Question

The accompanying data are the weights​ (kg) of poplar trees that were obtained from trees planted in a rich and moist region. The trees were given different treatments identified in the accompanying table. Also shown are partial results from using the Bonferroni test with the sample data. Complete parts​ (a) through​ (c).

No TreatmentNo Treatment	FertilizerFertilizer	IrrigationIrrigation	Fertilizer and IrrigationFertilizer and Irrigation
1.221.22	1.021.02	0.090.09	0.820.82
0.630.63	0.830.83	0.670.67	1.831.83
0.470.47	0.440.44	0.070.07	1.461.46
0.080.08	0.690.69	0.720.72	2.212.21
1.321.32	1.091.09	0.920.92	1.691.69

Bonferroni Results

		Mean
​ (I) TREATMENT	​ (J) TREATMENT	Difference​ (I-J)		Std. Error	Sig.
1.00	2.00	negative 0.0700−0.0700		0.271830.27183	1.0001.000
	3.00	0.25000.2500		0.271830.27183	1.0001.000
	4.00	negative 0.8580−0.8580		0.271830.27183	0.0400.040

c. Use the Bonferroni test procedure with a 0.10 significance level to test for a significant difference between the mean amount of the irrigation treatment group and the group treated with both fertilizer and irrigation. Identify the test statistic and the​ P-value. What do the results​ indicate?

The test statistic is

nothing.

​(Round to two decimal places as​ needed.)

Answer 1

Ho: μ1=μ2=μ3=μ4

Ha: At least one of the four population means is different from the others

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance	std dev
No	5	3.72	0.7440	0.2718	0.5214
F	5	4.07	0.8140	0.0685	0.2618
I	5	2.47	0.4940	0.1516	0.3894
F and I.	5	8.01	1.60	0.2653	0.5150
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	3.4432	3	1.1477	6.0626	0.0059	3.24
Within Groups	3.0290	16	0.1893
Total	6.4723	19

c)

test statistic = mean difference / √(MSE(1/ni+1/nj))      

test stat = -4.03 ; p value=0.0010 , result= significant different

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