Question

A buffet restaurant is considering increasing its current price for its luncheon buffet from $10 to $12. But first it wants to estimate, with 95% confidence, the true average cost of food eaten per customer. To estimate this, it randomly selected 25 customers eating lunch at its restaurant and determined how much it cost for food for each of these customers. Based on this sample, it calculated the average cost of food for these customers to be $6.75 and the standard deviation in costs per customer to be $2.25. Based on this information, determine the following:

• estimate, with 95% confidence, the true average cost of food per customer.

• what is the margin of error based on this sample?

• if the restaurant wanted the margin of error to be no greater than $0.50, what would the minimum sample size need to be?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 6.75

sample standard deviation = s = 2.25

sample size = n = 25

Degrees of freedom = df = n - 1 = 25 - 1 = 24

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t_0.025,24 = 2.064

Margin of error = E = t/2,df * (s /n)

= 2.064 * (2.25 / 25)

Margin of error = E = 0.93

The 95% confidence interval estimate of the population mean is,

  ± E  

= 6.75 ± 0.93

= ( $ 5.82, $ 7.68 )

Margin of error = E = t/2,df * (s /n)

= 2.064 * (2.25 / 25)

Margin of error = E = $ 0.93

margin of error = E = 0.50

sample size = n = [t/2,df* s / E]²

n = [2.064 * 2.25 / 0.50 ]²

n = 86.27

Sample size = n = 87