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

The 95% confidence interval is (-0.2451, 06912)

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{(pX\left (1-pX \right ))/(n_(1))+(pY\left (1-pY \right ))/(n_(2))}`

`(2)/(3) -(1)/(3) \pm 1.96 * \sqrt{((2)/(3) * \left (1-(2)/(3) \right ))/(12)+((1)/(3) * \left (1-(1)/(3) \right ))/(15)}`

Therefore, we have;

`\therefore 95\% \ CI = (1)/(3) \pm 0.3578454`

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