Question 1194707
**a) Hypothesis Testing**

* **Null Hypothesis (H0):** There is no significant difference in the mean exam scores of students from the four schools (μA = μB = μC = μD).
* **Alternative Hypothesis (H1):** At least one school's mean exam score is significantly different from the others.

**1. Calculate the necessary statistics:**

* **Calculate the mean and variance for each group:**
    * School A: Mean = 80.75, Variance = 8.6875
    * School B: Mean = 73.50, Variance = 10.1250
    * School C: Mean = 83.50, Variance = 4.6875
    * School D: Mean = 65.75, Variance = 66.1875

* **Calculate the overall mean:** 
    * Overall Mean = (Sum of all scores) / Total number of scores = 75.875

* **Calculate the between-groups sum of squares (SSB):**
    * SSB = Σ(n_i * (mean_i - overall_mean)^2) 
    * SSB = 4 * ((80.75 - 75.875)^2 + (73.50 - 75.875)^2 + (83.50 - 75.875)^2 + (65.75 - 75.875)^2) 
    * SSB = 4 * (23.890625 + 5.421875 + 58.007813 + 100.15625) 
    * SSB = 752.5

* **Calculate the within-groups sum of squares (SSW):**
    * SSW = Σ(n_i - 1) * variance_i
    * SSW = (4 - 1) * (8.6875 + 10.1250 + 4.6875 + 66.1875) 
    * SSW = 246

* **Calculate the degrees of freedom:**
    * Between-groups degrees of freedom (dfB) = k - 1 = 4 - 1 = 3
    * Within-groups degrees of freedom (dfW) = N - k = 16 - 4 = 12

* **Calculate the mean squares:**
    * Mean Square Between (MSB) = SSB / dfB = 752.5 / 3 = 250.83
    * Mean Square Within (MSW) = SSW / dfW = 246 / 12 = 20.5

* **Calculate the F-statistic:**
    * F = MSB / MSW = 250.83 / 20.5 = 12.23

**2. Determine the critical value:**

* Using an F-distribution table with dfB = 3 and dfW = 12, and α = 0.05, the critical value is approximately 3.49.

**3. Make a decision:**

* Since the calculated F-statistic (12.23) is greater than the critical value (3.49), we reject the null hypothesis.

**Interpretation:**

* There is sufficient evidence at the 0.05 significance level to conclude that there is a significant difference in the mean exam scores of students from the four schools.

**b) Determine the p-value:**

* Using statistical software (like R or Python), we can find the exact p-value associated with the calculated F-statistic (12.23) and the degrees of freedom (dfB = 3, dfW = 12). 
* The p-value will be very small (likely less than 0.001).

**c) 95% Confidence Interval for the Mean of School B:**

* **Calculate the standard error of the mean for School B:**
    * Standard Error (SE) = √(variance_B / n_B) = √(10.1250 / 4) = 1.586

* **Find the critical t-value:**
    * For a 95% confidence interval and df = n_B - 1 = 3, the critical t-value (from a t-distribution table) is approximately 3.182.

* **Calculate the margin of error:**
    * Margin of Error = t_critical * SE = 3.182 * 1.586 = 5.05

* **Calculate the confidence interval:**
    * Lower limit = Mean_B - Margin of Error = 73.50 - 5.05 = 68.45
    * Upper limit = Mean_B + Margin of Error = 73.50 + 5.05 = 78.55

* **95% Confidence Interval for School B:** (68.45, 78.55)

**Interpretation:**

We are 95% confident that the true mean exam score for students from School B lies between 68.45 and 78.55.

**Note:** This analysis assumes that the assumptions for ANOVA are met, including normality of the data within each group and homogeneity of variances. 

I hope this comprehensive analysis is helpful!