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 30 20 5 25 59 10 25 16 2 24 19 4 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: ?1 = ?2 = ?3; H1: Exactly two means are equal.Ho: ?1 = ?2 = ?3; H1: At least two means are equal. Ho: ?1 = ?2 = ?3; H1: All three means are different.Ho: ?1 = ?2 = ?3; H1: Not all the means are equal.
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 30 20 5 25 59 10 25 16 2 24 19 4 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: ?1 = ?2 = ?3; H1: Exactly two means are equal.Ho: ?1 = ?2 = ?3; H1: At least two means are equal. Ho: ?1 = ?2 = ?3; H1: All three means are different.Ho: ?1 = ?2 = ?3; H1: Not all the means are equal.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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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 |
30 | 20 | 5 |
25 | 59 | 10 |
25 | 16 | 2 |
24 | 19 | 4 |
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.
(b) Find SSTOT, SSBET, and SSW and check that SSTOT = SSBET + SSW. (Use 3 decimal places.)
Find d.f.BET, d.f.W, MSBET, and MSW. (Use 2 decimal places for MSBET, and 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.
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?
(e) Interpret your conclusion in the context of the application.
(f) Make a summary table for your ANOVA test.
State the null and alternate hypotheses.
Ho: ?1 = ?2 = ?3; H1: Exactly two means are equal.Ho: ?1 = ?2 = ?3; H1: At least two means are equal. Ho: ?1 = ?2 = ?3; H1: All three means are different.Ho: ?1 = ?2 = ?3; H1: Not all the 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 ? = 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 | ---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 |
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