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 differs significantly from the population mean and if the sample is above average, we can use the following steps: a. To determine if the sample mean differs significantly from the population mean, we can use the z-test. The formula for the z-test is: \[ z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}} \] where: - \( \bar{X} \) is the sample mean - \( \mu \) is the population mean - \( \sigma \) is the population standard deviation - \( n \) is the sample size Given: - \( \bar{X} = 80 \) - \( \mu = 75 \) - \( \sigma = 15 \) - \( n = 50 \) We can calculate the z-score and then compare it to the critical z-score for a 95% confidence level to determine if the sample mean differs significantly from the population mean. b. To determine if the sample is above average, we can compare the sample mean to the population mean. If the sample mean is greater than the population mean, it can be said that the sample is above average. Let's calculate the z-score and compare it to the critical z-score for a 95% confidence level. Calculate the value by following steps: - step0: Calculate: \(\frac{\left(80-75\right)}{\left(\frac{15}{\sqrt{50}}\right)}\) - step1: Remove the parentheses: \(\frac{80-75}{\frac{15}{\sqrt{50}}}\) - step2: Simplify the root: \(\frac{80-75}{\frac{15}{5\sqrt{2}}}\) - step3: Reduce the fraction: \(\frac{80-75}{\frac{3}{\sqrt{2}}}\) - step4: Subtract the numbers: \(\frac{5}{\frac{3}{\sqrt{2}}}\) - step5: Multiply by the reciprocal: \(5\times \frac{\sqrt{2}}{3}\) - step6: Multiply: \(\frac{5\sqrt{2}}{3}\) The z-score for the sample mean is approximately 2.357023. Next, we need to find the critical z-score for a 95% confidence level. The critical z-score for a 95% confidence level is approximately 1.96. Now, we can compare the calculated z-score to the critical z-score to determine if the sample mean differs significantly from the population mean. a. Since the calculated z-score (2.357023) is greater than the critical z-score (1.96), we can conclude that the sample mean differs significantly from the population mean. b. Since the sample mean (80) is greater than the population mean (75), it can be said that the sample is above average.