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The Wall Street Journal reported that 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 reported that 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return. The mean amount of deductions for this population of taxpayers was $17,190. Assume that the standard deviation is σ = $2,603. Use z-table.


 a. What is the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $166 of the population mean for each of the following sample sizes: 30, 50, 100, and 400? Round your answers to four decimals.


b. What is the advantage of a larger sample size when attempting to estimate the population mean? Round your answers to four decimals. 


A larger sample Select your answer - V the probability that the sample mean will be within a specified distance of the population mean. In the automobile insurance example, the probability of being within ±166 of μ ranges from for a sample of size 30 to _______  for a sample of size 400.

Solutions

Expert Solution

Given,

Mean = 17190

Standard deviation = 2603

a)

i)

sample n = 30

P(u-166 < xbar < u+166) = P(-166/(2603/sqrt(30)) < z < 166/(2603/sqrt(30)))

= P(-0.35 < z < 0.35)

= P(z < 0.35) - P(z < - 0.35)

= 0.6368307 - 0.3631693 [since from z table]

= 0.2737

ii)

sample n = 50

P(u-166 < xbar < u+166) = P(-166/(2603/sqrt(50)) < z < 166/(2603/sqrt(50)))

= P(- 0.45 < z < 0.45)

= P(z < 0.45) - P(z < - 0.45)

= 0.6736447 - 0.3263552 [since from z table]

= 0.3473

iii)

n = 100

P(u-166 < xbar < u+166) = P(-166/(2603/sqrt(100)) < z < 166/(2603/sqrt(100)))

= P(- 0.64 < z < 0.64)

= P(z < 0.64) - P(z < - 0.64)

= 0.7389137 - 0.2610863 [since from z table]

= 0.4778

(iv)

n = 400

P(u-166 < xbar < u+166) = P(-166/(2603/sqrt(400)) < z < 166/(2603/sqrt(400)))

= P(- 1.28 < z < 1.28)

= P(z < 1.28) - P(z < - 1.28)

= 0.8997274 - 0.1002726 [since from z table]

= 0.7995

b)

Here the larger sample the probability that the sample mean will be within a specified distance of the population mean. In the automobile insurance example, the probability of being within +/- 166 of ranges from  0.2737 for a sample of size 30 to 0.7995 for a sample of size 400.


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