Question

3. (a) Assume that 73% of American men favour a law that requires a police permit to buy a gun. A sample of 150 American men are randomly selected. Let P denote the proportion of this sample that are in favour of this law.
(i) Explain why P is an unbiased and consistent estimator of π, the population proportion of American men that are in favour of this law. (10 marks)
(ii) Use the central limit theorem to calculate an approximate value of Pr(P < 0.69). (10 marks)

(b) Suppose that a random sample of 400 American women were selected and 352 of them were in favour of this law.
(i) Calculate an approximate 99% confidence interval for the proportion of the population of American women that are in favour of this law. (10 marks)
(ii) Would a 95% confidence interval for π be wider or narrower than a 99% confidence interval? Explain your answer. (5 marks)

(c) Using the information given in part (a) and your answer to part (bi), does it appear that there are any differences in the proportions of American men and women that are in favour of this law? Explain your answer. (5 marks)

Answer 1

(i)

Here sample size is big. It is given that sample is randomly selected. So sample proportion will be an unbiased estimator of population proportion.

(ii)

The sampling distribution of sample proportion will be approximately normal with mean

Mp = p = 0.73

and standard deviation

ll-P) = \10.715027 = 0.0362

The z-score for P = 0.69 is

0.0362-=-1.10

The required probability using z table is

P(P < 0.69) = P(z < -1.10) = 0.1357

(b)(i)

Here we have -352, n=400, p=-=0.88 x Level of significance α-0.01 : Critical value: 2.58 The standard deviation is: ơ,-11--=sqrt( (0.88 *( 1-0.88)/400) Pn0.0162 The requried confidence interval is: ptŽe.Ơp (2.58*0.0162) -O.88 =0.88 ± 0.0418 (0.838,0.922) Excel function for critical value: NORMSINV(1-(0.01/2)) Hence, the required confidence interval is (0.838,0.922)

(ii)

The 95% confidence interval will be narrower because as the confidence level decreases critical value will also decreased.

(c)

The confidence interval for part a is :

Here we have n -150, p-0.73 Level of significance: α-001 -2.58 Critical value: The standard deviation is: P)sqrt(0.73*1-0.73)/150) = 0.0362 The requried confidence interval is: ptz, o 0.73+ 2.58 0.0362) =0.73 ± 0.0934 (0.637,0.823) Excel function for critical value: NORMSINV(1-(0.01/2)) Hence, the required confidence interval is (0.637,0.823)

Since confidence intervals for part a and part b do not overlap so it appears that there are differences in the proportions of American men and women that are in favour of this law.