Questions: Suppose a researcher wants to investigate the effect of the amount of fertilizer on the height of a common houseplant. More specifically, the researcher is interested in determining if there is a difference between the mean heights of the plants receiving one of three different fertilizer levels: high, medium, and low. Which of the following would be the correct null hypothesis? H0: μ1=μ2 H0: μ1=μ2=μ3 H0: μ1=μ2=μ3=μ4 HA: μ1=μ2=μ3

Solution Steps

The total variability in the data can be partitioned into two components: the variability between groups and the variability within groups.

• The sum of squares between groups is calculated as: $$SS_{between} = \sum_{i=1}^k n_i (\bar{X}_i - \bar{X})^2 = 91.2$$

• The sum of squares within groups is calculated as: $$SS_{within} = \sum_{i=1}^k \sum_{j=1}^{n_i} (X_{ij} - \bar{X}_i)^2 = 53.2$$

Next, we calculate the mean squares for both between and within groups.

• The mean square between groups is given by: $$MS_{between} = \frac{SS_{between}}{df_{between}} = \frac{91.2}{2} = 45.6$$

• The mean square within groups is given by: $$MS_{within} = \frac{SS_{within}}{df_{within}} = \frac{53.2}{12} \approx 4.4333$$

The F-statistic is calculated as the ratio of the mean square between groups to the mean square within groups: $$F = \frac{MS_{between}}{MS_{within}} = \frac{45.6}{4.4333} \approx 10.2857$$

The p-value is calculated based on the F-statistic and the degrees of freedom: $$P = 1 - F(F_{observed}; df_{between}, df_{within}) = 1 - F(10.2857; 2, 12) \approx 0.0025$$

Based on the calculated p-value, we compare it to the significance level ($\alpha = 0.05$):

• Since $P < \alpha$, we reject the null hypothesis.

Final Answer