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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