91Ó°ÊÓ

Suppose that $280$recent brides were interviewed as to where they spent their honeymoons. The information is given in Table $11.46$. Conduct a test of independence.

The null hypothesis is shown below
H_$0$: The choice of honeymoon hot spot is independent of the age of the bride.

Against the alternative hypothesis as shown below:

H_a : The choice of honeymoon hot spots is dependent of the age of the bride.

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 shown below:

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

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

$=\left(\right(B4−B13{)}^2)/B13$in cell B$22$and drag the same formula up to cell E$25$. Then, take the total of columns total and rows total. The table of the test statistics is shown below:

Hence, the test statistics are$15.70$

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

so, the p-value is$0.07$

g. The chi-square sketch is shown below:

i. Alpha: $0.05$

ii: Decision: Do not reject the null hypothesis H_$0$

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

iv. Conclusion: There is sufficient proof to ensure that the choice of honeymoon hot spots is independent of the age of the bride.