/*! 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 33 The following data represent the... [FREE SOLUTION] | 91Ó°ÊÓ

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The following data represent the repair cost for a low-impact collision in a simple random sample of mini- and micro-vehicles (such as the Chevrolet Aveo or Mini Cooper). $$ \begin{array}{lrlrr} \hline \$ 3148 & \$ 1758 & \$ 1071 & \$ 3345 & \$ 743 \\ \hline \$ 2057 & \$ 663 & \$ 2637 & \$ 773 & \$ 1370 \\ \hline \end{array} $$ (a) Draw a normal probability plot to determine if it is reasonable to conclude the data come from a population that is normally distributed. (b) Draw boxplot to check for outliers. (c) Construct and interpret a \(95 \%\) confidence interval for the population mean cost of repair. (d) Suppose you obtain a simple random sample of size \(n=10\) of a Mini Cooper that was in a low-impact collision and determine the cost of repair. Do you think a \(95 \%\) confidence interval would be wider or narrower? Explain.

Short Answer

Expert verified
1. Normal probability plot: use to check normality. 2. Boxplot: identify outliers. 3. 95% CI: compute interval. 4. Larger sample size: usually narrower CI.

Step by step solution

01

Organize the Data

Start by listing the given repair costs: 3148, 1758, 1071, 3345, 743, 2057, 663, 2637, 773, 1370.
02

Normal Probability Plot

To create a normal probability plot, use statistical software or graph paper. Plot the observed data on the vertical axis and the expected z-scores (corresponding to a standard normal distribution) on the horizontal axis. If the points approximately form a straight line, it suggests that the data may be normally distributed.
03

Draw Boxplot

Calculate the five-number summary of the data (minimum, Q1, median, Q3, maximum). Use these values to draw a boxplot. Identify potential outliers if any data points lie beyond 1.5 times the interquartile range (IQR) from the quartiles.
04

Calculate 95% Confidence Interval

First, compute the sample mean \(\bar{x}\) and sample standard deviation (s). Use the formula for the 95% confidence interval for the population mean:
05

Evaluate Sample Size Impact

With the new sample size \(n=10\), the 95% confidence interval would be narrower or wider based on sample variance and degrees of freedom. Generally, increasing sample size decreases standard error, leading to a narrower confidence interval.

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Key Concepts

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

normal probability plot
A normal probability plot helps us check if a set of data follows a normal distribution. To create this plot, we start by organizing the data in ascending order. We then plot the observed data points against the expected z-scores from a standard normal distribution.
When the points line up roughly in a straight line, it suggests that the data are normally distributed. However, significant deviations from the straight line indicate departures from normality.
This step is crucial because many statistical analyses assume normally distributed data. For our given repair costs, a good normal probability plot can be easily created using statistical software like R or Python.
boxplot
A boxplot is a graphical representation that shows the distribution of a dataset and highlights potential outliers. It's built using the five-number summary: minimum, first quartile (Q1), median, third quartile (Q3), and maximum.
To draw a boxplot for our repair costs:
  • Calculate the five-number summary of the data.
  • Draw a box from Q1 to Q3 with a line at the median.
  • Extend whiskers from the box to the minimum and maximum values.
Data points that fall far from the whiskers are potential outliers. Checking for outliers is essential as they can heavily influence measures like the mean and standard deviation.
confidence interval
A confidence interval provides a range of values that likely contain the population mean. For a 95% confidence interval:
  • Calculate the sample mean \(\bar{x}\) and sample standard deviation (s).
  • Use the formula: \[\bar{x} \pm t_{n-1} \cdot \frac{s}{\sqrt{n}}\] where \( t_{n-1} \) is the t-value from the t-distribution with n-1 degrees of freedom.
This formula assumes a normally distributed population. The confidence interval for our repair costs helps us understand the likely range for the average repair cost across all vehicles. It's useful in making informed economic decisions.
sample size effect
The size of the sample (n) affects the width of the confidence interval. A larger sample generally reduces the standard error, which is \( \frac{s}{\sqrt{n}} \), leading to a narrower interval. Conversely, a smaller sample size increases the standard error, resulting in a wider confidence interval.
This relationship shows the importance of drawing sufficiently large samples to make precise estimates. For example, if we had a larger sample size than n=10, the 95% confidence interval for the repair costs would be narrower, increasing the precision of our estimates.

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Most popular questions from this chapter

The data sets represent simple random samples from a population whose mean is \(100 .\) $$ \begin{array}{rrrrr} & {\text { Data Set I }} \\ \hline 106 & 122 & 91 & 127 & 88 \\ \hline 74 & 77 & 108 & & \end{array} $$ $$ \begin{array}{rrrrr} \quad{\text { Data Set II }} \\ \hline 106 & 122 & 91 & 127 & 88 \\ \hline 74 & 77 & 108 & 87 & 88 \\ \hline 111 & 86 & 113 & 115 & 97 \\ \hline 122 & 99 & 86 & 83 & 102 \end{array} $$ $$ \begin{array}{rrrrr} {\text { Data Set III }} \\ \hline 106 & 122 & 91 & 127 & 88 \\ \hline 74 & 77 & 108 & 87 & 88 \\ \hline 111 & 86 & 113 & 115 & 97 \\ \hline 122 & 99 & 86 & 83 & 102 \\ \hline 88 & 111 & 118 & 91 & 102 \\ \hline 80 & 86 & 106 & 91 & 116 \end{array} $$ (a) Compute the sample mean of each data set. (b) For each data set, construct a \(95 \%\) confidence interval about the population mean. (c) What effect does the sample size \(n\) have on the width of the interval? For parts \((d)-(e),\) suppose that the data value 106 was accidentally recorded as \(016 .\) (d) For each data set, construct a \(95 \%\) confidence interval about the population mean using the incorrectly entered data. (e) Which intervals, if any, still capture the population mean, 100? What concept does this illustrate?

