Question

The manager of a pizza chain in Albuquerque, New Mexico, wants to determine the average size of their advertised 19-inch pizzas. She takes a random sample of 29 pizzas and records their mean and standard deviation as 19.10 Inches and 120 inches, respectively. She subsequently computes the 99% confidence interval of the mean size of all pizzas as [18.53, 19.67). However, she finds this interval to be too broad to implement quality control and decides to reestimate the mean based on a bigger sample. Using the standard deviation estimate of 1.20 from her earlier analysis, how large a sample must she take if she wants the margin of error to be under 0.5 Inch? (You may find It useful to reference the z table. Round Intermediate calculations to at least 4 decimal places and "z" value to 3 decimal places. Round up your answer to the nearest whole number.) Sample size es

Answer 1

Solution :

Z_$\alpha$/2 = Z_0.005 = 2.576  

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

n = [ 2.576 * 1.20 / 0.5]²

n = 38.24

Sample size = n = 39