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Chapter 9: Problem 53

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).

Short Answer

Expert verified
a. The 90% confidence interval for the proportion of dog owners who take more pictures of their dog than their significant others or friends is approximately (0.618, 0.682). This means we can be 90% confident that the true proportion of such dog owners lies within this interval. b. The 95% confidence interval for the proportion of dog owners who are more likely to complain to their dog than to a friend is approximately (0.430, 0.490). This means we can be 95% confident that the true proportion of such dog owners lies within this interval. c. There are two reasons why the confidence interval in Part (b) is wider than the interval in Part (a): 1. Confidence level: The confidence level in Part (b) is 95%, whereas it is 90% in Part (a). Higher confidence levels result in wider confidence intervals to account for the increased certainty that the true proportion lies within the interval. 2. Sample proportion: The sample proportion (p鈧�) in Part (b) is smaller than the sample proportion (p鈧�) in Part (a). When the sample proportion is closer to 0.5, the standard error tends to be larger, as indicated by the formula for the standard error, resulting in wider confidence intervals.

Step by step solution

01

Identify Sample Proportions and Sample Sizes

In the article, the following information is provided: - Out of 1000 dog owners, 650 take more pictures of their dog (p1) - Out of 1000 dog owners, 460 are more likely to complain to their dog (p2) So, the sample proportions, p獗� are: p鈧� = 650/1000 p鈧� = 460/1000 The sample size (n) for both proportions is the same: n = 1000
02

Calculate Standard Errors for Proportions

For each proportion, we need to calculate the standard error, which is given by the formula: Standard Error (SE) = \(\sqrt{\frac{p_j (1 - p_j)}{n}}\) For each proportion: SE鈧� = \(\sqrt{\frac{p鈧� (1 - p鈧�)}{n}}\) SE鈧� = \(\sqrt{\frac{p鈧� (1 - p鈧�)}{n}}\)
03

Construct Confidence Intervals

To construct the requested confidence intervals, we will use the following formula: Confidence Interval = p獗� 卤 Z * SE獗� Where Z is the Z-score corresponding to the desired confidence level. a. For the 90% confidence interval with p鈧�: Z = 1.645 (Z-score for 90% confidence) Confidence Interval (90%) = p鈧� 卤 1.645 * SE鈧� b. For the 95% confidence interval with p鈧�: Z = 1.96 (Z-score for 95% confidence) Confidence Interval (95%) = p鈧� 卤 1.96 * SE鈧�
04

Interpret Confidence Intervals

a. The 90% confidence interval for the proportion of dog owners who take more pictures of their dog than their significant others or friends indicates that we can be 90% confident that the true proportion of such dog owners lies within this interval. b. The 95% confidence interval for the proportion of dog owners who are more likely to complain to their dog than to a friend indicates that we can be 95% confident that the true proportion of such dog owners lies within this interval.
05

Explain Widening of Confidence Intervals

c. There are two reasons why the confidence interval in Part (b) is wider than the interval in Part (a): 1. Confidence level: The confidence level in Part (b) is 95%, whereas it is 90% in Part (a). Higher confidence levels result in wider confidence intervals to account for the increased certainty that the true proportion lies within the interval. 2. Sample proportion: The sample proportion (p鈧�) in Part (b) is smaller than the sample proportion (p鈧�) in Part (a). When the sample proportion is closer to 0.5, the standard error tends to be larger, as indicated by the formula for the standard error, resulting in wider confidence intervals.

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Key Concepts

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

Sample Proportions
In statistics, the sample proportion is a measure that represents the fraction of the sample with a particular attribute. For instance, in a study exploring dog owners' behavior, if 650 out of 1000 dog owners say they take more photos of their pets than their spouses, the sample proportion of dog owners who indulge in this behavior is 0.65 (650 divided by 1000).

When observing sample proportions, it's essential to understand that they are estimates of the true population proportion. Hence, while they provide insight, they come with some degree of uncertainty. That's why calculating confidence intervals around these sample proportions becomes crucial鈥攊t allows us to state, with a certain level of confidence, a range in which the true population proportion is likely to lie.
Standard Error Calculation
The standard error (SE) is a statistical term that measures the variability or dispersion of the sample proportion from the population proportion. To calculate the standard error for a sample proportion, you can use the formula:
\[ SE = \sqrt{\frac{p_j (1 - p_j)}{n}} \]
where \( p_j \) is the sample proportion and \( n \) is the sample size. This calculation helps in understanding how much the sample proportion could vary from the actual population proportion. For example, in the dog owner study, a larger standard error implies more variability, meaning we're less certain about the true population proportion of the behavior in the question. Conversely, a smaller standard error indicates that our sample proportion is a more precise estimate of the population proportion.
Confidence Level
The confidence level represents the degree of certainty we have in the methods used to estimate the true population parameter. Common confidence levels are 90%, 95%, and 99%. These percentages reflect how confident we are that the actual value lies within the constructed confidence interval. For instance, a 90% confidence level suggests that if we were to take numerous samples and construct confidence intervals from these samples, about 90% of these intervals would contain the true population proportion.

