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For the year 2010, 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return ( The Wall Street Journal , October 25, 2012). The mean amount of deductions for this population of taxpayers was $16,642. Assume the standard deviation is σ = $2400. a. What is the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for each of the following sample sizes: 30, 50, 100, and 400? b. What is the advantage of a larger sample size when attempting to estimate the population mean?

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STATISTICS F/BUSINESS+ECONOMICS-TE...

13th Edition
Anderson
Publisher: CENGAGE L
ISBN: 9781305881884
BuyFind

STATISTICS F/BUSINESS+ECONOMICS-TE...

13th Edition
Anderson
Publisher: CENGAGE L
ISBN: 9781305881884

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Chapter
Section
Chapter 7.5, Problem 26E
Textbook Problem
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For the year 2010, 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return (The Wall Street Journal, October 25, 2012). The mean amount of deductions for this population of taxpayers was $16,642. Assume the standard deviation is σ = $2400.

  1. a. What is the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for each of the following sample sizes: 30, 50, 100, and 400?
  2. b. What is the advantage of a larger sample size when attempting to estimate the population mean?

Expert Solution

a.

To determine

Find the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for each of the sample sizes: 30, 50, 100, and 400.

Explanation of Solution

Calculation:

The given information is that the standard deviation is σ=$2,400 and the mean amount of deductions for this population of taxpayers is $16,642.

Sampling distribution of x¯:

The probability distribution all possible values of the sample mean x¯ is termed as the sampling distribution of x¯.

  • The expected value of x¯ is, E(x¯)=μ.
  • The standard deviation of x¯ is

    For finite population, σx¯=NnN1(σn)

    For infinite population, σx¯=σn

  • When the population size is infinite or finite and nN0.05 then the standard deviation of x¯ is σx¯=σn.

The expected value of x¯ is,

E(x¯)=μ=$16,642

Thus, the expected value of x¯ is $16,642.

For sample size 30:

The standard deviation is σx¯=σn

Substitute σ as $2,400 and n as 30 in the formula,

σx¯=σn=2,40030=2,4005.4772=438.18

Thus, the standard deviation of x¯ is $438.18.

Central limit theorem:

For a simple random sample of size n drawn from a population, the sampling distribution of the sample mean x¯ is approximately normal when the sample size is large.

Therefore, the sampling distribution of the sample mean x¯ is approximately normal with μ=$16,642 and σx¯=$438.18.

The probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for sample size 30 is denoted as P(x¯μ±200).

P(μ200x¯μ+200)=P[200σx¯x¯μσx¯200σx¯]=P(200438.18z200438.18)=P(0.46z0.46)=12P(z0.46)

From Table 1: Cumulative probabilities for the standard normal distribution,

  • Locate the value 0.4 in the first column.
  • Locate the value 0.06 in the first row corresponding to the value 0.4 in the first column.
  • The intersecting value of row and column is 0.6772.

P(μ200x¯μ+200)=12(0.6772)=11.3544=0.3544

Thus, the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for sample size 30 is 0.3544.

For sample size 50:

The standard deviation is σx¯=σn

Substitute σ as $2,400 and n as 50 in the formula,

σx¯=σn=2,40050=2,4007.0711=339.41

Thus, the standard deviation of x¯ is $339.41.

The sampling distribution of the sample mean x¯ is approximately normal with μ=$16,642 and σx¯=$339.41.

The probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for sample size 50 is denoted as P(x¯μ±200).

P(μ200x¯μ+200)=P[200σx¯x¯μσx¯200σx¯]=P(200339.41z200339

Expert Solution

b.

To determine

Explain the advantage of a larger sample size when attempting to estimate the population mean.

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