Question

suppose we know the population standard deviation σ = 2. we take a random sample of size 25 and calculate the sample mean = 10. what is the 99% confidence interval for the population mean based on this sample? hint: round the critical values to three decimal places 9.174, 10.826 9.216, 10.784 9.342, 10.658 8.97, 11.03

Explanation:

Step1: Find the critical value \( z_{\alpha/2} \)

For a 99% confidence interval, \( \alpha=1 - 0.99=0.01 \), so \( \alpha/2=0.005 \).
Looking up in the standard normal table (or using a calculator), \( z_{0.005}\approx 2.576 \)

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 \( \sigma = 2 \), \( n = 25 \), and \( z_{\alpha/2}=2.576 \)
\( E=2.576\times\frac{2}{\sqrt{25}}=2.576\times\frac{2}{5}=2.576\times0.4 = 1.0304 \)

Step3: Calculate the confidence interval

The confidence interval for the population mean \( \mu \) is \( \bar{x}-E<\mu<\bar{x} + E \)
Given \( \bar{x}=10 \)
\( 10-1.0304<\mu<10 + 1.0304 \)
\( 8.9696<\mu<11.0304 \approx[8.97,11.03] \) (rounded to three decimal places for the critical - value - based calculation as per the problem's hint intention, note that if we use more precise critical - value calculation:
If \( z_{0.005}\) is calculated more precisely (using a calculator with more decimal places, say \( z_{0.005}\approx2.5758\))
\( E = 2.5758\times\frac{2}{5}=2.5758\times0.4=1.03032\)
\(10 - 1.03032=8.96968\approx8.97\), \(10 + 1.03032 = 11.03032\approx11.03\))

Answer:

[8.97, 11.03]