Chapter 12.2, Problem 14E

A vehicle quality survey asked new owners a variety of questions about their recently purchased automobile. one question asked for the owner’s rating of the vehicle using categorical responses of average, outstanding, and exceptional. Another question asked for the owner’s education level with the categorical responses some high school, high school graduate, some college, and college graduate. Assume the sample data below are for 500 owners who had recently purchased an automobile.

Education

Quality Rating	Some HS	HS Grad	Some College	College Grad
Average	35	30	20	60
Outstanding	45	45	50	90
Exceptional	20	25	30	50

1. a. Use a .05 level of significance and a test of independence to determine if a new owner’s vehicle quality rating is independent of the owner’s education. What is the p-value and what is your conclusion?
2. b. Use the overall percentage of average, outstanding, and exceptional ratings to comment upon how new owners rate the quality of their recently purchased automobiles.

To determine

Perform a hypothesis test to check whether a new owner’s vehicle quality rating independent of the owner’s education at 0.05 level of significance.

Find the p-value and draw the conclusion of the study.

Answer to Problem 14E

The p-value is 0.3624.

The data provide sufficient evidence to conclude that a new owner’s vehicle quality rating independent of the owner’s education.

Explanation of Solution

Calculation:

The given information shows that the sample of 500 automobile owners provides information regarding the level of education with level of quality rating of owners.

State the test hypotheses:

Null hypothesis:

$H_0:\text{The two cataegorical variables are independent}$.

That is, quality rating is independent of the education of the owner.

Alternative hypothesis:

$H_a:\text{The two cataegorical variables are not independent}$.

That is, quality rating is not independent of the education of the owner.

The row and column total is tabulated below:

Quality rating	Some HS	HS Grad	Some College	College Grad	Total
Average	35	30	20	60	145
Outstanding	45	45	50	90	230
Exceptional	20	25	30	50	125
Total	100	100	100	200	500

The formula for expected frequency is given below:

$e_{ij}=\frac{\left(\text{Row }i\text{ Total}\right)\left(\text{Coloumn }j\text{ Total}\right)}{\text{Total Sample Size}}$.

The expected frequency for each category is calculated as follows:

Quality rating	Some HS	HS Grad	Some College	College Grad
Average	$\begin{array}{l}\frac{\left(145\right)\left(100\right)}{500}\\ =29\end{array}$	$\begin{array}{l}\frac{\left(145\right)\left(100\right)}{500}\\ =29\end{array}$	$\begin{array}{l}\frac{\left(145\right)\left(100\right)}{500}\\ =29\end{array}$	$\begin{array}{l}\frac{\left(145\right)\left(200\right)}{500}\\ =58\end{array}$
Outstanding	$\begin{array}{l}\frac{\left(230\right)\left(100\right)}{500}\\ =46\end{array}$	$\begin{array}{l}\frac{\left(230\right)\left(100\right)}{500}\\ =46\end{array}$	$\begin{array}{l}\frac{\left(230\right)\left(100\right)}{500}\\ =46\end{array}$	$\begin{array}{l}\frac{\left(230\right)\left(200\right)}{500}\\ =92\end{array}$
Exceptional	$\begin{array}{l}\frac{\left(125\right)\left(100\right)}{500}\\ =25\end{array}$	$\begin{array}{l}\frac{\left(125\right)\left(100\right)}{500}\\ =25\end{array}$	$\begin{array}{l}\frac{\left(125\right)\left(100\right)}{500}\\ =25\end{array}$	$\begin{array}{l}\frac{\left(125\right)\left(200\right)}{500}\\ =50\end{array}$

The formula for chi-square test statistic is given as,

${\chi }^2={\displaystyle \sum _{i=1}^r{\displaystyle \sum _{j=1}^c\frac{{\left(f_{ij}-e_{ij}\right)}^2}{e_{ij}}}}$ where r represents the r^th row and c represents the c^th column.

The value of chi-square test statistic is,

$\begin{array}{c}{\chi }^2=\left\{\begin{array}{l}\frac{{\left(35-29\right)}^2}{29}+\frac{{\left(30-29\right)}^2}{29}+\frac{{\left(20-29\right)}^2}{29}+\frac{{\left(60-58\right)}^2}{58}+\\ \frac{{\left(45-46\right)}^2}{46}+\frac{{\left(45-46\right)}^2}{46}+\frac{{\left(50-46\right)}^2}{46}+\frac{{\left(90-92\right)}^2}{92}+\\ \frac{{\left(20-25\right)}^2}{25}+\frac{{\left(25-25\right)}^2}{25}+\frac{{\left(30-25\right)}^2}{25}+\frac{{\left(50-50\right)}^2}{50}\end{array}\right\}\\ =1.24+0.03+2.79+0.07+0.02+0.02+0.35+0.04+1+0+1+0\\ =6.57\end{array}$

Thus, the chi-square test statistic is 6.57.

Degrees of freedom:

The degrees of freedom is $df=\left(r-1\right)\left(c-1\right)$ for the contingency table of r rows and c columns.

In the given problem $r=3$ and $c=4$.

Therefore,

$\begin{array}{c}df=\left(3-1\right)\left(4-1\right)\\ =6\end{array}$

Level of significance:

The given level of significance is $\alpha =0.05$.

p-value:

Software procedure:

Step -by-step software procedure to obtain p-value using MINITAB software is as follows:

• Select Graph > Probability distribution plot > view probability
• Select chi-square under distribution and enter 6 in degrees of freedom.
• Choose X-Value and Right Tail for the region of the curve to shade.
• Enter the X-value as 6.57 under shaded area.
• Select OK.
• Output using MINITAB software is given below:

From the MINITAB output, the p-value is 0.3624.

Rejection rule:

If the $p\text{-value}\le \alpha$, then reject the null hypothesis.

Conclusion:

Here, the p-value is greater than the level of significance.

That is, $p\text{-value}\left(=0.3624\right)>\alpha \left(=0.05\right)$

Thus, the decision is “fail to reject the null hypothesis”.

Therefore, the data do not provide sufficient evidence to conclude that the column variable is not independent of row variable.

That is, there is no association between column variable and row variable. In addition, quality ratings are not differs with the education of the owner.

Thus, the data provides sufficient evidence to conclude that the new owner’s vehicle quality rating independent of the owner’s education.

To determine

Find overall percentage of average, outstanding and exceptional ratings and comment how new owners rate the quality of their recently purchased automobiles.

Explanation of Solution

Calculation:

The overall percentage of average rating is,

$\frac{145}{500}\times 100=29\%$.

The overall percentage of outstanding rating is,

$\frac{230}{500}\times 100=46\%$.

The overall percentage of exceptional rating is,

$\frac{125}{500}\times 100=25\%$.

Thus, the new owners are satisfied with almost 50% outstanding quality rating and $46\%+25\%=69\%$ outstanding or exceptional rating for their new automobiles.