0.13 Ch 13: f distribution and anova

This section is a very brief overview of the $F$ distribution and two of its applications - One Way Analysis of Variance (ANOVA) and test of two variances. There are college courses which deal exclusively with these topics. ANOVA, particularly, is used regularly in industry.

Explanation of sum of squares, mean square, and the f ratio for anova

• $k$ = the number of different groups
• $n_j$ = the size of the $\text{jth}$ group
• $s_j$ = the sum of the values in the $\text{jth}$ group
• $N$ = the total number of all the values combined
• Total sample size: $\sum n_j$
• $x$ = one value: $\sum x=\sum s_j$
• Sum of squares of all values from every group combined: $\sum x^2$
• Between group variability: ${\text{SS}}_{\text{total}}=\sum x^2-\frac{{\left(\sum x\right)}^2}{N}$
• Total sum of squares: $\sum x^2-\frac{(\sum x{)}^2}{N}$
• Explained variation- sum of squares representing variation among the different samples ${\text{SS}}_{\text{between}}=\sum [\frac{(\text{sj}{)}^2}{n_j}]-\frac{(\sum s_j{)}^2}{N}$
• Unexplained variation- sum of squares representing variation within samples due to chance: ${\text{SS}}_{\text{within}}={\text{SS}}_{\text{total}}-{\text{SS}}_{\text{between}}$
• df's for different groups (df's for the numerator): ${\text{df}}_{\text{between}}=k-1$
• Equation for errors within samples (df's for the denominator): ${\text{df}}_{\text{within}}=N-k$
• Mean square (variance estimate) explained by the different groups: ${\text{MS}}_{\text{between}}=\frac{{\text{SS}}_{\text{between}}}{{\text{df}}_{\text{between}}}$
• Mean square (variance estimate) that is due to chance (unexplained): ${\text{MS}}_{\text{within}}=\frac{{\text{SS}}_{\text{within}}}{{\text{df}}_{\text{within}}}$
• F ratio or F statistic of two estimates of variance: $F=\frac{{\text{MS}}_{\text{between}}}{{\text{MS}}_{\text{within}}}$

Where...

• $(s_{-x}{)}^2=$ the variance of the sample means
• $n=$ the sample size of each group
• $(s_{\text{pooled}}{)}^2=$ the mean of the sample variances (pooled variance)
• ${\text{df}}_{\text{numerator}}=k-1$
• ${\text{df}}_{\text{denominator}}=k(n-1)=N-k$

These calculations are easily done with a graphing calculator or a computer program. We present the information in the chapter assuming some kind of technology will be used.

For ANOVA, the samples must come from normally distributed populations with the same variance, and the samples must be independent. The ANOVA test is right-tailed.

In a test of two variances, the samples must come from normal populations and must be independent of each other.

(one-way anova)

Three different diet plans are to be tested for average weight loss. For each diet plan, 4 dieters are selected and their weight loss (in pounds) in one month's time is recorded.

Plan 1	Plan 2	Plan 3
5	3.5	8
4.5	7	4
4	6	3.5
3	4	4.5

Is the average weight loss the same for each plan? Conduct an ANOVA test with a 1% level of significance.

Let ${\mu }_1$ , ${\mu }_2$ , and ${\mu }_3$ be the population means for the three diet plans.

• $H_o:{\mu }_1={\mu }_2={\mu }_3$
• $H_a:$ Not all pairs of means are equal.

• ${\text{df}}_{\text{numerator}}=3-1=2$
• ${\text{df}}_{\text{deno}\text{min}\text{ator}}=\text{12}-3=9$

The distribution for the test is $F_{\mathrm{2,9}}$

Using a calculator or computer, the test statistic is $F=0\text{.}\text{47}$ . The notation used for the $F$ statistic may also be $F\text{'}$ or $F_{\mathrm{2,9}}$ (like the distribution). The TI-83/84 series has the function ANOVA in STAT TESTS. Enter the lists of data separated by commas.

If you use the formulas for groups of the same size, the calculations are as follows:

Sample means are 4.13, 5.13, and 5, respectively. Sample standard deviations are 0.8539, 1.6250, and 2.0412, respectively.

$(s_{-x}{)}^2=0\text{.}\text{2956}$	The variance of the sample means
$(s_{\text{pooled}}{)}^2=2\text{.}\text{5416}$	The mean of the sample variances
$n=4$	The sample size of each group

$F=\frac{4\cdot 0\text{.}\text{2956}}{2\text{.}\text{5416}}$

Probability Statement: $P(F>0\text{.}\text{47})=0\text{.}\text{6395}$

Since -value, do not reject $H_o$ .

There is not sufficient evidence to conclude that the three diet plans are different. It appears that the three diet plans work equally well. The average weight loss is the same for all three plans.

Assign practice

Assign homework