Margin of error

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Violagirl

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Mar 9, 2008
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I'm confused on this problem:

As part of the 2000 census, the Census Bureau surveyed 700,000 households to study transportation to work. They reported that 76.3% drove alone to work, 11.2% carpooled, 5.1% took mass transity, 3.2% worked at home, 0.4% bicycled, and 3.8% took other means.

A) With such a large survey, explain why the margin of error for any of these values is extremely small.

I know that the MoE measures the accuaracy of a point estimate and that in most polls it's for a 95% confidence interval. However, I'm not sure how to derive from this in figure out why the values listed would be too small for the margin of error...Is it because it doesn't fit for what is normally found within a 95% confidence interval? :?
 


I'm thinking about the standard error of those reported percents, but I'm not sure what you've been taught in class.

Have you seen this formula before?

\(\displaystyle \text{Standard Error} \;=\; \sqrt{\frac{p(1 - p)}{n}}\)

\(\displaystyle \text{where p is a reported percent and n is the sample size}\)

For example, look at the reported percent for generalizing about how many workers in the entire country drive alone to work: 76.3%

The standard error of this survey result can be estimated using the formula above, IF the sample size of the survey is small compared to the whole population. (700,000 is fairly small compared to the total number of households in the US, so the formula above provides a good estimate of the error in this survey.)

p = 0.763

1 - p = 0.237

n = 700,000

Look at the formula, and think about the three numbers above. Do you "see" why the expression (i.e., the standard error for the 76.3% result) represents a very small number.

 
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