Question

High productivity and carbohydrate storage ability of the Jerusalem artichoke make it a promising agricultural crop. The article "Leaf Gas Exchange and Tuber Yield in Jerusalem Artichoke Cultivars" (Field Crops Research [1991]: \(241-252\) ) reported on various plant characteristics. Consider the accompanying data on chlorophyll concentration \(\left(\mathrm{gm} / \mathrm{m}^{2}\right)\) for four varieties of Jerusalem artichoke: $$\begin{array}{lccll} \text { Variety } & \text { BI } & \text { RO } & \text { WA } & \text { TO } \\\ \text { Sample mean } & .30 & .24 & .41 & .33 \end{array}$$ Suppose that the sample sizes were \(5,5,4\), and 6, respectively, and also that MSE \(=.0130 .\) Do the data suggest that true average chlorophyll concentration depends on the variety? State and test the appropriate hypotheses using a significance level of \(.05\).

Short answer

The answer cannot be precisely determined without performing the calculations. The calculations would allow comparison of the calculated F-statistic and the critical value from the F-distribution tables to make a final decision on accepting or rejecting the null hypothesis.

Step-by-step solution

Formulating Hypotheses

First, formulate the null hypothesis and alternative hypothesis: the null hypothesis, \(H_0: \mu_{BI} = \mu_{RO} = \mu_{WA} = \mu_{TO}\), assumes that the average chlorophyll concentrations are equal for all varieties. The alternative hypothesis, \(H_a\), states that at least one of the means is different, i.e., the average concentrations are not all the same.

Calculating F-value

Next step is to calculate F-statistic using the ANOVA formula. But first calculate sum of squares between groups (SSB) using the formula, SSB = \(\Sigma n_i * (\bar{X}_i - \bar{X})^2\), where \(n_i\) is the sample size, \(\bar{X}_i\) is the sample mean, and \(\bar{X}\) is the overall sample mean. Here, the overall mean \(\bar{X}\) is the weighted average of the sample means, \(\bar{X} = \Sigma (n_i/\Sigma n_i ) * \bar{X}_i\). Then, the degrees of freedom between groups (df_B) is \(k-1\), where \(k\) is the number of groups. This gives us the mean squares between groups (MSB) as MSB = SSB/df_B.

Mean Squares Within

Use the given mean squares within groups (or groups' mean square error) MSE = 0.013, and the degrees of freedom within groups (df_W) is \(\Sigma n_i - k = (5 + 5 + 4 + 6) - 4 = 16\).

F-statistic and Critical Region

Find the F-statistic as the ratio of MSB to MSE. F = MSB/MSE. Then, the critical region is found using the F-distribution table with df_B, df_W degrees of freedom at a significance level of 0.05. The null hypothesis is rejected if F > critical value.

Final Conclusion

Depending on the comparison of the calculated F-statistic and the critical value, the null hypothesis can either be accepted or rejected. If the F-statistic is greater than the critical value, we reject the null hypothesis and conclude that the chlorophyll concentration does vary among the different varieties. If the F-statistic is not greater than the critical value, we fail to reject the null hypothesis and conclude that the data does not provide enough evidence to suggest the chlorophyll concentration depends on the variety of the plant.