Question

A sample of size 36 from a normally distributed population variable with population standard deviation 20 had a sample mean of 88. find a 90% confidence interval for the population mean.

Answer 1

Given the sample mean is $\mu=88$ and population standard deviation is $\sigma=20$.

Now the standard deviation of the sample mean is $\sigma_X=\frac{\sigma}{\sqrt{n}} =\frac{20}{\sqrt{36}}=3.3333$

The $(1-\alpha )100\%$ confidence interval for population mean is

$\mu_X \pm z_{\alpha /2}\sigma_X$.

Here $\alpha =0.1,z_{\alpha /2}=-1.645$

The 90% CI for population mean is

$88 \pm 1.645 (3.3333)\\ (82.517,93.483)$