Question

# Where are the deer? Random samples of square-kilometer plots were taken in different ecological locations of...

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.

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

(b) Find SSTOT, SSBET, and SSW and check that SSTOT = SSBET + SSW. (Use 3 decimal places.)

 SSTOT = SSBET = SSW =

Find d.f.BET, d.f.W, MSBET, and MSW. (Use 2 decimal places for MSBET, and MSW.)

 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

a) level of significance =0.05

Ho: 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

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