We want to estimate the proportion of registered voters that plan to vote for ca

Question

We want to estimate the proportion of registered voters that plan to vote for candidate X in the upcoming election. At this point in the campaign, we are certain that the proportion of voters that favor X is between 10% and 30%. We plan to survey a random sample of registered voters and summarize our results with a 95% confidence interval for the proportion of people who plan to vote for X. How many people must we sample in order for the margin of error on or confidence interval to be a most 0.05?

Explanation / Answer

given that alpha = 0.05

z/2 = 1-0.05/2

z/2 value is  1.96

Standard deviation of sample is (p*(1-p)/n)

where n=sample size , p is being proportion

we know that

z*standard edviation = margin of error

given that margin of error=0.05

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

p*(1-p)/n = 0.00065

n=p*(1-p)/0.00065

For p = 0.1,

sample size(n) = 0.1*0.9/0.00065

= 138.46

= 139

For p = 0.3,

sample size(n) = 0.3*0.7/0.00065

= 323.07

= 324

Taking the highest value, sample size be at least n= 324

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