/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 Katrina wants to estimate the pr... [FREE SOLUTION] | 91Ó°ÊÓ

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Katrina wants to estimate the proportion of adult Americans who read at least 10 books last year. To do so, she obtains a simple random sample of 100 adult Americans and constructs a \(95 \%\) confidence interval. Matthew also wants to estimate the proportion of adult Americans who read at least 10 books last year. He obtains a simple random sample of 400 adult Americans and constructs a \(99 \%\) confidence interval. Assuming both Katrina and Matthew obtained the same point estimate, whose estimate will have the smaller margin of error? Justify your answer.

Short Answer

Expert verified
Matthew's estimate will have the smaller margin of error due to a larger sample size outweighing the higher confidence level.

Step by step solution

01

Understand the Concept of Margin of Error

The margin of error in a confidence interval measures the range within which the true population parameter is expected to lie. It depends on the sample size and the confidence level.
02

Recall the Formula for Margin of Error

The margin of error (E) for a proportion is given by \[ E = z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]where \(z\) is the z-score corresponding to the confidence level, \(\hat{p}\) is the sample proportion (point estimate), and \(n\) is the sample size.
03

Compare Sample Sizes

Katrina's sample size is 100 (\(n_1 = 100\)), while Matthew's sample size is 400 (\(n_2 = 400\)). A larger sample size reduces the margin of error, assuming the same point estimate and confidence level.
04

Compare Confidence Levels

Katrina's confidence level is 95%, and the z-score for this confidence level is approximately 1.96. Matthew's confidence level is 99%, and the z-score for this confidence level is approximately 2.576. A higher confidence level increases the margin of error.
05

Calculate the Margins of Error

For Katrina:\[ E_1 = 1.96 \sqrt{\frac{\hat{p}(1-\hat{p})}{100}} \]For Matthew:\[ E_2 = 2.576 \sqrt{\frac{\hat{p}(1-\hat{p})}{400}} \]
06

Justify the Comparison

Since Matthew has a larger sample size (400) compared to Katrina (100), and even though his confidence level is higher, the increase in sample size has a stronger effect to decrease the margin of error. Therefore, Matthew’s estimate will have the smaller margin of error.

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Key Concepts

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

Confidence Interval
A confidence interval provides a range of values which is likely to contain the population parameter with a certain level of confidence. When Katrina and Matthew construct a confidence interval for the proportion of adult Americans who read at least 10 books last year, they are essentially creating a range that is expected to include the true proportion for the entire population.

Confidence intervals have two main components: the sample statistic (e.g., sample proportion) and the margin of error. The sample statistic is the point estimate derived from the sample, while the margin of error accounts for how much the sample statistic is expected to vary from the true population parameter.

To construct a confidence interval, you need:
  • The sample proportion \( \hat{p} \)
  • The z-score corresponding to the desired confidence level
  • The sample size \( n \)
Using these, the confidence interval is typically formulated as:

\[ \hat{p} \pm E \] where \( E \) is the margin of error.
Sample Size
Sample size () is a critical factor in statistical analysis as it directly influences the margin of error and the reliability of the estimated population parameter. The larger the sample size, the more accurately it represents the population. In our exercise, Katrina's sample size is 100, while Matthew's is 400.

When it comes to margin of error, a larger sample size results in a smaller margin of error. This is because the variability or uncertainty in the estimate decreases with larger sample sizes. The relationship can be seen in the formula for the margin of error:

\[ E = z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]

Here, increasing \( n \) leads to a smaller value of \( E \), making the confidence interval narrower. Thus, Matthew’s estimate has a smaller margin of error compared to Katrina's due to his larger sample size.
Z-score
The z-score represents the number of standard deviations a data point is from the mean. In the context of confidence intervals, the z-score is used to determine the critical value that corresponds to the desired confidence level.

For common confidence levels, typical z-scores are:
  • 90% confidence level: z = 1.645
  • 95% confidence level: z = 1.96
  • 99% confidence level: z = 2.576
In our exercise, Katrina uses a 95% confidence level, giving her a z-score of 1.96. Matthew, on the other hand, uses a 99% confidence level, giving him a z-score of 2.576.

A higher z-score corresponds to a wider confidence interval because it reflects more certainty in the estimate. However, this also increases the margin of error. Although Matthew's z-score is higher, his larger sample size significantly reduces the margin of error, making his confidence interval more precise than Katrina's.

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Most popular questions from this chapter

Construct the appropriate confidence interval. A simple random sample of size \(n=25\) is drawn from a population that is normally distributed. The sample variance is found to be \(s^{2}=3.97\). Construct a \(95 \%\) confidence interval for the population standard deviation.

