Question

1. i randomly select 25 students math sat scores and find \\( \overline { x } \\) =600. i know that \\( \sigma \\) from this population is 50. find a 99% confidence interval and interpret.
2. i randomly select 100 students math sat scores and find \\( \overline { x } \\) =600. i know that \\( \sigma \\) from this population is 50. find a 99% confidence interval and interpret.

Explanation:

Step1: Find the z - value

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, \(z_{\alpha/2}=z_{0.005} = 2.576\)

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

The formula for the margin of error when the population standard deviation \(\sigma\) is known is \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\)

For \(n = 25\):

\(E=2.576\times\frac{50}{\sqrt{25}}=2.576\times\frac{50}{5}=2.576\times10 = 25.76\)
The confidence interval is \(\bar{x}-E<\mu<\bar{x} + E\)
Substituting \(\bar{x}=600\), we get \(600 - 25.76<\mu<600 + 25.76\)
\(574.24<\mu<625.76\)

For \(n = 100\):

\(E=2.576\times\frac{50}{\sqrt{100}}=2.576\times\frac{50}{10}=2.576\times5=12.88\)
The confidence interval is \(\bar{x}-E<\mu<\bar{x}+E\)
Substituting \(\bar{x} = 600\), we get \(600-12.88<\mu<600 + 12.88\)
\(587.12<\mu<612.88\)

Step3: Interpret the confidence intervals

For \(n = 25\):

We are 99% confident that the True population mean of Math SAT scores lies between \(574.24\) and \(625.76\)

For \(n = 100\):

We are 99% confident that the True population mean of Math SAT scores lies between \(587.12\) and \(612.88\)

Answer:

• For \(n = 25\): The 99% confidence interval is \((574.24,625.76)\)
• For \(n = 100\): The 99% confidence interval is \((587.12,612.88)\)