/*! 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 Samples of two different types o... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 33

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

Short Answer

Expert verified
The confidence interval \((49.4,50.6)\) for the second type of automobile is based on a larger sample size. This conclusion is reached by comparing the widths of the given confidence intervals.

Step by step solution

01

Understand the Confidence Intervals

The confidence intervals given for the two types of automobiles are \( (51.3,52.7) \) and \( (49.4,50.6) \). The width of these intervals gives us an idea of the precision of our estimate. The narrower the interval, the more precise our estimate is.
02

Compare the Width of the Confidence Intervals

To know which confidence interval is based on a larger sample size, we compare their width. The width of a confidence interval for the mean is largely determined by the standard deviation and the sample size. Because we are assuming the standard deviations to be identical, the interval with the smaller width should have a larger sample size. The width of the first interval is \(52.7 - 51.3 = 1.4\) and that of the second interval is \(50.6 - 49.4 = 1.2\)
03

Decide which sample size is larger

Since the width of the second interval (1.2) is less than the width of the first interval (1.4), this suggests that the second confidence interval is based on a larger sample size. This is because the greater the size of the sample, the lower the variability around the mean, resulting in a narrower confidence interval.

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Most popular questions from this chapter

The interval from \(-2.33\) to \(1.75\) captures an area of \(.95\) under the \(z\) curve. This implies that another largesample \(95 \%\) confidence interval for \(\mu\) has lower limit \(\bar{x}-2.33 \frac{\sigma}{\sqrt{n}}\) and upper limit \(\bar{x}+1.75 \frac{\sigma}{\sqrt{n}} .\) Would you recommend using this \(95 \%\) interval over the \(95 \%\) interval \(\bar{x} \pm 1.96 \frac{\sigma}{\sqrt{n}}\) discussed in the text? Explain. (Hint: Look at the width of each interval.)

The eating habits of 12 bats were examined in the article "Foraging Behavior of the Indian False Vampire Bat" (Biotropica \([1991]: 63-67) .\) These bats consume insects and frogs. For these 12 bats, the mean time to consume a frog was \(\bar{x}=21.9 \mathrm{~min}\). Suppose that the standard deviation was \(s=7.7 \mathrm{~min} .\) Construct and interpret a \(90 \%\) confidence interval for the mean suppertime of a vampire bat whose meal consists of a frog. What assumptions must be reasonable for the one-sample \(t\) interval to be appropriate?

Consumption of fast food is a topic of interest to researchers in the field of nutrition. The article "Effects of Fast-Food Consumption on Energy Intake and Diet Quality Among Children" (Pediatrics [2004]: \(112-118\) ) reported that 1720 of those in a random sample of 6212 U.S. children indicated that on a typical day they ate fast food. Estimate \(\pi\), the proportion of children in the U.S. who eat fast food on a typical day.

The report "2005 Electronic Monitoring \& Surveillance Survey: Many Companies Monitoring, Recording, Videotaping-and Firing-Employees" (American Management Association, 2005 ) summarized the results of a survey of 526 U.S. businesses. The report stated that 137 of the 526 businesses had fired workers for misuse of the Internet and 131 had fired workers for email misuse. For purposes of this exercise, assume that it is reasonable to regard this sample as representative of businesses in the United States. a. Construct and interpret a \(95 \%\) confidence interval for the proportion of U.S. businesses that have fired workers for misuse of the Internet. b. What are two reasons why a \(90 \%\) confidence interval for the proportion of U.S. businesses that have fired workers for misuse of email would be narrower than the \(95 \%\) confidence interval computed in Part (a).

The article "Sensory and Mechanical Assessment of the Quality of Frankfurters" (Journal of Texture Studies [1990]: \(395-409\) ) reported the following salt content (percentage by weight) for 10 frankfurters: \(\begin{array}{llllllllll}2.26 & 2.11 & 1.64 & 1.17 & 1.64 & 2.36 & 1.70 & 2.10 & 2.19 & 2.40\end{array}\) a. Use the given data to produce a point estimate of \(\mu\), the true mean salt content for frankfurters. b. Use the given data to produce a point estimate of \(\sigma^{2}\), the variance of salt content for frankfurters. c. Use the given data to produce an estimate of \(\sigma\), the standard deviation of salt content. Is the statistic you used to produce your estimate unbiased?
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