Let's test the hypothesis to see if the students did better. Step 1: State the Hypotheses The population mean mark (μ) is 42.3. We want to test if the students' mean mark (μ) is greater than 42.3. Null Hypothesis (H₀): μ ≤ 42.3 (The students did not do better) Alternative Hypothesis (H₁): μ > 42.3 (The students did better) This is a one-tailed test (specifically, a right-tailed test). Step 2: Identify Given Information and Significance Level Population mean (μ) = 42.3 Population variance (σ²) = 125.44 Population standard deviation (σ) = √(125.44) = 11.2 Sample size (n) = 15 Sample mean (x̄) = 49.8 Significance level (α) = 1% = 0.01 Step 3: Choose the Test Statistic Since the population standard deviation (σ) is known and the population is normally distributed, we use the z-test for a sample mean. The formula for the z-test statistic is: z = (x̄ - μ) / (σ / √n) Step 4: Calculate the Test Statistic Substitute the values into the formula: z = (49.8 - 42.3) / (11.2 / √15) z = 7.5 / (11.2 / 3.87298) z = 7.5 / 2.8919 z ≈ 2.593 Step 5: Determine the Critical Value For a one-tailed (right-tailed) test at a significance level of α = 0.01, we need to find the z-value that has 1% of the area in the right tail. This means the area to the left of the critical value is 1 - 0.01 = 0.99. Looking up 0.99 in the standard normal distribution table, the critical z-value is approximately 2.33. Step 6: Make a Decision Compare the calculated z-statistic with the critical z-value: Calculated z = 2.593 Critical z = 2.33 Since 2.593 > 2.33, the calculated z-statistic falls into the rejection region. Therefore, we reject the null hypothesis (H₀). Step 7: State the Conclusion At the 1% level of significance, there is sufficient evidence to conclude that the students from the particular school did better than the candidates in that GCE paper.