Watering Schedules and Crop Yield

Step 1: Graphs
Graph > Boxplot > with Groups, Graph variable=Yield, Categorical variable= Schedule

(Upper left panel)

There is a problem with outliers. Let's try the square root transform:
Calc > Calculator, Store in: Sqrt(Yield) Expression: SQRT('Yield')
Graph > Boxplot > with Groups, Graph variable=Sqrt(Yield), Categorical variable= Schedule
(Upper right panel)

This solves the problem with the outliers, but there is also evidence of a problem with equal variance (compare box of Schedule 4 with Schedule 6). Let's try a log transform:
Calc > Calculator, Store in: Log(Yield) Expression: LOGT('Yield')
Graph > Boxplot > with Groups, Graph variable=Log(Yield), Categorical variable= Schedule
(lower left panel)

This also fixes the problem of unequal variance. It uses numerical ordering, which here is no more meaningful than alphabetic ordering. It is better to change to some other ordering, say by means.
Stat > Basic Statistics > Display Descriptive Statistics, Variable=Log(Yield), By variables= Schedule
shows the order by means to be 4-5-8-1-3-7-6-2. Trying to change the order by
Select column Schedule by clicking c1, Editor > Column > Value Order
does not work because c1 is a numeric column. Turn it into a text column with
Data > Change data type > Numeric to text. Now change ordering
Redraw boxplot to get lower right panel

Step 2: Summary Statistics
Stat > Basic Statistics > Display Descriptive Statistics, Variable=Yield, By variables= Schedule
Schedule Sample Size Median IQR/1.35
1 6 674 1045
2 6 1410 1740
3 6 437 2137
4 6 5.21 12.3
5 6 104 102
6 6 2426 3362
7 6 864 2027
8 6 137 508

Step 3: Hypothesis Test: Stat > ANOVA > Oneway, Response=Log(Yield), Factor= Schedule, Graphs > Residual vs. Fits Plot and Normal Plot

both graphs are fine

1) α=0.05
2) H0: α1 = .. = α8=0 (no difference in the mean yields of the Schedules)
3) Ha: αi≠0 for some i (some differences in the mean yields of the Schedules)
4) p-value=0.000 < α
5) We reject H0, there are some differences in the mean yields of the Schedules

Step 4: Multiple Comparison
Stat > ANOVA > Oneway, Response=Log(Yield), Factor=Schedule, Comparisons > check Tukey


4 5 8 1 3 7 6 2
____________
____________
____
Interpretation: Schedule 2 is stat. signif. better than schedules 4 and 5. Other differences are not stat. signif., at least not at these sample sizes.