Question

Crash testing evaluates 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 proportions of small cars and large cars that were totaled.

Answer 1

Answer:

The  95% confidence interval for the difference between proportions is (-0.046, 0.713).

Step-by-step explanation:

We want to calculate the bounds of a 95% confidence interval.

For a 95% CI, the critical value for z is z=1.96.

The sample 1 (small cars), of size n1=12 has a proportion of p1=0.6667.

$p_1=X_1/n_1=8/12=0.6667$

The sample 2, of size n2=15 has a proportion of p2=0.3333.

$p_2=X_2/n_2=5/15=0.3333$

The difference between proportions is (p1-p2)=0.3333.

$p_d=p_1-p_2=0.6667-0.3333=0.3333$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{8+5}{12+15}=\dfrac{13}{27}=0.4815$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.4815*0.5185}{12}+\dfrac{0.4815*0.5185}{15}}\\\\\\s_{p1-p2}=\sqrt{0.0208+0.0166}=\sqrt{0.0374}=0.1935$

Then, the margin of error is:

$MOE=z \cdot s_{p1-p2}=1.96\cdot 0.1935=0.3793$

Then, the lower and upper bounds of the confidence interval are:

$LL=(p_1-p_2)-z\cdot s_{p1-p2} = 0.3333-0.3793=-0.046\\\\UL=(p_1-p_2)+z\cdot s_{p1-p2}= 0.3333+0.3793=0.713$

The  95% confidence interval for the difference between proportions is (-0.046, 0.713).