Question

The bad debt ratio for a financial institution is defined to the dollar value of loans defaulted divided by the total dollar value of all loans made. Suppose that a random sample of 7 Ohio banks is selected and that the bad debt ratios (written as percentages) for these banks are 6%, 8%, 5%, 9%, 7%, 5% and 8%. Banking officials claim that the mean debt ratio for all Midwestern banks is 3.5 percent and the mean bad debt ratio for Ohio banks is higher. Set up the null and alternative hypotheses needed to attempt to provide evidence supporting the claim that the mean bad debt ratio for Ohio banks exceed 3.5 percent. (a) H0: μ ≤ [ Select ] % versus Ha: μ > [ Select ] %. (b) Discuss the meanings of a Type I error and a Type II error in this situation. Type I : Conclude that Ohio's mean debt ratio is [ Select ] 3.5 % when it actually is 3.5%. Type II : Conclude that Ohio's mean debt ratio is [ Select ] 3.5 % when it actually is 3.5%.

Answer 1

Answer:

Step-by-step explanation:

From the information given:

Consider X to be the random variable denoting the bad debt ratios for Ohio Bank.

Then, $X \sim N ( \mu, \sigma ^2)$

Thus the null hypothesis and the alternative can be computed as:

Null hypothesis:

$H_o : \mu \leq3.5\%$

Alternative hypothesis

$H_1 : \mu > 3.5\%$

The type I and type II error is as follows:

Type I:

The mean bad debt ratio is > 3.5% when it is not

Type II:

The mean bad debt ratio is ≤ 3.5% when it is not.

The test statistics can be calculated by using the formula:

$t = \dfrac{\overline x - \mu}{\dfrac{\sigma}{\sqrt{n}}}$

where;

sample size n = 7

mean = 6+8+5+9+7+5+8 = 48

sample mean $\overline x =\dfrac{48}{7}$

$\overline x$ = 6.86

sample standard deviation is :

$s = \sqrt{\dfrac{\sum( x -\overline x)^2}{n-1}}$

$s = \sqrt{\dfrac{( 6 -6.86)^2+( 8-6.86)^2+ ( 5 -6.86)^2 + ...+( 7 -6.86)^2+ ( 5 -6.86)^2+( 8 -6.86)^2 }{7-1}}$

s = 1.573

population  mean $\mu = 3.5$

Therefore, the test statistics is :

$t = \dfrac{\overline x - \mu}{\dfrac{\sigma}{\sqrt{n}}}$

$t = \dfrac{6.86- 3.5}{\dfrac{1.573}{\sqrt{7}}}$

$t = \dfrac{3.36}{\dfrac{1.573}{2.65}}$

$t = \dfrac{3.36\times {2.65}}{{1.573}}$

t = 5.660

At significance level of 0.01

$t_{0.01} = 3.707$  

P - value = P(T > 5.66)

P - value = 1 - (T < 5.66)

P - value = 1 - 0.9993

P-value = 0.0007

Therefore, since $t_{0.01} < t$ , we reject the null hypothesis and conclude that the claim that the mean bad debt ratio for Ohio banks is higher than the mean for all financial institutions is true.