ARM Project(Sample)

8/7/2019 ARM Project(Sample)

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Summary of Data Used:When Sample size = n = 50

Eco BMS SI Finance

Mean 86.44 88.90 86.48 85.36

Standard Deviation 7.91 6.82 8.31 6.99

Maximum 98 99 99 95

Minimum 70 72 71 71

Reggression = r

When Sample size = n = 25

Eco BMS SI Finance

Mean 85.80 88.52 84.24 82.64

Standard Deviation 7.42 6.36 7.22 7.13

Maximum 98 97.00 96.00 95.00

Minimum 70 72.00 71.00 71.00

r b/w Eco and BMS 0.69

r b/w Eco and SI 0.71

r b/w Eco and Finance 0.61

r b/w BMS and SI 0.66

r

b/w BMS and Finance 0.62

r b/w SI and Finance 0.67


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Question No. 1

95% confidence intervals for single population means of marks in thefollowing subjects:

a)Using full sample of 50 (z-test)Note: is unknown and n30 thus we use z-test.A. ECONOMICS

Dx = 86.44

n = 50S = 7.42

= 0.05/2 = 0.025z /2 = 1.96

Dxs z/2 . S/n

= 86.44 (1.96 x 7.42/50)= 86.44 2.057

= 86.44 - 2.057 & = 86.44 + 2.057

= 84.383 = 88.497Confidence Interval(84.383 < < 88.497)

B. BUSINESS MATHS & STATISTICSDx = 88.90

n = 50

S = 6.82

= 0.05/2 = 0.025z /2 = 1.96

Dxs z/2 . S/n


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= 88.90 (1.96 x 6.82/50)= 88.90 1.891

= 88.90 1.891 & = 88.90 + 1.891


C. STATISTICAL INFERENCE

Dx = 86.48

n = 50

S = 8.31 = 0.05/2 = 0.025z /2 = 1.96

Dxs z/2 . S/n= 86.48 (1.96 x 8.31/50)= 86.48 2.30

= 86.48 2.30 & = 86.48 + 2.30= 84.18 = 88.78

Confidence Interval(84.18 < < 88.78)

D. FINANCEDx = 85.36

n = 50S = 6.99

= 0.05/2 = 0.025z /2 = 1.96

Dxs z/2 . S/n


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= 85.36 (1.96 x 6.99/50)= 85.36 1.94

= 85.36 1.94 & = 85.36 + 1.94



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b) Using first 25 as sample size (t-test)Note: is unknown and n


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= 85.89 = 91.15

Confidence Interval

(85.89 < < 91.15)C. STATISTICAL INFERENCE

Dx = 84.24n = 25S = 7.22 = 0.05/2 = 0.025

(n 1) = 24t0.025 (24) = 2.064_x t/2 (n 1) . S/n

= 84.24 (2.064 x 7.22/25)= 84.24 2.98= 84.24 2.98 & = 84.24 + 2.98= 81.26 = 87.22

Confidence Interval(81.26 < < 87.22)D. FINANCE

Dx = 82.64n = 25S = 7.13

= 0.05/2 = 0.025(n 1) = 24t0.025(24) = 2.064_x t/2 (n 1) . S/n

= 82.64 (2.064 x 7.13/25)


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= 82.64 2.94= 82.64 2.94 & = 82.64 + 2.94= 79.7 = 85.58



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Question No. 2

95% Confidence interval of difference of two population means of thefollowing subjects:

a)Using full sample of 50 (z-test)Note: is unknown and n1 & n2 30 thus we use z-test

A.ECONOMICS AND BUSINESS MATHEMATICSDx1 = 86.44 Dx2 = 88.90

n1 = 50 n2 = 50S1 = 7.91 S2 = 6.82

= 0.05/2 = 0.025z / 2 = z0.025 = 1.96

(Dx1 -Dx2) z/2 . (S12/n1 + S2

2/n2)= (86.44 88.90) 1.96 (7.912/50 + 6.822/50)= (-2.46) 2.895= -2.46 - 2.895 & = -2.46 + 2.895

= -5.355 = 0.435Confidence Interval(-5.355


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(Dx1 -Dx2) z/2 . (S12/n1 + S2

2/n2)= (86.44 - 86.48) 1.96 (7.912/50 + 8.312/50)= (-0.04) 3.18

= -0.04 - 3.18 & = -0.04 + 3.18= -3.22 = 3.14Confidence Interval(-3.22


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= 0.05/2 = 0.025z / 2 = z0.025 = 1.96

(Dx1 -Dx2) z/2 . (S12/n1 + S2

2/n2)

= (88.90 86.48) 1.96 (6.822/50 + 8.312/50)= 2.42 2.98= 2.42 2.98 & = 2.42 + 2.98= -0.56 = 5.4Confidence Interval

(-0.56


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F.STATISTICAL INFERENCE AND FINANCE

Dx1 = 86.48 Dx2 = 85.36

n1 = 50 n2 = 50

S1 = 8.31 S2 = 6.99= 0.05/2 = 0.025z / 2 = z0.025 = 1.96(Dx1 -Dx2) z/2 . (S1

2/n1 + S22/n2)

= (86.48 85.36) 1.96 (8.312/50 + 6.992/50)= 1.12 3.01

= 1.12 3.01 & = 1.12 + 3.01=-1.89 = 4.13Confidence Interval(-1.89


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b) Taking first 25 as sample size (t-test)Note: is unknown and n1 & n2 < 30 thus we use t-test

A.ECONOMICS AND BUSINESS MATHEMATICSDx1 = 85.80 Dx2 = 88.52

n1 = 25 n2 = 25S1 = 7.42 S2 = 6.36

= 0.05/2 = 0.025n1 + n2 2 = 48t0.025(48) = 2.011

(Dx1 -Dx2) t0.025(48) . Sp (1/n1 + 1/n2)

