The mean size of commercial loans by a bank has been $60,000 in the past. A rece

Question

The mean size of commercial loans by a bank has been $60,000 in the past. A recent change in the bank's credit policy allows larger amounts to be borrowed under the same terms. The credit manager now wishes to test whether the mean size of commercial loans made since the policy changes is larger than $60,000. The manager wishes to control the alpha risk at 0.01 when mu=$60,000. A random sample of n=144 loans made since the policy change yielded the following results x=$68,100, s=$45,000. i. Conduct the test. State the alternatives, the decision rule, the value standardised test statistic and the conclusion. ii. Calculate the p-value of the test and conduct the required test. Interpret its meaning here.

Explanation / Answer

1) H0:mu<=60,000 (mean chnage is atmost $60,000)

H1:mu>60,000 (mean chnage is larger than $60,000)

Perform 1-sample Z test.

Z=(xbar-mu)/(s/sqrt N-1), where xbar is sample mean, s is sample stnadrd deviation and N i ssample size.

=(68100-60000)/(45000/sqrt 144-1)

=2.15

Z critical at alpha=0.01=2.58

Reject null hypothesis if Z test statistic> Z critical.

The test statistic do not fall in critical region. Fail to reject H0, insufficient sample evidence to conclude that mean size of commericial loans made since policy change is larger than $60000.

ii). p value is 0.0157. The p value is not less than alpha=0.01. Fail to reject H0, insufficient sample evidence to conclude that mean size of commericial loans made since policy change is larger than $60000.

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