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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

For estimating a population characteristic, why is an unbiased statistic generally preferred over a biased statistic? Does unbiasedness alone guarantee that the estimate will be close to the actual value of the population characteristic? Explain.

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.

Consider taking a random sample from a population with \(p=0.40\). a. What is the standard error of \(\hat{p}\) for random samples of size \(100 ?\) b. Would the standard error of \(\hat{p}\) be greater for samples of size 100 or samples of size \(200 ?\) c. If the sample size were doubled from 100 to 200 , by what factor would the standard error of \(\hat{p}\) decrease?

Suppose that a city planning commission wants to know the proportion of city residents who support installing streetlights in the downtown area. Two different people independently selected random samples of city residents and used their sample data to construct the following confidence intervals for the population proportion: Interval 1:(0.28,0.34) Interval 2:(0.31,0.33) (Hint: Consider the formula for the confidence interval given on page 444.) a. Explain how it is possible that the two confidence intervals are not centered in the same place. b. Which of the two intervals conveys more precise information about the value of the population proportion? c. If both confidence intervals have a \(95 \%\) confidence level, which confidence interval was based on the smaller sample size? How can you tell? d. If both confidence intervals were based on the same sample size, which interval has the higher confidence level? How can you tell?

For each of the following choices, explain which one would result in a wider large-sample confidence interval for \(p:\) a. \(90 \%\) confidence level or \(95 \%\) confidence level b. \(n=100\) or \(n=400\)
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