Where Sp = ( (n1-1) s12 + (n2-1) s2

2)/n1+n2-2Sp = ( (25-1)7.422 + (25-1)6.362) / (25+25-2) = 6.91Thus,(Dx1 -Dx2) t0.025(48) . Sp (1/n1 + 1/n2)

= (85.80-88.52) 2.011x

6.91x

1/25 + 1/25

= (-2.72) 3.93= - 2.72 3.93 & = -2.72 + 3.93= -6.65 = 1.21Confidence Interval

(-6.65


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= 0.05/2 = 0.025n1 + n2 2 = 48t0.025(48) = 2.011

(Dx1 -Dx2) t0.025(48) . Sp (1/n1 + 1/n2)

Where Sp = ( (n1-1) s12 + (n2-1) s2

2)/n1+n2-2Sp = ( (25-1)7.422 + (25-1)7.222) / (25+25-2) = 7.32Thus,(Dx1 -Dx2) t0.025(48) . Sp (1/n1 + 1/n2)

= (85.80-84.24) 2.011x 7.32x 1/25 + 1/25

= 1.56 4.16= 1.56 4.16 & = 1.56 + 4.16= -2.6 = 5.72Confidence Interval

(-2.6


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Sp = ( (25-1)7.422 + (25-1)7.132) / (25+25-2) = 7.28Thus,

(Dx1 -Dx2) t0.025(48) . Sp (1/n1 + 1/n2) = (85.80-82.64) 2.011x 7.28 x 1/25 + 1/25

= 3.16 4.14= 3.16 4.14 & = 3.16 + 4.14= -0.98 =7.3Confidence Interval

(-0.98


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= 4.28 - 3.87 & = 4.28 + 3.87= 0.41 =8.15

Confidence Interval (0.41


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F.STATISTICAL INFERENCE AND FINANCEDx1 = 84.24 Dx2 = 82.64

n1 = 25 n2 = 25

S1 = 7.22 S2 = 7.13= 0.05/2 = 0.025n1 + n2 2 = 48t0.025(48) = 2.011

(x1 -Dx2) t0.025(48) . Sp (1/n1 + 1/n2)

Where Sp = ( (n1-1) s12 + (n2-1) s2

2)/n1+n2-2

Sp = ( (25-1)7.222 + (25-1)7.132) / (25+25-2) = 7.18Thus,(Dx1 -Dx2) t0.025(48) . Sp (1/n1 + 1/n2) = (84.24 82.64) 2.011x 7.18 x 1/25 + 1/25

= 1.6 4.08

= 1.6 4.08 & = 1.6 + 4.08= - 2.48 = 5.68Confidence Interval

(-2.48


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Question no.3

Hypothesis testing for single population means of the following subjects:

1.Using full sample of 50 (z-test)A.ECONOMICSDx = 86.44

S = 7.91n = 50


Ho: Q = 84.383Ha: Q > 84.383E = 0.05Critical Regionz > zEz > z0.05z > 1.65Computation

z = Dx - Qo

S / nz = 86.44 84.383

7.91/50z = 1.83Result

Since the calculated value of z lies inthe critical region, we may reject thenull hypothesis and conclude that thepopulation mean of economics isgreater than the minimum value of84.383. The lower limit of theconfidence interval is verified.

Ho: Q = 88.497Ha: Q < 88.497E = 0.05Critical Regionz < -zEz < -z0.05z < -1.65Computation

z = Dx - Qo

S / nz = 86.44 88.497

7.91/50z = - 1.83Result

Since the calculated value of z lies inthe critical region, we may reject thenull hypothesis and conclude that thepopulation mean of economics is lessthan the maximum value of 88.497.The upper limit of the confidenceinterval is verified.


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B.BUSINESS MATHS & STATISTICS

Dx = 88.90S = 6.82n = 50

Confidence Interval(87.009 < < 90.791)Ho: Q = 87.009Ha: Q > 87.009E = 0.05Critical Regionz > zE

z > z0.05z > 1.65Computationz = Dx - Qo

S / nz = 88.90 87.009

6.82/50z = 1.96ResultSince the calculated value of z lies inthe critical region, we may reject thenull hypothesis and conclude that thepopulation mean of BMS is greaterthan the minimum value of 87.009.The lower limit of the confidenceinterval is verified.

Ho: Q = 90.791Ha: Q < 90.791E = 0.05Critical Regionz < -zE

z < -z0.05z < -1.65Computationz = Dx - Qo

S / nz = 88.90 90.791

6.82/50z = - 1.96ResultSince the calculated value of z lies inthe critical region, we may reject thenull hypothesis and conclude that thepopulation mean of BMS is less thanthe maximum value of 90.791. Theupper limit of the confidence intervalis verified.


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C.STATISTICAL INFERENCE

Dx = 86.48S = 8.31n = 50



S / nz = 86.48 84.18

8.31/50z = 1.96ResultSince the calculated value of z lies inthe critical region, we may reject thenull hypothesis and conclude that thepopulation mean of SI is greater thanthe minimum value of 84.18. Thelower limit of the confidence intervalis verified.



S / nz = 86.48 88.78

8.31/50z = -1.96ResultSince the calculated value of z lies inthe critical region, we may reject thenull hypothesis and conclude that thepopulation mean of SI is less thanthe maximum value of 88.78. Theupper limit of the confidence intervalis verified.


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D.FINANCEDx = 85.36

S = 6.99n = 50



S / nz = 85.36 83.42

6.99/50z = 1.963ResultSince the calculated value of z lies inthe critical region, we may reject thenull hypothesis and conclude that thepopulation mean of finance is greaterthan the minimum value of 83.42.The lower limit of the confidenceinterval is verified.



