Answer:
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)
  ~ N(0,1)
where,  = sample proportion of small cars that were totaled =
 = sample proportion of small cars that were totaled =  = 0.67
 = 0.67
 = sample proportion of large cars that were totaled =
 = sample proportion of large cars that were totaled =  = 0.33
 = 0.33
 = sample of small cars = 12
 = sample of small cars = 12
 = sample of large cars = 15
 = sample of large cars = 15
 = population proportion of small cars that are totaled
 = population proportion of small cars that are totaled
 = population proportion of large cars that were totaled
 = population proportion of large cars that were totaled
<em>Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.</em>
So, 95% confidence interval for the difference between population population, ( ) is ;
) 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
 < 1.96) = 0.95
P(  <
 <  <
 <  ) = 0.95
 ) = 0.95
P(  <
 <  <
 <  ) = 0.95
 ) = 0.95
<u>95% confidence interval for</u>  = [
 = [ ,
 ,  ]
]
= [ ,
 ,  ]
]
= [-0.017 , 0.697]
Therefore, 95% confidence interval for the difference between proportions l and 2 is [-0.017 , 0.697].