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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

Check whether each of the conditions is met for calculating a confidence interval for the population proportion \(\bar{p}\). In the National AIDS Behavioral Surveys sample of 2673 adult heterosexuals, \(0.2 \%\) had both received a blood transfusion and had a sexual partner from a group at high risk of AlDS. We want to estimate the proportion \(p\) in the population who share these two risk factors.

Determine the point estimator you would use and calculate the value of the point estimate. How many pairs of shoes, on average, do female teens have? To find out, an AP Statistics class conducted a survey. They selected an SRS of 20 female students from their school. Then they recorded the number of pairs of shoes that each student reported having. Here are the data: $$\begin{array}{rrrrrrrrrr}50 & 26 & 26 & 31 & 57 & 19 & 24 & 22 & 23 & 38 \\\13 & 50 & 13 & 34 & 23 & 30 & 49 & 13 & 15 & 51\end{array}$$

A researcher plans to use a random sample of families to estimate the mean monthly family income for a large population. The researcher is deciding between a sample of size \(n=500\) and a sample of size \(n=1000 .\) Compared to using a sample size of \(n=500,\) a \(95 \%\) confidence interval based on a sample size of \(n=1000\) will be (a) narrower and would involve a larger risk of being incorrect. (b) wider and would involve a smaller risk of being incorrect. (c) narrower and would involve a smaller risk of being incorrect. (d) wider and would involve a larger risk of being incorrect. (e) narrower and would have the same risk of being incorrect.

A medical study finds that \(\bar{x}=114.9\) and \(s_{x}=9.3\) for the seated systolic blood pressure of the 27 members of one treatment group. What is the standard error of the mean? Interpret this value in context.

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.
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