Q 9RP. - Elementary Statistics

Question

New Car Passion. Edmunds.com publishes information on new car prices in Car Shopping Trends Report. During a recent year. Americans spent an average of $\backslash (30,803$ for a new car. Assume a standard deviation of $\backslash )10,200$

a. Identify the population and variable under consideration.

b. For samples of $50$ new car sales during the year in question, determine the mean and standard deviation of all possible sample mean prices.

c. Repeat part (b) for samples of size$100$

d. For samples of size $1000$ answer the following question without doing any computations: Will the standard deviation of all possible sample mean prices to be larger than, smaller than, or the same as that in part (c)? Explain your answer.

Step-by-Step Solution

Answer

Part (a) The variable is the amount spent by Americans on a new car, which fluctuates for each observational unit.

Part (b) The mean $\left({\mu }_x\right)$ of all possible sample mean prices for sample size $50$ is $\$30,803$ For sample size $50$ the standard deviation the standard deviation of all potential sample mean prices $\$1442.50$ is $\left({\sigma }_{\overline{x}}\right)$

Part (c) The mean $\left({\mu }_{\overline{x}}\right)$ of all possible sample mean prices for sample size $100$ is $\$30,803$For sample size $100$, the standard deviation of all potential sample mean prices $\$1,020$ is $\left({\sigma }_{\overline{x}}\right)$

Part (d) $\text{ The value of }{\sigma }_{\overline{x}}\text{ would be smaller when the sample size is increased as }{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\text{. }$

Part (a) Step 1: Given information

A new car cost the typical American $\$30,803$ dollars.

A $\$10,200$ standard deviation.

Part (a) Step 2: Explanation

Determine the population and variable in question.

The population is all new cars in the United States for the given year, according to the study. Furthermore, the variable is the amount spent by Americans on a new car, which fluctuates for each observational unit.

Part (b) Step 1: Calculation

Find the average $\left({\mu }_x\right)$ of $50$ new automobile sales samples.

It is assumed that the average $(\mu)$ cost of a new car is $\$30,803$

$$\begin{gathered} {\mu }_x=\mu \\ =\$30,803 \end{gathered}$$

Thus, with sample size $50$, the mean $\left({\mu }_x\right)$ of all potential sample mean prices is $\$30,803$

Part (b) Step 2: Calculation

Find the standard deviation of all potential sample mean prices for sample size $50$ using $\left({\sigma }_{\overline{x}}\right)$. The standard deviation $(sigma)$ is assumed to be $\$10,200$

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}} \\ =\frac{\$10,200}{\sqrt{50}} \\ =\frac{\$10,200}{7.0711} \\ =1,442.50 \end{gathered}$$

As a result, given sample size $50$, the standard deviation of all potential sample mean prices $\$1,442.50$ is $\left({\sigma }_{\overline{x}}\right)$

Part (c) Step 1: Calculation

Calculate the mean $\left({\mu }_{\overline{x}}\right)$ for $100$ new car sales samples.

It is assumed that the average $\left(\mu \right)$ cost of a new car is $\$30,803$

$$\begin{gathered} {\mu }_x=\mu \\ =\$30,803 \end{gathered}$$

The mean $\left({\mu }_{\overline{x}}\right)$ of all possible sample mean prices for sample size $100$ is $\$30,803$

Part (c) Step 2: Calculation

Find the standard deviation of all potential sample mean prices for sample size $100$ $\left({\sigma }_{\overline{x}}\right)$

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}} \\ =\frac{\$10,200}{\sqrt{100}} \\ =\frac{\$10,200}{10} \\ =\$1,020 \end{gathered}$$

For sample size $100$, the standard deviation of all potential sample mean prices $\$1,020$ is $\left({\sigma }_{\overline{x}}\right)$

Part (d) Step 1: Explanation

$\text{ The value of }{\sigma }_{\overline{x}}\text{ would be smaller when the sample size is increased as }{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\text{. }$