Question

construct a relative frequency marginal distribution for the given contingency table. round values to the nearest thousandth. x1 x2 x3 y1 15 30 15 y2 55 55 40

Explanation:

Step1: Calculate the total number of observations

$$\begin{align*} \text{Total}&=15 + 30+15+55+55+40\\ &=(15 + 30+15)+(55+55+40)\\ &=60 + 150\\ &=210 \end{align*}$$

Step2: Calculate row - relative frequencies

For row \(y1\):

$$\begin{align*} \text{Relative frequency of }y1&=\frac{15 + 30+15}{210}\\ &=\frac{60}{210}\approx0.286 \end{align*}$$

For row \(y2\):

$$\begin{align*} \text{Relative frequency of }y2&=\frac{55 + 55+40}{210}\\ &=\frac{150}{210}\approx0.714 \end{align*}$$

Step3: Calculate column - relative frequencies

For column \(x1\):

$$\begin{align*} \text{Relative frequency of }x1&=\frac{15 + 55}{210}\\ &=\frac{70}{210}\approx0.333 \end{align*}$$

For column \(x2\):

$$\begin{align*} \text{Relative frequency of }x2&=\frac{30 + 55}{210}\\ &=\frac{85}{210}\approx0.405 \end{align*}$$

For column \(x3\):

$$\begin{align*} \text{Relative frequency of }x3&=\frac{15 + 40}{210}\\ &=\frac{55}{210}\approx0.262 \end{align*}$$

Answer:

Row - relative frequencies: \(y1:0.286\), \(y2:0.714\)
Column - relative frequencies: \(x1:0.333\), \(x2:0.405\), \(x3:0.262\)