Step-by-step explanation:
From the given information:
For trained subjects:
sample size
= 1200
The sample mean
= 789
For non-trained subjects:
Sample size
= 1200
The sample mean = 632
For trained subjects, the proportion who repaid the loan is:
![\hat p_1 = \dfrac{x_1}{n_1}](/tpl/images/1008/1836/5e2a4.png)
![\hat p_1 = \dfrac{789}{1200}](/tpl/images/1008/1836/59e33.png)
![\hat p_1 = 0.6575](/tpl/images/1008/1836/5f108.png)
For non-trained loan takers, the proportion who repaid the loan was:
![\hat p_2 = \dfrac{x_2}{n_2}](/tpl/images/1008/1836/d7b8f.png)
![\hat p_2 = \dfrac{632}{1200}](/tpl/images/1008/1836/ea795.png)
![\hat p_2 = 0.5266](/tpl/images/1008/1836/dbd55.png)
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 ]](/tpl/images/1008/1836/1de83.png)
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} \times \sqrt{\dfrac{\hat p_1 (1- \hat p_1) }{n_1} + \dfrac{\hat p_2 (1- \hat p_2)}{n_2} }](/tpl/images/1008/1836/c9baf.png)
![=1.96 \times \sqrt{\dfrac{0.658 (1- 0.658) }{1200} + \dfrac{0.527 (1- 0.527)}{1200} }](/tpl/images/1008/1836/5b945.png)
![= 1.96 \times \sqrt{\dfrac{0.658 (0.342) }{1200} + \dfrac{0.527 (0.473)}{1200} }](/tpl/images/1008/1836/39f67.png)
![= 1.96 \times \sqrt{1.8753 \times 10^{-4}+2.07725833 \times 10^{-4} }](/tpl/images/1008/1836/fa411.png)
= 1.96 × 0.019881
≅ 0.039
The lower limit = ![( \hat p_1 - \hat p_2) - E](/tpl/images/1008/1836/e7eca.png)
= (0.658 - 0.527) - 0.0389
= 0.131 - 0.0389
= 0.092
The upper limit = ![( \hat p_1 - \hat p_2) + E](/tpl/images/1008/1836/4be56.png)
= (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](/tpl/images/1008/1836/0fe01.png)
For this study;
The null hypothesis is:
![H_o : p_1 -p_2 = 0](/tpl/images/1008/1836/0e832.png)
The alternative hypothesis is:
![H_a : p_1 -p_2 \ne 0](/tpl/images/1008/1836/1cd8f.png)
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
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.