S / nz = 85.36 87.3

6.99/50z = -1.963ResultSince the calculated value of z lies inthe critical region, we may reject thenull hypothesis and conclude that thepopulation mean of finance is lessthan the maximum value of 87.3.The upper limit of the confidenceinterval is verified.


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b)Using first 25 as sample size (t-test)

A. ECONOMICSDx = 85.80

S = 7.42n = 25Confidence Interval(82.74 < < 88.86)

Ho: Q = 82.74Ha: Q > 82.74E = 0.05

Critical Regiont > tE(n-1)t > t0.05(24)t > 1.711Computationt = Dx - Qo

S / nt = 85.80 82.74

7.42/25t = 2.062ResultSince the calculated value of t lies inthe critical region, we may reject thenull hypothesis and conclude that thepopulation mean of economics isgreater than the minimum value of82.74. The lower limit of theconfidence interval is verified.

Ho: Q = 88.86Ha: Q < 88.86E = 0.05

Critical Regiont < -tE(n-1)t < -t0.05(24)t < -1.711Computationt = Dx - Qo

S / nt= 85.80 88.86

7.42/25t = -2.062ResultSince the calculated value of t lies inthe critical region, we may reject thenull hypothesis and conclude that thepopulation mean of economics is lessthan the maximum value of 88.86.The upper limit of the confidenceinterval is verified.


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B. BUSINESS MATHS & STATISTICSDx = 88.52

S = 6.36n = 25

Confidence Interval(85.89 < < 91.15)Ho: Q = 85.89Ha: Q > 85.89E = 0.05Critical Regiont > tE(n-1)t > t0.05(24)

t > 1.711Computationt = Dx - Qo

S / nt = 88.52 85.89

6.36/25t = 2.068ResultSince the calculated value of z lies inthe critical region, we may reject thenull hypothesis and conclude that thepopulation mean of BMS is greaterthan the minimum value of 85.89.The lower limit of the confidenceinterval is verified.

Ho: Q = 91.15Ha: Q < 91.15E = 0.05Critical Regiont < -tE(n-1)t < -t0.05(24)

t < -1.711Computationt = Dx - Qo

S / nt = 88.52 91.15

6.36/25t = - 2.068ResultSince the calculated value of z lies inthe critical region, we may reject thenull hypothesis and conclude that thepopulation mean of BMS is less thanthe maximum value of 91.15. Theupper limit of the confidence intervalis verified.


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Dx = 84.24S = 7.22n = 25


Ho: Q = 81.26Ha: Q > 81.26E = 0.05Critical Regiont > t E (n-1)

t > t0.05(24)t > 1.711Computationt = Dx - Qo

S / nt = 84.24 81.26

7.22/25t = 2.064ResultSince the calculated value of z lies inthe critical region, we may reject thenull hypothesis and conclude that thepopulation mean of SI is greater thanthe minimum value of 81.26. Thelower limit of the confidence intervalis verified.

Ho: Q = 87.22Ha: Q < 87.22E = 0.05Critical Regiont < -t E (n-1)

t < -t0.05(24)t < -1.711Computationt = Dx - Qo

S / nt= 84.24 87.22

7.22/25t = -2.064ResultSince the calculated value of z lies inthe critical region, we may reject thenull hypothesis and conclude that thepopulation mean of SI is less thanthe maximum value of 87.22. Theupper limit of the confidence intervalis verified.


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D. FINANCEDx = 82.64

S = 7.13n = 25


Ho: Q = 79.7Ha: Q > 79.7E = 0.05Critical Regiont > tE(n-1)

t > t0.05(24)t > 1.711Computationt = Dx - Qo

S / nt = 82.64 79.7

7.13/25t = 2.062ResultSince the calculated value of z lies inthe critical region, we may reject thenull hypothesis and conclude that thepopulation mean of finance is greaterthan the minimum value of 79.7. Thelower limit of the confidence intervalis verified.

Ho: Q = 85.58Ha: Q < 85.58E = 0.05Critical Regiont < -tE(n-1)

t < -t0.05(24)t < -1.711Computationt = Dx - Qo

S / nt = 82.64- 85.58

7.13/25t = -2.062ResultSince the calculated value of z lies inthe critical region, we may reject thenull hypothesis and conclude that thepopulation mean of finance is lessthan the maximum value of 85.58.The upper limit of the confidenceinterval is verified.


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Question no.4Hypothesis testing for significant difference between avg. marks of thefollowing subjects:

1.Using full sample of 50 (z-test)

A.ECONOMICS and BUSINESS MATHS & STATISTICS

Dx1 = 86.44S1 = 7.91n1 = 50

Dx2 = 88.90S2 = 6.82n2 = 50

Ho: Q1 - Q2 = 0

Ha: Q1 - Q2 { 0

E = 0.05E/2 = 0.025Critical Regionz > zE/2z > z0.025z > 1.96

z < - zE/2z < - z0.025z < - 1.96

Computationz = (Dx1 - Dx2 ) (Q1 - Q2 )

( S12/n1 + S2

2/n2 )z = - 1.67Result

Since the calculated value of z lies in the accepted region, we may accept the nullhypothesis and conclude that there is no significant difference between the averagemarks of E and BMS.


