Question

The manager of a pizza chain in Albuquerque, New Mexico, wants to determine the average size of their advertised 20-inch pizzas. She takes a random sample of 28 pizzas and records their mean and standard deviation as 20.20 inches and 1.40 inches, respectively. She subsequently computes the 95% confidence interval of the mean size of all pizzas as [19.68, 20.72]. 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 1.40 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.)

Answer 1

Solution :

Given that,

Population standard deviation = = 1.40

Margin of error = E = 0.5

At 95% confidence level the z is,

= 1 - 95%

= 1 - 0.95 = 0.05

/2 = 0.025

Z/2 = Z0.025 = 1.96  

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

n = [ 1.96 * 1.40 / 0.5]2

n = 30.11

Sample size = n = 31