In: Statistics and Probability
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.
(ii) Use the central limit theorem to calculate an approximate
value of Pr(P < 0.69).
(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.
(ii) Would a 95% confidence interval for π be wider or narrower
than a 99% confidence interval? Explain your answer.
(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.
(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
and standard deviation
The z-score for P = 0.69 is
The required probability using z table is
P(P < 0.69) = P(z < -1.10) = 0.1357
(b)(i)
(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 :
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.