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, we can conduct a two-sample \( t \)-test to compare the means of the two classes. Since we assume equal population variances, we will use the pooled standard deviation.

### Step-by-Step Solution

1. **State the Hypotheses**:
   - Null Hypothesis (\( H_0 \)): The mean scores of the two classes are equal. \( \mu_A = \mu_B \).
   - Alternative Hypothesis (\( H_1 \)): The mean scores of the two classes are not equal. \( \mu_A \neq \mu_B \).

2. **Set the Significance Level**:
   - \( \alpha = 0.05 \).

3. **Calculate the Pooled Standard Deviation** \( (s_p) \):
   The formula for the pooled standard deviation is:
   \[
   s_p = \sqrt{\frac{(n_A - 1) \cdot s_A^2 + (n_B - 1) \cdot s_B^2}{n_A + n_B - 2}}
   \]
   where:
   - \( n_A = 13 \) (sample size for Class A),
   - \( n_B = 17 \) (sample size for Class B),
   - \( s_A = 6.0 \) (standard deviation for Class A),
   - \( s_B = 6.2 \) (standard deviation for Class B).

   Plugging in these values:
   \[
   s_p = \sqrt{\frac{(13 - 1) \cdot 6.0^2 + (17 - 1) \cdot 6.2^2}{13 + 17 - 2}}
   = \sqrt{\frac{12 \cdot 36 + 16 \cdot 38.44}{28}}
   = \sqrt{\frac{432 + 615.04}{28}}
   = \sqrt{\frac{1047.04}{28}}
   = \sqrt{37.39} \approx 6.11
   \]

4. **Calculate the Test Statistic** \( (t) \):
   The formula for the \( t \)-statistic in a two-sample \( t \)-test is:
   \[
   t = \frac{\bar{X}_A - \bar{X}_B}{s_p \cdot \sqrt{\frac{1}{n_A} + \frac{1}{n_B}}}
   \]
   where:
   - \( \bar{X}_A = 75.3 \) (mean for Class A),
   - \( \bar{X}_B = 80.1 \) (mean for Class B).

   Plugging in the values:
   \[
   t = \frac{75.3 - 80.1}{6.11 \cdot \sqrt{\frac{1}{13} + \frac{1}{17}}}
   = \frac{-4.8}{6.11 \cdot \sqrt{0.0769 + 0.0588}}
   = \frac{-4.8}{6.11 \cdot \sqrt{0.1357}}
   = \frac{-4.8}{6.11 \cdot 0.3684}
   = \frac{-4.8}{2.25} \approx -2.13
   \]

5. **Determine the Degrees of Freedom**:
   The degrees of freedom (\( df \)) for the test is:
   \[
   df = n_A + n_B - 2 = 13 + 17 - 2 = 28
   \]

6. **Find the Critical Value and Make a Decision**:
   - For a two-tailed test at \( \alpha = 0.05 \) with \( df = 28 \), the critical \( t \)-value is approximately \( \pm 2.048 \).
   - Since \( |t| = 2.13 \) is greater than \( 2.048 \), we reject the null hypothesis.

### Conclusion
At the 0.05 level of significance, there is sufficient evidence to conclude that the mean exam scores of the two classes are not equal.

Would you like further details on any steps? 

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Here are some related questions to explore:
1. How would the conclusion change if the significance level was 0.01 instead?
2. What if the sample sizes of the two classes were more unequal, say 10 and 20?
3. How does the assumption of equal variances affect the outcome of the test?
4. What are the effects of using a one-tailed test instead of a two-tailed test?
5. How would we approach this problem if the population variances were unknown and assumed unequal?

**Tip:** Always check the assumptions (like equal variances) before choosing a specific \( t \)-test formula.

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.

Conduct a two-sample t-test to determine if there is a significant difference in exam scores between two statistics classes. Class A has a mean score of 75.3 with a standard deviation of 6.0, and Class B has a mean score of 80.1 with a standard deviation of 6.2. At the 0.05 level of significance, calculate the t-statistic, pooled standard deviation, and determine if the null hypothesis should be rejected.

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