/*! 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 62 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 conversations 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
The 95% confidence interval for the mean number of lies per conversation for UVA students is approximately between 0.411 and 0.589. This does not necessarily imply that all students lie to their mothers despite the interval not including 0; rather, it suggests that in 95% of samples, the mean number of lies told per conversation is non-zero.

Step by step solution

01

Identify the Given Information

Identify the given information: sample size (n) is 77, sample mean (x̄) is 0.5, and standard deviation (σ) is 0.4. The Z statistic for a 95% confidence interval is approximately 1.96 (this value can be found in Z-tables that give the area under a standard normal curve).
02

Construct the Confidence Interval

Use the formula for a confidence interval: \(x̄ \pm Z * \frac{σ}{\sqrt{n}}\). Plug in the given values and perform the calculations: \(0.5 \pm 1.96 * \frac{0.4}{\sqrt{77}}\)
03

Calculate the Required Values

Calculate the required values: \(0.5 \pm 1.96 * \frac{0.4}{8.77}\) is approximately \(0.5 \pm 0.089\). Thus, the confidence interval is \(0.411, 0.589\)
04

Interpret the Confidence Interval

Interpret the confidence interval: We are 95% confident that the true population mean of lies told by UVA students to their mothers per conversation falls between 0.411 and 0.589.
05

Analyze the Results

The range does not include 0. This does not necessarily imply that all students lie to their mothers. Instead, it suggests that in 95% of samples, the mean number of lies per conversation will be non-zero. However, it does not preclude the possibility of a student not lying in a particular conversation.

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Key Concepts

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

Mean and Standard Deviation
In statistics, the mean and standard deviation are fundamental concepts that describe the central tendency and spread of a dataset, respectively. The mean, often referred to as the average, is calculated by summing all the values in a dataset and then dividing by the number of values. In our problem, the mean is given as 0.5, indicating the average number of lies students told their mothers per conversation.

The standard deviation, on the other hand, measures how spread out the numbers are in a dataset. A low standard deviation, like the 0.4 in this exercise, indicates that the majority of students' responses are close to the mean. By understanding these terms, one can easily evaluate the variation from the average number, helping to determine the reliability and consistency of the results in any given data.
Sample Size
Sample size is a crucial aspect of statistical analysis since it influences the accuracy and stability of the results. In this exercise, the sample size is 77, meaning data were collected from 77 students. A larger sample size generally leads to more reliable and accurate results because it better represents the entire population from which it is drawn, reducing variability and potential biases.

When constructing a confidence interval, the sample size affects the width of the interval. Larger sample sizes result in narrower intervals, suggesting more precise estimates of the population parameter. In our example, with a sample size of 77, the confidence interval provides a reliable approximation of the population mean.
Z-statistic
The Z-statistic is a value derived from the standard normal distribution used to standardize individual sample data and calculate confidence intervals. In this nature of problem, we use a Z-statistic of 1.96 for constructing a 95% confidence interval, a common level of confidence for statistical analysis.

To build a confidence interval using the Z-statistic, we apply this formula: \(x̄ \pm Z * \frac{σ}{\sqrt{n}}\) where \(x̄\) is the sample mean, \(σ\) is the standard deviation, and \(n\) is the sample size. The term \(\frac{σ}{\sqrt{n}}\) is known as the standard error, which adjusts the standard deviation for the sample size. In this case, the computed confidence interval suggests that with 95% confidence, the actual average number of lies for the entire student body lies between this range, allowing researchers to forecast with the degree of uncertainty they are willing to accept.

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Most popular questions from this chapter

Samples of two different models of cars were selected, and the actual speed for each car was determined when the speedometer registered \(50 \mathrm{mph}\). The resulting \(95 \%\) confidence intervals for mean actual speed were \((51.3,52.7)\) and \((49.4,50.6)\). Assuming that the two sample standard deviations are equal, which confidence interval is based on the larger sample size? Explain your reasoning.

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A random sample of \(n=12\) four-year-old red pine trees was selected, and the diameter (in inches) of each tree's main stem was measured. The resulting observations are as follows: \(\begin{array}{llllllll}11.3 & 10.7 & 12.4 & 15.2 & 10.1 & 12.1 & 16.2 & 10.5\end{array}\) \(\begin{array}{llll}11.4 & 11.0 & 10.7 & 12.0\end{array}\) a. Compute a point estimate of \(\sigma\), the population standard deviation of main stem diameter. What statistic did you use to obtain your estimate? b. Making no assumptions about the shape of the population distribution of diameters, give a point estimate for the population median diameter. What statistic did you use to obtain the estimate? c. Suppose that the population distribution of diameter is symmetric but with heavier tails than the normal distribution. Give a point estimate of the population mean diameter based on a statistic that gives some protection against the presence of outliers in the sample. What statistic did you use? d. Suppose that the diameter distribution is normal. Then the 90 th percentile of the diameter distribution is \(\mu+1.28 \sigma\) (so \(90 \%\) of all trees have diameters less than this value). Compute a point estimate

The article "Most Canadians Plan to Buy Treats. Many Will Buy Pumpkins, Decorations and/or Costumes" (Ipsos-Reid, October 24, 2005) summarized results from a survey of 1000 randomly selected Canadian residents. Each individual in the sample was asked how much he or she anticipated spending on Halloween during \(2005 .\) The resulting sample mean and standard deviation were \(\$ 46.65\) and \(\$ 83.70\), respectively. a. Explain how it could be possible for the standard deviation of the anticipated Halloween expense to be larger than the mean anticipated expense. b. Is it reasonable to think that the distribution of the variable anticipated Halloween expense is approximately normal? Explain why or why not. c. Is it appropriate to use the \(t\) confidence interval to estimate the mean anticipated Halloween expense for Canadian residents? Explain why or why not. d. If appropriate, construct and interpret a \(99 \%\) confidence interval for the mean anticipated Halloween expense for Canadian residents.

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