Chapter 11.1, Problem 10E

(a)

To determine

To State: The hypothesis and the claim.

Answer to Problem 10E

The null and alternative hypotheses are:

$H_0$: The distribution of the most popular colors for light trucks were as follows: white, 30%; black, 17%; red, 14%; silver, 12%; gray, 11%; blue, 8%; and other, 8%.

$H_1$: The distribution is not the same as stated in the null hypothesis.

The claim of the test is that at least one of the proportions differs from its original value.

Explanation of Solution

Given info:

The data shows the percentages of most popular colors for light trucks. The level of significance is $\alpha =0.05$.

Justification:

The survey wants to see that whether the proportions differ from the results of the survey or not. The null and alternative hypothesis can be defined as:

Null hypothesis:

$H_0$: The distribution of the most popular colors for light trucks were as follows: white, 30%; black, 17%; red, 14%; silver, 12%; gray, 11%; blue, 8%; and other, 8%.

Alternative hypothesis:

$H_1$: The distribution is not the same as stated in the null hypothesis.

The claim of the test is that at least one of the proportions differs from its original value.

(b)

To determine

The critical value.

Answer to Problem 10E

The required critical value is 12.592.

Explanation of Solution

Degrees of freedom

The number of colors is 7.

The degrees of freedom, is calculated as:

$\begin{array}{c}\text{d}\text{.f = Number of categories-1}\\ \text{=7-1}\\ \text{=6}\end{array}$

Critical value:

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

Therefore, the critical value is obtained from Table G: Chi-Square Distribution.

Procedure:

• Locate 6 in the column of degrees of freedom in the Table 6.
• Take the value corresponding to $\alpha =0.05$.

From the table, the critical value is 12.592.

Thus, the critical value using $\alpha =0.05$ is 12.592.

(c)

To determine

The value of the test statistic.

Answer to Problem 10E

The test statistic value is 10.9144.

Explanation of Solution

Calculation:

The expected frequency can be obtained as follows:

$\begin{array}{c}{E}_1=n{p}_1\\ =\left(180\right)\left(0.3\right)\\ =54\end{array}$

$\begin{array}{c}{E}_2=n{p}_2\\ =\left(180\right)\left(0.17\right)\\ =30.6\end{array}$

$\begin{array}{c}{E}_3=n{p}_3\\ =\left(180\right)\left(.14\right)\\ =25.2\end{array}$

$\begin{array}{c}{E}_4=n{p}_4\\ =\left(180\right)\left(.12\right)\\ =21.6\end{array}$

$\begin{array}{c}{E}_5=n{p}_5\\ =\left(180\right)\left(.11\right)\\ =19.8\end{array}$

$\begin{array}{c}{E}_6=n{p}_6\\ =\left(180\right)\left(.08\right)\\ =14.4\end{array}$

$\begin{array}{c}{E}_7=n{p}_7\\ =\left(180\right)\left(.08\right)\\ =14.4\end{array}$

Where $p_1$ is the probability of color of the truck is white, $p_2$ is the probability of color of the truck is black, $p_3$ is the probability of color of the truck is Red, $p_4$ is the probability of color of the truck is Silver, $p_5$ is the probability of color of the truck is Gray, $p_6$ is the probability of color of the truck is Blue and $p_7$ is the probability of color of the truck is Other.

The following table gives the observed and expected frequency:

Frequency	White	Black	Red	Silver	Gray	Blue	Other
Observed	45	32	30	30	22	15	6
Expected	54	30.6	25.2	21.6	19.8	14.4	14.4

Software procedure:

Step-by-step procedure to obtain the test statistic using the MINITAB software:

• Choose Stat> Tables> Chi-Square Goodness-of-Fit Test (one variable).
• Specify the “Observed count” as Observed and category names as Frequency.
• Under Test, select “Proportions specified by historical counts” as Expected.
• Click on OK.

Output using the MINITAB software is given below:

Therefore, the value of the test statistic is 10.9144.

(d)

To determine

To make: The decision.

Answer to Problem 10E

The null hypothesis will not be rejected.

Explanation of Solution

The test statistic is 10.9144.and the critical value is 12.592.

The chi-square statistic is less than the critical value.

That is, $10.9144<12.592$

Hence, it can be said that there is not enough evidence to reject the null hypothesis.

(e)

To determine

To summarize: The results.

Answer to Problem 10E

According to the obtained result, the claim is not true.

Explanation of Solution

The test statistic is 10.9144.and the critical value is 12.592.

The chi-square statistic is less than the critical value.

That is, $10.9144<12.592$

Thus, it can be said that there is not enough evidence to reject the null hypothesis.

Hence, the distribution is the same as stated in the null hypothesis.

That is, the claim is not true.