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.

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.