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 98% level of confidence.

a) With no prior research, what sample size should you gather in order to obtain a 0.035 margin of error? Round your answer up to the nearest whole number.

n = households

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 = 0.2 . Using this new information. what sample size should you gather in order to obtain a 0.035 margin of error?Round your answer up to the nearest whole number.

n = households

Answer #1

Solution :

Given that,

a)

= 0.5

1 - = 0.5

margin of error = E = 0.035

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_{/2}
= Z_{0.01} = 2.326

sample size = n = (Z_{
/ 2} / E )^{2} *
* (1 -
)

= (2.326 / 0.035)^{2} * 0.5 * 0.5

= 1104.14

**sample size = 1105**

**n = 1105 households**

b)

sample size = n = (Z_{
/ 2} / E )^{2} *
* (1 -
)

= (2.326 / 0.035)^{2} * 0.2 * 0.8

= 706.65

**sample size = 707**

**n =707 households**

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