Use the normal distribution to find a confidence interval for a proportion p giv
ID: 2949148 • Letter: U
Question
Use the normal distribution to find a confidence interval for a proportion p given the relevant sample results. Give the best point estimate for p , the margin of error, and the confidence interval. Assume the results come from a random sample. A 90% confidence interval for p given that p ^ = 0.9 and n = 110 . Round your answer for the point estimate to two decimal places, and your answers for the margin of error and the confidence interval to three decimal places.
Point estimate = .
Margin of error = ± .
The 90% confidence interval is to .
Explanation / Answer
solution:-
point estimate = 0.9
margin of error =
90% confidence for z is 1.645
given n = 110
formula = z * sqrt((p^ * q^)/n)
= 1.645 * sqrt((0.9 * 0.1)/110)
= 0.047
The 90% confidence interval is to .
confidence interval formula
p^ +/- z * sqrt((p^ * q^)/n)
=> 0.9 +/- 1.645 * sqrt((0.9 * 0.1)/110)
=> (0.853 , 0.947)
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