Could someone please help I will give brainiest to best answer but please show steps?
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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)
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
<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 ;
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
<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].