Question

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 $15,968. Assume that the standard deviation is $2,190. 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 $184 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 -increases or decreases) 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 +/-184 of u ranges from (blank) for a sample of size 30, to (blank) for a sample of size 400.

Answer 1

Solution:

We are given

Population mean = µ = $15,968

Population Standard deviation = σ = $2,190

Part a)

We have to find P(15968 – 184 < Xbar < 15968 + 184) = P(15784 < Xbar < 16152)

P(15784 < Xbar < 16152) =P(Xbar<16152) – P(Xbar<15784)

Z = (Xbar - µ) / [σ/sqrt(n)]

[All probability values for Z scores are calculated by using z-table or excel.]

For n = 30

First find P(Xbar<16152)

Z = (16152 – 15968)/[2190/sqrt(30)]

Z = 0.460187

P(Z<0.460187) = 0.677309

P(Xbar<16152) = 0.677309

Now find P(Xbar<15784)

Z = (15784 – 15968)/[2190/sqrt(30)]

Z = - 0.460187

P(Z<- 0.460187) = 0.322691

P(Xbar<15784) = 0.322691

P(15784 < Xbar < 16152) =P(Xbar<16152) – P(Xbar<15784)

P(15784 < Xbar < 16152) = 0.677309 - 0.322691 = 0.354618

Required probability = 0.354618

For n = 50

First find P(Xbar<16152)

Z = (16152 – 15968)/[2190/sqrt(50)]

Z = 0.594099

P(Z<0.594099) = 0.723777

P(Xbar<16152) = 0.723777

Now find P(Xbar<15784)

Z = (15784 – 15968)/[2190/sqrt(50)]

Z = - 0.594099

P(Z<- 0.594099) = 0.276223

P(Xbar<15784) = 0.276223

P(15784 < Xbar < 16152) = 0.723777 - 0.276223

P(15784 < Xbar < 16152) = 0.447554

Required probability = 0.447554

For n = 100

First find P(Xbar<16152)

Z = (16152 – 15968)/[2190/sqrt(100)]

Z = 0.840183

P(Z< 0.840183) = 0.799597

P(Xbar<16152) = 0.799597

Now find P(Xbar<15784)

Z = (15784 – 15968)/[2190/sqrt(100)]

Z = - 0.840183

P(Z<- 0.840183) = 0.200403

P(Xbar<15784) = 0.200403

P(15784 < Xbar < 16152) = 0.799597 - 0.200403

P(15784 < Xbar < 16152) = 0.599194

Required probability = 0.599194

For n = 400

First find P(Xbar<16152)

Z = (16152 – 15968)/[2190/sqrt(400)]

Z = 1.680365

P(Z< 1.680365) = 0.953557

P(Xbar<16152) = 0.953557

Now find P(Xbar<15784)

Z = (15784 – 15968)/[2190/sqrt(400)]

Z = - 1.680365

P(Z<- 1.680365) = 0.046443

P(Xbar<15784) = 0.046443

P(15784 < Xbar < 16152) = 0.953557 - 0.046443

P(15784 < Xbar < 16152) = 0.907114

Required probability = 0.907114

Part b)

A larger sample increases the probability that the sample mean will be within a specified distance of the population mean. In an automobile insurance example, the probability of being within +/- 184 ranges from 0.354618 for a sample of size 30, to 0.907114 for a sample of size 400.