Question

a manager of a technology start-up company wishes to estimate monthly expenses using a 98% confidence interval. she collected a random sample of 60 months and calculated average expenses of $2684 per month. assume σ=8 is known. what is the 98% confidence interval of the average monthly expense?

Answer 1

Given that, sample size ( n ) = 60

sample mean $(\bar x)$ = $2684

population standard deviation $(\sigma)$ = 8

A 98% confidence level has significance level $(\alpha)$ = 0.02 and critical value is, $Z_{\alpha/2} = 2.33$

98% confidence interval for population mean $(\mu)$ is,

$\bar x - Z_{\alpha/2} * \frac {\sigma}{\sqrt { n }} < \mu < \bar x &plus; Z_{\alpha/2} * \frac {\sigma}{\sqrt { n }}$

$2684 - 2.33 * \frac {8}{\sqrt { 60 }} < \mu < 2684 &plus; 2.33 * \frac {8}{\sqrt { 60 }}$

$2684 - 2.41 < \mu < 2684 &plus; 2.41$

$2681.59 < \mu < 2686.41$

$\$ 2681.59 < \mu < \$ 2686.41$

Therefore, the 98% confidence interval of the average monthly expense is between $2681.59 to $2686.41