91Ó°ÊÓ

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

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$\mathrm{Î\pm }$: 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

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

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

= $3×3$

$=9$

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 of the independence test is given below.

$\text{Test statistic}=\underset{(=),)}{∑}\frac{(O−E{)}^2}{E}$

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

$=\left(\right(B4−B14{)}^2)/B14\text{in cell B22 and drag the}$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

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

Hence, the p-value is$0.07$

The chi-square sketch is shown below

i. Alpha: $0.05$

ii: Decision: Do not reject the null hypothesis H₀

iii. Reason for decision: Because the p-value $>\mathrm{Î\pm }$

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.