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B. ECONOMICS and STATISTICAL INFERENCE

Dx1 = 86.44S1 = 7.91n1 = 50

Dx2 = 86.48S2 = 8.31n2 = 50

Ho: Q1 - Q2 = 0Ha: Q1 - Q2 { 0E = 0.05E/2 = 0.025Critical Regionz > zE/2z > z0.025

z > 1.96

z < - zE/2z < - z0.025

z < - 1.96

Computation

z = (Dx1 - Dx2 ) (Q1 - Q2 )( S1

2/n1 + S22/n2 )

z = -0.025

ResultSince the calculated value of z lies in the accepted region, we may accept the nullhypothesis and conclude that there is no significant difference between the averagemarks of E and SI.C. ECONOMICS and FINANCEDx1 = 86.44

S1 = 7.91n1 = 50

Dx2 = 85.36

S2 = 6.99n2 = 50

Ho: Q1 - Q2 = 0Ha: Q1 - Q2 { 0E = 0.05E/2 = 0.025


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Critical Regionz > zE/2z > z0.025z > 1.96

z < - zE/2z < - z0.025z < - 1.96


( S12/n1 + S2

2/n2 )z = 0.724

Result

Since the calculated value of z lies in the accepted region, we may accept the nullhypothesis and conclude that there is no significant difference between the averagemarks of E and F.

D. BUSINESS MATHS & STATISTICS and STATISTICAL INFERENCEDx1 = 88.90

S1 = 6.82n1 = 50

Dx2 = 86.48

S2 = 8.31n2 = 50

Ho: Q1 - Q2 = 0Ha: Q1 - Q2 { 0E = 0.05E/2 = 0.025Critical Regionz > zE/2z > z0.025z > 1.96

z < - zE/2z < - z0.025z < - 1.96

Computation

z = (Dx1 - Dx2 ) (Q1 - Q2 )( S1

2/n1 + S22/n2 )

z = 1.59


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ResultSince the calculated value of z lies in the accepted region, we may accept the nullhypothesis and conclude that there is no significant difference between the averagemarks of BMS and SI.

E. BUSINESS MATHS & STATISTICS and FINANCEDx1 = 88.90

S1 = 6.82n1 = 50

Dx2 = 85.36

S2 = 6.99n2 = 50

Ho: Q1 - Q2 = 0Ha: Q1 - Q2 { 0

E = 0.05E/2 = 0.025Critical Regionz > zE/2z > z0.025z > 1.96

z < - zE/2z < - z0.025z < - 1.96

Computation

z = (Dx1 - Dx2 ) (Q1 - Q2 )( S1

2/n1 + S22/n2 )

z = 2.56ResultSince the calculated value of z lies in the critical region, we may reject the nullhypothesis and accept the alternative and conclude that there is significantdifference between the average marks of BMS and F.


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F. STATISTICAL INFERENCE and FINANCEDx1 = 86.48

S1 = 8.31n1 = 50

Dx2 = 85.36

S2 = 6.99n2 = 50

Ho: Q1 - Q2 = 0Ha: Q1 - Q2 { 0E = 0.05E/2 = 0.025Critical Regionz > zE/2z > z0.025

z > 1.96

z < - zE/2z < - z0.025

z < - 1.96


( S12/n1 + S2

2/n2 )z = 0.73ResultSince the calculated value of z lies in the accepted region, we may accept the nullhypothesis and conclude that there is no significant difference between the averagemarks of SI and F.


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b) Using first 25 as sample size (t- test)

A.ECONOMICS and BUSINESS MATHS & STATISTICS

_Dx1 = 85.80S1 = 7.42n1 = 25

Dx2 = 88.52S2 = 6.36n2 = 25

Ho: Q1 - Q2 = 0Ha: Q1 - Q2 { 0E = 0.05E/2 = 0.025n1 + n2 2 = 48

Critical Regiont > tE/2 ( n1+n2 - 2)t > t0.025(48)t > 2.011

t < - tE/2 ( n1+n2 - 2)t < - t0.025(48)t < - 2.011

Computation

t = (Dx

1 -Dx

2 ) (Q1 - Q2 )Sp.( 1/n1 + 1/n2 )

Where Sp = (n1 - 1)S12 + (n2 1)S22n1 + n2 2

Sp = 6.91t = - 1.39

ResultSince the calculated value of t lies in the accepted region, we may accept the nullhypothesis and conclude that there is no significant difference between the averagemarks of E and BMS.


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B. ECONOMICS and STATISTICAL INFERENCEDx1 = 85.80

S1 = 7.42n1 = 25

Dx2 = 84.24

S2 = 7.22n2 = 25

Ho: Q1 - Q2 = 0Ha: Q1 - Q2 { 0E = 0.05E/2 = 0.025n1 + n2 2 = 48


t < - tE/2 ( n1+n2 - 2)t < - t0.025(48)t < - 2.011

Computationt = (Dx1 - Dx2 ) (Q1 - Q2 )

Sp.( 1/n1 + 1/n2 )

Where Sp =

(n1 - 1)S1

2

+ (n2 1)S22

n1 + n2 2

Sp = 7.32t = 0.75ResultSince the calculated value of t lies in the accepted region, we may accept the nullhypothesis and conclude that there is no significant difference between the averagemarks of E and SI.


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C. ECONOMICS and FINANCEDx1 = 85.80

S1 = 7.42n1 = 25

Dx2 = 82.64

S2 = 7.13n2 = 25

Ho: Q1 - Q2 = 0Ha: Q1 - Q2 { 0E = 0.05E/2 = 0.025n1 + n2 2 = 48Critical Regiont > tE/2 ( n1+n2 - 2)t > t0.025(48)t > 2.011

t < - tE/2 ( n1+n2 - 2)t < - t0.025(48)t < - 2.011


Sp.( 1/n1 + 1/n2 )

Where Sp = (n1 - 1)S12 + (n2 1)S2

2

n1 + n2 2Sp = 7.27t = 1.54ResultSince the calculated value of t lies in the accepted region, we may accept the nullhypothesis and conclude that there is no significant difference between the averagemarks of E and F.


