Question

Constructing Confidence Intervals, Part 1: Estimating Proportion Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level: In a random sample of 200 college students, 110 had part-time jobs. Find the margin of error for the 98% confidence interval used to estimate, for the entire population of college students, the percentage who have part-time jobs. Round your answer to three decimal places.

Constructing Confidence Intervals, Part 1: Estimating Proportion Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level: In a random sample of 200 college students, 110 had part-time jobs. Find the margin of error for the 98% confidence interval used to estimate, for the entire population of college students, the percentage who have part-time jobs. Round your answer to three decimal places. Your answer

Please answer with the explanation, especially about how 98% confidence level is used to be solved.

Thank you so much.

Answer 1

solution :- Given: - No.y favourable case (x) = 110 sample size (h) = 200. Point estimate () = 7 - 110 - 0.55. 200 At 98 % confidence level 121 is, by using '2' table, d = 1-98% x= 1-0.98 a = 0.02 9/2 = 0.02/2 = 0.01 20/= 2.326 Margin of Error (E) = 29/* Ê (1-6) = 2.326x10-55(1-0.55) 200 Margin of Error (E = 0.082 confidence interval at 98% cI; for Proportion is, point estimate I margin of Error ſ + E . 0.55 +0.082 (0.468, 0.632) At 98%, CI for Proportion is (0.468, 0.632)