Question

1. Online customer service is a key element to successful online retailing. According to a marketing survey, 6.8% of online customers take advantage of the online customer service. Random samples of 210 customers are selected. What is the standard error of all possible sample proportions?

2. A researcher wants to create a 90% confidence interval for the population proportion, with the margin of error of 4%. She has no idea what the population proportion is and she cannot take a trial sample. Calculate the required sample size.

Answer 1

Solution

Given that,

1) p = 0.068

1 - p = 1 - 0.068 = 0.932

n = 210

= p = 0.068

=  [p ( 1 - p ) / n] =   [(0.068 * 0.932) / 210 ] = 0.017

Given that,

2) =  1 - = 0.5

margin of error = E = 0.04

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

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

= (1.645 / 0.04)2 * 0.5 * 0.5

= 422.82

sample size = n = 423