Question

The bad debt ratio for a financial institution is defined to be 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%, 7%, 6%, 5%, 8%, 5%, and 8%.

(a-1) Banking officials claim that the mean bad debt ratio for all Midwestern banks is 3.5 percent and that 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 exceeds 3.5 percent. (Round your answers to 1 decimal place. Omit the "%" sign in your response.)

H0: μ < % versus Ha: μ > %.

(a-2) Discuss the meanings of a Type I error and a Type II error in this situation. Type I: Conclude that Ohio’s mean bad debt ratio is (Click to select)≤> 3.5% when it actually is ≤ 3.5%. Type II: Conclude that Ohio’s mean bad debt ratio is (Click to select)>≤ 3.5% when it actually is > 3.5%.

(b) Assuming that bad debt ratios for Ohio banks are approximately normally distributed, use critical values and the given sample information to test the hypotheses you set up in part a by setting α equal to .01. Also, interpret the p-value of 0.0004 for the test. (Round your answers to 3 decimal places.)

t

t.01

Since t.01 (Click to select)<> t, (Click to select)do not reject or reject H0 .

Answer 1

a-1) Claim: The mean bad ration for Ohio banks is higher that is exceeds 3.5%

That is, , the claim contains more than sign so it will be under the alternative hypothesis.

The null and alternative hypothesis are:

a-2) Meanings of type I and type II error.

Type I error: It is the probability of rejecting the null hypothesis actually it is true.

Type I error concludes that Ohio's mean bad debt ratio is > 3.5% when it actually is %.

Type II error: It is the probability of failing to reject the null hypothesis when it actually false.

Type II error conclude that Ohio's mean bad debt ratio is % when it actually is > 3.5%.

b)

Here the sample of size 7 is given and the population standard deviation is unknown so t-test is useful.

The null and alternative hypothesis is already written.

t-test statistics:

The formula of t-test statistics is,

Where,

s - sample standard deviation

The formula to find the sample standard deviation is,

T-test statistics = 6.090

t - critical value: The alternative hypothesis contains greater than sign, so the test is a right-tailed test.

Alpha = level of significance = 0.01 and degrees of freedom = n - 1 = 6

The t critical value using t table at alpha 0.01 in one-tailed with degrees of freedom 6 is 3.143

Decision rule: If t critical value > t-test statistics then fail to reject the null hypothesis otherwise reject the null hypothesis.

Here t critical value 3.143 is not > than t-test statistics 6.090, so reject the null hypothesis. That is

t_0.01 < t-test statistics so reject H₀.

Conclusion: Reject the null hypothesis, that is there is sufficient evidence to support the claim that the mean bad debt is higher than 3.5%.

P-value = 0.0004

P-value is the smallest level of significance lead to rejecting the null hypothesis. Here P-value 0.0004 is less than alpha that is 0.01, so reject the null hypothesis.