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The null hypothesis is shown below:

$H_0$: The ages and net worth is independent.

Against the alternative hypothesis as shown below:

$H_a:$The ages and net worth is dependent.

The degrees of freedom can be calculated by the formula given below:

$df=(numberofcolumns-1)(numberofrows-1)$

Therefore,

$df=(numberofcolumns-1)(numberofrows-1)$

$=(3-1)(2-1)$

$=2×1$

$=2$

From above calculation, it is clear that the distribution for the test is ${\mathrm{χ}}_2^2$.

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 independence test is given below:

$Teststatistic=\underset{(i×j)}{∑}\frac{(O-E{)}^2}{E}$

To calculate $\frac{(O-E{)}^2}{E}$apply formula $=\left((B4-B11{)}^2\right)/B11$in cell $B17$and drag the same formula up to cell $D18$. After that, take the total of columns total and rows total. The table of test statistic is shown below:

Hence, the test statistic is $1.76$.

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

Hence, the$p$value is$0.41$

The Chi-square sketch is given below

Alpha:$0.05$

Decision: Do not reject the null hypothesis $H_0$

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

Conclusion: There is sufficient evidence to insure that the ages and net worth is independent.