Question

According to a recent survey, the average earnings in a certain city were $41,500 in Educational Services and $77,800 in the Financial Services industry. a) Suppose the standard deviation of earnings in Educational Services for the whole of the city was $13,100 and we want the standard deviation of average eamings in our survey to be $1000. What sample size do we need to take? b) Suppose we used a sample of size 320 for our estimate of average earnings in Financial Services and the standard deviation of earnings in our survey was $15,300. What is the standard error of the average earnings in Financial Services? a) The sample size needed is |29450 Round up to the nearest whole number.)

Answer 1

a) Given that, standard deviation $(\sigma)$ = $13100

standard deviation of average $(\sigma_{\bar x})$ = $1000

we want to find, the sample size ( n ),

We know,

$\sigma_{\bar x} = \frac {\sigma}{\sqrt { n}}$

$=> \sqrt n = \frac {\sigma}{\sigma_{\bar x}}$

$=> n = (\frac {\sigma}{\sigma_{\bar x}})^2$

$=> n = (\frac {13100}{1000})^2$

$=> n = 171.61$

$=> n \approx 172$

Therefore, the sample size needed is 172

b) Given that, sample size ( n ) = 320

standard deviation ( S ) = $15300

we want to find, the standard error of the mean $(SE_{\bar x})$

$SE_{\bar x} = \frac {S}{\sqrt n } = \frac { 15300}{\sqrt { 320}}= 855.2960$

Therefore, the standard error of the average earnings in Financial Services is 855.2960