Question

Suppose a farmer wanted to test the effectiveness of three different types of fertilizer on strawberry production: no fertilizer, organic fertilizer, and non-organic fertilizer. Three sections of strawberry fields are selected, a random sample of 5 strawberry plants from each section are selected at the end of the growing season. The number of strawberries produced on each plant is counted. The purpose of the study is to determine if type of fertilizer affects strawberry production. Is there a significant effect of fertilizer on strawberry production?

a.Write out the null and alternative hypothesis for this study.

No Fertilizer                            Organic Fertilizer                    Non-organic fertilizer

5                                              8                                              6

4                                              7                                              9

4                                              4                                              4

7                                              7                                              4

3                                              9                                              5

b.  Conduct the appropriate computations to complete the ANOVA summary table. Calculate F using a non-directional test with an alpha level of .05.

c. Make a decision.
d. Interpret the findings.

Answer 1

The total sample size is N = 15.

Therefore, the total degrees of freedom are: df_total​ = 15−1 = 14

Also, the between-groups degrees of freedom are df_between​ = 3 − 1 = 2

The within-groups degrees of freedom are: df_within​ = df_total​ − df_between ​= 14 − 2 = 12

.

First, we need to compute the total sum of values and the grand mean.

The following is obtained

Also, the sum of squared values is

.

Based on the above calculations, the total sum of squares is computed as follows

The within sum of squares is computed as shown in the calculation below:

The between sum of squares is computed directly as shown in the calculation below:

.

Now that sum of squares are computed, we can proceed with computing the mean sum of squares:

.

Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows:

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1 = μ2 = μ3

Ha: Not all means are equal

The above hypotheses will be tested using an F-ratio for a One-Way ANOVA.

(2) Rejection Region

The significance level is α=0.05, and the degrees of freedom are df1​=2 and df2​=2, therefore, the rejection region for this F-test is R = { F : F > Fc ​= 3.885}.

(3) Test Statistics

(4) Decision about the null hypothesis

Since it is observed that F = 0.059 ≤ Fc ​= 3.885, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.9431, and since p = 0.9431 ≥ 0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that not all 3 population means are equal, at the α=0.05 significance level.

Therefore, there is not enough evidence to claim that there a significant effect of fertilizer on strawberry production at the α=0.05 significance level.

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