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 con

Question
Two-way tables
asked 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 \(\displaystyle{x}^{{{2}}}\) from each cell.
State and test the appropriate hypotheses using \(\displaystyle\alpha={0.05}\)
\(\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}&{a}\mp,\ {1}&{a}\mp,\ {2}&{a}\mp,\ {T}{o}{t}{a}{l}\backslash{h}{l}\in{e}{1}&{a}\mp,\ {35}&{a}\mp,\ {147}&{a}\mp,\ {182}\backslash{h}{l}\in{e}&{a}\mp,\ {25.48}&{a}\mp,\ {156.52}&{a}\mp,\backslash{h}{l}\in{e}{2}&{a}\mp,\ {101}&{a}\mp,\ {629}&{a}\mp,\ {730}\backslash{h}{l}\in{e}&{a}\mp,\ {102.20}&{a}\mp,\ {627.80}&{a}\mp,\backslash{h}{l}\in{e}{3}&{a}\mp,\ {28}&{a}\mp,\ {222}&{a}\mp,\ {250}\backslash{h}{l}\in{e}&{a}\mp,\ {35.00}&{a}\mp,\ {215.00}&{a}\mp,\backslash{h}{l}\in{e}{4}&{a}\mp,\ {4}&{a}\mp,\ {34}&{a}\mp,\ {38}\backslash{h}{l}\in{e}&{a}\mp,\ {5.32}&{a}\mp,\ {32.68}&{a}\mp,\backslash{h}{l}\in{e}{T}{o}{t}{a}{l}&{a}\mp,\ {168}&{a}\mp,\ {1032}&{a}\mp,\ {1200}\backslash{h}{l}\in{e}\)
\(\displaystyle{C}{h}{i}{s}{q}={a}\mp,\ {3.557}\ +\ {0.579}\ +\ {a}\mp,\ {0.014}\ +\ {0.002}\ +\ {a}\mp,\ {1.400}\ +\ {0.228}\ +\ {a}\mp,\ {0.328}\ +\ {0.053}={6.161}\)
\(\displaystyle{d}{f}={3}\)

Answers (1)

2020-11-24
Step 1
Testing for Independence - Lack of Association
When testing null hypothesis
\(\displaystyle{H}_{{{0}}}\ :\ {p}_{{{i}{j}}}={p}_{{{i}.}}\ \cdot\ {p}_{{.{j}}},\ {i}={1},\ {2},\ \cdots\ ,{I},\ {j}={1},\ {2},\ \cdots\ ,{J}\)
versus alternative hypothesis
\(\displaystyle{H}_{{\alpha}}\ :\ {H}_{{{0}}}\) is not true.
Let \(\displaystyle\hat{{{e}}}_{{{i}{j}}}\ \geq\ {5}\ {f}{\quad\text{or}\quad}\ {e}{a}{c}{h}\ {i},\ {j},\ {w}{h}{e}{r}{e}\ \hat{{{e}}}_{{{i}{j}}}\)
\(\displaystyle\hat{{{e}}}_{{{i}{j}}}={n}\ \cdot\ \hat{{{p}}}_{{{i}.}}\ \cdot\ \hat{{{p}}}_{{.{j}}}={n}\ \cdot\ {\frac{{{n}_{{{i}.}}}}{{{n}}}}\ \cdot\ {\frac{{{n}_{{.{j}}}}}{{{n}}}}={\frac{{{n}_{{{i}.}}\ \cdot\ {n}_{{.{j}}}}}{{{n}}}}\)
under regularity conditions, test statistic value is
\(\displaystyle{X}^{{{2}}}={\sum_{{{j}={1}}}^{{{I}}}}\ {\sum_{{{i}={1}}}^{{{J}}}}\ {\frac{{{\left({n}_{{{i}{j}}}\ -\ \hat{{{e}}}_{{{i}{j}}}\right)}^{{{2}}}}}{{\hat{{{e}}}_{{{i}{j}}}}}}=\ \sum_{{{a}{l}{l}\ {c}{e}{l}{l}{s}}}\ {\frac{{{o}{b}{s}{e}{r}{v}{e}{d}-{e}{s}{t}{i}{m}{a}{t}{e}{d}\ {\exp{{e}}}{c}{t}{e}{d}}}{^}}{\left\lbrace{2}\right\rbrace}\rbrace{\left\lbrace{e}{s}{t}{i}{m}{a}{t}{e}{d}\ {\exp{{e}}}{c}{t}{e}{d}\right\rbrace}\)
has approximated a chi-square distribution with (I - 1)(J - 1) degrees of freedom when \(\displaystyle{H}_{{{0}}}\) is true.
The P-value is corresponding area to the right of \(\displaystyle{X}^{{{2}}}\ {u}{n}{d}{e}{r}\ {t}{h}{e}\ {{X}_{{{\left({I}\ -\ {1}\right)}{\left({J}\ -\ {1}\right)}}}^{{{2}}}}\) curve.
The null hypothesis is
\(\displaystyle{H}_{{{0}}}\ :\) Television viewing and physical fitness are independent
versus alternative
\(\displaystyle{H}_{{\alpha}}\ :\) Television viewing and pshysical fitness are not independent.
Critical value, from the table in the appendix, is given by
\(\displaystyle{{X}_{{{0.05},\ {\left({4}\ -\ {1}\right)}{\left({2}\ -\ {1}\right)}}}^{{{2}}}}={7.815},\)
and the calculated \(\displaystyle{X}^{{{2}}}\) is given in the output as
\(\displaystyle{X}^{{{2}}}={3.557}\ +\ {0.579}\ +\ {0.014}\ +\ {0.002}\ +\ {1.400}\ +\ {0.228}\ +\ {0.328}\ +\ {0.053}={6.161},\)
thus, since
\(\displaystyle{{X}_{{{0.05},\ {3}}}^{{{2}}}}={7.815}\ {>}\ {6.161}={X}^{{{2}}},\)
do not reject null hypothesis at given significance level. The data indicates that there is no association between variables.
0

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