Chapter 15, Problem 4P

Data from the Department of Motor Vehicles indicate that 80% of all licensed drivers are older than age 25.

1. a. In a sample of n = 50 people who recently received speeding tickets, 33 were older than 25 years and the other 17 were age 25 or younger. Is the age distribution for this sample significantly different from the distribution for the population of licensed drivers? Use α = .05.
2. b. In a sample of n = 50 people who recently received parking tickets, 36 were older than 25 years and the other 14 were age 25 or younger. Is the age distribution for this sample significantly different from the distribution for the population of licensed drivers? Use α = .05.

To determine

To check: The age distribution for the given sample is significantly different from the distribution for the population of licensed drivers.

Answer to Problem 4P

The age distribution for the given sample is significantly different from the distribution for the population of licensed drivers.

Explanation of Solution

Given info:

In a sample of 50 people who received speeding tickets, 33 were of age 25 and older and 17 were of age 25 and younger. Data from the Department of Motor Vehicles indicate

that 80% of all licensed drivers are older than age 25. Use $\alpha =0.05$.

Calculations:

Step 1: Null hypothesis and Alternate hypothesis are:

$H_0:$ Age distribution for the given sample is not significantly different from the distribution for the population of licensed drivers

$H_1:$ Age distribution for the given sample is significantly different from the distribution for the population of licensed drivers.

Step 2: For the given sample, degrees of freedom equals:

$\begin{array}{c}df=(C-1) \text{where }C\text{ equals number of categories}\\ =2-1\\ =1\end{array}$

With $\alpha =0.05$ and $df=1$, the critical value (CV) is obtained from the ${\chi }^2-table$ as

${\chi }^2=3.841$

Step 3: ${\chi }^2-statistics$ is calculated as:

${\chi }^2={\displaystyle \sum \frac{{(f_o-f_e)}^2}{f_e}}$

The formula to calculate expected frequency is:

$f_e=p\times n$

Substituting respective values of proportion and $n=50$ in the above formula:

$\begin{array}{c}{f}_{e,25+}=0.8\times (50)\\ =40\\ f_{e,25-}=0.2\times (50)\\ =10\end{array}$

The contingency table is:

Frequency	Age 25+	Age 25-
Observed ( $f_o$)	33	17
Expected ( $f_e$)	40	10

Finally substitute the values in the ${\chi }^2$-statistics formula as:

$\begin{array}{c}{\chi }^2=\frac{{(33-40)}^2}{40}+\frac{{(17-10)}^2}{10}\\ =\frac{49}{40}+\frac{49}{10}\\ =6.125\end{array}$

Step 4: Rejection rule: Reject when ${\chi }^2-statistics>CV$.

Since ${\chi }^2-\text{statistics}(=6.125)<\text{critical}\_\text{value}(=3.841)$, reject the null hypothesis.

Step 5: Based on the results of hypothesis test, there is sufficient evidence to reject the null hypothesis at $\alpha =0.05$.

Hence, reject null and conclude that the age distribution for the given sample is significantly different from the distribution for the population of licensed drivers.

To determine

To check: The age distribution for the given sample is significantly different from the distribution for the population of licensed drivers.

Answer to Problem 4P

The age distribution for the given sample is not significantly different from the distribution for the population of licensed drivers.

Explanation of Solution

Given info:

In a sample of 50 people who received speeding tickets, 36 were of age 25 and older and 14 were of age 25 and younger. Data from the Department of Motor Vehicles indicate

that 80% of all licensed drivers are older than age 25. Use $\alpha =0.05$.

Calculations:

Step 1: Null Hypothesis and Alternate Hypothesis are:

$H_0:$ Age distribution for the given sample is not significantly different from the distribution for the population of licensed drivers

$H_1:$ Age distribution for the given sample is significantly different from the distribution for the population of licensed drivers.

Step 2: For the given sample, degrees of freedom equals:

$\begin{array}{c}df=(C-1) \text{where }C\text{ equals number of categories}\\ =2-1\\ =1\end{array}$

With $\alpha =0.05$ and $df=1$, the critical value (CV)  is obtained from the ${\chi }^2-table$ as

${\chi }^2=3.841$

Step 3: ${\chi }^2-\text{statistics}$ is calculated as:

${\chi }^2={\displaystyle \sum \frac{{(f_o-f_e)}^2}{f_e}}$

The formula to calculate expected frequency is:

$f_e=p\times n$

Substituting respective values of proportion and $n=50$ in the above formula:

$\begin{array}{c}{f}_{e,25+}=0.8\times (50)\\ =40\\ f_{e,25-}=0.2\times (50)\\ =10\end{array}$

The contingency table is:

Frequency	Age 25+	Age 25-
Observed ( $f_o$)	36	14
Expected ( $f_e$)	40	10

Finally substitute the values in the ${\chi }^2$-statistics formula as:

$\begin{array}{c}{\chi }^2=\frac{{(36-40)}^2}{40}+\frac{{(14-10)}^2}{10}\\ =\frac{16}{40}+\frac{16}{10}\\ =0.8\end{array}$

Step 4: Rejection rule:

Reject when ${\chi }^2-\text{statistics}>CV$.

Since ${\chi }^2-\text{statistics}(=0.8)<\text{critical}\_\text{value}(=3.841)$, fail to reject the null hypothesis.

Step 5: Based on the results of hypothesis test, there is no sufficient evidence to reject the null hypothesis at $\alpha =0.05$.

Hence, fail to reject null and conclude that the age distribution for the given sample is not significantly different from the distribution for the population of licensed drivers