Questions: Class Data again Use the data given in Question #1 a) What are the joint probabilities: P(Not ∩ JR) ? P(Car ∩ SR) ? b) Use probabilities to show that the events "Car" and "SO" are not independent.

Solution Steps

To find the joint probabilities, we calculate:

1. \( P(\text{Not} \cap \text{JR}) \): \[ P(\text{Not} \cap \text{JR}) = \frac{30}{110} = 0.2727 \]

2. \( P(\text{Car} \cap \text{SR}) \): \[ P(\text{Car} \cap \text{SR}) = \frac{25}{110} = 0.2273 \]

Next, we calculate the marginal probabilities:

1. \( P(\text{Car}) \): \[ P(\text{Car}) = \frac{60}{110} = 0.4545 \]

2. \( P(\text{SO}) \): \[ P(\text{SO}) = \frac{20 + 10}{110} = \frac{30}{110} = 0.2727 \]

3. \( P(\text{Car} \cap \text{SO}) \): \[ P(\text{Car} \cap \text{SO}) = \frac{10}{110} = 0.0909 \]

To check if the events "Car" and "SO" are independent, we compare \( P(\text{Car} \cap \text{SO}) \) with \( P(\text{Car}) \cdot P(\text{SO}) \):

\[ P(\text{Car}) \cdot P(\text{SO}) = 0.4545 \cdot 0.2727 = 0.1240 \]

Since \( P(\text{Car} \cap \text{SO}) = 0.0909 \) is not equal to \( P(\text{Car}) \cdot P(\text{SO}) = 0.1240 \), the events are not independent.

We perform a Chi-Square Test of Independence using the observed frequencies from the contingency table. The calculated Chi-Square statistic is:

\[ \chi^2 = 13.9683 \]

The critical value at \( \alpha = 0.05 \) for 2 degrees of freedom is:

\[ \chi^2_{\alpha, df} = 5.9915 \]

The p-value associated with the Chi-Square statistic is:

\[ P = 0.0009 \]

Since the p-value \( 0.0009 \) is less than \( 0.05 \), we reject the null hypothesis of independence. Therefore, the events "Car" and "SO" are not independent.

Final Answer