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D. BUSINESS MATHS & STATISTICS and STATISTICAL INFERENCEDx1 = 88.52

S1 = 6.36n1 = 25

Dx2 = 84.24

S2 = 7.22n2 = 25


t < - tE/2 ( n1+n2 - 2)t < - t0.025(48)t < - 2.011


Sp( 1/n1 + 1/n2 )

Where Sp =

(n1 - 1)S1

2

+ (n2 1)S22

n1 + n2 2

Sp = 6.81t = 2.22ResultSince the calculated value of t lies in the critical region thus we may reject the nullhypothesis and accept the alternative and conclude that there is significantdifference between the average means of BMS and SI.


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E. BUSINESS MATHS & STATISTICS and FINANCEDx1 = 88.52

S1 = 6.36n1 = 25

Dx2 = 82.64

S2 = 7.13n2 = 25

Ho: Q1 - Q2 = 0Ha: Q1 - Q2 { 0E = 0.05E/2 = 0.025n1 + n2 2 = 48


t < - tE/2 ( n1+n2 - 2)t < - t0.025(48)t < - 2.011


Sp.( 1/n1 + 1/n2 )

Where Sp =

(n1 - 1)S1

2

+ (n2 1)S22

n1 + n2 2

Sp = 6.76t = 3.075ResultSince the calculated value of t lies in the critical region, we may reject the nullhypothesis and conclude that there is significant difference between the averagemeans of BMS and F.


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F. STATISTICAL INFERENCE and FINANCEDx1 = 84.24

S1 = 7.22n1 = 25

Dx2 = 82.64

S2 = 7.13n2 = 25


t < - tE/2 ( n1+n2 - 2)t < - t0.025(48)t < - 2.011


Sp.( 1/n1 + 1/n2 )

Where Sp =

(n1 - 1)S1

2

+ (n2 1)S22

n1 + n2 2

Sp = 7.18t = 0.788ResultSince the calculated value of t lies in the accepted region, we may accept the nullhypothesis and conclude that there is no significant difference between the averagemeans of SI and F.


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E = 0.05E/2 = 0.025n 2 = 48

Critical Regiont > tE/2(n 2)t > t0.025(48)t > 2.011

t < - tE/2(n 2)t < - t0.025(48)t < -2.011

Computation

t = r(n 2) = 0.7148(1 r2) (1 - 0.712)

t = 6.98ResultSince the calculated value of t falls in the critical region, we may reject the nullhypothesis and conclude that there is correlation between E and SI.C. ECONOMICS and FINANCEn = 50

r = 0.61Ho: V = 0Ha: V{ 0E = 0.05E/2 = 0.025n 2 = 48Critical Region

t > tE/2(n 2)t > t0.025(48)t > 2.011

t < - tE/2(n 2)t < - t0.025(48)t < -2.011


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Computation

t = r(n 2) = 0.6148(1 r2) (1 - 0.612)

t = 5.33ResultSince the calculated value of t falls in the critical region, we may reject the nullhypothesis and conclude that there is correlation between E and F.D. BUSINESS MATHS & STATISTICS and STATISTICAL INFERENCEn = 50r = 0.66

Ho: V = 0Ha: V{ 0E = 0.05E/2 = 0.025n 2 = 48Critical Regiont > tE/2(n 2)t > t0.025(48)t > 2.011

t < - tE/2(n 2)t < - t0.025(48)t < -2.011

Computation

t = r(n 2) = 0.6648(1 r2) (1 - 0.662)

t = 6.09

ResultSince the calculated value of t falls in the critical region, we may reject the nullhypothesis and conclude that there is correlation between BMS and SI.


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E. BUSINESS MATHS & STATISTICS and FINANCEn = 50r = 0.62

Ho: V = 0Ha: V{ 0E = 0.05E/2 = 0.025n 2 = 48Critical Regiont > tE/2(n 2)

t > t0.025(48)t > 2.011

t < - tE/2(n 2)

t < - t0.025(48)t < -2.011

Computation

t = r(n 2) = 0.6248(1 r2) (1 - 0.622)

t = 6.98

ResultSince the calculated value of t falls in the critical region, we may reject the nullhypothesis and conclude that there is correlation between BMS and F.F. STATISTICAL INFERENCE and FINANCEn = 50r = 0.67

Ho: V = 0Ha: V{ 0E = 0.05E/2 = 0.025n 2 = 48


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Critical Regiont > tE/2(n 2)t > t0.025(48)

t > 2.011

t < - tE/2(n 2)t < - t0.025(48)

t < -2.011

Computation

t = r(n 2) = 0.6748(1 r2) (1 - 0.672)

t = 6.25Result

Since the calculated value of t falls in the critical region thus we reject the nullhypothesis and conclude that there is correlation between SI and F.


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Question.6Simple regression lines between the following subjects and hypothesis testingof their independence:

A. ECONOMICS and BUSINESS MATHS & STATISTICSn = 50X = 4322Y = 4445XY = 386054X2 = 376656

Y2 = 397439Dx = 86.44Dy = 88.9

Ho: = 0Ha: 0

= 0.05/2 = 0.025n 2 = 48

Critical Regiont > t/2(n 2)t > t0.025(48)t > 2.011

t < - t/2(n 2)t < - t0.025(48)t < - 2.011

Computation

b = nXY - X Y = 0.60nX2 - (X)2

a = Dy bDx = 37.036Regression line:y = 37.036 + 0.60x

SY.X = Y2 - a Y - b XY =4.96n 2

Sb = _____SY.X__________= 0.09

( X2- (X)2 )n


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t = b o =6.67Sb

Result

Since the calculated value of t lies in the critical region we may reject the null hypothesisand conclude that Economics and BMS are dependent on each other.B. ECONOMICS and STATISTICAL INFERENCEn = 50X = 4322Y = 4324XY = 376046X2 = 376656

