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Chapter 9: Problem 17

A large online retailer is interested in learning about the proportion of customers making a purchase during a particular month who were satisfied with the online ordering process. A random sample of 600 of these customers included 492 who indicated they were satisfied. For each of the three following statements, indicate if the statement is correct or incorrect. If the statement is incorrect, explain what makes it incorrect. Statement 1: It is unlikely that the estimate \(\hat{p}=0.82\) differs from the value of the actual population proportion by more than 0.016 . Statement 2: It is unlikely that the estimate \(\hat{p}=0.82\) differs from the value of the actual population proportion by more than 0.031 . Statement 3: The estimate \(\hat{p}=0.82\) will never differ from the value of the actual population proportion by more than 0.031 .

Short Answer

Expert verified
Statement 1: Incorrect. The margin of error is 0.031, which means the estimate can differ from the actual population proportion by up to that amount, greater than the 0.016 mentioned in this statement. Statement 2: Correct. The margin of error is 0.031, indicating that it is possible for the estimate to differ from the actual population proportion by up to 0.031. Statement 3: Incorrect. This statement incorrectly implies certainty about the margin of error, whereas the true margin of error is an estimate. The actual value might differ from the estimated proportion by more than 0.031, although it is unlikely.

Step by step solution

01

Calculate the Margin of Error

To determine if the given statements are correct or incorrect, we first need to calculate the margin of error. The margin of error is given by the formula: \[ME = Z \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\] With \(Z = 1.96\) for a 95% confidence interval, n = 600 (the number of customers in the sample), and \(\hat{p} = 0.82\) (the proportion of satisfied customers in the sample).
02

Calculate the values of \(ME\) using the given data

Now let's substitute the values to find the value of \(ME\): \[ME = 1.96 \times \sqrt{\frac{0.82 (1 - 0.82)}{600}}\] \[ME \approx 0.031\] The margin of error is approximately 0.031 for this dataset.
03

Analyze the statements

Now let's analyze each of the three statements given in the exercise: 1. Statement 1: It is unlikely that the estimate \(\hat{p}=0.82\) differs from the value of the actual population proportion by more than 0.016. Answer: Incorrect. The margin of error is 0.031, which means the estimate can differ from the actual population proportion by up to that amount, greater than the 0.016 mentioned in this statement. 2. Statement 2: It is unlikely that the estimate \(\hat{p}=0.82\) differs from the value of the actual population proportion by more than 0.031. Answer: Correct. As calculated in Step 2, the margin of error is 0.031, indicating that it is possible for the estimate to differ from the actual population proportion by up to 0.031. 3. Statement 3: The estimate \(\hat{p}=0.82\) will never differ from the value of the actual population proportion by more than 0.031. Answer: Incorrect. This statement incorrectly implies certainty about the margin of error, whereas the true margin of error is an estimate. The actual value might differ from the estimated proportion by more than 0.031, although it is unlikely.

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Key Concepts

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

Population Proportion Estimation
Population proportion estimation is a central concept in statistics, especially when trying to understand the preferences or behaviors of a large group based on a sample. In the context of our example, the population proportion refers to the true percentage of all customers at the online retailer who are satisfied with the ordering process. The retailer hopes that by examining a random sample of 600 customers, they can closely estimate this true population proportion.

To obtain this estimate, we use the sample proportion, denoted by \(\hat{p}\). In our case, \(\hat{p}\) is 492 satisfied customers out of 600, or 0.82. This estimate is expected to be close to the true population proportion, but because we used a sample and not the entire population, there is always some uncertainty. This is where margin of error and confidence intervals come into play, as they help us understand the potential discrepancy between the sample proportion and the true population proportion.
Confidence Interval Calculation
Confidence interval calculation provides a range of values that is likely to contain the population proportion. It reflects the reliability of the estimate and considers the variability within the sample. The confidence interval is built around the sample proportion and extends a certain distance on either side, depending on the desired level of confidence.

