A random sample of 256 credit unions that offer credit cards revealed that the a

Question

A random sample of 256 credit unions that offer credit cards revealed that the average annual fee charged by the credit union was $12.56 with a standard deviation of $2.33. A random sample of 225 federally chartered banks offering credit cards showed that the average annual fee was $22.48 with a standard deviation of $6.18. a) Construct a 90% and 95% confidence interval estimate for the true difference in means between the annual fees charged for credit cards by credit unions and federally chartered banks. b) Could the federally chartered banks be accused of having an average annual fee that is $10 more than that of credit unions? Support your answer using your knowledge of probability and statistics. c) Does the data clearly support the opinion that federally chartered banks have an average annual fee that is more tha $8 above that of credit unions? Use your knowledge of hypothesis testing when answering this question.

Explanation / Answer

Solution

Back-up Theory

Let X = annual fee charged by Credit Union, and Y = annual fee charged by Chartered Federal Banks.

We assume X ~ N(µ1, 12) and Y ~ N(µ2, 22) and 1 = 2 = , say which is unknown.

Suppose we have a sample of n1 observations on X and a sample of n2 observations on Y.

100(1 – ) % confidence interval for (1 - 2) is:

|Xbar – Ybar| ± (t(n1 + n2 - 2), /2)[s{(1/n1) + {(1/n2)}],………………………………..(1)

where Xbar = sample mean of X, Ybar = sample mean of Y,

s2 = {(n1 – 1)s12 + (n2 – 1)s22)/(n1 + n2 – 2),

s12 = sample variance of X = {1/(n - 1)}{sum over i = 1 to n of (xi – Xbar)2}

s22 = sample variance of Y = {1/(n - 1)}{sum over i = 1 to n of (yi – Ybar)2} and

t(n1 + n2 - 1), /2 = upper (/2) % point of t- Distribution with (n1 + n2 - 2) degrees of freedom.

Part (a) 90% CI

Given Xbar = 12.56, Ybar = 22.48, s12 = 2.332, s22 = 6.182, n1 = 256, n2 = 225, = 0.1 and using Excel Function, t479, 0.05 = 1.648

s2 = {(255 x 2.332) + (224 x 6.182)}/475 = 20.7505 and hence, s = 4.555,

90% confidence interval for (1 - 2) is:

(22.48 – 12.56) ± (1.648)[4.555{(1/256) + (1/225)}]

= 9.92 ± 0.686 ANSWER

Part (b) 95% CI

95% confidence interval for (1 - 2) is:

(22.48 – 12.56) ± (1.965)[4.555{(1/256) + (1/225)}]

= 9.92 ± 0.818 ANSWER

Part (c) [level of significance is taken to be 5%]

H0 : (1 - 2) = 10 Vs H0 : (1 - 2) < 10

Test Statistic: t = {(Xbar – Ybar) - 10}/[s{(1/n1) + {(1/n2)}] = - 0.08/0.4162 = - 0.192

Lower 5% point of t479 = - 1.965

Since calculated value of t > - 1.965, H0 is accepted. => there is evidence to suggest that

the federally chartered banks are having an average annual fee that is $10 more than that of credit unions ANSWER

Part (d) [level of significance is taken to be 5%]

H0 : (1 - 2) = 8 Vs H0 : (1 - 2) > 8

Test Statistic: t = {(Xbar – Ybar) - 8}/[s{(1/n1) + {(1/n2)}] = 1.92/0.4162 = 4.613

upper 5% point of t479 = 1.965

Since calculated value of t > 1.965, H0 is rejeted. => there is evidence to suggest that

the federally chartered banks are having an average annual fee that is more than $8 more than that of credit unions ANSWER

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