Question

10. A bank wonders whether omitting the annual credit card fee for customers who charge at least $2,500 in a year will increase the amount charged on its credit cards. The bank makes this offer to a simple random sample of 200 of its credit card customers. It then compares how much these customers charge this year with the amount they charged last year. The mean increase in the sample is $346 and the standard deviation is $112. Give a 90% confidence interval for the mean amount charges would have increased if this benefit had been extended to all such customers. Round CI limits to two decimal places.

Answer 1

Solution:

Given that, σ= $112, n= 200
x̄= $346
(1–α)%=90%
α=0.10
α/2 =0.05
Zα/2 =1.645 ....from standard normal table.
Margin of error=E=Zα/2 ×(σ/✓n)
=1.645 ×(112/✓200)
=1.645 × 7.9196
=13.0277
Margin of error=E=13.0277
90% confidence interval for the mean amount charges would have increased if this benefit had been extended to all such customers is given as,
x̄± Margin of error=(346 -13.0277,346+13.0277)
=(332.9723,359.0277)
=(332.97,359.03) ....upto 2 decimal place.
Lower limit =332.97
Upper limit=359.03
90% confidence interval for the mean amount charges would have increased if this benefit had been extended to all such customers is
(332.97,359.03)