Question

You work for a marketing firm that has a large client in the automobile industry. You have been asked to estimate the proportion of households in Chicago that have two or more vehicles. You have been assigned to gather a random sample that could be used to estimate this proportion to within a 0.035 margin of error at a 95% level of confidence

. a) With no prior research, what sample size should you gather in order to obtain a 0.035 margin of error?

b) Your firm has decided that your plan is too expensive, and they wish to reduce the sample size required. You conduct a small preliminary sample, and you obtain a sample proportion of ˆ p = 0.19 . Using this new information. what sample size should you gather in order to obtain a 0.035 margin of error?

Answer 1

Solution:

a)

Given that,

= 0.5

1 - = 0.5

margin of error = E = 0.035

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 / 0.035)2 * 0.5 * 0.5

= 784

Sample size = 784

b)

Given that,

= 0.19

1 - = 0.81

margin of error = E = 0.035

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 / 0.035)2 * 0.19 * 0.81

= 482.63

Sample size = 483