EXAMPLE 9.11

The chi-square significance test for health habits of college students. The first step in performing the significance test is to calculate the expected cell counts. Let’s start with the cell for students with low fruit consumption and low physical activity. Using the formula on page 533, we need three quantities: (1) the corresponding row total, 569, the number of students who have low fruit consumption; (2) the column total, 108, the number of students who have low physical activity; and (3) the total number of students, 1184. The expected cell count is, therefore,

Note that although any observed count of the number of students must be a whole number, an expected count need not be.

Calculations for the other eight cells in the table are performed in the same way. With these nine expected counts, we are now ready to use the formula for the statistic on page 534. The first term in the sum comes from the cell for students with low fruit consumption and low physical activity. The observed count is 69 and the expected count is 51.90. Therefore, the contribution to the statistic for this cell is

540

When we add the terms for each of the nine cells, the result is

Because there are levels of fruit consumption and levels of physical activity, the degrees of freedom for this statistic are

Under the null hypothesis that fruit consumption and physical activity are independent, the test statistic has a distribution. To obtain the P-value, look at the df = 4 row in Table F.

The calculated value lies between the critical points for probabilities 0.01 and 0.005. The P-value is, therefore, between 0.01 and 0.005. (Software gives the value as 0.0068.) There is strong evidence (, , ) that there is a relationship between fruit consumption and physical activity.