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Chapter 6: Problem 16

As discussed in Exercise 6.10, the General Social Survey reported a sample where about \(61 \%\) of US residents thought marijuana should be made legal. If we wanted to limit the margin of error of a \(95 \%\) confidence interval to \(2 \%\), about how many Americans would we need to survey?

Short Answer

Expert verified
Survey about 367 Americans.

Step by step solution

01

Understand the Formula for Margin of Error

The margin of error for a confidence interval is calculated using the formula \( ME = Z \times \sqrt{\frac{p(1-p)}{n}} \), where \( ME \) is the margin of error, \( Z \) is the Z-score for a given confidence level, \( p \) is the sample proportion (or probability, here 0.61), and \( n \) is the sample size.
02

Identify the Value to Use

Given the problem, we know that we need to find a sample size \( n \) to achieve a \( ME \) of 0.02 (2%) with a 95% confidence interval. For a 95% confidence interval, the Z-score \( Z \) is approximately 1.96.
03

Rearrange the Margin of Error Formula

Rearrange the formula to find \( n \):\[ n = \frac{Z^2 \times p(1-p)}{ME^2} \]Substitute the known values including \( Z = 1.96 \), \( p = 0.61 \), and \( ME = 0.02 \).
04

Calculate the Sample Size

Now, calculate the sample size using the rearranged formula:\[ n = \frac{1.96^2 \times 0.61 \times (1-0.61)}{0.02^2} \]\[ n = \frac{3.8416 \times 0.61 \times 0.39}{0.0025} \]\[ n = \frac{0.91560464}{0.0025} \]\[ n \approx 366.24 \]Round up to the nearest whole number: \( n = 367 \).
05

Final Conclusion

We need to survey approximately 367 Americans to limit the margin of error to 2% for a 95% confidence interval.

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Key Concepts

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

Margin of Error
The margin of error is a critical component in the world of statistics and surveys. It represents the range within which the true population parameter is expected to lie, and it reflects the uncertainty inherent in an estimate.
For a confidence interval, the margin of error ( ME ) provides the "buffer zone" around the sample proportion. In simpler terms, if we estimate that 61% of U.S. residents support legalizing marijuana, the margin of error tells us how confident we can be in this estimate given the sampled data.
A smaller margin of error requires a larger sample size, which means more data points are needed to increase the accuracy of the estimate. By controlling the margin of error, statisticians can influence the precision of the conclusions about the population. In our context, we aimed for a 2% margin of error to ensure our findings are as close as possible to reality within this tolerance.
Sample Size Calculation
Calculating the appropriate sample size is essential for obtaining reliable and valid results. It involves determining how many participants are needed to achieve a specific margin of error and confidence level.
Using the margin of error formula, we rearrange it to solve for the sample size:\[ n = \frac{Z^2 \times p(1-p)}{ME^2} \]"n" represents the sample size, "Z" is the Z-score for our desired confidence level, "p" is the estimated proportion of success (or likelihood of the event of interest, here 0.61), and "ME" is the margin of error, aimed here at 2%.
By plugging in the values for Z, p, and ME, we compute that approximately 367 participants are required to ensure the insights are within the desired range of precision. This ensures our estimates are statistically significant and dependable.
Z-score
A Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It is commonly used in determining the confidence interval's reliability.
In the context of a confidence interval, the Z-score determines how many standard deviations away from the mean the interval extends. For a typical 95% confidence interval, the Z-score is 1.96. This means we are 95% certain that the population parameter lies within the interval specified by our calculated margin of error.
Understanding Z-scores helps interpret how extreme or typical the data points are compared to the average findings. This, in turn, allows researchers to quantify the certainty regarding how representative the sample proportion is concerning the broader population.
Probability
Probability is a fundamental concept that underpins the entire practice of statistics. It provides a measure of how likely an event is to occur. In surveys and sampling, probability is used to predict patterns and inform decisions based on collected data.
In our context, probability refers to the proportion of people surveyed who support legalizing marijuana, set at 0.61. This probability determines the expected outcomes if the survey were repeated multiple times.
By combining probability with other statistical tools like the margin of error and Z-scores, statisticians can make informed predictions about the likelihood of certain outcomes. Probability thus aids in interpreting and validating survey results, adding another layer of confidence to the analysis of public opinion.

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Most popular questions from this chapter

Determine if the statements below are true or false. For each false statement, suggest an alternative wording to make it a true statement. (a) As the degrees of freedom increases, the mean of the chi-square distribution increases. (b) If you found \(\chi^{2}=10\) with \(d f=5\) you would fail to reject \(H_{0}\) at the \(5 \%\) significance level. (c) When finding the p-value of a chi-square test, we always shade the tail areas in both tails. (d) As the degrees of freedom increases, the variability of the chi-square distribution decreases.

Suppose that \(90 \%\) of orange tabby cats are male. Determine if the following statements are true or false, and explain your reasoning. (a) The distribution of sample proportions of random samples of size 30 is left skewed. (b) Using a sample size that is 4 times as large will reduce the standard error of the sample proportion by one-half. (c) The distribution of sample proportions of random samples of size 140 is approximately normal. (d) The distribution of sample proportions of random samples of size 280 is approximately normal.

Greece has faced a severe economic crisis since the end of \(2009 .\) A Gallup poll surveyed 1,000 randomly sampled Greeks in 2011 and found that \(25 \%\) of them said they would rate their lives poorly enough to be considered "suffering". (a) Describe the population parameter of interest. What is the value of the point estimate of this parameter? (b) Check if the conditions required for constructing a confidence interval based on these data are met. (c) Construct a \(95 \%\) confidence interval for the proportion of Greeks who are "suffering". (d) Without doing any calculations, describe what would happen to the confidence interval if we decided to use a higher confidence level. (e) Without doing any calculations, describe what would happen to the confidence interval if we used a larger sample.

A local news outlet reported that \(56 \%\) of 600 randomly sampled Kansas residents planned to set off fireworks on July \(4^{t h} .\) Determine the margin of error for the \(56 \%\) point estimate using a \(95 \%\) confidence level. \(^{10}\)

Suppose that \(8 \%\) of college students are vegetarians. Determine if the following statements are true or false, and explain your reasoning. (a) The distribution of the sample proportions of vegetarians in random samples of size 60 is approximately normal since \(n \geq 30\). (b) The distribution of the sample proportions of vegetarian college students in random samples of size 50 is right skewed. (c) A random sample of 125 college students where \(12 \%\) are vegetarians would be considered unusual. (d) A random sample of 250 college students where \(12 \%\) are vegetarians would be considered unusual. (e) The standard error would be reduced by one-half if we increased the sample size from 125 to 250 .
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