Question

Where are the deer? Random samples of square-kilometer plots were taken in different ecological locations of a national park. The deer counts per square kilometer were recorded and are shown in the following table.

Mountain Brush	Sagebrush Grassland	Pinon Juniper
34	15	1
31	58	7
24	15	10
25	26	13

Shall we reject or accept the claim that there is no difference in the mean number of deer per square kilometer in these different ecological locations? Use a 5% level of significance.

(a) What is the level of significance?


State the null and alternate hypotheses.

H_o: μ₁ = μ₂ = μ₃; H₁: Exactly two means are equal.H_o: μ₁ = μ₂ = μ₃; H₁: Not all the means are equal.    H_o: μ₁ = μ₂ = μ₃; H₁: All three means are different.H_o: μ₁ = μ₂ = μ₃; H₁: At least two means are equal.

(b) Find SS_TOT, SS_BET, and SS_W and check that SS_TOT = SS_BET + SS_W. (Use 3 decimal places.)

SSTOT	=
SSBET	=
SSW	=

Find d.f._BET, d.f._W, MS_BET, and MS_W. (Use 2 decimal places for MS_BET, and MS_W.)

dfBET	=
dfW	=
MSBET	=
MSW	=

Find the value of the sample F statistic. (Use 3 decimal places.)


What are the degrees of freedom?
(numerator)
(denominator)

(c) Find the P-value of the sample test statistic.

P-value > 0.1000.050 < P-value < 0.100    0.025 < P-value < 0.0500.010 < P-value < 0.0250.001 < P-value < 0.010P-value < 0.001

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?

Since the P-value is greater than the level of significance at α = 0.05, we do not reject H₀.Since the P-value is less than or equal to the level of significance at α = 0.05, we reject H₀.    Since the P-value is greater than the level of significance at α = 0.05, we reject H₀.Since the P-value is less than or equal to the level of significance at α = 0.05, we do not reject H₀.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance there is insufficient evidence to conclude that the means are not all equal.At the 5% level of significance there is sufficient evidence to conclude that the means are not all equal.    At the 5% level of significance there is insufficient evidence to conclude that the means are all equal.At the 5% level of significance there is sufficient evidence to conclude that the means are all equal.

(f) Make a summary table for your ANOVA test.

Source of Variation	Sum of Squares	Degrees of Freedom	MS	F Ratio	P Value	Test Decision
Between groups					---Select--- p-value > 0.100 0.050 < p-value < 0.100 0.025 < p-value < 0.050 0.010 < p-value < 0.025 0.001 < p-value < 0.010 p-value < 0.001	---Select--- Do not reject H0. Reject H0.
Within groups
Total

Answer 1

a) level of significance =0.05

H_o: μ₁ = μ₂ = μ₃; H₁: Not all the means are equal.   

b)APplying ANOVA on above data:

SSTOT	2536.917
SSBET	1148.167
SSW	1388.750
dfBET	2
dfW	9
MSBET	574.083
MSW	154.306
value of test statistic for factor A =				3.720
df(numerator) =		2
df(Denominator) =		9

c)

0.050 < P-value < 0.10

d)Since the P-value is greater than the level of significance at α = 0.05, we do not reject H₀

_e)At the 5% level of significance there is insufficient evidence to conclude that the means are not all equal.

f)

Source of Variation	SS	df	MS	F	P-value	Decision
Between Groups	1148.167	2	574.083	3.720	0.05< p<0.10	fail to reject Ho
Within Groups	1388.750	9	154.306
Total	2536.917	11