In-depth Notes on Analysis of Variance (ANOVA)

ANOVA is a statistical method used to test if there are any statistically significant differences between the means of three or more independent groups.
Depending on the hypothesis being tested, the null hypothesis can be represented as:
- $H0 : µ1 = µ2 = … = µk$ (means of populations are equal)
- Contrary hypothesis is $Ha : µi \neq µ_j$ for at least one pair i ≠ j.

Experiment Scenario: Participants are divided based on breakfast consumed and tested on aptitude.
- Groups:
- No breakfast
- Partial breakfast
- Full breakfast
Factors and Levels:
- Experimental units: Participants
- Factor: Breakfast (with 3 levels)
- Response variable: Test scores

If participants are also grouped by age in the breakfast example, this introduces a second factor:
- Age factor (young and old)
Specific combinations of factors are referred to as treatments (e.g., young with full breakfast)
One-way ANOVA focuses on a single factor, while two-way ANOVA includes multiple factors.

Testing multiple means using separate two-sample tests leads to potential issues of Type I error.
Example with 50 populations leads to 1225 tests, increasing the likelihood of incorrectly rejecting the null hypothesis due to chance.

To evaluate means across multiple populations, the F statistic is computed as:
$F = \frac{\text{variation between sample means}}{\text{variation within the samples}}$
This ratio tests the null hypothesis (means are equal) against the alternative that at least one mean is different.

Calculate sample means and variances:
- For each group, compute mean $\bar{x}i$ and variance $s^2i$ .
Overall mean: $\bar{x} = \frac{1}{N} \sum{i=1}^k ni \bar{x}_i$ where N is total sample size.
Sum of Squares Between Treatments (SST):
$SST = \sum{i=1}^k ni(\bar{x}_i - \bar{x})^2$
Sum of Squares Within Treatments (SSE):
$SSE = \sum{i=1}^k (ni - 1)s^2_i$
Total Sum of Squares (Total SS):
$Total~~SS = SST + SSE$
ANOVA Table: Fill in degrees of freedom (df), sums of squares (SS), mean squares (MS), and F statistic.

Scenario A: Sampling from different normal populations shows more clear mean differences than Scenario B.
ANOVA Results:
* Scenario A: SST = 720.82, Error = 43.75, F = 345.97
* Scenario B: SST = 849.18, Error = 1190.92, F = 14.97

Random sampling from normal populations with equal variances ( $σ^2$ ).
ANOVA is robust with regards to the normality assumption as long as sample sizes are similar and distributions are approximately normal.

Fill in degrees of freedom for error using Mean Square Error (MSE).
Calculate sums of squares between treatments and error based on total sum of squares.
Obtain F statistic.

Hypothesis: Test differences in calcium intake among groups (Normal, Osteopenia, Osteoporosis).
Compute sample means, variances, overall mean, SST and SSE to complete ANOVA table.

After identifying different means, use confidence intervals based on the Student’s t-distribution to analyze which means differ significantly.

Randomized block designs partition treatments into blocks that share similar characteristics, reducing variability in the response.
ANOVA Table for Block Design:
- Functional structure includes treatments and blocks with separate sum of squares calculations for treatments and blocks.

Hypothesis testing shows differences are primarily explained by usage levels rather than service providers. F statistics indicate significance of block factor (usage) overshadowed treatment differences (companies).

Ensure treatment and block design are suitable; block design only applies when treatment and block factors do not affect each other.

Even with complex designs like ANOVA, software can streamline calculations. Hypothesis testing principles remain similar, but ensure to understand all assumptions for validity.