Physical activity generally declines when students leave high school and enroll

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 1182 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.

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 =

Using the formula, we need three quantities:

(1) the corresponding row total, 569, 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, 1182.

The expected cell count is therefore the following.

= 51.99

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 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 2 statistic.

2 =

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 51.99. Therefore, the contribution to the 2 statistic for this cell is the following.

= 5.57

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

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 2 has a

2

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

df =

row in Table F. The calculated value

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 0.0071.) There  ---Select--- is or is not strong evidence at = 0.01 that there is a relationship between fruit consumption and physical activity.

Physical activity Fruit consumption Low Moderate Vigorous Total Low 69 207 293 569 Medium 25 125 170 320 High 14 110 169 293 Total 108 442 632 1182

Explanation / Answer

Here

H0: There is no significant relationship between fruit consumption and physical activity.

Ha : There is significant relationship between fruit consumption and physical activity.

The expected table is

The chi - square table by the given formula is given below

X2 = (observed - Expected)2 / Expected

X2= 14.06275

Here dF = 2 *2 = 4 and alpha = 0.05

X2(critical) = 9.4877

Here, X2 > X2(critical) so we shall reject the null hypothesis ..

There is very strong evidence at = 0.01 that there is a relationship between fruit consumption and physical activity.

Physical activity Fruit consumption Low Moderate Vigorous Total Low 51.99 212.77 304.24 569 Medium 29.24 119.66 171.10 320 High 26.77 109.57 156.66 293 Total 108 442 632 1182

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