A professor is concerned that the two sections of applied statistics 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 class. In Class A, the mean exam score for 13 students is 75.3 with a standard deviation of 6.0. In Class B, the mean exam score for 17 students is 80.1 with a standard deviation of 6.2. Test the professor's claim at the 0.05 level of significance. Assume that the population variances are equal.

To test the professor's claim that the two classes perform at different levels, we can perform a two-sample t-test assuming equal variances. Here’s how we proceed:

### Step 1: Set Up the Hypotheses

- **Null Hypothesis (\(H_0\))**: There is no difference in mean exam scores between Class A and Class B (\(\mu_A = \mu_B\)).
- **Alternative Hypothesis (\(H_a\))**: There is a difference in mean exam scores (\(\mu_A \neq \mu_B\)).

### Step 2: Gather the Sample Statistics

- Class A:
  - Sample size (\(n_A\)) = 13
  - Mean (\(\bar{x}_A\)) = 75.3
  - Standard deviation (\(s_A\)) = 6.0

- Class B:
  - Sample size (\(n_B\)) = 17
  - Mean (\(\bar{x}_B\)) = 80.1
  - Standard deviation (\(s_B\)) = 6.2

### Step 3: Calculate the Test Statistic

First, we calculate the pooled standard deviation (\(s_p\)):
\[
s_p = \sqrt{\frac{(n_A - 1) s_A^2 + (n_B - 1) s_B^2}{n_A + n_B - 2}}
\]

\[
s_p = \sqrt{\frac{(13 - 1)(6.0^2) + (17 - 1)(6.2^2)}{13 + 17 - 2}}
\]

Calculating each component:

- \(s_A^2 = 36\)
- \(s_B^2 = 38.44\)

Now substituting into the formula:
\[
s_p = \sqrt{\frac{(12)(36) + (16)(38.44)}{28}} = \sqrt{\frac{432 + 615.04}{28}} = \sqrt{\frac{1047.04}{28}} \approx \sqrt{37.64} \approx 6.14
\]

Now, we can calculate the t-statistic:
\[
t = \frac{\bar{x}_A - \bar{x}_B}{s_p \sqrt{\frac{1}{n_A} + \frac{1}{n_B}}}
\]
\[
t = \frac{75.3 - 80.1}{6.14 \sqrt{\frac{1}{13} + \frac{1}{17}}}
\]
Calculating the denominator:
\[
\sqrt{\frac{1}{13} + \frac{1}{17}} = \sqrt{\frac{17 + 13}{221}} = \sqrt{\frac{30}{221}} \approx 0.48
\]
Thus:
\[
t \approx \frac{-4.8}{6.14 \times 0.48} \approx \frac{-4.8}{2.95} \approx -1.63
\]

### Step 4: Determine the Critical t-Value

With a significance level (\(\alpha = 0.05\)) and degrees of freedom (\(df = n_A + n_B - 2 = 28\)), we look up the critical t-value for a two-tailed test:

Using a t-table, the critical t-value for \(df = 28\) at \(\alpha = 0.05\) is approximately \( \pm 2.048\).

### Step 5: Make a Decision

The calculated t-statistic is approximately -1.63. Since -1.63 is within the range of -2.048 and 2.048, we fail to reject the null hypothesis.

### Conclusion

At the 0.05 level of significance, there is not enough evidence to support the claim that the mean exam scores of the two classes are significantly different.

---

If you have any questions or need further details, feel free to ask! Here are some follow-up questions you might consider:

1. What assumptions are made in performing a two-sample t-test?
2. How would the conclusion change if we had a one-tailed test instead?
3. What would happen if we did not assume equal variances?
4. How can we visually represent the results of this test?
5. What are the implications of this test for the professor's teaching methods? 

**Tip**: Always check the assumptions of the statistical test you are using to ensure the validity of your results!

Explore the two-sample t-test used to compare the mean exam scores of two applied statistics classes, with Class A scoring 75.3 and Class B scoring 80.1. This analysis covers hypothesis testing at the 0.05 significance level, assuming equal population variances.

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