A department store manager wants to estimate with a 99% confidence interval the mean amount spent by all customers at the store. How large of a sample should be taken if the manager is willing to tolerate an error of $3. Assume the population standard deviation is $31.
Solution :
Given that,
standard deviation =s = =31
Margin of error = E = 3
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* / E] 2
n = ( 2.576* 31 / 3 )2
n =708.55
n=709
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