Questions: In your own words briefly describe when you would use a post hoc test.

Transcript text: In your own words briefly describe when you would use a post hoc test.

Solution

Solution Steps

Step 1: Calculate Sum of Squares Between Groups

The sum of squares between groups is calculated as follows:

\[ SS_{between} = \sum_{i=1}^k n_i (\bar{X}_i - \bar{X})^2 = 1210.0 \]

Step 2: Calculate Sum of Squares Within Groups

The sum of squares within groups is calculated as:

\[ SS_{within} = \sum_{i=1}^k \sum_{j=1}^{n_i} (X_{ij} - \bar{X}_i)^2 = 4065.6 \]

Step 3: Calculate Mean Squares

The mean square between groups is given by:

\[ MS_{between} = \frac{SS_{between}}{df_{between}} = \frac{1210.0}{2} = 605.0 \]

The mean square within groups is calculated as:

\[ MS_{within} = \frac{SS_{within}}{df_{within}} = \frac{4065.6}{12} = 338.8 \]

Step 4: Calculate F-Statistic

The F-statistic is calculated using the mean squares:

\[ F = \frac{MS_{between}}{MS_{within}} = \frac{605.0}{338.8} = 1.7857 \]

Step 5: Calculate P-Value

The p-value is determined using the F-distribution:

\[ P = 1 - F(F_{observed}; df_{between}, df_{within}) = 1 - F(1.7857; 2, 12) = 0.2095 \]

Step 6: Interpret Results

The degrees of freedom between groups is \(df_{between} = 2\) and the degrees of freedom within groups is \(df_{within} = 12\). The calculated F-statistic is \(F = 1.7857\) and the p-value is \(P = 0.2095\).

Since the p-value \(0.2095\) is greater than the significance level \(\alpha = 0.05\), we conclude that the ANOVA test is not significant. Therefore, there is no need for a post hoc test.

Final Answer

\(\boxed{\text{The ANOVA test is not significant. No need for a post hoc test.}}\)

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