Question

Credit scores are used by banks, financial institutions, and retailers to determine one's trustworthiness, and their ability to pay their credit card, when he/she is given a credit card. The higher the credit score, the better the credit (and the more financially trustworthy) a consumer is. In light of the current economic downturn, an economist claims that the credit score of Canadians between the ages of 25 to 40 has remained the same. Prior to the current economic downturn, the mean credit score for Canadians between the ages of 25 to 40 was 681681.

A random sample of ?=24n=24 Canadians between the ages of 25 to 40 was taken, the mean was found to be 688, with a standard deviation of 16.6.


(a) Choose the correct statistical hypotheses.

A. ?0:?⎯⎯⎯⎯⎯=681,??:?⎯⎯⎯⎯⎯≠681H0:X¯=681,HA:X¯≠681
B. ?0:?=681,??:?≠681H0:μ=681,HA:μ≠681
C. ?0:?>681,??:?<681H0:μ>681,HA:μ<681
D. ?0:?⎯⎯⎯⎯⎯≥681,??:?⎯⎯⎯⎯⎯<681H0:X¯≥681,HA:X¯<681
E. ?0:?⎯⎯⎯⎯⎯>681,??:?⎯⎯⎯⎯⎯≤681H0:X¯>681,HA:X¯≤681
F. ?0:?≥681??:?<681H0:μ≥681HA:μ<681

(b) Find a 93% confidence interval for ?μ. Use at least 4 digits after the decmial if rounding.

Minimum lower bound =   when ?=7%
Maximum upper bound =    when ?=7%

Answer 1

Solution : -

( a )

The null and alternative hypothesis is ,

H0 :   = 681

Ha : >  681

This is the right tailed test .

( b )

Given that,

Point estimate = sample mean = = 688

Population standard deviation =    = 16.6

Sample size = n = 24

At 93% confidence level

= 1 - 93%  

= 1 - 0.93 = 0.07

/2 = 0.035

Z/2 = Z0.035 = 1.812

Margin of error = E = Z/2 * ( /n)

= 1.812 * ( 16.6 /  24 )

= 6.1399

At 93% confidence interval estimate of the population mean is,

- E < < + E

688 - 6.1399 <   < 688 + 6.1399

681.8601 <   < 694.1399

( 681.8601 , 694.1399 )

The 93% confidence interval estimate of the population mean is :

Minimum lower bound = 681.8601
Maximum upper bound = 694.1399