/*! 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 21 Twenty-five adult citizens of th... [FREE SOLUTION] | 91Ó°ÊÓ

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Twenty-five adult citizens of the United States were asked to estimate the average income of all U.S. households. The mean estimate was \(x^{-} \bar{x}=\$ 65,000\) and \(s=\$ 15,000\). (Note: The actual average household income at the time of the study was about \(\$ 79,000\).) Assume the 25 adults in the study can be considered an SRS from the population of all adult citizens of the United States. A 95\% confidence interval for the mean estimate of the average income of all U.S. households is (a) \(\$ 62,000\) to \(\$ 68,000\). (b) \(\$ 58,808\) to \(\$ 71,192\). (c) \(\$ 72,808\) to \(\$ 85,192\).

Short Answer

Expert verified
The correct confidence interval is (b) $58,808 to $71,192.

Step by step solution

01

Identify the question

We need to find the 95% confidence interval for the mean estimate of the average income based on the given sample of adult citizens.
02

Understand the given data

Based on the problem statement, we have: the sample mean \( \bar{x} = 65,000 \), the sample standard deviation \( s = 15,000 \), and the sample size \( n = 25 \).
03

Identify the critical value

Since the sample size is 25, we check a t-distribution table for a 95% confidence interval with \( n-1 = 24 \) degrees of freedom to find \( t^* \), which is approximately \( 2.064 \).
04

Calculate the standard error

The standard error (SE) is given by \( SE = \frac{s}{\sqrt{n}} = \frac{15000}{\sqrt{25}} = 3000 \).
05

Compute the margin of error

The margin of error is found using \( t^* \times SE = 2.064 \times 3000 = 6192 \).
06

Determine the confidence interval

The confidence interval is calculated by \( \bar{x} \pm \text{margin of error} \). Thus, the interval is \( 65000 \pm 6192 \), which results in \( 58808 \) to \( 71192 \).
07

Match with given options

Compare the calculated interval \( 58808 \) to \( 71192 \) with the given choices. Option (b) matches this interval.

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Key Concepts

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

Confidence Interval
When we want to estimate a population parameter, such as the mean income of households, it's essential to provide not just an estimate, but also a range in which we are confident that the true parameter lies. This range is called a confidence interval. In this exercise, the 95% confidence interval signifies that we are 95% confident that the true average income of U.S. households falls within this range. The number "95%" is the confidence level, indicating the probability that the interval calculated from the sample data includes the true parameter value. Typically, higher confidence levels lead to wider intervals. This is because more certainty about containing the true parameter value requires accounting for more variability. Calculating the interval involves determining the margin of error and adjusting the sample mean by this amount.
Mean Estimate
The mean estimate is essentially the average value obtained from a sample and is used as a point estimate for the population parameter. In this exercise, 25 adult citizens provided income estimates, and their mean estimate was $65,000. A mean estimate is crucial as it represents our best guess of the population average based on the sample data. However, it’s important to remember that a mean estimate taken from a sample is only an approximation of the true population mean. Therefore, we use statistical methods to assess how close our sample mean might be to the real mean, guiding decisions and predictions more accurately.
T-distribution
The t-distribution is a vital concept in statistics, especially when dealing with sample sizes less than 30 or when the population standard deviation is unknown. It resembles the normal distribution but has heavier tails, which means it accounts for more variability and uncertainty. In our case, with a sample size of 25 and unknown population variance, the degrees of freedom (df) is 24, calculated as sample size minus one. This degrees of freedom is used to find the critical value from the t-distribution table, which provides how many standard deviations we extend on either side of the mean for the confidence interval. The t-distribution becomes very similar to the normal distribution as sample size increases, hence why it's pivotal for smaller samples.
Standard Error
The standard error (SE) measures the variability of the sample mean estimate. It essentially tells us how much we expect the sample mean to differ from the actual population mean if we repeatedly took samples from the population. Calculated as the sample standard deviation divided by the square root of the sample size, \[ SE = \frac{s}{\sqrt{n}} \]where in this case, SE comes out to be $3,000. The standard error is what we multiply by the critical t-value to find the margin of error. The margin of error tells us how far from the sample mean the true population mean is likely to be, giving us both ends of the confidence interval when added and subtracted from the sample mean. It's crucial for constructing confidence intervals and assessing the precision of our estimate.

