Answer to Question #164911 in Statistics and Probability for Sowmya

Question

A university administrator would like to estimate the true mean salary of all professors at the university. She uses $4000 as the population standard deviation of the salary of all professors at the university. What sample size would be required so that a 90% confidence interval for 𝜇 has a length of $2000?

Answer 1

Given,

Standard deviation $\sigma=$ $$4000$

Mean salary $\mu=$ $$2000$

Confidence level=$90$ %

Let $n$ be the sample size

The value of z at 90% confidence level is 1.645

As we know, $z=\dfrac{x-\mu}{\sigma}$

$1.645=\dfrac{x-2000}{4000}$

$x=1.64\times4000+2000=8560$ $

Also The confidence interval is given by-

$x=\mu\pm z\dfrac{\sigma}{\sqrt{n}}$

$\implies 8560=2000\pm 1.645\times \dfrac{4000}{\sqrt{n}}$

$\implies \dfrac{4000}{\sqrt{n}}=\dfrac{8560-2000}{1.645}$

$\implies \sqrt{n}=\dfrac{4000\times 1.645}{6560}=1.003$

$\implies n=1.01$