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.