The following small data set represents a simple random sample from a population whose mean is \(50 .\) $$ \begin{array}{llllll} \hline 43 & 63 & 53 & 50 & 58 & 44 \\ \hline 53 & 53 & 52 & 41 & 50 & 43 \\ \hline \end{array} $$ (a) A normal probability plot indicates that the data could come from a population that is normally distributed with no outliers. Compute a \(95 \%\) confidence interval for this data set. (b) Suppose that the observation, 41 , is inadvertently entered into the computer as 14 . Verify that this observation is an outlier (c) Construct a \(95 \%\) confidence interval on the data set with the outlier. What effect does the outlier have on the confidence interval? (d) Consider the following data set, which represents a simple random sample of size 36 from a population whose mean is 50\. Verify that the sample mean for the large data set is the same as the sample mean for the small data set from part (a). $$ \begin{array}{llllll} \hline 43 & 63 & 53 & 50 & 58 & 44 \\ \hline 53 & 53 & 52 & 41 & 50 & 43 \\ \hline 47 & 65 & 56 & 58 & 41 & 52 \\ \hline 49 & 56 & 57 & 50 & 38 & 42 \\ \hline 59 & 54 & 57 & 41 & 63 & 37 \\ \hline 46 & 54 & 42 & 48 & 53 & 41 \\ \hline \end{array} $$ (e) Compute a \(95 \%\) confidence interval for the large data set. Compare the results to part (a). What effect does increasing the sample size have on the confidence interval? (f) Suppose that the last observation, 41 , is inadvertently entered as 14 . Verify that this observation is an outlier. (g) Compute a \(95 \%\) confidence interval for the large data set with the outlier. Compare the results to part (e). What effect does an outlier have on a confidence interval when the data set is large?

The trade magazine QSR routinely checks the drive-through service times of fast-food restaurants. A \(90 \%\) confidence interval that results from examining 607 customers in Taco Bell's drive-through has a lower bound of 161.5 seconds and an upper bound of 164.7 seconds. What does this mean?

Travelers pay taxes for flying, car rentals, and hotels. The following data represent the total travel tax for a 3-day business trip in eight randomly selected cities. Note: Chicago has the highest travel taxes in the country at 101.27 dollar. In Problem 32 from Section \(9.2,\) it was verified that the data are normally distributed and that \(s=12.324\) dollars. Construct and interpret a \(90 \%\) confidence interval for the standard deviation travel tax for a 3 -day business trip. $$ \begin{array}{llll} \hline 67.81 & 78.69 & 68.99 & 84.36 \\ \hline 80.24 & 86.14 & 101.27 & 99.29 \\ \hline \end{array} $$

Researchers Havar Brendryen and Pal Kraft conducted a study in which 396 subjects were randomly assigned to either an experimental smoking cessation program or control group. The experimental program consisted of the Internet and phone-based Happy Ending Intervention, which lasted 54 weeks and consisted of more than 400 contacts by e-mail, web pages, interactive voice response, and short message service (SMS) technology. The control group received a self- help booklet. Both groups were offered free nicotine replacement therapy. Abstinence was defined as "not even a puff of smoke, for the last 7 days," and assessed by means of Internet surveys or telephone interviews. The response variable was abstinence after 12 months. Of the participants in the experimental program, \(22.3 \%\) reported abstinence; of the participants in the control group, \(13.1 \%\) reported abstinence. (a) What type of experimental design is this? (b) What is the treatment? How many levels does it have? (c) What is the response variable? (d) What are the statistics reported by the authors? (e) An odds ratio is the ratio of the odds of an event occurring in one group to the odds of it occurring in another group. These groups might be men and women, an experimental group and a control group, or any other dichotomous classification. If the probabilities of the event in each of the groups are \(p\) (first group) and \(q\) (second group), then the odds ratio is $$ \frac{\frac{p}{(1-p)}}{\frac{q}{(1-q)}}=\frac{p(1-q)}{q(1-p)} $$ An odds ratio of 1 indicates that the condition or event under study is equally likely in both groups. An odds ratio greater than 1 indicates that the condition or event is more likely in the first group. And an odds ratio less than 1 indicates that the condition or event is less likely in the first group. The odds ratio must be greater than or equal to zero. As the odds of the first group approach zero, the odds ratio approaches zero. As the odds of the second group approach zero, the odds ratio approaches positive infinity. Verify that the odds ratio for this study is \(1.90 .\) What does this mean? (f) The authors of the study reported a \(95 \%\) confidence interval for the odds ratio to be lower bound: 1.12 and upper bound 3.26. Interpret this result. (g) Write a conclusion that generalizes the results of this study to the population of all smokers.

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