Y2 = 377322Dx = 86.44Dy = 86.48

Ho: = 0Ha: 0 = 0.05/2 = 0.025n 2 = 48Critical Regiont > t/2(n 2)

t > t0.025(48)t > 2.011

t < - t/2(n 2)

t < - t0.025(48)t < - 2.011

Computation

b = nXY - X Y = 0.75nX2 - (X)2

a = Dy bDx = 21.65

Regression line:

y = 21.65+ 0.75x

SY.X = Y2 - a Y - b XY = 5.9n 2


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Sb = _____SY.X__________= 0.11( X2- (X)2 )

nt = b o =6.82

SbResultSince the calculated value of t lies in the critical region we may reject the null hypothesisand conclude that Economics and SI are dependent on each other.C. ECONOMICS and FINANCEn = 50

X = 4322Y = 4268XY = 370589X2 = 376656

Y2 = 366714

Dx

= 86.44Dy = 85.36

Ho: = 0Ha: 0 = 0.05/2 = 0.025

n 2 = 48Critical Regiont > t/2(n 2)t > t0.025(48)t > 2.011

t < - t/2(n 2)t < - t0.025(48)t < - 2.011

Computation

b = nXY - X Y = 0.543nX2 - (X)2

a = Dy bDx = 38.42 Regression line:y = 38.42 + 0.543x


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SY.X = Y2 - a Y - b XY = 7.39n 2

Sb = _____SY.X__________= 0.10

(X2

- (X)2

)nt = b o =5.43

SbResultSince the calculated value of t lies in the critical region we may reject the null hypothesisand conclude that Economics and finance are dependent of each other.

D. BUSINESS MATHS & STATISTICS and STATISTICAL INFERENCEn = 50X = 4445Y = 4324XY = 386238X2 = 397439

Y2 = 377322Dx = 88.9Dy = 86.48

Ho: = 0Ha: 0

= 0.05/2 = 0.025n 2 = 48Critical Regiont > t/2(n 2)t > t0.025(48)t > 2.011

t < - t/2(n 2)t < - t0.025(48)t < - 2.011

Computation

b = nXY - X Y = 0.81nX2 - (X)2

a = Dy bDx = 14.47


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Regression line:y = 14.47+ 0.81x

SY.X =

Y2 - a Y - b XY = 6.29

n 2Sb = _____SY.X__________= 0.13

( X2- (X)2 )n

t = b o =6.23

SbResult

Since the calculated value of t lies in the critical region we may reject the null hypothesisand conclude that BMS and SI are dependent on each other.E. BUSINESS MATHS & STATISTICS and FINANCEn =50X = 4445Y = 4268XY = 380878

X

2

= 397439

Y2 = 366714Dx = 88.9Dy = 85.36

Ho: = 0Ha: 0 = 0.05/2 = 0.025n 2 = 48Critical Regiont > t

/2(n

2)t > t0.025(48)t > 2.011

t < - t/2(n

2)t < - t0.025(48)t < - 2.011

Computation

b = nXY - X Y = 0.64nX2 - (X)2


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a = Dy bDx = 28.464 Regression line:

y = 28.464 + 0.64x

SY.X = Y2 - a Y - b XY = 5.53n 2

Sb = _____SY.X__________= 0.12

(X2- (X)2 )n

t = b o =5.33

SbResultSince the calculated value of t lies in the critical region we may reject the null hypothesisand conclude that BMS and finance are dependent on each other.F. STATISTICAL INFERNECE and FINANCEn = 50

X = 4324Y = 4268XY = 371017X2 = 377322

Y

2

= 366714Dx = 86.48Dy = 85.36

Ho: = 0Ha: 0 = 0.05/2 = 0.025

n 2 = 48Critical Regiont > t/2(n 2)t > t0.025(48)t > 2.011

t < - t/2(n 2)t < - t0.025(48)t < - 2.011


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Question no. 7Confidence intervals for the population variances of the following subjects:A. ECONOMICS

1.95 % confidence intervaln = 25S = 7.42E = 0.05E/2 = 0.0251 - E/2 = 0.975n - 1 = 24

G2E/2(n-1) = G

20.025(24) = 39.364

G2(1 - E/2)(n-1) = G

20.975(24) =12.401

(n 1)S2 < H2 < (n 1)S2 .G2E/2(n-1) G

2(1 - E/2)(n-1)

(33.57 < H2 < 106.55)

At 5% significance level value forH2 lies between 33.57 and 106.55.

2.99 % confidence interval

n = 25S = 7.42E = 0.01

E/2 = 0.0051 - E/2 = 0.995n - 1 = 24G2E/2(n-1) = G

20.005(24) = 45.558


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G2(1 - E/2)(n-1) = G2

0.995(24) =9.886(n 1)S2 < H2 < (n 1)S2 .G2E/2(n-1) G

2(1 - E/2)(n-1)

(29 < H2 < 133.66)

At 1% significance level value forH2 lies between 29 and 133.66

B. BUSINESS MATHS & STATISTICS

1.95 % confidence intervaln = 25S = 6.36E = 0.05E/2 = 0.0251 - E/2 = 0.975n - 1 = 24G2E/2(n-1) = G

20.025(24) =39.364

G2(1 - E/2)(n-1) = G20.975(24) =12.401(n 1)S2 < H2 < (n 1)S2 .G2E/2(n-1) G

2(1 - E/2)(n-1)

(24.66 < H2 < 78.28)



n = 25S = 6.36E = 0.01E/2 = 0.005


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1 - E/2 = 0.995n - 1 = 24G2E/2(n-1) = G

20.005(24) = 45.558

G2(1 - E/2)(n-1) = G

20.995(24) =9.886

(n 1)S2 < H2 < (n 1)S2 .G2E/2(n-1) G

2(1 - E/2)(n-1)

(21.31 < H2 < 98.2)