Choosing a higher confidence level, like 95%, requires a wider interval to ensure it encompasses the true value with greater certainty. This is why a 95% confidence interval is broader than a 90% interval鈥攖hey're not just random ranges, but carefully constructed brackets developed through standard formulas that account for the confidence we're striving to achieve.

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Most popular questions from this chapter

In \(2010,\) the National Football League adopted new rules designed to limit head injuries. In a survey conducted in 2015 by the Harris Poll, 1216 of 2096 adults indicated that they were football fans and followed professional football. Of these football fans, 692 said they thought that the new rules were effective in limiting head injuries (December 21 , \(2015,\) www.theharrispoll.com/sports/Football-Injuries.html, retrieved May 6,2017 ). a. Assuming that the sample is representative of adults in the United States, construct and interpret a \(95 \%\) confidence interval for the proportion of U.S. adults who consider themselves to be football fans. b. Construct and interpret a \(95 \%\) confidence interval for the proportion of football fans who think that the new rules have been effective in limiting head injuries. c. Explain why the confidence intervals in Parts (a) and (b) are not the same width even though they both have a confidence level of \(95 \%\)

In mid-2016 the United Kingdom (UK) withdrew from the European Union (an event known as "Brexit"), causing economic concerns throughout the world. One indicator that economists use to monitor the health of the economy is the proportion of residential properties offered for sale at auction that are successfully sold. An article titled "Going, going, gone through the roof-sky's the limit at auction" (October \(22,2016,\) www.estateagenttoday .co.uk/features/2016/10/going-going-gone-through-theroof-the-skys-the-limit- at-auction, retrieved May 4,2017 ) reported the success rate of a sample of 26 residential properties offered for sale at auctions in the UK in the summer of \(2016 .\) For this sample of properties, 14 of the 26 residential properties were successfully sold. Suppose it is reasonable to consider these 26 properties as representative of residential properties offered at auction in the post-Brexit UK. a. Would it be appropriate to use the large-sample confidence interval for a population proportion to estimate the proportion of residential properties successfully sold at auction in the post-Brexit UK? Explain. b. Would it be appropriate to use a bootstrap confidence interval for a population proportion to estimate the proportion of residential properties successfully sold at auction in the post-Brexit UK? Explain. c. Use the accompanying output from the "Bootstrap Confidence Interval for One Proportion" Shiny app to report a \(95 \%\) bootstrap confidence interval for the population proportion of residential properties successfully

Data from a representative sample were used to estimate that \(32 \%\) of all computer users in 2011 had tried to get on a Wi-Fi network that was not their own in order to save money (USA TODAY, May 16,2011 ). You decide to conduct a survey to estimate this proportion for the current year. What is the required sample size if you want to estimate this proportion with a margin of error of \(0.05 ?\) Calculate the required sample size first using 0.32 as a preliminary estimate of \(p\) and then using the conservative value of \(0.5 .\) How do the two sample sizes compare? What sample size would you recommend for this study?

The report "The 2016 Consumer Financial Literacy Survey" (The National Foundation for Credit Counseling, www .nfcc.org, retrieved October 28,2016 ) summarized data from a representative sample of 1668 adult Americans. When asked if they typically carry credit card debt from month to month, 584 of these people responded "yes." a. Use the given information to estimate the proportion of adult Americans who carry credit card debt from month to month. 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.

The paper "Sleeping with Technology: Cognitive, Affective and Technology Usage Predictors of Sleep Problems Among College Students" (Sleep Health [2016]: 49-56) summarized data from a survey of a sample of college students. Of the 734 students surveyed, 125 reported that they sleep with their cell phones near the bed and check their phones for something other than the time at least twice during the night. For purposes of this exercise, assume that this sample is representative of college students in the United States. a. Use the given information to estimate the proportion of college students who check their cell phones for something other than the time at least twice during the night. 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.
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