/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 Seventy-seven students at the Un... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Seventy-seven students at the University of Virginia were asked to keep a diary of a conversation with their mothers, recording any lies they told during these conversations (San 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
Our \(95\% \) confidence interval for the average number of lies told per conversation is calculated around the sample mean of \(0.5\) lies. Although 0 is not included in our confidence interval, this does not mean every student lies to their mothers. This only implies that the average number of lies per conversation for the whole student population is likely to be greater than 0.

Step by step solution

01

Identify Given Values

Identify the sample size \(n = 77\), the sample mean \(\overline{x} = 0.5\), and the sample standard deviation \(s = 0.4\). The confidence level is \(95\% \).
02

Determine the Confidence Level in Z-Scores

For a \(95\% \) confidence level, the z-value is found using the Standard Normal (Z) Table or commonly used probability values for the normal distribution. The z-value for \(95\% \) confidence level is approximately \(1.96\). This is found by learning that \(95\% \) of the area under the standardized normal curve lies within 1.96 standard deviations of the mean.
03

Calculate Confidence Interval

Use the formula to calculate the confidence interval. The formula for confidence interval is: \(\overline{x} \pm z \dfrac{s}{\sqrt{n}}\). Substituting in values, the calculation becomes \(0.5 \pm 1.96 \dfrac{0.4}{\sqrt{77}}\). Calculate the upper and lower confidence interval values.
04

Interpretation of Part a

The result is the \(95\% \) confidence interval for the average number of lies per conversation reported by the students at that University.
05

Interpretation of Part b

The interval not including 0 means that it is statistically unlikely (less than \(5\% \) chance) that the true average number of lies per conversation is 0. However, this doesn't imply that all students lie to their mothers. It only means that, on average, the students in this sample reported telling some lies. There could be many students who didn't tell any lies at all, so long as there are others who told more than one lie to balance it.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Statistics Education
Understanding statistics is essential for interpreting research findings, making informed decisions based on data, and even becoming a critical consumer of information presented in the media. A reliable statistics education entails learning about various concepts, including data collection, analysis, interpretation, and presentation. When studying confidence intervals, as in our exercise, students learn to estimate the range within which a population parameter is likely to lie with a given level of certainty. This is a key concept in inferential statistics, which allows us to make predictions or inferences about a population based on sample data.
Normal Distribution
A normal distribution, often represented by a bell curve, is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. In the context of creating a confidence interval, when the sampling distribution of the mean is normally distributed or the sample size is large, the normal distribution can be used to determine how likely it is that the sample mean will fall within a certain range around the actual population mean. This information, combined with the concept of z-scores, lets us calculate the boundaries of the confidence interval for the mean number of lies told by students to their mothers.
Sample Standard Deviation
The sample standard deviation is a measure of the amount of variability or dispersion of a set of values. It reflects how much the elements in the sample differ from the sample mean. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates the values are spread out over a wider range. When computing the confidence interval in our exercise, the sample standard deviation is used to estimate the standard deviation of the population, helping to determine the margin of error and the width of our confidence interval.
z-scores
Z-scores are the number of standard deviations away from the mean a data point is. In the context of confidence intervals, z-scores are used to identify the critical values that correspond to a designated level of confidence. These z-scores, when multiplied by the standard deviation and adjusted for the sample size, provide the margin of error for the interval. For example, a z-score of 1.96 is commonly used for a 95% confidence interval. This tells us that there is a 95% chance that the population mean will fall between 1.96 standard deviations below and above the sample mean.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The article "National Geographic, the Doomsday Machine," which appeared in the March 1976 issue of the Journal of Irreproducible Results (yes, there really is a journal by that name -it's a spoof of technical journals!) predicted dire consequences resulting from a nationwide buildup of National Geographic magazines. The author's predictions are based on the observation that the number of subscriptions for National Geographic is on the rise and that no one ever throws away a copy of National Geographic. A key to the analysis presented in the article is the weight of an issue of the magazine. Suppose that you were assigned the task of estimating the average weight of an issue of National Geographic. How many issues should you sample to estimate the average weight to within \(0.1 \mathrm{oz}\) with \(95 \%\) confidence? Assume that \(\sigma\) is known to be 1 oz.

Why is an unbiased statistic generally preferred over a biased statistic for estimating a population characteristic? Does unbiasedness alone guarantee that the estimate will be close to the true value? Explain. Under what circumstances might you choose a biased statistic over an unbiased statistic if two statistics are available for estimating a population characteristic?

The authors of the paper "Short-Term Health and Economic Benefits of Smoking Cessation: Low Birth Weight" (Pediatrics [1999]: \(1312-1320\) ) investigated the medical cost associated with babies born to mothers who smoke. The paper included estimates of mean medical cost for low-birth-weight babies for different ethnic groups. For a sample of 654 Hispanic low-birth- weight babies, the mean medical cost was \(\$ 55,007\) and the standard error \((s / \sqrt{n})\) was \(\$ 3011 .\) For a sample of 13 Native American low-birth- weight babies, the mean and standard error were \(\$ 73,418\) and \(\$ 29,577\), respectively. Explain why the two standard errors are so different.

Given a variable that has a \(t\) distribution with the specified degrees of freedom, what percentage of the time will its value fall in the indicated region? a. \(10 \mathrm{df}\), between \(-1.81\) and \(1.81\) b. \(10 \mathrm{df}\), between \(-2.23\) and \(2.23\) c. \(24 \mathrm{df}\), between \(-2.06\) and \(2.06\) d. \(24 \mathrm{df}\), between \(-2.80\) and \(2.80\) e. 24 df, outside the interval from \(-2.80\) to \(2.80\) f. \(24 \mathrm{df}\), to the right of \(2.80\) g. \(10 \mathrm{df}\), to the left of \(-1.81\)

In an AP-AOL sports poll (Associated Press, December 18,2005 ), 394 of 1000 randomly selected U.S. adults indicated 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 that consider themselves to be baseball fans. b. Construct a \(95 \%\) confidence interval for the proportion of those who consider themselves to be baseball fans that 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 \(95 \%\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.