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Most popular questions from this chapter

Which of these settings does not allow use of a matched pairs \(t\) procedure? (a) You interview both the instructor and one of the students in each of 20 introductory statistics classes and ask each how many hours per week homework assignments require. (b) You interview a sample of 15 instructors and another sample of 15 students and ask each how many hours per week homework assignments require. (c) You interview 40 students in the introductory statistics course at the beginning of the semester and again at the end of the semester and ask how many hours per week homework assignments require.

Use software or Table \(C\) to find (a) the critical value for a one-sided test with level \(\alpha=0.01\) based on the \(t_{2}\) distribution. (b) the critical value for a \(90 \%\) confidence interval based on the \(t_{28}\) distribution.

Researchers claim that women speak significantly more words per day than men. One estimate is that a woman uses about 20,000 words per day while a man uses about 7000 . To investigate such claims, one study used a special device to record the conversations of male and female university students over a four- day period. From these recordings, the daily word counts of the 20 men in the study were determined. Here are their daily word counts: \({ }^{23}\) \(\begin{array}{rrrrr}28,408 & 10,084 & 15,931 & 21,688 & 37,786 \\ 10,575 & 12,880 & 11,071 & 17,799 & 13,182 \\ 8,918 & 6,495 & 8,153 & 7,015 & 4,429 \\\ 10,054 & 3,998 & 12,639 & 10,974 & 5,255\end{array}\) (a) Examine the data. Is it reasonable to use the \(t\) procedures (assume these men are an SRS of all male students at this university)? (b) If your conclusion in part (a) is Yes, do the data give convincing evidence that the mean number of words per day of men at this university differs from 7000 ?

Fortunately, we aren't really interested in the number of seeds velvetleaf plants produce (see Exercise \(20.41\) ). The velvetleaf seed beetle feeds on the seeds and might be a natural weed control. Here are the total seeds, seeds infected by the beetle, and percent of seeds infected for 28 velvetleaf plants: $$ \begin{array}{l|rrrrrrrrrr} \hline \text { Seeds } & 2450 & 2504 & 2114 & 1110 & 2137 & 8015 & 1623 & 1531 & 2008 & 1716 \\ \text { Infected } & 135 & 101 & 76 & 24 & 121 & 189 & 31 & 44 & 73 & 12 \\ \text { Percent } & 5.5 & 4.0 & 3.6 & 2.2 & 5.7 & 2.4 & 1.9 & 2.9 & 3.6 & 0.7 \\\ \hline \text { Seeds } & 721 & 863 & 1136 & 2819 & 1911 & 2101 & 1051 & 218 & 1711 & 164 \\ \text { Infected } & 27 & 40 & 41 & 79 & 82 & 85 & 42 & 0 & 64 & 7 \\ \text { Percent } & 3.7 & 4.6 & 3.6 & 2.8 & 4.3 & 4.0 & 4.0 & 0.0 & 3.7 & 4.3 \\\ \hline \text { Seeds } & 2228 & 363 & 5973 & 1050 & 1961 & 1809 & 130 & 880 & & \\ \text { Infected } & 156 & 31 & 240 & 91 & 137 & 92 & 5 & 23 & & \\ \text { Percent } & 7.0 & 8.5 & 4.0 & 8.7 & 7.0 & 5.1 & 3.8 & 2.6 & & \\ \hline \end{array} $$ Do a complete analysis of the percent of seeds infected by the beetle. Include a \(90 \%\) confidence interval for the mean percent infected in the population of all velvetleaf plants. Do you think that the beetle is very helpful in controlling the weed? Why is analyzing percent of seeds infected more useful than analyzing number of seeds infected?

The Trial Urban District Assessment (TUDA) is a government-sponsored study of student achievement in large urban school districts. TUDA gives a mathematics test scored from 0 to 500 . A score of 262 is a "basic" mathematics level and a score of 299 is "proficient." Scores for a random sample of 1100 eighth- graders in Dallas had \(x \bar{x}=271\) with standard error 1.3. 16 (a) We don't have the 1100 individual scores, but use of the \(t\) procedures is surely safe. Why? (b) Give a \(99 \%\) confidence interval for the mean score of all Dallas eighthgraders. (Be careful: the report gives the standard error of \(x^{-} \bar{x}\), not the standard deviation s.) (c) Urban children often perform below the basic level. Is there good evidence that the mean for all Dallas eighth-graders is more than the basic level?

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