Answer:
The 95% confidence interval is (-0.2451, 06912)
Step-by-step explanation:
From the question, we have;
The number of small cars in the sample of small cars, n₁ = 12
The number of small cars that were totaled, x = 8
The number of large cars in the sample of small cars, n₂ = 15
The number of large cars that were totaled, y = 5
Therefore, the proportion of small cars that were totaled, pX = x/n₁
∴ pX = 8/12 = 2/3
The proportion of large cars that were totaled, pY = y/n₁
∴ pY = 5/15 = 1/3
The 95% confidence interval for the difference pX - pY is given as follows;
![pX-pY\pm z^{*}\sqrt{\dfrac{pX\left (1-pX \right )}{n_{1}}+\dfrac{pY\left (1-pY \right )}{n_{2}}}](https://tex.z-dn.net/?f=pX-pY%5Cpm%20z%5E%7B%2A%7D%5Csqrt%7B%5Cdfrac%7BpX%5Cleft%20%281-pX%20%20%5Cright%20%29%7D%7Bn_%7B1%7D%7D%2B%5Cdfrac%7BpY%5Cleft%20%281-pY%20%20%5Cright%20%29%7D%7Bn_%7B2%7D%7D%7D)
![\dfrac{2}{3} -\dfrac{1}{3} \pm 1.96 \times \sqrt{\dfrac{\dfrac{2}{3} \times \left (1-\dfrac{2}{3} \right )}{12}+\dfrac{\dfrac{1}{3} \times \left (1-\dfrac{1}{3} \right )}{15}}](https://tex.z-dn.net/?f=%5Cdfrac%7B2%7D%7B3%7D%20-%5Cdfrac%7B1%7D%7B3%7D%20%5Cpm%201.96%20%5Ctimes%20%5Csqrt%7B%5Cdfrac%7B%5Cdfrac%7B2%7D%7B3%7D%20%5Ctimes%20%5Cleft%20%281-%5Cdfrac%7B2%7D%7B3%7D%20%20%20%5Cright%20%29%7D%7B12%7D%2B%5Cdfrac%7B%5Cdfrac%7B1%7D%7B3%7D%20%5Ctimes%20%5Cleft%20%281-%5Cdfrac%7B1%7D%7B3%7D%20%20%20%5Cright%20%29%7D%7B15%7D%7D)
Therefore, we have;
![\therefore 95\% \ CI = \dfrac{1}{3} \pm 0.3578454](https://tex.z-dn.net/?f=%5Ctherefore%2095%5C%25%20%5C%20%20CI%20%3D%20%5Cdfrac%7B1%7D%7B3%7D%20%5Cpm%200.3578454)
The 95% confidence interval, CI = (-0.2451, 06912)