96.8k views
4 votes
6.Suppose the Gallup Organization wants to estimate the population proportion of those who think there should be a law that would ban the possession of handguns. In a previous study of 1012 randomly chosen respondents, 374 said that there should be such a law. How large a sample size is needed to be 95% confident with a margin of error of E

User Ehz
by
8.4k points

1 Answer

3 votes

Answer:

A sample of
n = ((1.96√(0.3696*0.6304))/(E))^2 is needed, in which E is the desired margin of error, as a proportion. If we find a decimal value, we round up to the next whole number.

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
\pi, and a confidence level of
1-\alpha, we have the following confidence interval of proportions.


\pi \pm z\sqrt{(\pi(1-\pi))/(n)}

In which

z is the zscore that has a pvalue of
1 - (\alpha)/(2).

The margin of error is of:


M = z\sqrt{(\pi(1-\pi))/(n)}

In a previous study of 1012 randomly chosen respondents, 374 said that there should be such a law.

This means that
n = 1012, \pi = (374)/(1012) = 0.3696

95% confidence level

So
\alpha = 0.05, z is the value of Z that has a pvalue of
1 - (0.05)/(2) = 0.975, so
Z = 1.96.

How large a sample size is needed to be 95% confident with a margin of error of E?

A sample size of n is needed, and n is found when M = E.


M = z\sqrt{(\pi(1-\pi))/(n)}


E = 1.96\sqrt{(0.3696*0.6304)/(n)}


E√(n) = 1.96√(0.3696*0.6304)


√(n) = (1.96√(0.3696*0.6304))/(E)


(√(n))^2 = ((1.96√(0.3696*0.6304))/(E))^2


n = ((1.96√(0.3696*0.6304))/(E))^2

A sample of
n = ((1.96√(0.3696*0.6304))/(E))^2 is needed, in which E is the desired margin of error, as a proportion. If we find a decimal value, we round up to the next whole number.

User Nahid Bin Azhar
by
8.1k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.