Answer:
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}](https://tex.z-dn.net/?f=%5Chat%20p_1%20%3D%20%5Cdfrac%7Bx_1%7D%7Bn_1%7D)
![\hat p_1 = \dfrac{789}{1200}](https://tex.z-dn.net/?f=%5Chat%20p_1%20%3D%20%5Cdfrac%7B789%7D%7B1200%7D)
![\hat p_1 = 0.6575](https://tex.z-dn.net/?f=%5Chat%20p_1%20%3D%200.6575)
For non-trained loan takers, the proportion who repaid the loan was:
![\hat p_2 = \dfrac{x_2}{n_2}](https://tex.z-dn.net/?f=%5Chat%20p_2%20%3D%20%5Cdfrac%7Bx_2%7D%7Bn_2%7D)
![\hat p_2 = \dfrac{632}{1200}](https://tex.z-dn.net/?f=%5Chat%20p_2%20%3D%20%5Cdfrac%7B632%7D%7B1200%7D)
![\hat p_2 = 0.5266](https://tex.z-dn.net/?f=%5Chat%20p_2%20%3D%200.5266)
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 ]](https://tex.z-dn.net/?f=%5B%20%28%20%5Chat%20p_1%20-%20%5Chat%20p_2%20%29%20-%20E%20%5C%20%2C%20%5C%20%28%5Chat%20p_1%20-%20%5Chat%20p_2%20%29%20%2B%20E%20%20%5D)
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} }](https://tex.z-dn.net/?f=Z_%7B%5Calpha%2F2%7D%20%5Ctimes%20%5Csqrt%7B%5Cdfrac%7B%5Chat%20p_1%20%281-%20%5Chat%20p_1%29%20%7D%7Bn_1%7D%20%2B%20%5Cdfrac%7B%5Chat%20p_2%20%281-%20%5Chat%20p_2%29%7D%7Bn_2%7D%20%7D)
![=1.96 \times \sqrt{\dfrac{0.658 (1- 0.658) }{1200} + \dfrac{0.527 (1- 0.527)}{1200} }](https://tex.z-dn.net/?f=%3D1.96%20%5Ctimes%20%5Csqrt%7B%5Cdfrac%7B0.658%20%281-%200.658%29%20%7D%7B1200%7D%20%2B%20%5Cdfrac%7B0.527%20%281-%200.527%29%7D%7B1200%7D%20%7D)
![= 1.96 \times \sqrt{\dfrac{0.658 (0.342) }{1200} + \dfrac{0.527 (0.473)}{1200} }](https://tex.z-dn.net/?f=%3D%201.96%20%5Ctimes%20%5Csqrt%7B%5Cdfrac%7B0.658%20%280.342%29%20%7D%7B1200%7D%20%2B%20%5Cdfrac%7B0.527%20%280.473%29%7D%7B1200%7D%20%7D)
![= 1.96 \times \sqrt{1.8753 \times 10^{-4}+2.07725833 \times 10^{-4} }](https://tex.z-dn.net/?f=%3D%201.96%20%5Ctimes%20%5Csqrt%7B1.8753%20%5Ctimes%2010%5E%7B-4%7D%2B2.07725833%20%5Ctimes%2010%5E%7B-4%7D%20%7D)
= 1.96 × 0.019881
≅ 0.039
The lower limit = ![( \hat p_1 - \hat p_2) - E](https://tex.z-dn.net/?f=%28%20%5Chat%20p_1%20-%20%5Chat%20p_2%29%20-%20E)
= (0.658 - 0.527) - 0.0389
= 0.131 - 0.0389
= 0.092
The upper limit = ![( \hat p_1 - \hat p_2) + E](https://tex.z-dn.net/?f=%28%20%5Chat%20p_1%20-%20%5Chat%20p_2%29%20%2B%20E)
= (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](https://tex.z-dn.net/?f=%3D0.092%20%5Cle%20p_1-p_2%20%5Cle%200.167)
For this study;
The null hypothesis is:
![H_o : p_1 -p_2 = 0](https://tex.z-dn.net/?f=H_o%20%3A%20p_1%20-p_2%20%3D%200)
The alternative hypothesis is:
![H_a : p_1 -p_2 \ne 0](https://tex.z-dn.net/?f=H_a%20%3A%20p_1%20-p_2%20%5Cne%200)
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.