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This is a Chi-square goodness-of-fit test problem, where the goal is to test if the observed distribution of grades differs from the claimed distribution at a 1% significance level.

### Step-by-step solution:

1. **Claimed (expected) proportions**:
   - A: 20%
   - B: 25%
   - C: 35%
   - D: 10%
   - F: 10%

2. **Total sample size (sum of observed frequencies)**:
   \[
   75 + 52 + 115 + 12 + 35 = 289
   \]
   
3. **Expected frequencies**: For each category, the expected frequency is calculated by multiplying the claimed proportion by the total sample size.

   - Expected frequency for A: \( 0.20 \times 289 = 57.8 \)
   - Expected frequency for B: \( 0.25 \times 289 = 72.25 \)
   - Expected frequency for C: \( 0.35 \times 289 = 101.15 \)
   - Expected frequency for D: \( 0.10 \times 289 = 28.9 \)
   - Expected frequency for F: \( 0.10 \times 289 = 28.9 \)

4. **Chi-square test statistic**: The formula for the test statistic is
   \[
   \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}
   \]
   where \( O_i \) is the observed frequency and \( E_i \) is the expected frequency for each category.

   Now we calculate the contributions to the chi-square statistic for each category:

   - For A: \( \frac{(75 - 57.8)^2}{57.8} = 4.9845 \)
   - For B: \( \frac{(52 - 72.25)^2}{72.25} = 5.6767 \)
   - For C: \( \frac{(115 - 101.15)^2}{101.15} = 1.9282 \)
   - For D: \( \frac{(12 - 28.9)^2}{28.9} = 9.5516 \)
   - For F: \( \frac{(35 - 28.9)^2}{28.9} = 1.2775 \)

   Summing these values gives the chi-square statistic:
   \[
   \chi^2 = 4.9845 + 5.6767 + 1.9282 + 9.5516 + 1.2775 = 23.4185
   \]

5. **Degrees of freedom**: The degrees of freedom (df) for this test is the number of categories minus 1:
   \[
   df = 5 - 1 = 4
   \]

6. **p-value**: Using the chi-square distribution table or a calculator, we compare the chi-square statistic (\( \chi^2 = 23.4185 \)) to the critical value at a 1% significance level and 4 degrees of freedom.

   The critical value for \( \alpha = 0.01 \) and \( df = 4 \) is approximately 13.277. Since \( 23.4185 > 13.277 \), the p-value is less than 0.01.

### Conclusion:

- Since the test statistic exceeds the critical value, we reject the null hypothesis.
- **There is enough evidence to conclude that the teacher's claim about the grade distribution is not supported at the 1% significance level.**

Would you like further details on any step or calculations? Here are some follow-up questions you may find useful:

1. How do we determine the expected frequencies for different categories?
2. Why do we use the chi-square distribution for this test?
3. How does changing the significance level affect the test's outcome?
4. What does the p-value represent in this context?
5. What are the assumptions underlying the chi-square goodness-of-fit test?

**Tip**: Always make sure your expected frequencies are based on a valid probability distribution to avoid errors in hypothesis testing.

A statistics teacher claims that, on average, 20% of students in his class get an A, 25% get a B, 35% get a C, 10% get a D and the rest get an F. The grades of a random sample of his students over the years is recorded. Test the claim at 1% significance. Round to the fourth as needed.

Test a statistics teacher's claim about student grade distribution using a Chi-Square Goodness-of-Fit test at a 1% significance level. This problem involves calculating expected frequencies, determining test statistics, and using hypothesis testing.

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