test statistic \( z \) is \(-1.86\), rounded to two decimal places.

Would you like further details or clarification on this calculation?

Here are 5 related questions for practice:
1. How would you determine the critical value for this hypothesis test at a 98% confidence level?
2. What would the decision rule be for rejecting or not rejecting the null hypothesis in this scenario?
3. How would you calculate the p-value associated with this test statistic?
4. What changes in the test would occur if the population standard deviations were unknown?
5. How could the sample size affect the test statistic and its interpretation?

**Tip:** Always check whether you are performing a one-tailed or two-tailed test, as this impacts your critical values and conclusion.

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 88.9. He knows the population standard deviation for the evening classes to be 2.5 points. A random sample of 250 students from morning classes results in a mean test score of 89.4. He knows the population standard deviation for the morning classes to be 3.2 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. Compute the value of the test statistic. Round your answer to two decimal places.

Test the professor's claim that evening class students have lower mean test scores than morning class students using a Z-test for two independent samples. Data includes sample sizes of 200 and 250, with population standard deviations of 2.5 and 3.2, respectively. Calculate the test statistic and assess the claim at a 98% confidence level.

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