Question

A nonprofit bank that provides loans to small entrepreneurs in developing countries is considering organizing a mandatory training on good financial practices for all its loan-takers. To verify whether attending the course affects the chance that borrowers repay their short-term debt, two years ago the bank randomly split 2400 loan applicants into a group of 1200 subjects that received the mandatory training and a group of 1200 subjects that didn't receive it.

The repayments of these 2400 loans were due last month. The bank is now looking at the following table, that indicates how many loan-takers in the two groups have repaid/not repaid their loan:

Repaired Not repaid
Trained 789 411
Non- trained 632 568

a. Of trained subjects, the proportion who repaid the loan was __________
b. Of non-trained loan-takers, the proportion who repaid the loan was ______
c. What is a 95% confidence interval for the difference between the proportion of trained and non- trained loan-takers who repaid the loan?
d. Based on your previous answer, if we conduct a test of the null hypothesis that the proportion of trained and non-trained loan-takers who repaid the loan are the same against the alternative hypothesis that they differ, what decision would we make?

Answer 1

Explanation:

From the given information:

For trained subjects:

sample size
`n_1` = 1200

The sample mean
`x_1` = 789

For non-trained subjects:

Sample size
`n_2` = 1200

The sample mean = 632

For trained subjects, the proportion who repaid the loan is:

`\hat p_1 = (x_1)/(n_1)`

`\hat p_1 = (789)/(1200)`

`\hat p_1 = 0.6575`

For non-trained loan takers, the proportion who repaid the loan was:

`\hat p_2 = (x_2)/(n_2)`

`\hat p_2 = (632)/(1200)`

`\hat p_2 = 0.5266`

The confidence interval for the difference between the given proportion is:

=
`[ ( \hat p_1 - \hat p_2 ) - E \ , \ (\hat p_1 - \hat p_2 ) + E ]`

where;

Level of significance = 1 - C.I

= 1 - 0.95

= 0.05

Z - Critical value at ∝ = 0.05 is 1.96

The Margin of Error (E) =
`Z_(\alpha/2) * \sqrt{(\hat p_1 (1- \hat p_1) )/(n_1) + (\hat p_2 (1- \hat p_2))/(n_2) }`

`=1.96 * \sqrt{(0.658 (1- 0.658) )/(1200) + (0.527 (1- 0.527))/(1200) }`

`= 1.96 * \sqrt{(0.658 (0.342) )/(1200) + (0.527 (0.473))/(1200) }`

`= 1.96 * \sqrt{1.8753 * 10^(-4)+2.07725833 * 10^(-4) }`

= 1.96 × 0.019881

≅ 0.039

The lower limit =
`( \hat p_1 - \hat p_2) - E`

= (0.658 - 0.527) - 0.0389

= 0.131 - 0.0389

= 0.092

The upper limit =
`( \hat p_1 - \hat p_2) + E`

= (0.658 - 0.527) + 0.0389

= 0.131 + 0.0389

= 0.167

Thus, 95% C.I for the difference between the proportion of trained and non-trianed loan takers who repaired the loan is:

`=0.092 \le p_1-p_2 \le 0.167`

For this study;

The null hypothesis is:

`H_o : p_1 -p_2 = 0`

The alternative hypothesis is:

`H_a : p_1 -p_2 \\e 0`

Since the C.I lie between (0.092, 0.17);

And the null hypothesis value does not lie within the interval (0.092, 0.17).

∴

we reject the null hypothesis
`H_o` at ∝(0.05).

Conclusion: We conclude that there is enough evidence to claim that the proportion of trained and non-trained loan takers who repaired the loan are different.