A researcher used a developed problem-solving test to randomly selected 50 Grade 6 pupils. In this sample, \( \bar{X}=80 \) ands \( =10 \). The mean \( \mu \) and the standard deviation of the population used in the standardization of the test were 75 and 15 , respectively. Use \( 95 \% \) confidence level to answer the following: a. Does the sample mean differ significantly from the population mean? b. Can it be said that the sample is above average?

To determine if the sample mean significantly differs from the population mean, we can use a one-sample z-test. With a sample mean (\( \bar{X} \)) of 80, a population mean (\( \mu \)) of 75, a sample size (\( n \)) of 50, and a population standard deviation (\( \sigma \)) of 15, we calculate the z-score as follows: First, find the standard error (SE) using \( SE = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{50}} \approx 2.12 \). Next, compute the z-score \( z = \frac{\bar{X} - \mu}{SE} = \frac{80 - 75}{2.12} \approx 2.36 \). For a 95% confidence level, the critical z-value is approximately 1.96. Since 2.36 > 1.96, we conclude the sample mean differs significantly from the population mean. Now, regarding whether we can say the sample is above average, the sample mean of 80 is indeed greater than the population mean of 75. Since our z-score calculation showed a significant difference, we can confidently claim the sample reflects a performance that is statistically above average within the context of this test. Keep celebrating those brainy Grade 6 pupils!