A poll is taken in which 342 out of 500 randomly selected voters
indicated their preference for a certain candidate.
(a) Find a 98% confidence interval for p.
≤ p ≤
(b) Find the margin of error for this 98% confidence interval for
p.
solution:-
given that n = 500 , x = 342
proportion p = x/n = 342/500 = 0.684
(a) the value of 98% confidence from z table is 2.33
confidence interval formula
=> p +/- z* sqrt(p*(1-p)/n)
=> 0.684 +/- 2.33 * sqrt(0.684*(1-0.684)/500)
=> 0.636 ≤ p ≤ 0.732 rounded to three decimals
(b) margin of error formula
=> z* sqrt(p*(1-p)/n)
=> 2.33 * sqrt(0.684*(1-0.684)/500)
=> 0.048 rounded to three decimals
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