Question

Q. Suppose a consumer advocacy group would like to conduct a survey to find the proportion p of consumers who bought the newest generation of mobile phone were happy with their purchase.
(a) How large a sample n should they take to estimate p with 2% margin of error and 90% confidence?
(b) The advocacy group took a random sample of 1000 consumers who recently purchased this mobile phone and found that 400 were happy with their purchase. Find
a 95% confidence interval for p.

Answer 1

a)

When prior estimate for proportion is not specified then p = 0.50

Sample size = Z2/2 * p ( 1 - p) / E2

= 1.64492 * 0.5 * 0.5 / 0.022

= 1691.06

n = 1692 (Rounded up to nearest integer)

b)

p̂ = X / n = 400/1000 = 0.4

95% confidence interval is
p̂ ± Z(α/2) √( (p * q) / n)
0.4 ± Z(0.05/2) √( (0.4 * 0.6) / 1000)
Z(α/2) = Z(0.05/2) = 1.96
Lower Limit = 0.4 - Z(0.05) √( (0.4 * 0.6) / 1000) = 0.370
upper Limit = 0.4 + Z(0.05) √( (0.4 * 0.6) / 1000) = 0.430
95% Confidence interval is ( 0.370 , 0.430 )