Question

A simple random sample (SRS) of 100 of a certain popular car in 1997 found that 20 had a certain minor defect in the brakes. An SRS of 400 of the same car in 1998 found that 50 had the minor defect in the brakes. Let p1 and p2 be the proportion of all cars of this model in 1997 and 1998, respectively,that contained the defect. A 90% confidence interval for p1−p2 is:

Answer 1

Answer: The required interval is (0.679, 0.821).

Step-by-step explanation:

Since we have given that

$n_1=100\\\\p_1=20\\\\n_2=400\\\\p_2=50$

So, the proportions would be

$\hat{p_1}=\dfrac{p_1}{n_1}=\dfrac{20}{100}=0.2\\\\\hat{p_2}=\dfrac{p_2}{n_2}=\dfrac{50}{400}=0.125$

So, At 90% confidence interval, the z score value would be

z = 1.65

and the hypothesis are :

$H_0:\hat{p_1}-\hat{p_2}=0\\\\H_1:\hat{p_1}-\hat{p_2}\neq 0$

So, the 90% confidence interval would be

$(\hat{p_1}-\hat{p_2})\pm z\sqrt{\dfrac{\hat{p_1}(1-\hat{p_1}}{n_1}+\dfrac{\hat{p_2}(1-\hat{p_2})}{n_2}}\\\\=(0.2-0.125)\pm 1.65\sqrt{\dfrac{0.2\times 0.8}{100}+\dfrac{0.125\times 0.875}{400}}\\\\=0.075\pm 1.65\times 0.0433\\\\=(0.75-0.0714,0.75+0.0714)\\\\=(0.679,0.821)$

Hence, the required interval is (0.679, 0.821).