A simple random sample of 100 of a certain popular model car in 2003 found that 20 had a certain minor defect in the brakes.A simple random sample of 400 of this model car in 2004 found that 50 had the minor defect in the brakes.Let p₁ and p₂ be the proportion of all cars of this model in 2003 and 2004,respectively,that actually have the defect.What is a 90% confidence interval for p₁ - p₂? A)0.075 ± 0.071 B)0.075 ± 0.085 C)0.075 ± 0.112 D)0.079 ± 0.071

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Quizplus · Accepted Answer

The confidence interval for the difference between two proportions $p_1 - p_2$ can be calculated using the formula:$$\hat{p}_1 - \hat{p}_2 \pm z \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}$$where $\hat{p}_1 = \frac{20}{100} = 0.20$, $\hat{p}_2 = \frac{50}{400} = 0.125$, $n_1 = 100$, $n_2 = 400$, and $z$ is the z-score corresponding to the desired confidence level (for 90% confidence, $z \approx 1.645$).Plugging in the values:$$0.20 - 0.125 \pm 1.645 \sqrt{\frac{0.20(0.80)}{100} + \frac{0.125(0.875)}{400}}$$$$0.075 \pm 1.645 \sqrt{\frac{0.16}{100} + \frac{0.109375}{400}}$$$$0.075 \pm 1.645 \sqrt{0.0016 + 0.0002734375}$$$$0.075 \pm 1.645 \sqrt{0.0018734375}$$$$0.075 \pm 1.645 \times 0.0433$$$$0.075 \pm 0.071$$Therefore, the correct answer is 0.075 ± 0.071.

A Simple Random Sample of 100 of a Certain Popular