Question

QUESTION 05 (10 points) - Sample Size Determination in Estimating Population Mean In this hypothetical scenario, you are working for a consumer protection agency. Your boss assigned you the task of conducting a survey on cheese burgers at a national restaurant chain. In your report, you would need to construct a 90% confidence interval for the mean weight of cheese burgers. Based on past surveys on cheese burgers at this restaurant chain, the standard deviation of weight is 0.21 pound. It is required that the maximum likely sampling error of mean weight be 0.05 pound. A cheese burger at the restaurant chain costs $1.29. What is the required budget in dollars (2 decimals) to have the cheese burgers in your sample, assuming no sales tax

Answer 1

Answer:

$ 37.32 is the required budget in dollars (2 decimals) to have the cheese burgers in your sample

Step-by-step explanation:

Here

Z∝ =1.64

Standard Deviation=S= 0.21

Sampling Error= SE= 0.05

As Sampling Error is found by

SE= z * s/√n

√n= z*s/ SE

n= (z* s/ SE)²

We use the above formula  to calculate N

N= ( Z * S/SE)²

N= (1.64 * 0.21/0.05)²

N= 28.9296= 28.93

Cost of N number of burgers

= Cost *N

= $1.29 * 28.93= $ 37.32 is the required budget in dollars (2 decimals) to have the cheese burgers in your sample