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Chapter 9: Problem 15

A researcher wants to estimate the proportion of city residents who favor spending city funds to promote tourism. Would the standard error of the sample proportion \(\hat{p}\) be smaller for random samples of size \(n=100\) or random samples of size \(n=200 ?\)

Short Answer

Expert verified
The standard error of the sample proportion \(\hat{p}\) would be smaller for random samples of size \(n=200\).

Step by step solution

01

Understand the formula for standard error

The formula to calculate the standard error of the sample proportion \(\hat{p}\) is \(\sqrt{\frac{p(1 - p)}{n}}\), where \(p\) is the sample proportion and \(n\) is the sample size.
02

Apply the formula for \(n=100\)

To find out whether the standard error of the sample proportion would be smaller for \(n=100\) or \(n=200\), apply this formula for both. Although the value of \(p\) is not given, it won't affect the comparison since it would be the same for both calculations. For \(n=100\), the standard error would be \(\sqrt{\frac{p(1 - p)}{100}}\).
03

Apply the formula for \(n=200\)

Applying the formula for \(n=200\), the standard error would be \(\sqrt{\frac{p(1 - p)}{200}}\). Notice that this value would be smaller than the standard error for \(n=100\) because \(n\) has a larger value in the denominator, making the whole fraction and therefore the standard error smaller.
04

Compare the standard errors

From Step 2 and Step 3, it can be concluded that the standard error of the sample proportion \(\hat{p}\) would be smaller for random samples of size \(n=200\) when compared to random samples of size \(n=100\). This is because the standard error decreases as the sample size \(n\) increases.

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Most popular questions from this chapter

In a survey on supernatural experiences, 722 of 4,013 adult Americans reported that they had seen a ghost (鈥淲hat Supernatural Experiences We've Had," USA Today, February 8,2010 ). Assume that this sample is representative of the population of adult Americans. a. Use the given information to estimate the proportion of adult Americans who would say they have seen a ghost. b. Verify that the conditions for use of the margin of error formula to be appropriate are met. c. Calculate the margin of error. d. Interpret the margin of error in context. e. Construct and interpret a \(90 \%\) confidence interval for the proportion of all adult Americans who would say they have seen a ghost. f. Would a \(99 \%\) confidence interval be narrower or wider than the interval calculated in Part (e)? Justify your answer.

Suppose that 935 smokers each received a nicotine patch, which delivers nicotine to the bloodstream at a much slower rate than cigarettes do. Dosage was decreased to 0 over a 12 -week period. Of these 935 people, 245 were still not smoking 6 months after treatment. Assume this sample is representative of all smokers. a. Use the given information to estimate the proportion of all smokers who, when given this treatment, would refrain from smoking for at least 6 months. b. Verify that the conditions needed in order for the margin of error formula to be appropriate are met. c. Calculate the margin of error. d. Interpret the margin of error in the context of this problem.

Will \(\hat{p}\) from a random sample from a population with \(60 \%\) successes tend to be closer to 0.6 for a sample size of \(n=400\) or a sample size of \(n=800 ?\) Provide an explanation for your choice.

In an AP-AOL sports poll (Associated Press, December 18,2005\(), 394\) of 1,000 randomly selected U.S. adults indicated that they considered themselves to be baseball fans. Of the 394 baseball fans, 272 stated that they thought the designated hitter rule should either be expanded to both baseball leagues or eliminated. a. Construct and interpret a \(95 \%\) confidence interval for the proportion of U.S. adults who consider themselves to be baseball fans. b. Construct and interpret a \(95 \%\) confidence interval for the proportion of baseball fans who think the designated hitter rule should be expanded to both leagues or eliminated. c. Explain why the confidence intervals of Parts (a) and (b) are not the same width even though they both have a confidence level of \(95 \%\).

The formula used to calculate a large-sample confidence interval for \(p\) is $$ \hat{p} \pm(z \text { critial value }) \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} $$ What is the appropriate \(z\) critical value for each of the following confidence levels? a. \(95 \%\) b. \(98 \%\) c. \(85 \%\)
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