91Ó°ÊÓ

Given in the question that, we have presented a contingency table that gives a cross-classification of a random sample of values for two variables, $x$and $y$, of a population.

We need to find the expected frequencies.

The values of two variables $x$and $y$picked from a random sample of a population are classified in the table below.

localid="1653738509387" $\begin{array}{|lll|}\hline y& & x\\ & A& B\\ a& 10& 20\\ b& 40& 30\\ \hline\end{array}$

The anticipated frequency is calculated as localid="1653738515030" $E=\frac{R·C}{n},$where localid="1653738523116" $R$represents the row total, localid="1653738529573" $C$represents the column total, and localid="1653738583207" $n$represents the sample size.

Determine the totals for each row and column:

localid="1653738588542" $\begin{array}{|llll|}\hline y& & x& \\ & A& B& \text{Total}\\ a& 10& 20& 30\\ b& 40& 30& 70\\ \text{Total}& 50& 50& 100\\ \hline\end{array}$

Using the equation localid="1653738593262" $E=\frac{R·C}{100}$the following table shows the computed anticipated frequencies:

$\begin{array}{|llll|}\hline y& & x& \\ & A& B& \text{Total}\\ a& 15& 15& 30\\ b& 35& 35& 70\\ \text{Total}& 50& 50& 100\\ \hline\end{array}$

Given in the question that, we have presented a contingency table that gives a cross-classification of a random sample of values for two variables, $x$and $y$, of a population.

We need to find the value of the chi-square statistic .

The statistics of the chi-square test

${\mathrm{χ}}^2=∑\frac{(O-E{)}^2}{E},$

observed frequency is $O$

Calculate the test statistics as follows:

${\mathrm{χ}}^2=(10-15{)}^2/15+(40-35{)}^2/35+(20-15{)}^2/15+(30-35{)}^2/35$

=4.7619

Given in the question that, we have presented a contingency table that gives a cross-classification of a random sample of values for two variables, $x$and $y$, of a population.

We need to find that whether the data provide sufficient evidence to conclude that the two variables are associated at the$5\%$ significance level

If the test statistic value is greater than ${\mathrm{χ}}_{\mathrm{Î\pm }}^2$with degrees of freedom, reject the null hypothesis.

$df=(r-1)(c-1)$

Establish the hypotheses:

$H_0$:The variables $x$and $y$do not have any relationship.

$Ha$: The variables $x$and $y$have a relationship

$$\begin{gathered} \mathrm{Î\pm }=5\% \\ =0.05 \end{gathered}$$

Table can be used to determine the crucial value.$\mathrm{V}$with $df=(2-1)(2-1)=1$

${\mathrm{χ}}_{0.05}^2=3.841$

The value of the calculated test statistics

$4.7619>3.841⇒\text{Reject}{H}_0$

At a $5$percent significance level, there is sufficient evidence to indicate the presence of an association between the variables $x$ and $y$.