The bad debt ratio for a financial institution is defined to be the dollar value

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 7%, 6%, 9%, 8%, 9%, 8%, and 6%.

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.)

Discuss the meanings of a Type I error and a Type II error in this situation.

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 equal to .01. Also, interpret the p-value of 0.0001 for the test. (Round your answers to 3 decimal places.)

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 7%, 6%, 9%, 8%, 9%, 8%, and 6%.

Explanation / Answer

Question a-1)

H0: µ 3.5% versus Ha: µ > 3.5%

Question a -2)

Type I Conclude the mean bad debt ratio is > when it actually is not.

Type II Conclude the mean bad debt is <= 3.5% when it is actually not.

Question b )

x (%)

7

6

9

8

9

8

6

We go to excel and use =average( ) and =stdev() formula to get the sample mean and sample standard deviation.

X bar = 7.5714

Sample standard deviation ( s ) = 1.2724

Test Statistics:

t = ( x bar – Mew)/(s/sqrt(n))

= (7.5714-3.5)/( 1.2724/sqrt(7))

= 8.466

t = 8.466

t .01 = 3.143

Since t.01 is < reject H0

p-value = .0001; p-value is less than the level of significance (.01); we reject the null hypothesis.

x (%)

7

6

9

8

9

8

6

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