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

Solution :

Given that,

n = 90

x = 61

Point estimate = sample proportion = = x / n = 61/90=0.678

1 -   = 1- 0.678 =0.322

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E = Z/2 * ((( * (1 - )) / n)

= 1.96 (((0.678*0.322) / 90)

= 0.0965

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.678-0.0965 < p <0.678+ 0.0965

0.5815< p < 0.7745

The 95% confidence interval for the population proportion p is : 0.5815, 0.7745

(B)

Solution :

Given that,

= 0.68

1 - = 1 - 0.68 = 0.32

margin of error = E = +/-0.02

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96

Sample size = n = (Z/2 / E)2 * * (1 - )

= (1.96 / 0.02)2 * 0.68 * 0.32

= 2089.8

Sample size =2090