Question

An opinion poll based on a sample of 50 subjects estimated p, the proportion of the population in favor of the proposition, as 0.72.
(i) Estimate the true proportion, θ, with a 95% confidence interval. State any assumptions you may have to make in answering this question.

(ii) If the true population proportion is suspected to be θ =0.8, and the estimate from an opinion poll is to be determined to within ±0.05 with 95% confidence, how
many people, n, should be sampled?

(iii) If the proportion is to be estimated to within the same margin of ±0.05, but with 90% confidence, what is the value of n required? Comment on the effect that
reducing the confidence level has on the sample size, n required to achieve the desired precision.

Answer 1

Given: Sample size = n = 50, sample proportion = p = 0.72

(i) The random variable has a binomial distribution with probability of success as P. As the sample size is sufficiently large, the normal approximation to binomial distribution can be used.

Thus, the required 95% confidence interval is { p - 1.96*sqrt(p*(1-p)/n) , p +1.96*sqrt(p*(1-p)/n) }

i.e. {0.72 - 1.96*sqrt(0.72*0.28/50), 0.72 + 1.96*sqrt(0.72*0.28/50)}

i.e.(0.72-0.1245 , 0.72+0.1245), i.e. (0.5955,0.8445)

(ii) According to the given condition,

1.96*sqrt(p*(1-p)/n) <= 0.05

i.e. 1.96*sqrt(0.8*0.2/n) <= 0.05

i.e. n >= 0.16*(1.96/0.05)^2

i.e. n >= 245.8624

Thus the sample should be of size at least 246.

(iii) 1.64*sqrt(p*(1-p)/n) <= 0.05

i.e. n >= 0.16*(1.64/0.05)^2

i.e. n >= 172.1344

Thus the sample should be of size at least 173.

Thus, for lower confidence, a smaller sample size suffices.

Hope this helps!