Question

1. suppose you administer a certain aptitude test to a random sample of 9 students in your school, and that the average score is 105. we want to determine the mean μ of the population of all students in the school. assume a standard deviation of σ = 15 for the test. perform a 99% ci and interpret.

Explanation:

Step1: Find the z - score

For a 99% confidence interval, the significance level \(\alpha=1 - 0.99 = 0.01\). Then \(\alpha/2=0.005\). Looking up in the standard normal table (z - table), the z - score \(z_{\alpha/2}=z_{0.005} = 2.576\)

Step2: Calculate the margin of error \(E\)

The formula for the margin of error when the population standard deviation \(\sigma\) is known is \(E = z_{\alpha/2}\times\frac{\sigma}{\sqrt{n}}\)
Given \(n = 9\), \(\sigma=15\), and \(z_{\alpha/2}=2.576\)
\(E=2.576\times\frac{15}{\sqrt{9}}=2.576\times\frac{15}{3}=2.576\times5 = 12.88\)

Step3: Calculate the confidence interval

The formula for the confidence interval for the population mean \(\mu\) when \(\sigma\) is known is \(\bar{x}-E<\mu<\bar{x} + E\)
Given \(\bar{x}=105\)
\(105-12.88<\mu<105 + 12.88\)
\(92.12<\mu<117.88\)

Interpretation:

We are 99% confident that the True population mean \(\mu\) of the scores of all students in the school on the aptitude test lies between \(92.12\) and \(117.88\)

Answer:

The 99% confidence interval for the population mean \(\mu\) is \((92.12,117.88)\)