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Chapter 22: Problem 27

Exercise \(22.25\) describes a Harris Poll survey of smokers in which 848 of a sample of 1010 smokers agreed that smoking would probably shorten their lives. Harris announces a margin of error of \(\pm 3\) percentage points for all samples of about this size. Opinion polls announce the margin of error for \(95 \%\) confidence. (a) What is the actual margin of error (in percent) for the large-sample confidence interval from this sample? (b) The margin of error is largest when \(\mathrm{p}^{\wedge \hat{p}}=0.5\). What would the margin of error (in percent) be if the sample had resulted in \(\mathrm{p}^{\wedge \hat{p}}=0.5\) ? (c) Why do you think that Harris announces a \(\pm 3\) margin of error for all samples of about this size?

Short Answer

Expert verified
(a) 2.33%, (b) 3.08%, (c) It's a worst-case scenario assumption for sample proportions.

Step by step solution

01

Calculate the Sample Proportion

First, we need to find the sample proportion \( \hat{p} \). This is calculated by dividing the number of smokers who agreed that smoking would shorten their life by the total number of smokers surveyed.\[ \hat{p} = \frac{848}{1010} \approx 0.8396. \]
02

Find the Standard Error

Next, we find the standard error (SE), which is calculated using the formula: \[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \] where \( \hat{p} \approx 0.8396 \) and \( n = 1010 \). Thus, \[ SE = \sqrt{\frac{0.8396 \times (1 - 0.8396)}{1010}} \approx 0.0119. \]
03

Determine the Margin of Error for 95% Confidence Interval

For a 95% confidence interval, the critical value (z*) is approximately 1.96. The margin of error (ME) is given by \[ ME = z^* \times SE. \] Thus, \[ ME = 1.96 \times 0.0119 \approx 0.0233, \] or 2.33% when expressed in percentage form.
04

Calculate Margin of Error for \( \hat{p} = 0.5 \)

If \( \hat{p} = 0.5 \), the standard error is maximized. This is calculated as:\[ SE = \sqrt{\frac{0.5 \times (1 - 0.5)}{1010}} = \sqrt{\frac{0.25}{1010}} \approx 0.0157. \] The margin of error at 95% confidence is: \[ ME = 1.96 \times 0.0157 \approx 0.0308, \] or 3.08% when expressed as a percentage.
05

Explanation for Announced Margin of Error

The margin of error is declared as \( \pm3\% \) because it represents the worst-case scenario when the sample proportion \( \hat{p} = 0.5 \). This is a standard practice in polling to cover all possible distributions and ensure the stated confidence level is upheld across all samples.

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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 crucial element in statistics that indicates how much the sample results may vary from the actual population values. It provides a way to express the uncertainty within survey results. - In our exercise, the margin of error helps to estimate how closely the opinions of the 1010 smokers surveyed reflect those of all smokers. - The margin of error depends on the confidence level you choose, which is commonly set at 95% in polling. - In the context of the solution, for the actual data, we calculated it to be approximately 2.33%, suggesting the survey results may differ by this percentage. - For a sample proportion of 0.5 (where variability is highest), the margin of error increases to 3.08%.
Sample Proportion
Sample Proportion (\(\hat{p}\)) reflects the fraction of your sample that has the particular attribute of interest. In our exercise, it's the ratio of smokers who believe smoking shortens life to the total number of smokers surveyed.- We calculated the sample proportion as \(\hat{p} = \frac{848}{1010} \approx 0.8396\), indicating that around 83.96% of those surveyed think smoking is harmful.- \(\hat{p}\) is essential in computing the Standard Error and the margin of error.- A smaller or larger sample proportion can dramatically affect survey accuracy and representativeness.
Standard Error
Standard Error (SE) gives us an idea of how much variability exists in our sample. It is derived from the sample proportion and allows for estimating the margin of error.- The formula for Standard Error is: \[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\], where \(\hat{p}\) is the sample proportion, and \(n\) is the sample size.- In our given scenario, \(\hat{p} = 0.8396\) and \(n = 1010\), leading to \(SE \approx 0.0119\).- Lower SE indicates less variability and more reliable results. However, it is influenced by both the sample proportion and the sample size.
95% Confidence Level
A 95% Confidence Level signifies the degree of certainty we have that the sample result contains the true population parameter.- For any given confidence interval, the margin of error is calculated using the choice of confidence level. Commonly, in statistics, we use a critical value (\(z^*\)) of approximately 1.96 for a 95% confidence level.- In context, our study's confidence interval assures that if we were to take 100 different samples, approximately 95 of them will contain the true population parameter.- The choice of 95% is conventional for opinion polls because it balances reliability and precision effectively. Choosing a higher confidence level would result in a larger margin of error, while a lower one may increase the risk of inaccuracy.

