Question

The Wall Street Journal reports 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 $16,642. Assume the standard deviation is

σ = $2,400.

(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: 20, 50, 100, and 300? (Round your answers to four decimal places.)

sample size n = 20sample size n = 50sample size n = 100sample size n = 300

(b)

What is the advantage of a larger sample size when attempting to estimate the population mean?

A larger sample has a standard error that is closer to the population standard deviation.A larger sample increases the probability that the sample mean will be a specified distance away from the population mean.    A larger sample increases the probability that the sample mean will be within a specified distance of the population mean.A larger sample lowers the population standard deviation.

Answer 1

a)

for n = 20

Here, μ = 16642, σ = 536.6563, x1 = 16442 and x2 = 16842. We need to compute P(16442<= X <= 16842). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (16442 - 16642)/536.6563 = -0.37
z2 = (16842 - 16642)/536.6563 = 0.37

Therefore, we get
P(16442 <= X <= 16842) = P((16842 - 16642)/536.6563) <= z <= (16842 - 16642)/536.6563)
= P(-0.37 <= z <= 0.37) = P(z <= 0.37) - P(z <= -0.37)
= 0.6443 - 0.3557
= 0.2886

Here, μ = 16642, σ = 339.4113, x1 = 16442 and x2 = 16842. We need to compute P(16442<= X <= 16842). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (16442 - 16642)/339.4113 = -0.59
z2 = (16842 - 16642)/339.4113 = 0.59

Therefore, we get
P(16442 <= X <= 16842) = P((16842 - 16642)/339.4113) <= z <= (16842 - 16642)/339.4113)
= P(-0.59 <= z <= 0.59) = P(z <= 0.59) - P(z <= -0.59)
= 0.7224 - 0.2776
= 0.4448

Here, μ = 16642, σ = 240, x1 = 16442 and x2 = 16842. We need to compute P(16442<= X <= 16842). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (16442 - 16642)/240 = -0.83
z2 = (16842 - 16642)/240 = 0.83

Therefore, we get
P(16442 <= X <= 16842) = P((16842 - 16642)/240) <= z <= (16842 - 16642)/240)
= P(-0.83 <= z <= 0.83) = P(z <= 0.83) - P(z <= -0.83)
= 0.7967 - 0.2033
= 0.5934

Here, μ = 16642, σ = 138.5641, x1 = 16442 and x2 = 16842. We need to compute P(16442<= X <= 16842). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (16442 - 16642)/138.5641 = -1.44
z2 = (16842 - 16642)/138.5641 = 1.44

Therefore, we get
P(16442 <= X <= 16842) = P((16842 - 16642)/138.5641) <= z <= (16842 - 16642)/138.5641)
= P(-1.44 <= z <= 1.44) = P(z <= 1.44) - P(z <= -1.44)
= 0.9251 - 0.0749
= 0.8502

b)

A larger sample increases the probability that the sample mean will be within a specified distance of the population mean.