Question

7. Pete’s Thai Bistro in Dubuque collected data from a sample of 93 customers to estimate the average amount spent per customer for dinner. The sample average was $48. The sample standard deviation was s = $6.

(a) Working at the 90% confidence level, what is the margin of error for the population average?

(b) Working at the 90% confidence level, what is the confidence interval for the population average?

Answer 1

Solution :

Given that,

= 48

s = 6

n = 93

Degrees of freedom = df = n - 1 = 93 - 1 = 92

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t /2,df = t0.05,92 =1.662

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

= 1.662* (6 / 93)

= 1.03

Margin of error = 1.03

b ) The 90% confidence interval estimate of the population mean is,

- E < < + E

48 - 1.03 < < 48 + 1.03

46.97 < < 49.03

(46.97, 49.03 )