Concerned about campus safety, college officials wish to estimate the proportion

Question

Concerned about campus safety, college officials wish to estimate the proportion of students who carry a gun, knife or other such weapon. Assuming that there is no available information that can be used to estimate the percentage of college students carrying weapons, (a) how many randomly selected students must be surveyed in order to be 95% confident that the sample percentage has a margin of error of 3%?. After the study has started, a paper is published describing a similar survey at another comparable college which found that 7% of college students carried weapons. Being provided with this information, (b) what sample size would you now recommend to meet the aforementioned criteria?

Explanation / Answer

This question is related to margin of error

we know that margin of error is given as

MOE = Z*sqrt(p*(1-p)/n)

a) in the first question we are given moe = 0.03 and 95% confidence , so the z value is 1.96 from the z table ,(use the z table for this)

also as the proportion is not given , so we shall assuma p = 0.5 , as p*(1-p) would be maximum at p = 0.5

putting the values in above equation and solving for n

0.03= 1.96*sqrt(0.5*(0.5)/n)

n = 0.25/(0.03/1.96)^2 = 1067.11 == 1067 approx

b) in the second case we are given p = 0.07 , agian using the same formula and putting the values as

0.03= 1.96*sqrt(0.07*(0.93)/n)

n = (0.07*0.93)/(0.03/1.96)^2 = 277.87 = 278 approx ...

Hope this helps !!

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