Question

use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round expected frequency to two decimal places car manufacturers are interested in whether there is a relationship between the size of the car an individual drives and the number of people in the driver’s family (that is, whether car size and family size are independent).To test this, suppose that $800$car owners were randomly surveyed with the results in Table $11.44$. Conduct a test of independence.

Family Size	Sub & Compact	Mid-size	Full-size	Van & Truck
$1$	$20$	$35$	$40$	$35$
$2$	$20$	$50$	$70$	$80$
$3-4$	$20$	$50$	$100$	$90$
$5+$	$20$	$30$	$70$	$70$

Table 11.44

Short answer

The null hypothesis is shown below:

H_$0$ : The size of the car an individual drive is independent of the number of people in the driver's family.

Decision: Do not reject the null hypothesis H₀

Conclusion: There is sufficient evidence to ensure that the size of the car an individual driver is independent of the number of people in the driver's family.

Step-by-step solution

Given Information

To test this, suppose that $800$ car owners were randomly surveyed with the results in Table $11.44$. Conduct a test of independence.

Hypotheses Test

The null hypothesis is shown below:

H_$0$ : The size of the car an individual drive is independent of the number of people in the driver's family.

Against the alternative hypothesis as shown below:
H$\alpha$: The size of the car an individual drive is not independent of the number of people in the driver's family.

The degree of freedom can be calculated by the formula given below

$df=(\text{number of columns}-1)(\text{number of rows}-1)$

Therefore

$df=(\text{number of columns}-1)(\text{number of rows}-1)$

=$(4-1)(4-1)$

= $3\times 3$

$=9$

Excel Worksheet

From the above calculation, it is clear that the distribution for the test is,$x_9^2$.

The observed value table is already given in the textbook. calculate the expected frequencies by using the formula shown below,

$E=\frac{\left(\text{row total}\right)\left(\text{column total}\right)}{\text{overall total}}$

All calculations can be done in an excel worksheet. Hence the expected (E) values table is shown below.

The test statistic

The test statistic of the independence test is given below.

$\text{Test statistic}=\sum _{(=),)}\frac{(O-E{)}^2}{E}$

$\text{To calculate}\frac{(O-E{)}^2}{E}\text{apply formula}$

$=\left(\right(B4-B14{)}^2)/B14\text{in cell B22 and drag the}$same formula up to cell E. After that, take the total of columns total and rows total. The table of test statistics is shown below.

Hence, the test statistic is 15.82

Explanation

f. The p-value can be calculated in excel by using the CHIDIST ( ) formula as shown below:

Hence, the p-value is$0.07$

Step 6: The chi-square sketch

The chi-square sketch is shown below

Conclusion

i. Alpha: $0.05$

ii: Decision: Do not reject the null hypothesis H₀

iii. Reason for decision: Because the p-value $>\alpha$

iv. Conclusion: There is adequate evidence to ensure that the size of the car an individual drives is independent of the number of people in the driver's family.