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# A sample of 250 observations is selected from a normal population for which the population standard deviation is known to be 25. The sample mean is 20. a. Determine the standard error of the mean. b. Explain why we can use formula to determine the 95 percent confidence interval. c. Determine the 95 percent confidence interval for the population mean

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A sample of $n=250$ observations is taken from a normal population with a standard deviation of $\sigma=25$. Sample mean is $\bar{X}=20$.

$\textbf{a. Standard error of the mean}$

\begin{align*} \sigma_{\bar{X}}&=\sigma/ \sqrt{n}\\ &=25 / \sqrt{250}\\ &=25/15.81\\ &=1.58\\ \end{align*}

$\textbf{b. }$The central limit theorem says that the sampling distribution of the mean becomes a normal distribution as sample size increases. A sample of 250 observations is large enough to assume that the sampling distribution will follow the normal distribution.

$\textbf{c. 95 percent Confidence Interval }$

For a 95 percent level of confidence, the value of $z$ is obtained by locating $0.95/2=0.4750$ in the table given in Appendix B1. The value of $z=1.96$.

\begin{align*} \bar{X} \pm z\frac{\sigma}{\sqrt{n}}&=20 \pm 1.96 \times \frac{25}{\sqrt{250}}\\ &=20 \pm 1.96 \times 1.58\\ &=20 \pm 3.097 \end{align*}

So, $X_1 =20-3.097=16.903$ and $X_2=20+3.097=23.097$. So, $95$ percent confidence interval for population mean is $(16.903, 23.097)$.

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