Chapter 29.8, Problem 20SRE

To determine

Obtain the chance that between 9% and 11% of the sample families will not own cars.

Answer to Problem 20SRE

The chance that between 9% and 11% of the sample families will not own carsis 0.8064.

Explanation of Solution

Calculation:

Based on the given information, it is clear that the population size is 25,000. 10% of them have no cars. The opinion survey includes the simple random sample of 1,500.

The box should contain 25,000, which represents the population of 25,000 people.

Let 1 represent a person who does not have a car and 0 represent a person who has a car.

In this case, the researcher is interested in drawing 1,500 tickets from a box containing 90% of populations containing 22,200 zeros $\left(=0.90\times 25,000\right)$ and 2500 ones $\left(=0.10\times 25,000\right)$.

The number of draws is 1500.

Expected value of the box:

$\begin{array}{c}\text{Expected value}=\frac{\text{Sum of all values}}{\text{Number of tickets}}\\ =\frac{0\left(22,500\right)+1\left(2,500\right)}{25,000}\\ =\frac{2,500}{25,000}\\ =\frac{1}{10}\\ =0.1\end{array}$

Standard deviation of the box:

$\begin{array}{c}SD=\left(\text{Big number}-\text{small number}\right)\sqrt{\left(\begin{array}{l}\text{Fraction with big number}\times \\ \text{Fraction with small number}\end{array}\right)}\\ =\left(1-0\right)\sqrt{0.10\times 0.90}\\ =0.3\end{array}$

Expected value of sum:

$\begin{array}{c}\text{Expected value of sum}=\left\{\begin{array}{l}\text{Number of draws}\times \\ \text{Expected value of box}\end{array}\right\}\\ =1,500\times \left(0.1\right)\\ =150\end{array}$

Standard error of the sum:

$\begin{array}{c}\text{SE sum}=\sqrt{\text{Number of draws}}\times \text{SD box}\\ \text{=}\sqrt{1,500}\times 0.3\\ =11.6170\end{array}$

Expected value of the percentage:

$\begin{array}{c}\text{Expected value of percentage}=\frac{\text{Expected value sum}}{\text{Number of draws}}\times 100\%\\ =\frac{150}{1,500}\times 100\%\\ =10\%\end{array}$

Standard error of the percentage:

$\begin{array}{c}\text{SE percentage}=\frac{\text{SE for number}}{\text{Number of draws}}\times 100\%\\ =\frac{11.6170}{1,500}\times 100\%\\ =0.0077\times 100\%\\ =0.77\%\end{array}$

The known values are given below:

$\begin{array}{l}\mu =\text{Mean}=10\\ \sigma =0.77\\ x=9\text{ and }11\end{array}$

The formula for z-score is as follows:

$z=\frac{x-\mu }{\sigma }$

The z-score is obtained as given below:

At $x=9$,

$\begin{array}{c}z=\frac{9-\left(10\right)}{\text{0}\text{.77}}\\ =-1.30\end{array}$

At $x=11$,

$\begin{array}{c}z=\frac{11-\left(10\right)}{\text{0}\text{.77}}\\ =1.30\end{array}$

The chance that between 9% and 11% of the sample families will not own carsis given below:

$\begin{array}{c}P\left(9<p<1\right)=P\left(-1.30<Z<1.30\right)\\ =P\left(Z<1.30\right)-P\left(Z<-1.30\right)\end{array}$

The value corresponding to $P\left(Z<-1.30\right)$ is 0.0968 and $P\left(Z<1.30\right)$ is 0.9032 using the standard normal table.

Remaining calculation:

$\begin{array}{c}P\left(9<p<1\right)=P\left(Z<1.30\right)-P\left(Z<-1.30\right)\\ =0.9032-0.0968\\ =0.8064\end{array}$

Therefore, the chance that between 9% and 11% of the sample families will not own cars is 0.8064.