Question

Physical activity generally declines when students leave high school and enroll in college. This suggests that college is an ideal setting to promote physical activity. One study examined the level of physical activity and other health-related behaviors in a sample of 1183 college students. Let's look at the data for physical activity and consumption of fruits. We categorize physical activity as low, moderate, or vigorous and fruit consumption as low, medium, or high. Here is the two-way table that summarizes the data.

	Physical activity
Fruit consumption	Low	Moderate	Vigorous	Total
Low	68	207	293	568
Medium	25	126	170	321
High	14	110	170	294
Total	107	443	633	1183

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. Use the following formula for expected cell counts.expected count =

row total ✕ column total

n

Using the formula, we need three quantities:

(1) the corresponding row total, 568, the number of students who have low fruit consumption,
(2) the column total,  , the number of students who have low physical activity, and
(3) the total number of students, 1183.

The expected cell count is therefore the following.

(568)

1183

= 51.37Note 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 3 ✕ 3 table are performed in the same way. With these nine expected counts we are now ready to use the following formula for the  χ² statistic.χ² =

	(observed − expected count)2 expected count

The first term in the sum comes from the cell for students with low fruit consumption and low physical activity. The observed count is  and the expected count is 51.37. Therefore, the contribution to the χ ² statistic for this cell is the following.

− 51.37 2

51.37

= 5.38When we add the terms for each of the nine cells, the result is the following. (Round your answer to two decimal places.)

χ ² =

Because there are

r = 3

levels of fruit consumption and

c =

levels of physical activity, the degrees of freedom for this statistic are the following.df = (r − 1)(c − 1) = (3 − 1)

  − 1

= Under the null hypothesis that fruit consumption and physical activity are independent, the test statistic χ ² has a χ ²

distribution. To obtain the P-value, look at the

df =

row in Table F. The calculated value

χ ² =

lies between the critical points for probabilities 0.005 and 0.01. The P-value is therefore between 0.005 and 0.01. (Software gives the value, rounded to four decimal places, as 0.0081.) There  ---Select--- is is not strong evidence at α = 0.01 that there is a relationship between fruit consumption and physical activity.

Answer 1

Using the expected count formula, we calculate the expected frequencies and they are given below.

	Low	Moderate	Vigorous	Total
Low	51.37	212.70	303.93	568
Medium	29.03	120.21	171.76	321
High	26.59	110.09	157.31	294
Total	107	443	633	1183

The observed frequencies are = O = (68, 25, 14, 207, 126, 110, 293, 170, 170).
The corresponding expected frequencies are = E = (51.37, 29.03, 26.59, 212.70, 120.21, 110.09, 303.93, 171.76, 157.31).

Now, the test statistic is = = .
The contribution to the test statistic is given in the following table.

	Low	Moderate	Vigorous
Low	5.38	0.15	0.39
Medium	0.56	0.28	0.02
High	5.96	0.00	1.02

The test statistic value = 13.77, which follows under H0.
The p-value = = 0.0081. Since the p-value is less than 0.01, we reject H0.
There is strong evidence at α = 0.01 that there is a relationship between fruit consumption and physical activity.