Question

the monthly earnings of a group of business students are normally distribution with a standard devition of 345 dollars. a researcher wants to estimate the mean monthly earnings of all business students. find the sample size needed to have a confidence level of 89% and a margin of error of 118 dollars.

The monthly earnings of a group of business students are normally distributed with a standard deviation of 345 dollars. A researcher wants to estimate the mean monthly earnings of all business students. Find the sample size needed to have a confidence level of 89% and a margin of error of 118 dollars. Select one: a. 39 b.52 C. 22 O 2.42

Answer 1

Option C.

22

The sample size is given by,

n = ( Z$\alpha$/2* $\sigma$ / E) 2​​​​​​

Z$\alpha$​​​​​​​/2 : Z value from table for confidence level

$\alpha$ = level of significance

=1 - confidence level

= 1 - 0.89

= 0.11

E : margin of error

$\sigma$ : standard deviation

n = (Z0.055 * 345 / 118) 2​​​​​​

= ( 1.59 * 345 / 118) 2​​​​​​

n = 21.61

n = 22