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&nbsp;(x2)&nbsp;from each cell.State and test the appropriate hypotheses using&nbsp;α=0.05&nbsp;12Total135147182&nbsp;25.48156.52&nbsp;2101629730&nbsp;102.20627.80&nbsp;328222250&nbsp;35.00215.00&nbsp;443438&nbsp;5.3232.68&nbsp;Total16810321200Chisq=3.557&nbsp;+&nbsp;0.579&nbsp;+&nbsp;0.014&nbsp;+&nbsp;0.002&nbsp;+&nbsp;1.400&nbsp;+&nbsp;0.228&nbsp;+&nbsp;0.328&nbsp;+&nbsp;0.053=6.161df=3

Question

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&amp;nbsp;(x2)&amp;nbsp;from each cell.State and test the appropriate hypotheses using&amp;nbsp;α=0.05&amp;nbsp;12Total135147182&amp;nbsp;25.48156.52&amp;nbsp;2101629730&amp;nbsp;102.20627.80&amp;nbsp;328222250&amp;nbsp;35.00215.00&amp;nbsp;443438&amp;nbsp;5.3232.68&amp;nbsp;Total16810321200Chisq=3.557&amp;nbsp;+&amp;nbsp;0.579&amp;nbsp;+&amp;nbsp;0.014&amp;nbsp;+&amp;nbsp;0.002&amp;nbsp;+&amp;nbsp;1.400&amp;nbsp;+&amp;nbsp;0.228&amp;nbsp;+&amp;nbsp;0.328&amp;nbsp;+&amp;nbsp;0.053=6.161df=3

Demi-Leigh Barrera · Accepted Answer

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 observed−estimated 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 &amp;gt; 6.161=X2, do not reject null hypothesis at given significance level. The data indicates that there is no association between variables.

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).

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