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 2020-11-23 Answered

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 \(\displaystyle{x}^{{{2}}}\) from each cell.
State and test the appropriate hypotheses using \(\displaystyle\alpha={0.05}\)
\(\begin{array}{|c|c|}\hline & 1 & 2 & Total \\ \hline 1 & 35 & 147 & 182 \\ \hline & 25.48 & 156.52 & \\ \hline 2 & 101 & 629 & 730 \\ \hline & 102.20 & 627.80 & \\ \hline 3 & 28 & 222 & 250 \\ \hline & 35.00 & 215.00 & \\ \hline 4 & 4 & 34 & 38 \\ \hline & 5.32 & 32.68 & \\ \hline Total & 168 & 1032 & 1200 \\ \hline \end{array}\)
\(Chisq= 3.557\ +\ 0.579\ +\ 0.014\ +\ 0.002\ +\ 1.400\ +\ 0.228\ +\ 0.328\ +\ 0.053=6.161\)
\(\displaystyle{d}{f}={3}\)

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Expert Answer

Demi-Leigh Barrera
Answered 2020-11-24 Author has 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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