Question

Weights​ (kg) of poplar trees were obtained from trees planted in a rich and moist region. The trees were given different treatments identified in the accompanying table. Use a

0.05 significance level to test the claim that the four treatment categories yield poplar trees with the same mean weight. Is there a treatment that appears to be most​ effective?

No TreatmentNo Treatment	FertilizerFertilizer	IrrigationIrrigation	Fertilizer and IrrigationFertilizer and Irrigation
1.13	0.82	0.07	0.91
0.52	0.82	0.62	1.45
0.53	0.38	0.08	1.04
0.08	0.42	0.87	1.59
1.23	1.08	0.89	1.11

Determine the null and alternative hypotheses.

Upper H 0H0​:

▼

mu 1 greater than mu 2 greater than mu 3 greater than mu 4μ1>μ2>μ3>μ4

mu 1 not equals mu 2 not equals mu 3 not equals mu 4μ1≠μ2≠μ3≠μ4

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

At least two of the population means are equal.At least two of the population means are equal.

Not all of the population means are equal.Not all of the population means are equal.

Exactly two of the population means are equal.Exactly two of the population means are equal.

Upper H 1H1​:

▼

At least two of the population means are equal.At least two of the population means are equal.

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

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.

Exactly two of the population means are different from the others.Exactly two of the population means are different from the others.

mu 1 greater than mu 2 greater than mu 3 greater than mu 4μ1>μ2>μ3>μ4

mu 1 not equals mu 2 not equals mu 3 not equals mu 4μ1≠μ2≠μ3≠μ4

Find the F test statistic.

Fequals=nothing

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

Find the​ P-value using the F test statistic.

​P-valueequals=nothing

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

What is the conclusion for this hypothesis​ test?

A.

RejectReject

Upper H 0H0.

There is

sufficientsufficient

evidence to warrant rejection of the claim that the four different treatments yield the same mean poplar weight.

B.

RejectReject

Upper H 0H0.

There is

insufficientinsufficient

evidence to warrant rejection of the claim that the four different treatments yield the same mean poplar weight.

C.

Fail to rejectFail to reject

Upper H 0H0.

There is

insufficientinsufficient

evidence to warrant rejection of the claim that the four different treatments yield the same mean poplar weight.

D.

Fail to rejectFail to reject

Upper H 0H0.

There is

sufficientsufficient

evidence to warrant rejection of the claim that the four different treatments yield the same mean poplar weight.

Is there a treatment that appears to be most​ effective?

A.

The​ 'No Treatment' method seems to be most effective.

B.

The​ 'Irrigation' method seems to be most effective.

C.

The 'Fertilizer and Irrigation'The 'Fertilizer and Irrigation'

method seems to be most effective.

D.

The​ 'Fertilizer' method seems to be most effective.

E.

No one treatmentNo one treatment

method seems to be most effective.

Answer 1

One-way ANOVA: No Treatment, Fertilizer, Irrigation, ... and Irrigation

Method

Null hypothesis	All means are equal
Alternative hypothesis	Not all means are equal
Significance level	α = 0.05

Equal variances were assumed for the analysis.

Factor Information

Factor	Levels	Values
Factor	4	No Treatment, Fertilizer, Irrigation, Fertilizer and Irrigation

Analysis of Variance

Source	DF	Adj SS	Adj MS	F-Value	P-Value
Factor	3	1.406	0.4686	3.3171	0.0468
Error	16	2.260	0.1413
Total	19	3.666

Model Summary

S	R-sq	R-sq(adj)	R-sq(pred)
0.375859	38.35%	26.79%	3.67%

Means

Factor	N	Mean	StDev	95% CI
No Treatment	5	0.698	0.477	(0.342, 1.054)
Fertilizer	5	0.704	0.297	(0.348, 1.060)
Irrigation	5	0.506	0.408	(0.150, 0.862)
Fertilizer and Irrigation	5	1.220	0.287	(0.864, 1.576)

Pooled StDev = 0.375859

Tukey Pairwise Comparisons

Tukey Simultaneous Tests for Differences of Means

Difference of Levels	Difference of Means	SE of Difference	95% CI	T-Value	Adjusted P-Value
Fertilizer - No Treatment	0.006	0.238	(-0.675, 0.687)	0.03	1.000
Irrigation - No Treatment	-0.192	0.238	(-0.873, 0.489)	-0.81	0.850
Fertilizer and Irrigation - No Treatment	0.522	0.238	(-0.159, 1.203)	2.20	0.167
Irrigation - Fertilizer	-0.198	0.238	(-0.879, 0.483)	-0.83	0.838
Fertilizer and Irrigation - Fertilizer	0.516	0.238	(-0.165, 1.197)	2.17	0.174
Fertilizer and Irrigation - Irrigation	0.714	0.238	(0.033, 1.395)	3.00	0.038

Tukey Simultaneous 95% CIs

1)

H0​: μ1=μ2=μ3=μ4

H1: At least one of the four population means is different from the others.

2)

F= 3.3171

p-value = 0.0468

since p-value is less than 0.05

Reject H0. There is sufficient evidence to warrant rejection of the claim that the four different treatments yield the same mean poplar weight.

Is there a treatment that appears to be most​ effective?

The 'Fertilizer and Irrigation' method seems to be most effective. which is concluded from Tukey's simultaneous intervals.