Question

A researcher conducted an Independence test by using data consisting of 2 categorical variables: Zip code and Diet. Her data can be organized into a 3 by 4 contingency table. If she found the test statistic 19.96:

Answer 1

The degrees of freedom of the chi-squared statistic is 6 and the p-value of the independence test is 0.057. Since the p-value is greater than 0.05, the researcher cannot conclude that zip code and diet are dependent on one another.

Therefore, the final answer is: Zip code and diet are independent of one another.

To calculate the degrees of freedom (df) of the chi-squared statistic, we use the following formula:

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

where r is the number of rows in the contingency table and c is the number of columns.

In the given example, the contingency table is 4 by 3, so the degrees of freedom are:

df = (4 - 1) * (3 - 1) = 6

To calculate the p-value of the independence test, we use the following formula:

p-value = P(χ² ≥ χ²_obs)

where χ²_obs is the observed chi-squared statistic.

We can use a chi-squared table to find the p-value for a given chi-squared statistic and degrees of freedom. In the given example, the observed chi-squared statistic is 10.78 and the degrees of freedom are 6. The p-value for this combination is approximately 0.057.