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=(\sigma)/(√(n)) =(20)/(√(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)`