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

a. Use a

0.100.10

significance level to test the claim that the different treatments result in the same mean weight.

Determine the null and alternative hypotheses.

Upper H 0H0​:

mu 1 equals mu 2 equals mu 3 equals mu 4μ1=μ2=μ3=μ4

Upper H 1H1​:

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

b. Find the test statistic.

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

Answer 1

a) Ho: μ1=μ2=μ3=μ4

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

b)

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

Test statistic = 6.06