Question

You want to obtain a sample to estimate how much parents spend on their kids birthday parties. Based on previous study, you believe the population standard deviation is approximately o = 45.1 dollars. You would like to be 98% confident that your estimate is within 2.5 dollar(s) of average spending on the birthday parties. How many parents do you have to sample? Get help: Video Written Example

Answer 1

Solution :

Given that,

standard deviation =  $\sigma$ =45.1

Margin of error = E = 2.5

At 98% confidence level the z is ,

$\alpha$ = 1 - 98% = 1 - 0.98 = 0.02

$\alpha$ / 2 = 0.02/ 2 = 0.01

Z$\alpha$/2 = Z0.01 = 2.326 ( Using z table    )

sample size = n = [Z$\alpha$/2* $\sigma$ / E] 2

n = ( 2.326*45.1 / 2.5 )2

n =1760.72

Sample size = n =1761

( when you have z value 2 decimal Z0.01 = 2.33 when n=1766.78=1767)