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 0.03 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 0.03 margin of error? Round your answer up to the nearest whole number.

Answer 1

Answer:

The sample size must be approximately 1067 to get a margin of error of 0.03.

Step-by-step explanation:

We are given the following in the question:

Margin of error = 0.03

Significance level = 5%

Margin of error =

$z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

Since, no prior proportions are given, we take

$\hat{p} = 0.5$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Putting values, we get,

$0.03 = 1.96\sqrt{\dfrac{0.5(1-0.5)}{n}}\\\\\sqrt{n} = \dfrac{1.96\times 0.5}{0.03}\\\\\\sqrt{n} = 32.67\\n = 1067.3289\approx 1067$

Thus, the sample size must be approximately 1067 to get a margin of error of 0.03.