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 500? (Round your answers to four decimal places.)
b. What is the advantage of a larger sample size when attempting to estimate the population mean?
a. A larger sample increases the probability that the sample mean will be a specified distance away from the population mean.
b. A larger sample increases the probability that the sample mean will be within a specified distance of the population mean.
c. A larger sample lowers the population standard deviation.
d. A larger sample has a standard error that is closer to the population standard deviation.

Answer 1

Answer:

(a)                             n :      20           50          100         500

P (-200 < X - μ < 200) : 0.2886    0.4444    0.5954    0.9376

(b) The correct option is (b).

Step-by-step explanation:

Let the random variable X represent the amount of deductions for taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return.

The mean amount of deductions is, μ = $16,642 and standard deviation is, σ = $2,400.

Assuming that the random variable X follows a normal distribution.

(a)

Compute 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 as follows:

• For a sample size of n = 20

$P(\mu-200$

                                           $=P(-0.37$

• For a sample size of n = 50

$P(\mu-200$

                                           $=P(-0.59$

• For a sample size of n = 100

$P(\mu-200$

                                           $=P(-0.83$

• For a sample size of n = 500

$P(\mu-200$

                                           $=P(-1.86$

                                 n :      20           50          100         500

P (-200 < X - μ < 200) : 0.2886    0.4444    0.5954    0.9376

(b)

The law of large numbers, in probability concept, states that as we increase the sample size, the mean of the sample ($\bar x$) approaches the whole population mean ($\mu_{x}$).

Consider the probabilities computed in part (a).

As the sample size increases from 20 to 500 the probability that the sample mean is within $200 of the population mean gets closer to 1.

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

Thus, the correct option is (b).