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

$P(\mu-200<X<\mu+200)=P(\frac{(16642-200)-16642}{2400/\sqrt{20}}<\frac{X-\mu}{\sigma/\sqrt{n}}<\frac{(16642+200)-16642}{2400/\sqrt{20}})$

                                           $=P(-0.37<Z<0.37)\\\\=P(Z<0.37)-P(Z<-0.37)\\\\=0.6443-0.3557\\\\=0.2886$

• For a sample size of n = 50

$P(\mu-200<X<\mu+200)=P(\frac{(16642-200)-16642}{2400/\sqrt{50}}<\frac{X-\mu}{\sigma/\sqrt{n}}<\frac{(16642+200)-16642}{2400/\sqrt{50}})$

                                           $=P(-0.59<Z<0.59)\\\\=P(Z<0.59)-P(Z<-0.59)\\\\=0.7222-0.2778\\\\=0.4444$

• For a sample size of n = 100

$P(\mu-200<X<\mu+200)=P(\frac{(16642-200)-16642}{2400/\sqrt{100}}<\frac{X-\mu}{\sigma/\sqrt{n}}<\frac{(16642+200)-16642}{2400/\sqrt{100}})$

                                           $=P(-0.83<Z<0.83)\\\\=P(Z<0.83)-P(Z<-0.83)\\\\=0.7977-0.2023\\\\=0.5954$

• For a sample size of n = 500

$P(\mu-200<X<\mu+200)=P(\frac{(16642-200)-16642}{2400/\sqrt{500}}<\frac{X-\mu}{\sigma/\sqrt{n}}<\frac{(16642+200)-16642}{2400/\sqrt{500}})$

                                           $=P(-1.86<Z<1.86)\\\\=P(Z<1.86)-P(Z<-1.86)\\\\=0.9688-0.0312\\\\=0.9376$

                                 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).

Answer 2

Using the normal distribution and the central limit theorem, it is found that:

a)

0.2886 = 28.86% probability that a sample size of 20 has a sample mean within $200 of the population mean.

0.4448 = 44.48% probability that a sample size of 50 has a sample mean within $200 of the population mean.

0.5934 = 59.34% probability that a sample size of 100 has a sample mean within $200 of the population mean.

0.9372 = 93.72% probability that a sample size of 500 has a sample mean within $200 of the population mean.

b)

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

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.  
• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
• By the Central Limit Theorem, the standard deviation of the sampling distribution of sample sizes of n is: $s = \frac{\sigma}{\sqrt{n}}$.

Hence, the probability of sample of size n having a sample mean within 200 is the p-value of $Z = \frac{200}{s}$ subtracted by the p-value of $Z = -\frac{200}{s}$

In this problem:

• The standard deviation is $\sigma = 2400$.

Item a:

For samples of 20, $n = 20$, and thus:

$s = \frac{2400}{\sqrt{20}}$

Then

$Z = \frac{200}{s} = \frac{200}{\frac{2400}{\sqrt{20}}} = 0.37$

• Z = 0.37 has a p-value of 0.6443.
• Z = -0.37 has a p-value of 0.3557.

0.6443 - 0.3557 = 0.2886.

0.2886 = 28.86% probability that a sample size of 20 has a sample mean within $200 of the population mean.

For samples of 50, $n = 50$, and thus:

$s = \frac{2400}{\sqrt{50}}$

Then

$Z = \frac{200}{s} = \frac{200}{\frac{2400}{\sqrt{50}}} = 0.59$

• Z = 0.59 has a p-value of 0.7224.
• Z = -0.59 has a p-value of 0.2776.

0.7224 - 0.2776 = 0.4448.

0.4448 = 44.48% probability that a sample size of 50 has a sample mean within $200 of the population mean.

For samples of 100, $n = 100$, and thus:

$s = \frac{2400}{\sqrt{100}}$

Then

$Z = \frac{200}{s} = \frac{200}{\frac{2400}{\sqrt{100}}} = 0.83$

• Z = 0.83 has a p-value of 0.7967.
• Z = -0.83 has a p-value of 0.2033.

0.7967 - 0.2033 = 0.5934

0.5934 = 59.34% probability that a sample size of 100 has a sample mean within $200 of the population mean.

For samples of 500, $n = 500$, and thus:

$s = \frac{2400}{\sqrt{500}}$

Then

$Z = \frac{200}{s} = \frac{200}{\frac{2400}{\sqrt{500}}} = 1.86$

• Z = 1.86 has a p-value of 0.9686.
• Z = -1.86 has a p-value of 0.0314.

0.9686 - 0.0314 = 0.9372

0.9372 = 93.72% probability that a sample size of 500 has a sample mean within $200 of the population mean.

Item b:

A larger sample size reduces the standard error, as $s = \frac{\sigma}{\sqrt{n}}$, that is, the standard error is inversely proportional to the square root of the sample size n, meaning that values are closer to the mean. Thus, the correct option is:

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

A similar problem is given at https://brainly.com/question/24663213