Who owns a home? What is the relationship between educational

achievement and home ownership? A random sample of 500 U.S. adults was selected. Each member of the sample was identified as a high school graduate (or not) and as a homeowner (or not). The two-way table summarizes the data.

Are the events “homeowner” and “high school graduate” independent? Justify your answer.

Data in a two-way table for the relationship between educational achievement and home ownership:

Two events are independent if the probability of one event's occurrence has no bearing on the probability of the other event's occurrence.

Conditional probability states that

$P(\mathrm{B}\mid \mathrm{A})=\frac{P(\mathrm{A}\cap \mathrm{B})}{P(\mathrm{A})}=\frac{P(\mathrm{A}\text{and}\mathrm{B})}{P(\mathrm{A})}$

Two events are independent if the probability of one event's occurrence has no bearing on the probability of the other event's occurrence.

Conditional probability states that

$P(\mathrm{B}\mid \mathrm{A})=\frac{P(\mathrm{A}\cap \mathrm{B})}{P(\mathrm{A})}=\frac{P(\mathrm{A}\text{and}\mathrm{B})}{P(\mathrm{A})}$

Let,

G: Graduate
H: Homeowner

The table contains data on $500$adults in the United States.

The information about $500$U.S adults is provided in the table.

There are $500$different outcomes to choose from.

Also, note that

In the table, 310 of the 500 U.S adults are graduates.

Thus,

The number of favorable outcomes is $310$.

When the number of favorable outcomes is divided by the number of possible outcomes, we get the probability.

$P\left(\mathrm{G}\right)=\frac{\text{Numberoffavourable outcomes}}{\text{Numberof possible outcomes}}=\frac{310}{500}=0.62$

Now,
Note that
In the table, 340 of the $500$U.S adults are homeowners. In this case, the number of favorable outcomes is $340$and the number of possible outcomes is $500$:

$P\left(\mathrm{H}\text{and}\mathrm{G}\right)=\frac{\text{Numberoffavourable outcomes}}{\text{Number of possible outcomes}}=\frac{221}{500}$

Use the conditional probability formula:

$P(\mathrm{G}\mid \mathrm{H})=\frac{P\left(\mathrm{GandH}\right)}{P\left(\mathrm{H}\right)}=\frac{221/500}{340/500}=\frac{221}{340}=\frac{13}{20}=0.65$

For events $G$and $H$to be independent,

$P(\mathrm{G}\mid \mathrm{H})=P\left(\mathrm{G}\right)$

And, $P(\mathrm{H}\mid \mathrm{G})=P\left(\mathrm{H}\right)$

But,

In this case,

Note that $P(\mathrm{G}\mid \mathrm{H})=0.65$

$P\left(\mathrm{G}\right)=0.62$

Both probabilities are not the same.

Therefore, the two events are not independent.