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

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

Step-by-step 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. =  ~ N(0,1)

where, = sample proportion of small cars that were totaled = = 0.67

= sample proportion of large cars that were totaled = = 0.33

= sample of small cars = 12

= sample of large cars = 15

= population proportion of small cars that are totaled

= 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, () 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 < < 1.96) = 0.95

P( < < ) = 0.95

P( < < ) = 0.95

95% confidence interval for = [ , ]

= [ , ]

= [-0.017 , 0.697]

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

1,150 give brainliest

step-by-step explanation

2r= 2300

r =1150

f is the midpoint of