Question

Suppose a random sample of 36 is selected from a population with a standard deviation of 12. If the sample mean is 98, the 99% confidence Interval to estimate the population mean is what.

Answer 1

Sample size = n = 36

Sample mean = $\bar{x}$ = 98

Population mean = $\mu$

Population standard deviation = $\sigma$ = 12

100(1-$\alpha$)% confidence interval to estimate $\mu$ is calculated as follows :-

$\therefore$ 100(1-$\alpha$)% confidence interval to estimate $\mu$ is given as :-  $\left [ \bar{x}-\frac{\sigma}{\sqrt{n}} Z_{\alpha/2}\;,\;\bar{x}+\frac{\sigma}{\sqrt n} Z_{\alpha/2} \right ]$

Given:-  100(1-$\alpha$)% = 99% $\Rightarrow$$\alpha$ = 0.01 $\Rightarrow$$\alpha$/2 = 0.005

$\therefore$ 99% confidence interval to estimate the population mean is given as :-  

  $\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\left [ \bar{x}-\frac{\sigma}{\sqrt{n}} Z_{0.005}\;,\;\bar{x}+\frac{\sigma}{\sqrt n} Z_{0.005} \right ]=\left [ 98-\frac{12}{\sqrt {36}}\times2.576\;,\;98+\frac{12}{\sqrt {36}}\times2.576 \right ]=[92.848\;,103.152]$

[The Z-scores are obtained from the Standard Normal table]