The marks of a GCE paper were normally distributed with mean 42.3 and variance 125.44. As a practical examination 15 students from a particular School set for the examination and their mean Mean was 49.8. Test at the 1% level of significance whether the students did better than did better than the candidate in that GCE paper.

Question

Asked on March 31, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-03-31T08:47:35.785Z

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.

Accepted Answer · 2026-03-31T08:47:35.785Z

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.

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