Question

A fast food restaurant executive wishes to know how many fast food meals adults eat each week. They want to construct a 90%90% confidence interval with an error of no more than 0.070.07. A consultant has informed them that a previous study found the mean to be 7.47.4 fast food meals per week and found the variance to be 0.810.81. What is the minimum sample size required to create the specified confidence interval? Round your answer up to the next integer.

Answer 1

Solution :

Given that,

standard deviation =   =0.81=0.9

Margin of error = E = 0.07

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

sample size = n = [Z/2* / E] 2

n = ( 1.645 * 0.9 / 0.07 )2

n =447.3225

Sample size = n =448