Question

Crash testing is a highly expensive procedure to evaluate the ability of an automobile to withstand a serious accident. A simple random sample of 12 small cars were subjected to a head-on collision at 40 miles per hour. Of them 8 were "totaled," meaning that the cost of repairs is greater than the value of the car. Another sample of 15 large cars were subjected to the same test, and 5 of them were totaled. Let pX represent the population proportion of small cars that are totaled and let pY represent the population proportion of large cars that are totaled. Find a 95% confidence interval for the difference pX – pY . Round the answers to four decimal places. The 95% confidence interval is ( , ).

Answer 1

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;

$pX-pY\pm z^{*}\sqrt{\dfrac{pX\left (1-pX \right )}{n_{1}}+\dfrac{pY\left (1-pY \right )}{n_{2}}}$

$\dfrac{2}{3} -\dfrac{1}{3} \pm 1.96 \times \sqrt{\dfrac{\dfrac{2}{3} \times \left (1-\dfrac{2}{3} \right )}{12}+\dfrac{\dfrac{1}{3} \times \left (1-\dfrac{1}{3} \right )}{15}}$

Therefore, we have;

$\therefore 95\% \ CI = \dfrac{1}{3} \pm 0.3578454$

The 95% confidence interval, CI = (-0.2451, 06912)