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Most popular questions from this chapter

Sample surveys usually contact large samples, so we can use the large-sample confidence interval if the sample design is close to an SRS. Scientific studies often use smaller samples that require the plus four method. For example, Familial Adenomatous Polyposis (FAP) is a rare inherited disease characterized by the development of an extreme number of polyps early in life and colon cancer in virtually \(100 \%\) of patients before the age of 40 . A group of 14 people suffering from FAP being treated at the Cleveland Clinic drank black raspberry powder in a slurry of water every day for nine months. The numbers of polyps were reduced in 11 out of 14 of these patients. \({ }^{18}\) (a) Why can't we use the large-sample confidence interval for the proportion \(p\) of patients suffering from FAP that will have the number of polyps reduced after nine months of treatment? (b) The plus four method adds four observations-two successes and two failures. What are the sample size and the number of successes after you do this? What is the plus four estimate \(\mathrm{p}^{-\bar{p}}\) of \(p\) ? (c) Give the plus four \(90 \%\) confidence interval for the proportion of patients suffering from FAP who will have the number of polyps reduced after nine months of treatment.

You are planning a survey of students at a large university to determine what proportion favor an increase in student fees to support an expansion of the student newspaper. Using records provided by the registrar, you can select a random sample of students. You will ask each student in the sample whether he or she is in favor of the proposed increase. Your budget will allow a sample of 100 students. (a) For a sample of size 100, construct a table of the margins of error for \(95 \%\) confidence intervals when \(\mathrm{p}^{\wedge \hat{p}}\) takes the values \(0.1,0.3,0.5,0.7\), and \(0.9\). (b) A former editor of the student newspaper offers to provide funds for a sample of size 500 . Repeat the margin of error calculations in part (a) for the larger sample size. Then write a short thank-you note to the former editor describing how the larger sample size will improve the results of the survey.

A random digit dialing telephone survey of 880 drivers asked, "Recalling the last 10 traffic lights you drove through, how many of them were red when you entered the intersections?" Of the 880 respondents, 171 admitted that at least one light had been red. \({ }^{24}\) (a) Give a \(95 \%\) confidence interval for the proportion of all drivers who ran one or more of the last 10 red lights they met. (b) Nonresponse is a practical problem for this survey-only \(21.6 \%\) of calls that reached a live person were completed. Another practical problem is that people may not give truthful answers. What is the likely direction of the bias: do you think more or fewer than 171 of the 880 respondents really ran a red light? Why?

Explain whether we can use the \(z\) test for a proportion in these situations. (a) You toss a coin 10 times in order to test the hypothesis \(H_{0}: p=0.5\) that the coin is balanced. (b) A local candidate contacts an SRS of 900 of the registered voters in his district to see if there is evidence that more than half support the bill he is sponsoring. (c) A college president says, " \(99 \%\) of the alumni support my firing of Coach Boggs." You contact an SRS of 200 of the college's 15,000 living alumni to test the hypothesis \(H_{0}: p=0.99\).

The value of the \(z\) statistic for the Exercise \(22.22\) is \(2.53\). This test is (a) not significant at either \(\alpha=0.05\) or \(\alpha=0.01\). (b) significant at \(\alpha=0.05\) but not at \(\alpha=0.01\). (c) significant at both \(\alpha=0.05\) and \(\alpha=0.01\).
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