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 manager assigned you the task of conducting a survey on cheese burgers at a regional 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 0.17 pound. It is required that the maximum likely sampling error of mean weight be 0.016 pound. A cheese burger at the restaurant chain costs $1.49. What is the budget in dollars (2 decimals) to have the cheese burgers in your sample, assuming no sales tax?

Answer 1

Answer:

$454.45

Explanation:

Given that,

Confidence interval = 90%

From the Z-table,

$Z_{0.90} =1.645$

Standard deviation of weight (SD) = 0.17 pound

Sampling error of mean weight (SE) = 0.016 pound

Therefore,

$n=(\frac{Z_{0.90}\times SD }{SE} )^{2}$

$n=(\frac{1.645\times 0.17 }{0.016} )^{2}$

$=(17.478125)^{2}$

= 305.4

n ≅ 305 (approx)

Thus, the needed sample size is 305.

Budget in dollars = 305 × $1.49

                            = $454.45