56.8k views
4 votes
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.

1 Answer

7 votes


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)

User Jeffrey Wilges
by
8.7k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories