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.

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).

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

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)

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

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.

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).

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.