Question

s defined to be the dollar value of loans defaulted divided by the total dollar value of all loans made. Banking officials claim that the mean bad debt ratio for all Midwestern banks is 3.5%, but that Ohio’s is different. Suppose that a random sample of seven Ohio banks is selected and that the bad debt ratios for these banks are: 7%, 4%, 6%, 3%, 5%, 4%, and 2%. Assuming that bad debt ratios are approximately normal, test at the 0.05 level of significance whether the mean bad debt ratio for Ohio banks is different than the Midwestern average.H0: Mu = 3.5Ha: Mu neq 3.5Hint: data <- c(7, 4, 6, 3, 5, 4, 2) # you can find xbar=mean(data) and s=sd(data) of the sample in R.a) What type of hypothesis test do you need to use? (3 points)b) Determine the test statistic? (3 points)c) Determine the critical value? Stat your conclusion. (5 points)d) Determine the p-value? State your conclusion. (4 points)

Answer 1

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the mean bad debt ratio for Ohio banks is different than the Midwestern average (3.5%).

Test statistic t = 1.431

Critical values tc = ±2.447

P-value = 0.203

Explanation:

This is a hypothesis t-test for the population mean.

The claim is that the mean bad debt ratio for Ohio banks is different than the Midwestern average (3.5%).

Then, the null and alternative hypothesis are:

H_0: \mu=3.5\\\\H_a:\mu\\eq 3.5

The significance level is 0.05.

The sample has a size n=7.

We calculate the sample mean and standard deviation as:

`M=(1)/(n)\sum_(i=1)^n\,x_i\\\\\\M=(1)/(7)(7+4+6+3+5+4+2)\\\\\\M=(31)/(7)\\\\\\M=4.43\\\\\\`

`s=\sqrt{(1)/(n-1)\sum_(i=1)^n\,(x_i-M)^2}\\\\\\s=\sqrt{(1)/(6)((7-4.43)^2+(4-4.43)^2+(6-4.43)^2+. . . +(2-4.43)^2)}\\\\\\s=\sqrt{(17.71)/(6)}\\\\\\s=√(2.95)=1.72\\\\\\`

The sample mean is M=4.43.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1.72.

The estimated standard error of the mean is computed using the formula:

`s_M=(s)/(√(n))=(1.72)/(√(7))=0.65`

Then, we can calculate the t-statistic as:

`t=(M-\mu)/(s/√(n))=(4.43-3.5)/(0.65)=(0.93)/(0.65)=1.431`

The degrees of freedom for this sample size are:

`df=n-1=7-1=6`

This test is a two-tailed test, with 6 degrees of freedom and t=1.431, so the P-value for this test is calculated as (using a t-table):

`\text{P-value}=2\cdot P(t>1.431)=0.203`

As the P-value (0.203) is bigger than the significance level (0.05), the effect is not significant.

If we use the critical value approach, for this level of confidence, the critical values are tc = ±2.447. The test statistic is within the bounds of the critical values and falls within the acceptance region.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the mean bad debt ratio for Ohio banks is different than the Midwestern average (3.5%).