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. Find a 95% confidence interval for the difference in the prop

Answer 1

95% confidence interval for the difference in the proportion is [-0.017 , 0.697].

Explanation:

We are given that 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.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

P.Q. =
`\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } }` ~ N(0,1)

where,
`\hat p_1` = sample proportion of small cars that were totaled =
`(8)/(12)` = 0.67

`\hat p_2` = sample proportion of large cars that were totaled =
`(5)/(15)` = 0.33

`n_1` = sample of small cars = 12

`n_2` = sample of large cars = 15

`p_1` = population proportion of small cars that are totaled

`p_2` = population proportion of large cars that were totaled

Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.

So, 95% confidence interval for the difference between population population, (
`p_1-p_2`) is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 <
`\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } }` < 1.96) = 0.95

P(
`-1.96 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } }` <
`{(\hat p_1-\hat p_2)-(p_1-p_2)}` <
`1.96 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } }` ) = 0.95

P(
`(\hat p_1-\hat p_2)-1.96 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } }` <
`p_1-p_2` <
`(\hat p_1-\hat p_2)+1.96 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } }` ) = 0.95

95% confidence interval for
`p_1-p_2` = [
`(\hat p_1-\hat p_2)-1.96 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } }` ,
`(\hat p_1-\hat p_2)+1.96 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } }`]

= [
`(0.67-0.33)-1.96 * {\sqrt{(0.67(1-0.67))/(12)+(0.33(1-0.33))/(15) } }` ,
`(0.67-0.33)+1.96 * {\sqrt{(0.67(1-0.67))/(12)+(0.33(1-0.33))/(15) } }`]

= [-0.017 , 0.697]

Therefore, 95% confidence interval for the difference between proportions l and 2 is [-0.017 , 0.697].