Statistic using R

Question

In order to adjust the values ​​of the food vouchers of the 9.443 employees of your company, a study will be carried out with the objective of evaluating the average monthly expenditure in supermarkets and bakeries (places that generally accept food vouchers) of its employees. It is known that the standard deviation of this monthly expense for all company employees is $11.42.

How many employees should be selected so that the estimate of the average monthly expenditure in supermarkets and bakeries does not deviate from the real average by more than $1.68 with 98% confidence?

Answer 1

Error = 1.68 and the Z_level for 98% confidence

is the z-score whose probability is 0.99 for 2-tails.

So Z_level = 2.326348

Error = Z_level * stndDev/ sqrt(N)

Error * sqrt(N) = Z_level * stndDev

sqrt(N) = Z_level* stndDev / Error

N = [ Z_level * stndDev / Error ] ^2

= [ 2.326348 * 11.42 / 1.68 ] ^2

= 250 +

just over 250, or 251