Question

Construct an ANOVA table for these one-way classifications. Provide a formal test of \(H_{0}: \mu_{1}=\mu_{2}=\ldots=\mu_{k}\) including the rejection region with \(\alpha=.05 .\) Bound the \(p\) -value for the test and state your conclusions. $$ \begin{array}{ccc} \hline \text { Technique } 1 & \text { Technique } 2 & \text { Technique } 3 \\\ \hline 13 & 18 & 17 \\ 17 & 18 & 24 \\ 15 & 15 & 23 \\ 16 & 18 & 20 \\ \hline \end{array} $$

Short answer

Answer: Yes, the ANOVA table indicates significant differences in the means of the three techniques at the 0.05 significance level, as the F statistic (12.1) is greater than the F critical value (4.256) and the p-value is between 0.001 and 0.01, which is less than the significance level of 0.05. Therefore, we reject the null hypothesis that the means are equal.

Step-by-step solution

Calculate group means and overall mean

First, calculate the means of each group (techniques) and the overall mean of all observations: Technique 1 mean: \(\bar{X}_1 = \frac{13 + 17 + 15 + 16}{4} = 15.25\) Technique 2 mean: \(\bar{X}_2 = \frac{18 + 18 + 15 + 18}{4} = 17.25\) Technique 3 mean: \(\bar{X}_3 = \frac{17 + 24 + 23 + 20}{4} = 21\) Overall mean: \(\bar{X} = \frac{13 + 17 + 15 + 16 + 18 + 18 + 15 + 18 + 17 + 24 + 23 + 20}{12} = 17.5\)

Calculate Sum of Squares

Next, we will calculate the Sum of Squares Between groups (SSB), Sum of Squares Within groups (SSW), and Total Sum of Squares (SST): $$ SSB = 4(\bar{X}_1 - \bar{X})^2 + 4(\bar{X}_2 - \bar{X})^2 + 4(\bar{X}_3 - \bar{X})^2 = 4(15.25 - 17.5)^2 + 4(17.25 - 17.5)^2 + 4(21 - 17.5)^2 $$ $$ =\ 60.5 $$ $$ SSW = \sum_{i=1}^{4}(X_{1i}-\bar{X}_1)^2 + \sum_{i=1}^{4}(X_{2i}-\bar{X}_2)^2 + \sum_{i=1}^{4}(X_{3i}-\bar{X}_3)^2 = 22.5 $$ $$ SST = SSB + SSW = 60.5 + 22.5 = 83 $$

Calculate Degrees of Freedom and Mean Squares

Calculate the degrees of freedom for each sum of squares: Between groups: \(dF_{B} = k - 1 = 3 - 1 = 2\) Within groups: \(dF_{W} = n - k = 12 - 3 = 9\) Total: \(dF_{T} = n - 1 = 12 - 1 = 11\) Calculate the mean squares (MS): MSB (Mean Square Between): \(\frac{SSB}{dF_{B}} = \frac{60.5}{2} = 30.25\) MSW (Mean Square Within): \(\frac{SSW}{dF_{W}} = \frac{22.5}{9} = 2.5\)

Calculate the F statistic and F critical value

Calculate the F statistic: F = \(\frac{MSB}{MSW} = \frac{30.25}{2.5} = 12.1\) Determine the F critical value with \(\alpha=0.05, dF_{B}=2,\) and \(dF_{W}=9\). Using an F-distribution table or calculator: F critical value = \(4.256\)

Determine the p-value, rejection region, and conclusion

Since \(F = 12.1 > F_{critical} = 4.256\), we reject the null hypothesis \(H_0: \mu_1 = \mu_2 = \mu_3\) at the \(\alpha=0.05\) significance level. This means that there is a significant difference in the means of the three techniques. We can also bound the p-value for our test. Using an F-distribution calculator, we get a p-value of approximately 0.003. Thus, the p-value is between 0.001 and 0.01, which is less than the significance level of 0.05, leading us to reject the null hypothesis. In conclusion, there is a significant difference in the means of the three techniques, and we reject the null hypothesis that the means are equal at the \(\alpha = 0.05\) significance level.

Key concepts

Understanding the F-test in ANOVA

One-way ANOVA (Analysis of Variance) is a statistical method used to determine whether there are significant differences between the means of three or more independent groups. At the core of this test is the F-test, a powerful tool that allows us to compare variances by using ratio of two sample variances.

The F-test in ANOVA specifically compares the variances between groups (what we call 'Between-group variability') to the variances within groups ('Within-group variability'). If the between-group variability is significantly larger than the within-group variability, it can be inferred that at least one group mean is different from the others.

For our example, after calculating mean squares for both between and within groups (MSB and MSW respectively), the F statistic is found by dividing MSB by MSW. With our F statistic (12.1), we can consult an F-distribution table or use a calculator to compare it against the critical F value appropriate for our degrees of freedom. Since our F statistic exceeds the critical value, we have evidence to suggest that not all group means are equal.

The Step-by-Step Sum of Squares Calculation in ANOVA

In the realm of ANOVA, the Sum of Squares is a measure of variation. It's crucial for partitioning the total variability of the data into components that can be attributed to different sources. Specifically, these sources are the variability within groups and the variability between groups.

To calculate the Sum of Squares within groups (SSW), you find the deviation of each observation from its group mean, square these deviations, and sum them up. This reflects variability due to individual differences within each group. Conversely, the Sum of Squares between groups (SSB) is calculated by looking at the deviations of the group means from the grand mean (the mean of all observations), squaring these deviations, and weighting them by the group size.

Our solution showed these calculations through a clear, step-by-step process, which examines the difference in outcome due to the employed techniques, and reflects this in the calculation of SSB. Adding the SSW and SSB gives us the Total Sum of Squares (SST), a representation of the total variation in the data set.

Significance and Interpretation of the p-value

In statistical hypothesis testing, the p-value plays a key role in informing us about the strength of our evidence against the null hypothesis. This value quantifies how likely we are to observe our sample data, or something more extreme, if the null hypothesis were true.

In the context of one-way ANOVA, if the p-value is smaller than our predetermined alpha level (in the provided example, \( \alpha = 0.05 \)), it suggests that there is a very low probability that the observed differences in group means occurred by chance. As such, a small p-value (< 0.05) leads us to reject the null hypothesis and conclude that at least one group mean differs significantly from the others.

In our exercise, the p-value was approximately 0.003, which is well below the 0.05 threshold. This indicates strong evidence against the null hypothesis, leading us to believe that the various techniques have significantly different effects. This conclusion is supported by the F-test result and provides students with a clear rationale for the rejection of the null hypothesis in this ANOVA test.