Question

Where are the deer? Random samples of square-kilometer plots were taken in different ecological locations of Mesa Verde National Park. The deer counts per square kilometer were recorded and are shown in the following table.

Mountain Brush	Sagebrush Grassland	Pinon Juniper
31	20	8
32	56	3
22	17	6
27	19	5

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: ?₁ = ?₂ = ?₃; H₁: Not all the means are equal.

H_o: ?₁ = ?₂ = ?₃; H₁: Exactly two means are equal.     

H_o: ?₁ = ?₂ = ?₃; H₁: At least two means are equal.

H_o: ?₁ = ?₂ = ?₃; H₁: All three means are different.

(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. (Use 4 decimal places.)

(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 ? = 0.05, we do not reject H₀.

Since the P value is less than or equal to the level of significance at ? = 0.05, we reject H₀.     

Since the P value is greater than the level of significance at ? = 0.05, we reject H₀.

Since the P value is less than or equal to the level of significance at ? = 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						-Do not reject H0/Reject H0.
Within groups
Total

Answer 1

Random samples of square-kilometer plots were taken in different ecological locations of Mesa Verde National Park. The deer counts per square kilometer were recorded and are given.

The 3 ecological locations are Mountain Brush, Sagerbrush and Pinon Junpler.

We are to test if the mean number of deers in these 3 ecological locations per square kilometer vary or not.

Clearly, the data provided is classified according to only one factor , i.e. the ecological location.

Hence, the appropriate design to be used to test our subject of concern is One Way Classified Data , also known as Completely Randomised Design.

We are given to use a 5% level of significance.

a.) So the level of significance is 5% or 0.05.

     The corresponding null and alternative hypothesis to conduct our test shall be :

                  H_o: ?₁ = ?₂ = ?₃            ; H₁: Not all the means are equal.

b.) Accordingly our model for study shall be :

                              

           where ,    represents the j-th observation from the i-th ecological location

                         is the general mean number of deers irrespective of the location

                         is the effect due to the ith location

                         eij is the error

Hence, it can be written as ;

SSTOT (Total Sum of Squares)

= SSBET (Sum of Squares between the locations) + SSW(Sum of Squares Within which is the error)

By formula,

                                   

                                   

                                    

Hence, on calculating, The sum of squares are as follows:

                                                      SSTOT = 2475

                                                      SSBET = 1350

                                                      SSW = 1125

Clearly, SSTOT=SSBET + SSW

Now, the degrees of freedom for the Sum of Squares between and Sum of squares within are (3-1=2 since there were 3 classifications) and (12-3=9 since there were 12 observations) respectively.

The Mean sum of squares are obtained by dividing the Sum of squares by the degrees of freedom.

MSBET= 1350/2 = 675

MSW = 1125/9 = 125

The sample F statistic is obtained as :

                                         F = MSBET / MSW

                                            = 675/125

                                            = 5.4

c.) The p-value of the test statistic is the probability of obtaining such a value under the alternate   hypothesis and its value here is 0.02878

d.) It is known that if the obtained p-value is less than our provided level of significance then we reject our null hypothesis otherwise we accept it.

     Here, 0.02878 < 0.05 (our level of significance )

   So, we reject our null hypothesis.

e.) From our study we can thus conclude that,

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

f.) The summary table is :

    

ANOVA
Source of Variation	SS	df	MS	F	P-value
Between Groups	1350.0	2	675	5.4	0.02878
Within Groups	1125	9	125
Total	2475	11