In a survey of 700 community college students, 481 indicated that they have read a book for personal enjoyment during the school year (based on data from the Community College Survey of Student Engagement). (a) Determine a \(90 \%\) confidence interval for the proportion of community college students who have read a book for personal enjoyment during the school year. (b) Determine a \(95 \%\) confidence interval for the proportion of community college students who have read a book for personal enjoyment during the school year. (c) What is the impact of increasing the level of confidence on the margin of error?

The American Society for Microbiology (ASM) and the Soap and Detergent Association (SDA) jointly commissioned two separate studies, both of which were conducted by Harris Interactive. In one of the studies, 1001 adults were interviewed by telephone and asked about their handwashing habits. In the telephone interviews, 921 of the adults said they always wash their hands in public restrooms. In the other study, the hand-washing behavior of 6076 adults was inconspicuously observed within public restrooms in four U.S. cities and 4679 of the 6076 adults were observed washing their hands. (a) In the telephone survey, what is the variable of interest? Is it qualitative or quantitative? (b) What is the sample in the telephone survey? What is the population to which this study applies? (c) Verify that the requirements for constructing a confidence interval for the population proportion of adults who say they always wash their hands in public restrooms are satisfied. (d) Using the results from the telephone interviews, construct a \(95 \%\) confidence interval for the proportion of adults who say they always wash their hands in public restrooms. (e) In the study where hand-washing behavior was observed, what is the variable of interest? Is it qualitative or quantitative? (f) We are told that 6076 adults were inconspicuously observed, but were not told how these adults were selected. We know randomness is a key ingredient in statistical studies that allows us to generalize results from a sample to a population. Suggest some ways randomness might have been used to select the individuals in this study. (g) Verify the requirements for constructing a confidence interval for the population proportion of adults who actually washed their hands while in a public restroom. (h) Using the results from the observational study, construct a \(95 \%\) confidence interval for the proportion of adults who wash their hands in public restrooms. (i) Based on your findings in parts (a) through (h), what might you conclude about the proportion of adults who say they always wash their hands versus the proportion of adults who actually wash their hands in public restrooms? (j) Cite some sources of variability in both studies.

A recent Gallup poll asked Americans to disclose the number of books they read during the previous year. Initial survey results indicate that \(s=16.6\) books. (a) How many subjects are needed to estimate the number of books Americans read the previous year within four books with \(95 \%\) confidence? (b) How many subjects are needed to estimate the number of books Americans read the previous year within two books with \(95 \%\) confidence? (c) What effect does doubling the required accuracy have on the sample size? (d) How many subjects are needed to estimate the number of books Americans read the previous year within four books with \(99 \%\) confidence? Compare this result to part (a). How does increasing the level of confidence in the estimate affect sample size? Why is this reasonable?

Researchers Havar Brendryen and Pal Kraft conducted a study in which 396 subjects were randomly assigned to either an experimental smoking cessation program or control group. The experimental program consisted of the Internet and phone-based Happy Ending Intervention, which lasted 54 weeks and consisted of more than 400 contacts by e-mail, web pages, interactive voice response, and short message service (SMS) technology. The control group received a self- help booklet. Both groups were offered free nicotine replacement therapy. Abstinence was defined as "not even a puff of smoke, for the last 7 days," and assessed by means of Internet surveys or telephone interviews. The response variable was abstinence after 12 months. Of the participants in the experimental program, \(22.3 \%\) reported abstinence; of the participants in the control group, \(13.1 \%\) reported abstinence. (a) What type of experimental design is this? (b) What is the treatment? How many levels does it have? (c) What is the response variable? (d) What are the statistics reported by the authors? (e) An odds ratio is the ratio of the odds of an event occurring in one group to the odds of it occurring in another group. These groups might be men and women, an experimental group and a control group, or any other dichotomous classification. If the probabilities of the event in each of the groups are \(p\) (first group) and \(q\) (second group), then the odds ratio is $$ \frac{\frac{p}{(1-p)}}{\frac{q}{(1-q)}}=\frac{p(1-q)}{q(1-p)} $$ An odds ratio of 1 indicates that the condition or event under study is equally likely in both groups. An odds ratio greater than 1 indicates that the condition or event is more likely in the first group. And an odds ratio less than 1 indicates that the condition or event is less likely in the first group. The odds ratio must be greater than or equal to zero. As the odds of the first group approach zero, the odds ratio approaches zero. As the odds of the second group approach zero, the odds ratio approaches positive infinity. Verify that the odds ratio for this study is \(1.90 .\) What does this mean? (f) The authors of the study reported a \(95 \%\) confidence interval for the odds ratio to be lower bound: 1.12 and upper bound 3.26. Interpret this result. (g) Write a conclusion that generalizes the results of this study to the population of all smokers.

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