d and f only do not understand how to calculate 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 35 17 2 30 53 6 20 19 2 26 19 8 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: All three means are different.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: 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 ofVariation Sum ofSquares Degrees ofFreedom MS FRatio P Value TestDecision 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

d and f only do not understand how to calculate 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 35 17 2 30 53 6 20 19 2 26 19 8 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: All three means are different.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: 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 ofVariation Sum ofSquares Degrees ofFreedom MS FRatio P Value TestDecision 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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MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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d and f only do not understand how to calculate

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
35	17	2
30	53	6
20	19	2
26	19	8

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: All three means are different.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: 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.)

SS_TOT	=
SS_BET	=
SS_W	=

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

df_BET	=
df_W	=
MS_BET	=
MS_W	=

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

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