Question

suppose we know the population standard deviation $sigma = 2$. we take a random sample of size 25 and calculate the sample mean $overline{x}=10$. what is the 95% 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 8.968, 11.032

Explanation:

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

For a 95% confidence interval, \( \alpha=1 - 0.95=0.05 \), and \( \alpha/2=0.025 \). Using the standard normal distribution table or a calculator, \( z_{0.025}\approx1.960 \) (rounded to three decimal places as \( 1.960 \)).

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 \( z_{\alpha/2}=1.960 \), we have \( E=1.960\times\frac{2}{\sqrt{25}}=1.960\times\frac{2}{5}= 0.784 \).

Step3: Calculate the confidence interval

The confidence interval for the population mean \( \mu \) is \( \bar{X}-E<\mu<\bar{X} + E \). Given \( \bar{X}=10 \), we get \( 10 - 0.784<\mu<10 + 0.784 \), which is \( 9.216<\mu<10.784 \).

Answer:

[9.216, 10.784]