Question

1. In a survey, 90 people were asked if they received at least some of their news from social media. 61 responded Yes.

(a) What is the 95% confidence interval for the proportion of people in the population who receive at least some of their news from social media?

(b) Working at the 95% confidence level, how large a sample would be necessary to estimate the population proportion with a margin of error of ±0.02? Assume a planning value of p^∗ = 0.68

Answer 1

Answer)

A)

N = 90

P = 61/90

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 61

N*(1-p) = 29

Both the conditions are met so we can use standard normal z table to estimate the interval

Critical value z from z table for 95% confidence level is 1.96

Margin of error (MOE) = Z*√{P*(1-P)}/√N

MOE = 0.09637497139

Interval is given by

P-MOE < P < P+MOE

0.58362502860 < P < 0.77637497139

B)

0.02 = 1.96*√{0.68*(1-0.68)}/√n

N = 2090