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Chapter 9: Problem 62

Seventy-seven students at the University of \(V\) irginia were asked to keep a diary of conversations with their mothers, recording any lies they told during these converations iSan Luis Obispo Telegram-Tribune. August 16,1995\() .\) It was reported that the mean number of lies per conversation was \(0.5 .\) Suppose that the standard deviation (which was not reported) was \(0.4 .\) a. Suppose that this group of 77 is a random sample from the population of students at this university. Construct a \(95 \%\) confidence interval for the mean number of lies per conversation for this population. b. The interval in Part (a) does not include \(0 .\) Does this imply that all students lie to their mothers? Explain.

Short Answer

Expert verified
The 95% confidence interval for the population mean is approximately \(0.411, 0.589\). This does not imply that all students lie to their mothers, but on average, students are likely to tell some lies in their conversations.

Step by step solution

01

Identifying the data

Identify and confirm the data required to solve the problem: sample size \(n = 77\), sample mean \(x虅 = 0.5\), and sample standard deviation \(s = 0.4\).
02

Establishing the confidence level

Establish and confirm the confidence level for the interval, which in this case is \(95%\). This corresponds to a z-value (a theoretical value from a standard normal distribution table) of \(1.96\) (for a two-tailed test).
03

Computing the standard error of the mean

Compute the standard error of the mean, which is given by the formula \(\sigma_{x虅} = s/\sqrt{n}\). Substituting the values in gives \(\sigma_{x虅} = 0.4/\sqrt{77} \approx 0.0456\).
04

Computing the confidence interval

Compute the confidence interval using the formula \(x虅 卤 (Z * \sigma_{x虅})\). Substituting the values in gives \(0.5 卤 (1.96 * 0.0456) \approx 0.5 卤 0.089\).
05

Interpreting the results

The 95% confidence interval for the population mean is approximately \(0.411, 0.589\). This interval gives us the range that we are 95% sure contains the true population mean. As the interval does not contain zero, we can say that on average, students are likely to tell some lies in their conversation with their mothers. However, it does not imply that all students lie to their mothers. Some may not lie at all, while others may lie more than average. This is just an average tendency.

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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 is a range that estimates an unknown population parameter, like the mean in our case, with a certain level of confidence. For example, with a 95% confidence interval, we are 95% certain that the interval includes the true population parameter.
To calculate a confidence interval for a mean, we use the sample mean and the standard error derived from the sample data. The purpose of constructing a confidence interval is to give an estimate of where the actual mean of the entire group might lie based on our sample.
In this exercise, we constructed a 95% confidence interval around the sample mean of 0.5, with the result being an interval from approximately 0.411 to 0.589. This interval suggests where the true average number of lies per conversation for all students usually falls.
Sample Mean
The sample mean, often represented as \( \bar{x} \), is the average of all data points in a sample. It serves as an estimate of the population mean when a complete census is not possible. By using the sample mean, we can make informed inferences about the larger population.
In the given exercise, the sample mean is 0.5. This indicates that, on average, students reported telling 0.5 lies per conversation in the sample.
It is crucial because it acts as the center point of our confidence interval. Therefore, any statistical analysis on a sample often starts by calculating the sample mean to use as a benchmark for other calculations or estimations.
Standard Deviation
Standard deviation (SD) measures the amount of variation or dispersion of a set of values from the mean. A low standard deviation means the data points are closer to the mean, while a high standard deviation indicates that they are spread out over a broader range of values.
In our exercise, the standard deviation is 0.4. This value reflects how much the number of lies varies from one student to another within the sample.
It's essential to understanding data distributions and variability, as a high variability may suggest that the sample does not uniformly represent the population's behavior. With confidence intervals, a smaller standard deviation will generally produce a narrower interval, indicating more precise estimates of the population mean.
Z-Value
A z-value, or z-score, is a measure of how many standard deviations an element is from the mean. It is used in statistics to determine the probability of a statistic occurring within a standard normal distribution.
In this problem, a z-value of 1.96 is used for constructing a 95% confidence interval. This value comes from the standard normal distribution and tells us the number of standard errors we should go out from the mean to capture 95% of the possible sample means.
The choice of the z-value is critical in determining the accuracy and reliability of the confidence interval. For different confidence levels, different z-values are used. Typically, \( \approx 1.645 \) for 90%, \( \approx 1.96 \) for 95%, and \( \approx 2.576 \) for 99% confidence levels, respectively.

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Most popular questions from this chapter

The article "Nine Out of Ten Drivers Admit in Survey to Having Done Something Dangerous" (Knight Ridder Newspapers, July 8. 2005 ) reported the results of a survey of 1100 drivers. Of those surveyed, 990 admitted to careless or aggressive driving during the previous 6 months. Assuming that it is reasonable to regard this sample of 1100 as representative of the population of drivers, use this information to construct a \(99 \%\) confidence interval to estimate \(p\), the proportion of all drivers who have engaged in careless or aggressive driving in the previous 6 months.

In a survey on supernatural experiences, 722 of 4013 adult Americans surveyed reported that they had seen or been with a ghost ( 7 What Supernatural Experiences We've Had," USA Today. February 8. 2010). a. What assumption must be made in order for it to be appropriate to use the formula of this section to construct a confidence interval to estimate the proportion of all adult Americans who have seen or been with a ghost? b. Construct and interpret a \(90 \%\) confidence interval for the proportion of all adult Americans who have scen or been with a ghost. c. Would a \(99 \%\) confidence interval be narrower or wider than the interval compured in Part (b)? Justify your answer.

The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21,2006\()\) reported that \(37 \%\) of college freshmen and \(48 \%\) of college seniors carry a credit card balance from month to month. Suppose that the reported percentages were based on random samples of 1000 college freshmen and 1000 college seniors. a. Construct a \(90 \%\) confidence interval for the proportion of college freshmen who carry a credit card balance from month to month. b. Construct a \(90 \%\) confidence interval for the proportion of college seniors who carry a credit card balance from month to month. c. Explain why the two \(90 \%\) confidence intervals from Parts (a) and (b) are not the same width.

In an AP-AOL sports poll (Associated Press, December 18,2005\(), 394\) of 1000 randomly selected U.S. adults indicared that they considered themselves to be baseball fans. Of the 394 baseball fans, 272 stated that they thought the designated hitter rule should either be expanded to both baseball leagues or eliminated. a. Construct a \(95 \%\) confidence interval for the proportion of U.S. adults who consider themselves to be bascball fans. b. Construct a \(95 \%\) confidence interval for the proportion of those who consider themselves to be baseball fans who think the designated hitter rule should be expanded to both leagues or eliminated. c. Explain why the confidence intervals of Parts (a) and (b) are not the same width even though they both have a confidence level of \(95 \%\).

Authors of the news release titled "Major Gaps Still Exist Between the Perception and the Reality of Americans' Internet Security Protections, Study Finds" (The National Cyber Security Alliance) estimated the proportion of Americans who claim to have a firewall installed on their computer to protect them from computer hackers to be .80 based on a survey conducted by the Zogby market research firm. They also estimated the proportion of those who actually have a firewall installed to be .42, based on checkups performed by Norton's \(\mathrm{PC}\) Help software. The following quote is from the news release: For the study, NCSA commissioned a Zogby survey of more than 3000 Americans and Symantec conducted checkups of 400 Americans' personal computers performed by \(\mathrm{PC}\) Help by Norton (www.norton.com/tuneup). The Zogby poll has a margin of error of \(+1-1.6 \%\) and the checkup has a margin of error of \(+1-5 \%\). Explain why the margins of error for the two estimated proportions are different.
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