To calculate this interval, we start with the sample proportion. We then add and subtract the margin of error from this proportion to define the range. For the online retailer's sample, a confidence interval at the 95% level means we have a 95% chance that the range captures the true population proportion. The addition of the margin of error to the sample proportion gives the upper bound of the interval, while subtracting it gives the lower bound. Our example showed that with a margin of error of approximately 0.031, we can be confident that the true satisfaction rate lies within the interval of 0.789 to 0.851.
Sampling Error Analysis
Sampling error analysis assesses the degree to which the sample might differ from the population. It acknowledges that the sample result is only an estimate and that different samples can yield different results. A critical part of this analysis is the margin of error, which quantifies the possible variation between the sample proportion and the population proportion.

The margin of error in our retailer's study, calculated as approximately 0.031, offers insight into the precision of the estimate \(\hat{p}=0.82\). This value implies that if we were to take many samples of the same size from the population, the proportion of satisfied customers would likely fluctuate within the range defined by this margin of error. It is essential to recognize that the margin of error does not provide a guarantee but rather a probability statement about the extent of potential sampling error.
Hypothesis Testing
Hypothesis testing is a foundational methodology in statistics used to make inferences about populations based on sample data. It involves making an assumption (the null hypothesis) about a population parameter and then determining whether the sample data provides enough evidence to reject this assumption in favor of an alternative hypothesis.

In the example of the online retailer, if they hypothesized that the true satisfaction rate was 0.85, they would use hypothesis testing to assess whether their sample proportion (0.82) significantly differs from this hypothesized rate. The margin of error and confidence intervals are used to determine if the observed difference could reasonably occur due to sampling variability or if it is too large, suggesting that the null hypothesis should be rejected.

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Most popular questions from this chapter

A survey of a representative sample of 478 U.S. employers found that 359 ranked stress as their top health and productivity concern (June \(29,2016,\) www.globenewswire .com/news-release/2016/06/29/852338/0/en/Seventy-five- percent-of-U-S-employers-say-stress-is-their-number-one-workplace-health- concern.html?print=1, retrieved May 4,2017) a. Use the accompanying output from the "Bootstrap Confidence Interval for One Proportion" Shiny app to report a \(95 \%\) bootstrap confidence interval for the proportion of all U.S. employers who would rank stress at their top health and productivity concern. Interpret the confidence interval in context. b. A number of international employers were also surveyed. If the international employers had a similar rate of identifying stress as their top health and productivity concern, and if the results from international employers were included in the sample, would the width of the resulting confidence interval remain the same, decrease, or increase? Explain your reasoning.

The article "Most Dog Owners Take More Pictures of Their Pet Than Their Spouse" (August \(22,2016,\) news .fastcompany.com/most-dog-owners-take-more- pictures-oftheir-pet-than-their-spouse-4017458, retrieved May 6,2017 ) indicates that in a sample of 1000 dog owners, 650 said that they take more pictures of their dog than their significant others or friends, and 460 said that they are more likely to complain to their dog than to a friend. Suppose that it is reasonable to consider this sample as representative of the population of dog owners. a. Construct and interpret a \(90 \%\) confidence interval for the proportion of dog owners who take more pictures of their dog than of their significant others or friends. b. Construct and interpret a \(95 \%\) confidence interval for the proportion of dog owners who are more likely to complain to their dog than to a friend. c. Give two reasons why the confidence interval in Part (b) is wider than the interval in Part (a).

A random sample will be selected from the population of all adult residents of a particular city. The sample proportion \(\hat{p}\) will be used to estimate \(p,\) the proportion of all adult residents who are registered to vote. For which of the following situations will the estimate tend to be closest to the actual value of \(p ?\) I. \(\quad n=1000, p=0.5\) II. \(\quad n=200, p=0.6\) III. \(n=100, p=0.7\)

Based on data from a survey of 1200 randomly selected Facebook users (USA TODAY, March 24, 2010), a \(98 \%\) confidence interval for the proportion of all Facebook users who say it is OK to ignore a coworker's "friend" request is \((0.35,0.41) .\) What is the meaning of the confidence level of \(98 \%\) that is associated with this interval?

The article "Should Canada Allow Direct-to-Consumer Advertising of Prescription Drugs?" (Canadian Family Physician [2009]: \(130-131\) ) calls for the legalization of advertising of prescription drugs in Canada. Suppose you wanted to conduct a survey to estimate the proportion of Canadians who would allow this type of advertising. How large a random sample would be required to estimate this proportion with a margin of error of \(0.02 ?\)
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