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Chapter 7: Problem 24

Use the data and confidence level to construct a confidence interval estimate of \(p,\) then address the given question. In a survey of 1002 people, \(70 \%\) said that they voted in a recent presidential election (based on data from ICR Research Group). Voting records show that \(61 \%\) of eligible voters actually did vote. a. Find a \(98 \%\) confidence interval estimate of the proportion of people who say that they voted. b. Are the survey results consistent with the actual voter turnout of \(61 \% ?\) Why or why not?

Short Answer

Expert verified
The 98% confidence interval is (0.6669, 0.7331). The survey results are not consistent with the actual voter turnout.

Step by step solution

01

- Identify given data

Identify and note down the given data. The sample size, n, is 1002. The sample proportion, \(\hat{p}\), is 0.70. The confidence level is 98%, which corresponds to a z-score of approximately 2.33.
02

- Calculate the standard error

Use the formula for the standard error of the proportion: \(SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\). Substituting the values, \(SE = \sqrt{\frac{0.70(1-0.30)}{1002}} = 0.0142\).
03

- Calculate the margin of error

Use the z-score for the 98% confidence level and the standard error. Margin of error (ME) = z * SE = 2.33 * 0.0142 = 0.0331.
04

- Determine the confidence interval

Calculate the confidence interval using the formulas: \(\hat{p} - ME\) and \(\hat{p} + ME\). Therefore, lower limit is 0.70 - 0.0331 = 0.6669 and upper limit is 0.70 + 0.0331 = 0.7331. The 98% confidence interval is (0.6669, 0.7331).
05

- Compare with actual voter turnout

Compare the actual voter turnout proportion (0.61) with the confidence interval. Since 0.61 lies outside the interval (0.6669, 0.7331), the survey results are not consistent with the actual voter turnout.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Proportion
The sample proportion, denoted as \(\backslash hat\backslash p\), represents the proportion of a specific outcome in a sample. In the context of voter turnout from the exercise, if out of 1002 surveyed individuals, 70% reported they had voted, we denote \(\backslash hat\backslash p = 0.70\). This indicates the fraction of the sample that participated in voting. Understanding sample proportion is crucial as it serves as a fundamental building block in determining confidence intervals, letting us assess broader population behaviors from our sample.
Standard Error
The standard error (SE) provides an estimate of how much a sample proportion \(\backslash hat\backslash p\) may fluctuate around the true population proportion. It is calculated using the formula:
\[SE = \backslash sqrt\backslash\frac\backslash (\backslash hat\backslash p(1 - \backslash hat\backslash p)}\backslash n}\]
where n is the sample size. For our example, with \(\backslash hat\backslash p = 0.70\) and n = 1002, the standard error:
\[SE = \backslash sqrt\backslash\frac{0.70(1 - 0.30)}\backslash{1002} = 0.0142\]
This helps gauge the expected variability in the sample proportion, a critical component in constructing a confidence interval.
Margin of Error
The margin of error (ME) defines the range within which we expect our sample proportion to lie with a certain confidence level. It is calculated by multiplying the standard error by the z-score (associated with the confidence level). For a 98% confidence interval, the z-score is approximately 2.33. Thus:
\[ME = z \backslash * SE = 2.33 \backslash * 0.0142 = 0.0331\]
This means the range \(\backslash pm ME\) from the sample proportion provides an interval that is expected to contain the true population proportion 98% of the time.
Voter Turnout
Voter turnout represents the percentage of eligible voters who cast a ballot in an election. Actual voter turnout, as seen in the exercise at 61%, can substantially differ from reported or perceived turnout (70% reported in the survey). Comparing these proportions using a confidence interval (e.g., 66.69% to 73.31% from the exercise) helps validate the consistency of survey results. Here, the actual voter turnout of 61% falls outside this interval, suggesting the survey results are not consistent with the true turnout.

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Most popular questions from this chapter

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