At 1% significance level value forH2 lies between 21.31 and 98.2


1.95 % confidence intervaln = 25S = 7.22E = 0.05

E/2 = 0.0251 - E/2 = 0.975n - 1 = 24G2E/2(n-1) = G

20.025(24) = 39.364

G2(1 - E/2)(n-1) = G

20.975(24) =12.401

(n 1)S2 < H2 < (n 1)S2 .G2E/2(n-1) G2(1 - E/2)(n-1)

(31.78 < H2 < 100.89)



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n = 25S = 7.22

E = 0.01E/2 = 0.0051 - E/2 = 0.995n - 1 = 24G2E/2(n-1) = G

20.005(24) = 45.558

G2(1 - E/2)(n-1) = G

20.995(24) =9.886

(n 1)S2 < H2 < (n 1)S2 .G2E/2(n-1) G

2(1 - E/2)(n-1)

( 27.46 < H2 < 126.55)


D. FINANCE

1.95 % confidence intervaln = 25S = 7.13E = 0.05E/2 = 0.0251 - E/2 = 0.975n - 1 = 24

G2E/2(n-1) = G20.025(24) = 39.364

G2(1 - E/2)(n-1) = G

20.975(24) =12.401


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(n 1)S2 < H2 < (n 1)S2 .G2E/2(n-1) G

2(1 - E/2)(n-1)

(30.99 < H2 < 98.39)



n = 25S = 7.13E = 0.01E/2 = 0.005

1 - E/2 = 0.995n - 1 = 24G2E/2(n-1) = G

20.005(24) = 45.558

G2(1 - E/2)(n-1) = G

20.995(24) =9.886

(n 1)S2 < H2 < (n 1)S2 .

G2E/2(n-1) G

2(1 - E/2)(n-1)

(26.78< H2 < 123.42)



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Question no. 8

Hypothesis testing of the population variances for the following subjectstaking extreme values of their confidence intervals:A. ECONOMICS

n = 25S = 7.42

Confidence Interval(33.57 < H2 < 106.55)

Ho: H2 = 33.57Ha: H2 > 33.57

E = 0.05 ,1 - E = 0.95n - 1 = 24Critical Region

G2 >G2E(n-1)G2 >G20.05(24)G2 >36.415

ComputationG2 = (n 1)S2

H02

G2 = 39.36

Result

Since the calculated value ofG2 liesin the critical region, we may rejectthe null hypothesis and conclude thatthe population variance of economicsis greater than the minimum value of33.57. The lower limit of theconfidence interval is verified.

Ho: H2 = 106.55Ha: H2 < 106.55


G2 < G2(1 - E)(n-1)G2 < G20.95(24)G2 < 13.848

ComputationG2 = (n 1)S2

H02

G2 = 12.40

Result

Since the calculated value ofG2 liesin the critical region, we may rejectthe null hypothesis and conclude thatthe population variance of economicsis less than the maximum value of106.55. The upper limit of theconfidence interval is verified.


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B. BUSINESS MATHS & STATISTICSn = 25S = 6.36

Confidence Interval

(24.66 < H2

< 78.28)Ho: H2 = 24.66Ha: H2 > 24.66


G2 > G2E(n-1)G2 > G20.05(24)G2 > 36.415ComputationG2 = (n 1)S2

H02

G2 = 39.37

Result

Since the calculated value ofG2 liesin the critical region, we may rejectthe null hypothesis and conclude thatthe population variance of BMS isgreater than the minimum value of24.66.The lower limit of the

confidence interval is verified.

Ho: H2 = 78.28Ha: H2 < 78.28


G2 < G2(1 - E)(n-1)G2 < G20.95(24)G2 < 13.848ComputationG2 = (n 1)S2

H02

G2 = 12.40

Result

Since the calculated value ofG2 liesin the critical region, we may rejectthe null hypothesis and conclude thatthe population variance of BMS isless than the maximum value of78.28. The upper limit of the

confidence interval is verified.


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n = 25S = 7.22

Confidence Interval

(31.78 < H2

< 100.89)Ho: H2 = 31.78Ha: H2 > 31.78

E = 0.05 ,1 - E = 0.95n - 1 = 24

Critical RegionG2 > G2E(n-1)G2 > G20.05(24)G2 > 36.415ComputationG2 = (n 1)S2

H02G2 = 39.37

Result

Since the calculated value ofG2 liesin the critical region, we may rejectthe null hypothesis and conclude thatthe population variance of Statistical

Inference is greater than theminimum value of31.78.The lowerlimit of the confidence interval isverified.

Ho: H2 = 100.89Ha: H2 < 100.89

E = 0.051 - E = 0.95n - 1 = 24

Critical RegionG2 < G2(1 - E)(n-1)G2 < G20.95(24)G2 < 13.848ComputationG2 = (n 1)S2

H02G2 = 12.40

Result

Since the calculated value ofG2 liesin the critical region, we may rejectthe null hypothesis and conclude thatthe population variance of Statistical

Inference is less than the maximumvalue of100.89. The upper limit ofthe confidence interval is verified.


