A political consultant wants to sample residents of a county to estimate what proportion of all the county's residents support a certain candidate. They will use the sample data to construct a one-sample $z$ interval for the proportion, and they are considering two sample sizes: a large sample of $n=900$ for more accurate results, or a small sample of $n=100$ to save time and money.

Assuming the sample proportion is the same in each sample, what is true about the margins of error from these two samples?

The margin of error from the smaller sample will be about select ratio the margin of error from the larger sample.

Question

A political consultant wants to sample residents of a county to estimate what proportion of all the county's residents support a certain candidate. They will use the sample data to construct a one-sample $z$ interval for the proportion, and they are considering two sample sizes: a large sample of $n=900$ for more accurate results, or a small sample of $n=100$ to save time and money.

Assuming the sample proportion is the same in each sample, what is true about the margins of error from these two samples?

The margin of error from the smaller sample will be about select ratio the margin of error from the larger sample.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To compare the margins of error for the two sample sizes, we need to understand the formula for the margin of error in a one-sample $z$ interval for a proportion. ‖ ‖ step 2 ⋮ The margin of error (ME) for a one-sample $z$ interval for a proportion is given by $ME = z \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$, where $z$ is the $z$-score corresponding to the desired confidence level, $\hat{p}$ is the sample proportion, and $n$ is the sample size. ‖ ‖ step 3 ⋮ Since the sample proportion $\hat{p}$ is assumed to be the same for both samples, the margin of error is inversely proportional to the square root of the sample size $n$. ‖ ‖ step 4 ⋮ The ratio of the margins of error for the two sample sizes can be found by taking the square root of the inverse ratio of the sample sizes. That is, $\frac{ME_{small}}{ME_{large}} = \sqrt{\frac{n_{large}}{n_{small}}}$. ‖ ‖ step 5 ⋮ Substituting the given sample sizes, we get $\frac{ME_{small}}{ME_{large}} = \sqrt{\frac{900}{100}} = \sqrt{9} = 3$. ‖ ∻Answer∻ ⚹ The margin of error from the smaller sample will be about 3 times the margin of error from the larger sample. ⚹ ∻Key Concept∻ ⚹Margin of Error and Sample Size Relationship⚹ ∻Explanation∻ ⚹The margin of error for a proportion is inversely proportional to the square root of the sample size, meaning that a larger sample size will result in a smaller margin of error.⚹◊What is the formula for calculating the margin of error in a sample proportion using the $z$ interval?⍭ Generate me a similar question◊

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