Question

A randomized block design has \(k=3\) treatments, \(b=6\) blocks, with \(S S T=11.4, S S B=17.1\), and Total \(S S=42.7 . \bar{T}_{A}=21.9\) and \(\bar{T}_{B}=24.2 .\) Construct an ANOVA table showing all sums of squares, mean squares, and pertinent \(F\) -values. Then use this information to answer the questions. Do the data provide sufficient evidence to indicate differences among the treatment means? Test using \(\alpha=.05 .\)

Short answer

Answer: No, we cannot conclude that there are significant differences among the treatment means at the 5% level of significance.

Step-by-step solution

Calculate the Sum of Squares for Error (SS_E)

We can find the SS_E using the following formula: $$ SS_E = SS_{total} - SS_T - SS_B $$ Plugging in the given values, $$ SS_E = 42.7 - 11.4 - 17.1 = 14.2 $$

Calculate the degrees of freedom for each source of variation

Using the formulas for degrees of freedom for treatments (df_T), blocks (df_B), and error (df_E), we can obtain the following values: - \(df_T=k-1=3-1=2\) - \(df_B=b-1=6-1=5\) - \(df_E=(k-1)(b-1)=(3-1)(6-1)=10\)

Calculate the Mean Squares for Treatments (MST), Blocks (MSB), and Error (MSE)

Now we calculate the mean squares using the given formulas: - \(MST = \frac{SST}{df_T}=\frac{11.4}{2}=5.7\) - \(MSB = \frac{SSB}{df_B}=\frac{17.1}{5}=3.42\) - \(MSE = \frac{SSE}{df_E}=\frac{14.2}{10}=1.42\)

Form the ANOVA table and Calculate the F-statistics

Now we form the ANOVA table: | Source | Sum of Squares | Degrees of Freedom | Mean Square | F-value | |------------|--------------|------------------|-----------|-------| | Treatments | 11.4 | 2 | 5.7 | F | | Blocks | 17.1 | 5 | 3.42 | | | Error | 14.2 | 10 | 1.42 | | | Total | 42.7 | 17 | | | The F-value can be calculated as follows: $$ F = \frac{MST}{MSE} = \frac{5.7}{1.42} = 4.01 $$

Perform the Hypothesis Test

The null and alternative hypotheses are: - H0: All treatment means are equal - H1: At least one treatment mean is different Using α = 0.05 level of significance and the F-distribution table, we find the critical F-value for df_T = 2 and df_E = 10: $$ F_{0.05, (2, 10)} = 4.10 $$ Since our calculated F-value is 4.01, which is less than the critical F-value of 4.10, we fail to reject the null hypothesis at the 5% level of significance.

Conclusion

The data does not provide sufficient evidence to indicate differences among the treatment means. Therefore, we cannot conclude that there are significant differences among the treatments at the 5% level of significance.

Key concepts

Sum of Squares

When we talk about sum of squares (SS) in the context of an ANOVA in a randomized block design, we’re discussing a way to measure variation. It's a critical component to analyze differences between groups. In simple terms, think of it as a tally of how much data points deviate from their group mean, and in turn, how much the groups deviate from the overall mean. The total sum of squares (SS_total) combines the variation within each group and between the groups.

In the given exercise, the SS_total is the total variability in the data, whereas the treatment sum of squares (SS_T) and the block sum of squares (SS_B) represent variability due to treatments and blocks respectively. The error sum of squares (SS_E) is the variability that's not explained by the treatments or blocks - it's essentially the 'leftover' variation in the experiment. Calculating it, as shown in Step 1 of our exercise, helps us separate this unexplained variation from the total variation captured in SS_total.

Mean Squares

You can think of mean squares as a form of 'averaged' sum of squares. After we have our SS values, we divide them by their respective degrees of freedom (df) to obtain the mean squares (MS).

The calculation of mean squares normalizes the SS values, allowing them to be compared directly, even when the degrees of freedom differ. In our randomized block design ANOVA example, we calculate mean squares for treatments (MST), for blocks (MSB), and for error (MSE). By doing so, we can compare the variance within treatments to the variance within error, which is crucial for determining if our treatments have a significant effect.

F-statistics

The F-statistic is the ratio of the mean square value for treatments to the mean square value for error, and it's the key metric in conducting our hypothesis test. Intuitively, the F-statistic tells us how much the treatment means deviate from the overall mean, relative to the variability within the treatment groups.

A higher F-statistic indicates a greater disparity between groups compared to the variation within the groups. The critical point to note in the exercise is the direct calculation of the F-statistic in Step 4. It serves as the basis for deciding whether our treatments have a statistically significant effect. However, we must always compare this calculated F-value against the critical F-value from established F-distribution tables to reach a conclusion.

Degrees of Freedom

In statistics, the degrees of freedom refer to the number of independent values that can vary in an analysis without breaking any constraints. It's essential in calculating mean squares and critical values for various statistical tests.

For instance, in our exercise, degrees of freedom help us adjust the sum of squares to account for the size of each sample. They help ensure our statistical measures, like mean squares, are scaled appropriately. The degrees of freedom act as a balancing factor - taking into account both the number of treatments and blocks in our design.

Hypothesis Testing

The whole idea behind hypothesis testing in an ANOVA is to determine whether the differences among group means are statistically significant. We start by proposing two contrasting hypotheses: the null hypothesis (H₀) suggests that there is no effect or difference, while the alternative hypothesis (H₁) claims there is.

In our case, the null hypothesis states that all treatment means are equal. By comparing our calculated F-statistic to a critical F-value from statistical tables, we're essentially putting our null hypothesis to the test. If the F-statistic is large enough to surpass the critical F-value, we reject the null hypothesis, concluding that not all treatment means are equal. As our final step in the exercise showed, the F-statistic wasn't large enough to reach this threshold, leading us to retain the null hypothesis for now.