The accompanying two-way table was constructed using data in the article “Television Viewing and Physical Fitness in Adults” (Research Quarterly for Exercise and Sport, 1990: 315–320).

smileycellist2

smileycellist2

Answered question

2020-11-23

The accompanying two-way table was constructed using data in the article “Television Viewing and Physical Fitness in Adults” (Research Quarterly for Exercise and Sport, 1990: 315–320). The author hoped to determine whether time spent watching television is associated with cardiovascular fitness. Subjects were asked about their television-viewing habits and were classified as physically fit if they scored in the excellent or very good category on a step test. We include MINITAB output from a chi-squared analysis. The four TV groups corresponded to different amounts of time per day spent watching TV (0, 1–2, 3–4, or 5 or more hours). The 168 individuals represented in the first column were those judged physically fit. Expected counts appear below observed counts, and MINITAB displays the contribution to (x2) from each cell.
State and test the appropriate hypotheses using α=0.05
 12Total135147182 25.48156.52 2101629730 102.20627.80 328222250 35.00215.00 443438 5.3232.68 Total16810321200
Chisq=3.557 + 0.579 + 0.014 + 0.002 + 1.400 + 0.228 + 0.328 + 0.053=6.161
df=3

Answer & Explanation

Demi-Leigh Barrera

Demi-Leigh Barrera

Skilled2020-11-24Added 97 answers

Step 1
Testing for Independence - Lack of Association
When testing null hypothesis
H0 : pij=pi.  p.j, i=1, 2,  ,I, j=1, 2,  ,J
versus alternative hypothesis
Hα : H0 is not true.
Let e^ij  5 for each i, j, where e^ij
e^ij=n  p^i.  p^.j=n  ni.n  n.jn=ni.  n.jn
under regularity conditions, test statistic value is
X2=j=1I i=1J (nij  e^ij)2e^ij= all cells observedestimated expected2estimated expected
has approximated a chi-square distribution with (I - 1)(J - 1) degrees of freedom when H0 is true.
The P-value is corresponding area to the right of X2 under the X(I  1)(J  1)2 curve.
The null hypothesis is
H0 : Television viewing and physical fitness are independent
versus alternative
Hα : Television viewing and pshysical fitness are not independent.
Critical value, from the table in the appendix, is given by
X0.05, (4  1)(2  1)2=7.815,
and the calculated X2 is given in the output as
X2=3.557 + 0.579 + 0.014 + 0.002 + 1.400 + 0.228 + 0.328 + 0.053=6.161,
thus, since
X0.05, 32=7.815 > 6.161=X2,
do not reject null hypothesis at given significance level. The data indicates that there is no association between variables.

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