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Chapter 22: Problem 31

A random digit dialing telephone survey of 880 drivers asked, "Recalling the last 10 traffic lights you drove through, how many of them were red when you entered the intersections?" Of the 880 respondents, 171 admitted that at least one light had been red. \({ }^{24}\) (a) Give a \(95 \%\) confidence interval for the proportion of all drivers who ran one or more of the last 10 red lights they met. (b) Nonresponse is a practical problem for this survey-only \(21.6 \%\) of calls that reached a live person were completed. Another practical problem is that people may not give truthful answers. What is the likely direction of the bias: do you think more or fewer than 171 of the 880 respondents really ran a red light? Why?

Short Answer

Expert verified
(a) The 95% confidence interval is approximately (0.168, 0.221). (b) Likely fewer admitted to running a red light; actual number might be higher.

Step by step solution

01

Determine Sample Proportion

First, calculate the sample proportion \( \hat{p} \) of respondents who admitted to running at least one red light. The proportion is given by \( \hat{p} = \frac{171}{880} \approx 0.1943 \).
02

Calculate Standard Error

Use the sample proportion to calculate the standard error \( SE \): \[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.1943 \times 0.8057}{880}} \approx 0.0134 \].
03

Determine Critical Value

For a 95% confidence interval, the critical value \( z^* \) (approximately) is 1.96 for a normal distribution.
04

Construct Confidence Interval

The confidence interval is given by \( \hat{p} \pm z^* \times SE \). Substitute the known values: \( 0.1943 \pm 1.96 \times 0.0134 = (0.1943 \pm 0.0263) \). Thus, the confidence interval is approximately \( (0.168, 0.221) \).
05

Address Bias in Responses

Nonresponse and respondent honesty can affect results. Since only 21.6% completed the survey, nonrespondents are assumed to have different characteristics. Additionally, respondents might underreport running red lights due to social desirability bias, likely causing fewer admissions than actual incidents.

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Key Concepts

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

Sample Proportion
When conducting a survey or experiment, the sample proportion is essential for determining the attributes of a population based on a sample. In the context of the traffic light survey, it is the ratio of respondents admitting to running a red light at least once out of the total respondents. By calculating the sample proportion, you'll have a foundational figure to carry out further statistical analysis, such as a confidence interval. The formula to find the sample proportion is straightforward:
  • Expressed mathematically as: \( \hat{p} = \frac{X}{n} \), where \( X \) is the number of favorable responses (in our case, 171) and \( n \) is the total number of respondents surveyed (880).
  • By substituting the numbers: \( \hat{p} = \frac{171}{880} \approx 0.1943 \).
This means approximately 19.43% of the respondents admitted to running at least one red light in their recent experience. By understanding the sample proportion, you gauge the behavior or attribute of the larger population based on the sample itself.
Standard Error
The standard error measures the variability or uncertainty of a sample statistic, such as the sample proportion. It helps in determining how much the sample proportion is expected to fluctuate from the actual population proportion. In statistical analyses like surveys or experiments, a smaller standard error indicates more precise estimates.Here's how you determine it:
  • Use the formula: \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \).
  • Bar this, \( \hat{p} \) is 0.1943 (from our sample proportion calculation), and \( n \) is 880.
  • So, the standard error is: \( SE = \sqrt{\frac{0.1943 \times 0.8057}{880}} \approx 0.0134 \).
This calculates how much the sample proportion might differ from the true population proportion, guiding us to construct a reliable confidence interval.
Nonresponse Bias
Nonresponse bias occurs when a portion of sampled individuals fails to respond to the survey, possibly leading to unrepresentative results. In our traffic light survey example, only 21.6% of those contacted by phone completed the survey. This low response rate provides cause for concern:
  • The drivers who did not respond could have different habits or circumstances than those who did.
  • They might have had a higher or lower tendency to run red lights, skewing the overall perception derived from the sample that did respond.
  • As a result, your findings might not accurately reflect the behavior of the general population related to the survey's subject.
By acknowledging and trying to address nonresponse bias, researchers aim to create more accurate, comprehensive results that reflect the whole population's behavior.
Social Desirability Bias
Social desirability bias occurs when respondents alter their answers to appear more favorable in the eyes of others. In the traffic light survey, respondents might either underreport or overreport running red lights. Here鈥檚 why this bias impacts survey results:
  • Drivers might not be fully truthful about running red lights to appear law-abiding.
  • This desire to present themselves in a better light can lead to fewer admissions of running red lights than what actually occurred.
  • Consequently, the reported proportion might be lower than the true proportion鈥攃ausing a misrepresentation of the observed behavior.
Recognizing social desirability bias is essential, as it helps researchers adjust their methodologies or interpret results more accurately, maintaining study reliability.

