Question

the financial report of a business association indicates that 29% of businesses in a particular sector claim deductions on specific expenditures. the mean amount of deductions for these businesses is $15,676 with a standard deviation of $sigma = $2,200.
a. what is the probability that a random sample of businesses in this sector will show a sample mean within $250 of the population mean for the following sample sizes? (round your answers to four decimal places.)
$n = 10$
$n = 50$
$n = 150$
$n = 400$
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 within a specified distance of the population mean.
b. a larger sample size decreases the population standard deviation.
c. a larger sample size increases the probability that the sample mean will be further away from the population mean.
d. a larger sample size diminishes variability completely.

Explanation:

Step1: Recall the formula for z - score and probability for normal distribution

We use the formula $z=\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}$, and we want to find $P(|\bar{X}-\mu|\leq250)$. This is equivalent to $P(- \frac{250}{\sigma/\sqrt{n}}\leq Z\leq\frac{250}{\sigma/\sqrt{n}})= \Phi(\frac{250}{\sigma/\sqrt{n}})-\Phi(-\frac{250}{\sigma/\sqrt{n}})=2\Phi(\frac{250}{\sigma/\sqrt{n}})- 1$, where $\Phi(z)$ is the cumulative - distribution function of the standard normal distribution and $\sigma = 2200$.

Step2: Calculate for $n = 10$

First, calculate the $z$ - value: $z=\frac{250}{2200/\sqrt{10}}\approx\frac{250}{2200/3.1623}\approx\frac{250}{695.69}\approx0.36$. Then $P(|\bar{X}-\mu|\leq250)=2\Phi(0.36)-1$. Looking up in the standard - normal table, $\Phi(0.36) = 0.6406$. So $P(|\bar{X}-\mu|\leq250)=2\times0.6406 - 1=0.2812$.

Step3: Calculate for $n = 50$

Calculate the $z$ - value: $z=\frac{250}{2200/\sqrt{50}}\approx\frac{250}{2200/7.0711}\approx\frac{250}{311.12}\approx0.80$. Then $P(|\bar{X}-\mu|\leq250)=2\Phi(0.80)-1$. Looking up in the standard - normal table, $\Phi(0.80)=0.7881$. So $P(|\bar{X}-\mu|\leq250)=2\times0.7881 - 1 = 0.5762$.

Step4: Calculate for $n = 150$

Calculate the $z$ - value: $z=\frac{250}{2200/\sqrt{150}}\approx\frac{250}{2200/12.2474}\approx\frac{250}{179.64}\approx1.39$. Then $P(|\bar{X}-\mu|\leq250)=2\Phi(1.39)-1$. Looking up in the standard - normal table, $\Phi(1.39)=0.9177$. So $P(|\bar{X}-\mu|\leq250)=2\times0.9177 - 1 = 0.8354$.

Step5: Calculate for $n = 400$

Calculate the $z$ - value: $z=\frac{250}{2200/\sqrt{400}}=\frac{250}{2200/20}=\frac{250}{110}\approx2.27$. Then $P(|\bar{X}-\mu|\leq250)=2\Phi(2.27)-1$. Looking up in the standard - normal table, $\Phi(2.27)=0.9884$. So $P(|\bar{X}-\mu|\leq250)=2\times0.9884 - 1 = 0.9768$.

For part b, the correct answer is: A larger sample size increases the probability that the sample mean will be within a specified distance of the population mean.

Answer:

When $n = 10$, the probability is $0.2812$.
When $n = 50$, the probability is $0.5762$.
When $n = 150$, the probability is $0.8354$.
When $n = 400$, the probability is $0.9768$.
For part b: A. A larger sample size increases the probability that the sample mean will be within a specified distance of the population mean.