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Based on the following values: Mathematical expectation: E(X) = ∑Xn1 = 1204 = 30 E(Y) = ∑Yn2 = 1884 = 47 E(Z) = ∑Zn3 = 1784 = 44.5 Variance: var(X) = 1n1-1∑X2-1n∑X2 = 14-13678-12024 =26 var(Y) = 1n2-1∑Y2-1n∑Y2 = 14-18890-18824 =18 var(Z) = 1n3-1∑Z2-1n3∑Z2 = 14-18322-17824 =133.6667 Root mean square deviation sd(X)= var(X) =26 = 5.0990 sd(Y)= var(Y) =18 = 4.2426 sd(Z)= var(Z) =133.6667 = 11.5614 Correlation coefficients X Y Z     25 44 28   28 43 48   37 49 55   30 52 47 1. Expectation 120 188 178 2. Variance 26 18 133.6667 3. root mean square deviation 5.099 4.2426 11.5614    Check the hypotheses in the image below, in accordance with the tablic values.

Based on the following values: Mathematical expectation: E(X) = ∑Xn1 = 1204 = 30 E(Y) = ∑Yn2 = 1884 = 47 E(Z) = ∑Zn3 = 1784 = 44.5 Variance: var(X) = 1n1-1∑X2-1n∑X2 = 14-13678-12024 =26 var(Y) = 1n2-1∑Y2-1n∑Y2 = 14-18890-18824 =18 var(Z) = 1n3-1∑Z2-1n3∑Z2 = 14-18322-17824 =133.6667 Root mean square deviation sd(X)= var(X) =26 = 5.0990 sd(Y)= var(Y) =18 = 4.2426 sd(Z)= var(Z) =133.6667 = 11.5614 Correlation coefficients X Y Z     25 44 28   28 43 48   37 49 55   30 52 47 1. Expectation 120 188 178 2. Variance 26 18 133.6667 3. root mean square deviation 5.099 4.2426 11.5614    Check the hypotheses in the image below, in accordance with the tablic values.

MATLAB: An Introduction with Applications
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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Based on the following values:

Mathematical expectation:

E(X) = ∑Xn1 = 1204 = 30

E(Y) = ∑Yn2 = 1884 = 47

E(Z) = ∑Zn3 = 1784 = 44.5

Variance:

var(X) = 1n1-1∑X2-1n∑X2 = 14-13678-12024 =26

var(Y) = 1n2-1∑Y2-1n∑Y2 = 14-18890-18824 =18

var(Z) = 1n3-1∑Z2-1n3∑Z2 = 14-18322-17824 =133.6667

Root mean square deviation

sd(X)= var(X) =26 = 5.0990

sd(Y)= var(Y) =18 = 4.2426

sd(Z)= var(Z) =133.6667 = 11.5614

Correlation coefficients


X

Y

Z

 

 

25

44

28

 

28

43

48

 

37

49

55

 

30

52

47

1. Expectation

120

188

178

2. Variance

26

18

133.6667

3. root mean square deviation

5.099

4.2426

11.5614

 

 Check the hypotheses in the image below, in accordance with the tablic values. 

 

 

α = 0.05 T(a/2, v = 2) = 4.3027 T(a/2, v = 3) = 3.1825 T(a/2, v = 4) = 2.7765 T(a/2, v = 6) = 2.4469 F(a/2, 4, 4) = 9.6045 F(a/2,3,3)= 15.4392 F(a/2,3,4)= 9.9792 F(a/2, 1,4)= 7.708 F(a/2, 4, 3) = 9.12 G(a,3,3): = 0.7977 x² = x²(2,3)= 0.2158 x² = x² (1-2, 3) = T(α, 4) = 1.71 = 9.348
Transcribed Image Text:α = 0.05 T(a/2, v = 2) = 4.3027 T(a/2, v = 3) = 3.1825 T(a/2, v = 4) = 2.7765 T(a/2, v = 6) = 2.4469 F(a/2, 4, 4) = 9.6045 F(a/2,3,3)= 15.4392 F(a/2,3,4)= 9.9792 F(a/2, 1,4)= 7.708 F(a/2, 4, 3) = 9.12 G(a,3,3): = 0.7977 x² = x²(2,3)= 0.2158 x² = x² (1-2, 3) = T(α, 4) = 1.71 = 9.348
Ho: 02 = 03: H1:03 + 0 осии Ho: 03 = 02; H1: 02 +02 сии Ho: 0 = 02; H1:02 + 02 # o
Transcribed Image Text:Ho: 02 = 03: H1:03 + 0 осии Ho: 03 = 02; H1: 02 +02 сии Ho: 0 = 02; H1:02 + 02 # o
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Thank you, could you please test the two hypotheses below as well? 

Thanks, 

 

4.Verification of hypothesis of uniformity of many variances Họ: đải = đẳz = đẳs; H: Gải toả + đảo 5. Testing a hypothesis that the mathematical expectation is equal to a given constant HO: mx = 31; H1: mx # 31 |
Transcribed Image Text:4.Verification of hypothesis of uniformity of many variances Họ: đải = đẳz = đẳs; H: Gải toả + đảo 5. Testing a hypothesis that the mathematical expectation is equal to a given constant HO: mx = 31; H1: mx # 31 |
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