Question

4. (Exam 3, Fall 2018) An article presents a study that investigated the effect of varying the type of fertilizer on the height of certain Mediterranean woody tree species. In one experiment, three samples, each consisting of ten trees, were grown with three different fertilizers. One, the control group, was grown with a standard fertilizer. Another was grown with a fertilizer containing only half the nutrients of the standard fertilizer. The third was grown with the standard fertilizer to which a commercial slow-release fertilizer had been added. Following are the heights of the trees after one year. Fertilizer Control17.912.2 14.9 13.8 26.115.4 20.316.920.814.8 Deficient | 7.0 | 6.9 | 13.3 | 11.1 | 11.0 | 16.5 | 12.7 | 12.4 | 17.1 | 9.0 Height Slow-release 19.820.3 16.17.912.4 12.5 17.4 19.9 27.314.4 Can you conclude that the heights differ among the types of fertilizer? Use α = 0.05. You are given that SSTreatments-229.74067, SST-669.12967. Be sure to show all your work including stating the appropriate hypotheses, constructing the ANOVA table to obtain the test statistic, obtaining the critical value and stating your conclusion in the contert of the problem.

Answer 1

H0: Null Hypothesis: $\mu 1=\mu 2=\mu 3$ (The heights do not differ among the types of fertilizer)

HA:Alternative Hypothesis: $\mu 1\neq \mu 2\neq \mu 3$ (At least one mean is different from other 2 means)

From the given data, the following Table is calculated:

	Control	Deficient	Slow - release	Total
N	10	10	10	30
$\sum X$	173.1	117	178	468.1
Mean	173.1/10=17.31	117/10=11.7	178/10=17.8	468.1/30=15.603
$\sum X^{2}$	3149.45	1478.42	3345.18	7973.05
Std. Dev.	4.1243	3.4884	4.432	4.8035

From the above Table, ANOVA Table is calculated as follows:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Sum of Squares	F
Between treatments	229.7407	2	229.7407/2 = 114.8703	114.8703/16.2737 = 7.0587
Within treatments	439.389	27	439.389/27 = 16.2737
Total	669.1297	29

F - Ratio value = 114.8703/16.2737 = 7.0587

Degrees of Freedom for Numerator = 2

Degrees of Freedom for Denominator = 27

$\alpha$ = 0.05

From Table:

Critical value of F = 3.354

Since the calculated value of F = 7.0587 is greater than critical value of F = 3.354, the difference is significant. Reject null hypothesis.

Conclusion:
The data support the claim that The heights differ among the types of fertilizer

By Technology, p - value =0.0034.

Since p - value = 0.0034 is less than $\alpha$ = 0.05, the difference is significant. Reject null hypothesis.

Conclusion:
The data support the claim that The heights differ among the types of fertilizer