i.Analysis of Variance (ANOVA)

Analysis of variance (ANOVA) is used to determine whether there are any statistically significant differences between the means of two or more independent unrelated groups.

A Research scenario for the Analysis of Variance ANOVA is described below:

A clinical trial is run to compare weight loss programs. Three popular weight loss programs are considered (Low calorie, Low fat and Low carbohydrate). A total of 15 patients agree to participate in the study and are randomly assigned to one of the three diets. The data on the weight loss of the 15 patients under different weight loss program are given below:

\[\begin{array}{|c|c|}\hline Low \ Calorie & Low \ fat & Low \ carbohydrate \\ \hline 2 & 2 & 2 \\ \hline 6 & 5 & 1 \\ \hline 5 & 6 & 5 \\ \hline 1 & 3 & 2 \\ \hline 4 & 2 & 6 \\ \hline \end{array}\]

Is there a statistically significant difference in the mean weight loss among the three diets?

Step 1 - Setting of hypothesis and determination of level of significance

\(\displaystyle{H}_{{{0}}}:\mu_{{{1}}}=\mu_{{{2}}}=\mu_{{{3}}}\)

\(\displaystyle{H}_{{{1}}}:\) All means are not equal

Level of significance a=0.05

Step 2 - Selection of appropriate test statistics

\(\displaystyle{F}={\frac{{{M}{S}{B}}}{{{M}{S}{E}}}}\)

Step 3 - Setting up of Decision rule:

Degrees of freedom.

\(\displaystyle{d}{f}_{{{1}}}={k}-{1}\)

=3-1

= 2

\(\displaystyle{d}{f}_{{{2}}}={N}-{k}\)

=15-3

=12

The critical value of F-statistics for degrees of freedoms 2,12 at 0.05 level of significance is 3.885

Therefore, Decision rule: Reject \(\displaystyle{H}_{{{0}}}:{\quad\text{if}\quad}\ {F}{>}{3.885}\)

Step 4 - computation of test statistics:

\[\begin{array}{|c|c|}\hline & Low \ Calorie & Low \ fat & Low \ carbohydrate \\ \hline n & 5 & 5 & 5 \\ \hline Group \ mean & 3.6 & 3.6 & 3.2 \\ \hline \end{array}\]

The overvall mean is obtained as \(\displaystyle\overline{{{X}}}={3.46}\)

Therefore,

\(\displaystyle{S}{S}{B}=\sum{n}_{{{j}}},{\left(\overline{{{X}_{{{j}}}}}-\overline{{{X}}}\right)}^{{{2}}}\)

\(\displaystyle={5}{\left({3.6}-{3.46}\right)}^{{{2}}}+{5}{\left({3.6}-{3.46}\right)}^{{{2}}}+{5}{\left({3.2}-{3.46}\right)}^{{{2}}}\)

=0.534

\(\displaystyle{S}{S}{E}=\sum\sum{\left({X}-\overline{{{X}}}_{{{j}}}\right)}^{{{2}}}\)

=(17.2+13.2+18.8)

= 49.2

The required ANOVA table is given as follows:

\[\begin{array}{|c|c|}\hline Sources \ of \ variation & Sum \ of \ squares (SS) & Degrees \ of \ freedom(df) & Means \ Square \ (MS) & F \\ \hline Between \ treatment & 0.534 & 3-1=2 & 0.534/2=0.267 & 0.267/4.1=0.065 \\ \hline Error & 49.2 & 15-3=12 & 49.2/12=4.1 & \\ \hline Total & 49.734 & 15-1=14 & & \\ \hline \end{array}\]

Conclusion:

Since \(\displaystyle{F}{\left({0.065}\right)}{<}{3.885}\)</span>, Therefore, there is no evidence to reject the null hypothesis and we can conclude that there is no statistically significant difference in the mean weight loss among the three diets.

Analysis of variance (ANOVA) is used to determine whether there are any statistically significant differences between the means of two or more independent unrelated groups.

A Research scenario for the Analysis of Variance ANOVA is described below:

A clinical trial is run to compare weight loss programs. Three popular weight loss programs are considered (Low calorie, Low fat and Low carbohydrate). A total of 15 patients agree to participate in the study and are randomly assigned to one of the three diets. The data on the weight loss of the 15 patients under different weight loss program are given below:

\[\begin{array}{|c|c|}\hline Low \ Calorie & Low \ fat & Low \ carbohydrate \\ \hline 2 & 2 & 2 \\ \hline 6 & 5 & 1 \\ \hline 5 & 6 & 5 \\ \hline 1 & 3 & 2 \\ \hline 4 & 2 & 6 \\ \hline \end{array}\]

Is there a statistically significant difference in the mean weight loss among the three diets?

Step 1 - Setting of hypothesis and determination of level of significance

\(\displaystyle{H}_{{{0}}}:\mu_{{{1}}}=\mu_{{{2}}}=\mu_{{{3}}}\)

\(\displaystyle{H}_{{{1}}}:\) All means are not equal

Level of significance a=0.05

Step 2 - Selection of appropriate test statistics

\(\displaystyle{F}={\frac{{{M}{S}{B}}}{{{M}{S}{E}}}}\)

Step 3 - Setting up of Decision rule:

Degrees of freedom.

\(\displaystyle{d}{f}_{{{1}}}={k}-{1}\)

=3-1

= 2

\(\displaystyle{d}{f}_{{{2}}}={N}-{k}\)

=15-3

=12

The critical value of F-statistics for degrees of freedoms 2,12 at 0.05 level of significance is 3.885

Therefore, Decision rule: Reject \(\displaystyle{H}_{{{0}}}:{\quad\text{if}\quad}\ {F}{>}{3.885}\)

Step 4 - computation of test statistics:

\[\begin{array}{|c|c|}\hline & Low \ Calorie & Low \ fat & Low \ carbohydrate \\ \hline n & 5 & 5 & 5 \\ \hline Group \ mean & 3.6 & 3.6 & 3.2 \\ \hline \end{array}\]

The overvall mean is obtained as \(\displaystyle\overline{{{X}}}={3.46}\)

Therefore,

\(\displaystyle{S}{S}{B}=\sum{n}_{{{j}}},{\left(\overline{{{X}_{{{j}}}}}-\overline{{{X}}}\right)}^{{{2}}}\)

\(\displaystyle={5}{\left({3.6}-{3.46}\right)}^{{{2}}}+{5}{\left({3.6}-{3.46}\right)}^{{{2}}}+{5}{\left({3.2}-{3.46}\right)}^{{{2}}}\)

=0.534

\(\displaystyle{S}{S}{E}=\sum\sum{\left({X}-\overline{{{X}}}_{{{j}}}\right)}^{{{2}}}\)

=(17.2+13.2+18.8)

= 49.2

The required ANOVA table is given as follows:

\[\begin{array}{|c|c|}\hline Sources \ of \ variation & Sum \ of \ squares (SS) & Degrees \ of \ freedom(df) & Means \ Square \ (MS) & F \\ \hline Between \ treatment & 0.534 & 3-1=2 & 0.534/2=0.267 & 0.267/4.1=0.065 \\ \hline Error & 49.2 & 15-3=12 & 49.2/12=4.1 & \\ \hline Total & 49.734 & 15-1=14 & & \\ \hline \end{array}\]

Conclusion:

Since \(\displaystyle{F}{\left({0.065}\right)}{<}{3.885}\)</span>, Therefore, there is no evidence to reject the null hypothesis and we can conclude that there is no statistically significant difference in the mean weight loss among the three diets.