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Most popular questions from this chapter

The Monterey Bay Aquarium, founded in 1984 , is situated on the beautiful coast of Monterey Bay in the historic Cannery Row district. In 1985 , the aquarium began a survey program that involved randomly sampling visitors as they exit for the day. The survey includes visitor demographic information, use of social media, and opinions on their aquarium visit. In 2015, the survey included 356 visitors over age 65 , of which 52 used a mobile device such as an Android phone or iPad during their visit. \({ }^{11}\) (a) What is the margin of error of the large-sample \(95 \%\) confidence interval for the proportion of visitors over 65 who used a mobile device during their visit? (b) How large a sample is needed to get the common \(\pm 3\) percentage point margin of error? Use the 2015 sample as a pilot study to get \(p^{*}\).

Some shrubs have the useful ability to resprout from their roots after their tops are destroyed. Fire is a particular threat to shrubs in dry climates because it can injure the roots as well as destroy the aboveground material. One study of resprouting took place in a dry area of Mexico. \({ }^{32}\) The investigators clipped the tops of samples of several species of shrubs. In some cases, they also applied a propane torch to the stumps to simulate a fire. Of 12 specimens of the shrub Krameria cytisoides, five resprouted after fire. Estimate with \(90 \%\) confidence the proportion of all shrubs of this species that will resprout after fire.

The Internal Revenue Service plans to examine an SRS of individual federal income tax returns from each state. One variable of interest is the proportion of returns claiming itemized deductions. The total number of tax returns in a state varies from almost 30 million in California to approximately 500,000 in Wyoming. (a) Will the margin of error for estimating the population proportion change from state to state if an SRS of 2000 tax returns is selected in each state? Explain your answer. (b) Will the margin of error change from state to state if an SRS of \(1 \%\) of all tax returns is selected in each state? Explain your answer.

Although more than \(50 \%\) of American adults believe the maxim that breakfast is the most important meal of the day, only about \(30 \%\) eat breakfast daily. \({ }^{6}\) A cereal manufacturer contacts an SRS of 1500 American adults and calculates the proportion \(\mathrm{p}^{\wedge \hat{p}}\) in this sample who eat breakfast daily. (a) What is the approximate distribution of \(\mathrm{p}^{\wedge \hat{p}}\) ? (b) If the sample size were 6000 rather than 1500 , what would be the approximate distribution of \(\mathrm{p}^{\wedge \hat{p}}\) ?

A husband and wife, Stan and Lucretia, share a digital music player that has a feature that randomly selects which song to play. A total of 2643 songs have been loaded into the player, some by Stan and the rest by Lucretia. They are interested in determining whether they have loaded different proportions of songs into the player. Suppose that when the player was in the random- selection mode, 27 of the first 40 songs selected were songs loaded by Lucretia. Let \(p\) denote the proportion of songs that were loaded by Lucretia. (a) State the null and alternative hypotheses to be tested. How strong is the evidence that Stan and Lucretia have loaded different proportions of songs into the player? Make sure to check the conditions for the use of this test. (b) Are the conditions for the use of the large-sample confidence interval met? If so, estimate with \(95 \%\) confidence the proportion of songs that were loaded by Lucretia.
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