The following table gives the yields of wheat from 16 plots, all of approximately equal fertility, where 4 varieties of wheat were cultivated in a completely randomized fashion. Test the hypothesis that the varieties are not significantly different.
The correct answer and explanation is :
To test the hypothesis that the four varieties of wheat do not significantly differ in yield, we use a one-way Analysis of Variance (ANOVA). The null hypothesis ((H_0)) states that all four varieties have the same mean yield, while the alternative hypothesis ((H_1)) suggests that at least one variety differs.
Steps:
- State the Hypotheses:
- (H_0): The mean yields of all four varieties are equal.
- (H_1): At least one variety has a different mean yield.
- Compute the ANOVA Table Components:
- Calculate the Total Sum of Squares (SST):
[
SST = \sum{(X_{ij} – \bar{X})^2}
] - Calculate the Between-group Sum of Squares (SSB):
[
SSB = \sum{n_i(\bar{X}_i – \bar{X})^2}
] - Calculate the Within-group Sum of Squares (SSW):
[
SSW = SST – SSB
]
- Determine Degrees of Freedom (df):
- df for between groups = (k – 1 = 4 – 1 = 3)
- df for within groups = (N – k = 16 – 4 = 12)
- Calculate Mean Squares (MS):
- (MSB = SSB / df_B)
- (MSW = SSW / df_W)
- Compute the F-statistic:
[
F = \frac{MSB}{MSW}
] - Compare with the Critical Value:
- Find the critical value from the F-table for (df_B = 3) and (df_W = 12) at a chosen significance level (e.g., 0.05).
- If (F) is greater than the critical value, reject (H_0).
Conclusion:
- If (H_0) is rejected, the varieties have significantly different yields.
- If (H_0) is not rejected, the differences are due to random chance.