Question

the financial report of a business association indicates that 32% of businesses in a particular sector claim deductions on specific expenditures. the mean amount of deductions for these businesses is $18,144 with a standard deviation of $sigma = 32,400$. a. what is the probability that a random sample of businesses in this sector will show a sample mean within $300 of the population mean for the following sample sizes? (round your answers to four decimal places.) $n = 10$: $n = 50$: $n = 150$: $n = 600$: b. what is the impact of choosing a larger sample size when estimating the population mean? choose the best answer below: a. a larger sample size increases the probability that the sample mean will be further away from the population mean. b. a larger sample size decreases the population standard deviation. c. a larger sample size diminishes variability completely. d. a larger sample size increases the probability that the sample mean will be within a specified distance of the population mean.

Explanation:

Step1: Recall the formula for the standard error of the mean

The standard error of the mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, and we use the z - score formula $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$. We want to find $P(|\bar{X}-\mu|\leq300)$. First, find the z - scores for $\bar{x}_1=\mu - 300$ and $\bar{x}_2=\mu + 300$. The z - score is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$.

Step2: Calculate z - scores for each sample size

For $n = 10$:
The standard error $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{32400}{\sqrt{10}}\approx10247.84$.
$z_1=\frac{- 300}{\frac{32400}{\sqrt{10}}}\approx - 0.03$ and $z_2=\frac{300}{\frac{32400}{\sqrt{10}}}\approx0.03$.
$P(-0.03\leq Z\leq0.03)=\Phi(0.03)-\Phi(-0.03)=2\Phi(0.03)-1$.
Using the standard normal table, $\Phi(0.03) = 0.5120$, so $P(-0.03\leq Z\leq0.03)=2\times0.5120 - 1=0.0240$.
For $n = 50$:
The standard error $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{32400}{\sqrt{50}}\approx4582.58$.
$z_1=\frac{-300}{\frac{32400}{\sqrt{50}}}\approx - 0.07$ and $z_2=\frac{300}{\frac{32400}{\sqrt{50}}}\approx0.07$.
$P(-0.07\leq Z\leq0.07)=2\Phi(0.07)-1$.
Using the standard normal table, $\Phi(0.07)=0.5279$, so $P(-0.07\leq Z\leq0.07)=2\times0.5279 - 1 = 0.0558$.
For $n = 150$:
The standard error $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{32400}{\sqrt{150}}\approx2635.28$.
$z_1=\frac{-300}{\frac{32400}{\sqrt{150}}}\approx - 0.11$ and $z_2=\frac{300}{\frac{32400}{\sqrt{150}}}\approx0.11$.
$P(-0.11\leq Z\leq0.11)=2\Phi(0.11)-1$.
Using the standard normal table, $\Phi(0.11)=0.5438$, so $P(-0.11\leq Z\leq0.11)=2\times0.5438 - 1=0.0876$.
For $n = 600$:
The standard error $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{32400}{\sqrt{600}}\approx1326.34$.
$z_1=\frac{-300}{\frac{32400}{\sqrt{600}}}\approx - 0.23$ and $z_2=\frac{300}{\frac{32400}{\sqrt{600}}}\approx0.23$.
$P(-0.23\leq Z\leq0.23)=2\Phi(0.23)-1$.
Using the standard normal table, $\Phi(0.23)=0.5910$, so $P(-0.23\leq Z\leq0.23)=2\times0.5910 - 1 = 0.1820$.

Answer:

For $n = 10$: $0.0240$
For $n = 50$: $0.0558$
For $n = 150$: $0.0876$
For $n = 600$: $0.1820$
For part b: D. A larger sample size increases the probability that the sample mean will be within a specified distance of the population mean.