Question

A corn farmer wants to determine a 99% confidence interval estimate for the mean number of pounds of corn that he harvest each day. Assuming that the farmer reports that the standard deviation of his harvests is 100 pounds, how many days should he sample so that the margin of error will be 40.0 pounds or less?

Answer 1

Solution :

Given that,

standard deviation =  σ =100

Margin of error = E = 40.0

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2  = 2.576

sample size = n = [ Z/2 *  σ/ E]2

n = ( 2.576* 100 / 40.0 )2

n =41.47

Sample size = n =42