Given the contingency table shown here, does the decision to retire appear independent of the employee type? Survey question: Do you plan on retiring or keep working when you turn 65? $$\begin{array} { | l | c | c | c | }
\hline\text { Employee } & \text { Retire } ( R ) & \text { Work } ( W ) & \text { Total }\
\hline \text { Management } ( M ) & 13 & 18 &31 \
\hline \text { Line worker } ( L ) & 39 & 54 & 93 \
\hline \text { Total } & 52 & 72 & 124\
\hline
\end{array}$$
A)Yes.
B)No.

Question

Given the contingency table shown here, does the decision to retire appear independent of the employee type? Survey question: Do you plan on retiring or keep working when you turn 65? $$\begin{array} { | l | c | c | c | } 
\hline\text { Employee } & \text { Retire } ( R ) & \text { Work } ( W ) & \text { Total }\
\hline  \text { Management } ( M ) & 13 & 18 &31 \
\hline  \text { Line worker } ( L ) & 39 & 54 & 93 \
\hline  \text { Total } & 52 & 72 & 124\
\hline
\end{array}$$
A)Yes.
B)No.

Quizplus · Accepted Answer

The expected counts for each cell are:$$E_{MR} = \frac{31 \times 52}{124} = 13.2$$$$E_{MW} = \frac{31 \times 72}{124} = 17.8$$$$E_{LR} = \frac{93 \times 52}{124} = 38.8$$$$E_{LW} = \frac{93 \times 72}{124} = 54.2$$The chi-square statistic is:$$\chi^2 = \frac{(13 - 13.2)^2}{13.2} + \frac{(18 - 17.8)^2}{17.8} + \frac{(39 - 38.8)^2}{38.8} + \frac{(54 - 54.2)^2}{54.2} = 0.03$$The p-value for this chi-square statistic is 0.98, which is greater than 0.05. Therefore, we fail to reject the null hypothesis and conclude that the decision to retire is independent of the employee type.

Given the Contingency Table Shown Here, Does the Decision to Retire