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Chapter 8: Problem 50

Most people can roll their tongues, but many can't. The ability to roll the tongue is genetically determined. Suppose we are interested in determining what proportion of students can roll their tongues. We test a simple random sample of 400 students and find that 317 can roll their tongues. The margin of error for a \(95 \%\) confidence interval for the true proportion of tongue rollers among students is closest to (a) 0.0008 . (c) 0.03 . (c) 0.05 . (b) 0.02 (d) 0.04

Short Answer

Expert verified
The margin of error is closest to \(0.04\).

Step by step solution

01

Determine the sample proportion (p虃)

The sample proportion \(\hat{p}\) is calculated by dividing the number of students who can roll their tongues by the total number of students in the sample. Thus, \(\hat{p} = \frac{317}{400} = 0.7925\).
02

Identify the critical value for a 95% confidence level

For a 95% confidence interval, the critical value \(z^*\) is approximately 1.96. This value is common for 95% confidence intervals in a standard normal distribution.
03

Calculate the standard error (SE)

The standard error of the proportion is calculated using the formula: \(SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\), where \(n\) is the sample size. Substituting the values, \(SE = \sqrt{\frac{0.7925 \times 0.2075}{400}} \approx 0.0203\).
04

Compute the margin of error (ME)

The margin of error is calculated by multiplying the standard error by the critical value: \(ME = z^* \times SE = 1.96 \times 0.0203 \approx 0.0398\).

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Key Concepts

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

Sample Proportion
In statistics, the sample proportion is a straightforward and crucial concept used to estimate population characteristics. It represents the fraction or percentage of the sample that possesses a particular attribute of interest.
For instance, in the exercise about tongue rolling among students, the sample proportion is calculated as the number of students who can roll their tongues divided by the total number of students sampled. Specifically, here's how you determine it:
  • Identify the total number of successes (in our case, 317 students who can roll their tongues).
  • Divide this number by the sample size (400 students in total).
This formula gives:\[ \hat{p} = \frac{317}{400} = 0.7925 \]This result means approximately 79.25% of the sample can roll their tongues. This provides a useful estimate when making inferences about the population.
Standard Error
The standard error measures the variability of the sample proportion. It essentially tells us how much we expect the sample proportion to vary from sample to sample due to random chance.
In the tongue-rolling example, the standard error is obtained using the formula:
  • Formula: \[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]
  • Substitute values: \[ SE = \sqrt{\frac{0.7925 \times 0.2075}{400}} \approx 0.0203 \]
This calculation suggests that the sample proportion (\(0.7925\)) might vary by roughly \(0.0203\) in repeated sampling due to natural sampling variability. Understanding the standard error is crucial for creating confidence intervals and assessing the precision of the sample proportion.
Margin of Error
Margin of error constitutes the range of uncertainty around the sample proportion. It combines the standard error with the critical value to quantify the confidence in the estimate.
The formula to find the margin of error (ME) is:
  • Multiply the critical value by the standard error:
  • \[ ME = z^* \times SE \]
  • Using our numbers: \[ ME = 1.96 \times 0.0203 \approx 0.0398 \]
The calculation implies the sample proportion of students who can roll their tongues could be \(0.7925 \pm 0.0398\). This indicates our estimate of the population proportion is relatively precise and provides insights into how results might generalize to the entire student population.
Critical Value
The critical value is a key element in statistics, particularly for constructing confidence intervals. It reflects the confidence level we want in our results.
For a 95% confidence interval, which is common, the critical value from a standard normal distribution is typically \(z^* = 1.96\). This value stems from the properties of the normal distribution:
  • It indicates that there is a 95% probability the true population parameter lies within this range on either side of the sample estimate.
  • Provides a measure of how 'confident' we can be about the derived confidence interval.
By using \(1.96\) as the critical value, the margin of error becomes suitable for accurately estimating the population parameter with the desired confidence level.

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Most popular questions from this chapter

PTC is a substance that has a strong bitter taste for some people and is tasteless for others. The ability to taste \(\mathrm{PTC}\) is inherited. About \(75 \%\) of Italians can taste \(\mathrm{PTC}\), for example. You want to estimate the proportion of Americans who have at least one Italian grandparent and who can taste PTC. (a) How large a sample must you test to estimate the proportion of PTC tasters within 0.04 with \(90 \%\) confidence? Answer this question using the \(75 \%\) estimate as the guessed value for \(\hat{p}\). (b) Answer the question in part (a) again, but this time use the conservative guess \(\hat{p}=0.5 .\) By how much do the two sample sizes differ?

A New York Times/CBS News Poll asked a random sample of U.S. adults the question, "Do you favor an amendment to the Constitution that would permit organized prayer in public schools?" Based on this poll, the \(95 \%\) confidence interval for the population proportion who favor such an amendment is (0.63,0.69) (a) Interpret the confidence interval. (b) What is the point estimate that was used to create the interval? What is the margin of error? (c) Based on this poll, a reporter claims that more than two-thirds of U.S. adults favor such an amendment. Use the confidence interval to evaluate this claim.

Check whether each of the conditions is met for calculating a confidence interval for the population proportion \(\bar{p}\). Latoya wants to estimate what proportion of the seniors at her boarding high school like the cafeteria food. She interviews an SRS of 50 of the 175 seniors living in the dormitory. She finds that 14 think the cafeteria food is good.

Losing weight A Gallup Poll asked a random sample of U.S. adults, "Would you like to lose weight?" Based on this poll, the \(95 \%\) confidence interval for the population proportion who want to lose weight is (0.56,0.62) (a) Interpret the confidence interval. (b) What is the point estimate that was used to create the interval? What is the margin of error? (c) Based on this poll, Gallup claims that more than half of U.S. adults want to lose weight. Use the confidence interval to evaluate this claim.

The news article goes on to say: "The theoretical errors do not take into account \(\cdots\) additional error resulting from the various practical difficulties in taking any survey of public opinion." List some of the "practical difficulties" that may cause errors which are not included in the 卤3 percentage point margin of error.
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