Wiki Page, Math 13 - Perez Sandbox : ANSWER(S): Module 17 - Distribution of Sample Proportions (3 of 6 discussion 1)

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2026-01-17T21:14:48-08:00

Here is an example of a correct response to the previous discussion prompt.

1. Simulation: Set p = 0.30 and n = 100. Check the “Show standard deviation bar.”

2. Based on this simulation, the average amount of expected error is 0.0458, about 4.6%.

3. Based on this simulation, the margin of error is about 0.0916 (0.0458 x 2 = 0.0916), about 9.2%.

4. Based on this simulation, a sample proportion of 0.20 is highly unlikely. There are two ways to think about this:

1. A sample proportion of 0.20 has an error of 0.10 (0.30 - 0.20 = 0.10). This error is more than 2 standard errors (more than 0.092) from the mean.

2. Sample proportions within 2 standard errors of the mean lie between about 0.21 (0.301 – 0.092 = 0.209, about 0.21) and 0.39 (0.301 + 0.092 = 0.393, about 0.39) A sample proportion of 0.20 lies outside of this interval; therefore, it is highly unlikely to occur.

Here is what all of this means in context: Researchers estimated the percentage of all adult smokers in the U.S. who started smoking before the age of 16. If this percentage is 30%, then we expect the percentage from most random samples of 100 smokers to be within a margin of error of 9.2%.

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