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!*

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