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D. FINANCE

n = 25S = 7.13

Confidence Interval

(30.99 < H2

< 98.39)Ho: H2 = 30.99Ha: H2 > 30.99


G2 > G2E(n-1)G2 > G20.05(24)G2 > 36.415ComputationG2 = (n 1)S2

H02

G2 = 39.37

Result

Since the calculated value ofG2 liesin the critical region, we may rejectthe null hypothesis and conclude thatthe population variance of finance isgreater than the minimum value of

30.99. The lower limit of theconfidence interval is verified.

Ho: H2 = 98.39Ha: H2 < 98.39


G2 < G2(1 - E)(n-1)G2 < G20.95(24)G2 < 13.848ComputationG2 = (n 1)S2

H02

G2 = 12.40

Result

Since the calculated value ofG2 liesin the critical region, we may rejectthe null hypothesis and conclude thatthe population variance of finance isless than the maximum value of

98.39. The upper limit of theconfidence interval is verified.


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Question no.9Hypothesis testing for significant difference between variance marks offollowing subjects:

A.ECONOMICS and BUSINESS MATHS & STATISTICSS1 = 7.42 2 = 6.36n1 = 25 2 = 25

Ho: H1

2 = H22

Ha: H12{H2

2

E = 0.05E/2 = 0.025

1 - E/2 = 0.975Critical RegionV1 = n1 1 = 24V2 = n2 1 = 24F0.975 (24,24) = ____1_____ = __1__ = 0.4

F0.025 (24,24) 2.2693F > FE/2 (V1,V2)

F > F0.025 (24,24)F > 2.2693

F < F(1 - E/2) (V1,V2)

F < F0.975 (24,24)F < 0.441

Computation

F = S1

2__= 7.42

2 = 1.36S2

2 6.362

Result

Since the calculated value of F lies in the accepted region, we may accept the nullhypothesis and conclude that there is no significant difference between thepopulation variances of E and BMS.


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B.ECONOMICS and STATISTICAL INFERENCE

S1 = 7.42 2 = 7.22n1 = 25 2 = 25

Ho: H12

= H22

Ha: H12{H2

2

E = 0.05E/2 = 0.0251 - E/2 = 0.975Critical RegionV1 = n1 1 = 24V

2= n

21 = 24

F0.975 (24,24) = ____1_____ = __1__ = 0.4

F0.025 (24,24) 2.2693F > FE/2 (V1,V2)F > F0.025 (24,24)F > 2.2693

F < F(1 - E/2) (V1,V2)F < F0.975 (24,24)F < 0.441

Computation

F = S1

2__= 7.42

2 = 1.056S2

2 7.222

Result

Since the calculated value of F lies in the accepted region, we may accept the nullhypothesis and conclude that there is no significant difference between thepopulation variances of E and SI.


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C.BUSINESS MATHS & STATISTICS and STATISTICAL INFERENCE

S1 = 6.36 2 = 7.22n1 = 25 2 = 25

Ho: H12

= H22

Ha: H12{H2

2

E = 0.05E/2 = 0.0251 - E/2 = 0.975Critical RegionV1 = n1 1 = 24V

2= n

21 = 24

F0.975 (24,24) = ____1_____ = __1__ = 0.4

F0.025 (24,24) 2.2693F > FE/2 (V1,V2)F > F0.025 (24,24)F > 2.2693

F < F(1 - E/2) (V1,V2)F < F0.975 (24,24)F < 0.441

Computation

F = S1

2= 6.36

2 = 0.776S2

2 7.222

Result

Since the calculated value of F lies in the accepted region, we may accept the nullhypothesis and conclude that there is no significant difference between thepopulation variances of BMS and SI.


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Question no.10Contingency tables and testing of independence of the following variables:

A. Gender & CGPA

CGPA2 2.99 3 - 4 Total

GenderMale 5 32 37

Female 0 13 13Total 5 45 50

Ho: Gender and CGPA are independent of each other.

Ha: Gender and CGPA are dependent on each other.E = 0.05r 1 = 1c 1 = 1Critical RegionG2 > G2E(r 1)(c 1)G2 > G20.05(1)G2 > 3.841

Computation

O E O-E (O-E)/E5 3.7 1.3 0.4632 33.3 -1.3 0.050 1.3 -1.3 1.3013 11.7 1.3 0.14

(O-E)/E 1.95G2 = (O E)2 = 1.95

EResultSince the value ofG2 falls in the acceptance region, we may accept the null hypothesisand conclude that gender and CGPA are independent of each other.


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C.Marks in Eco, BMS, Finance and SI & background in Math

Economics2 -2.99 3 - 4 Total

Background inMath

Yes 10 32 42No 2 6 8

Total 12 38 50Ho: Marks in Economics & background in Math are independent of each other.Ha: Marks in Economics & background in Math are dependent on each other.E = 0.05r 1 = 1c 1 = 1Critical RegionG2 > G2E(r 1)(c 1)G2 > G20.05(1)G2 > 3.841Computation

O E O-E (O-E)/E10 10.08 -0.08 0.0006

2 1.92 0.08 0.003332 31.92 0.08 0.0002

6 6.08 -0.08 0.0011

(O-E)/E 0.0052

G2 = (O E)2 = 0.0052

EResult

Since the value ofG2falls in the acceptance region, we may accept the null hypothesisand conclude that Marks in Economics and background in Math are independent of eachother.


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BMS

2 -2.99 3 - 4 Total

Background inMath

Yes 5 37 42No 2 6 8

Total 7 43 50Ho: Marks in BMS & background in Math are independent of each other.Ha: Marks in BMS & background in Math are dependent on each other.E = 0.05r 1 = 1c 1 = 1Critical Region

G2 > G2E(r 1)(c 1)G2 > G20.05(1)G2 > 3.841Computation

O E O-E (O-E)/E

5 5.88 -0.88 0.13

2 1.12 0.88 0.69

37 36.12 0.88 0.02

6 6.88 -0.88 0.11

(O-E)/E 0.96G2 = (O E)2 = 0.96

EResult

Since the value ofG

2

falls in the acceptance region, we may accept the null hypothesisand conclude that Marks in BMS and background in Math are independent of each other


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SI

2 -2.99 3 - 4 Total

Background inMath

Yes 8 34 42No 2 6 8

Total 10 40 50Ho: Marks in SI & background in Math are independent of each other.Ha: Marks in SI & background in Math are dependent on each other.E = 0.05r 1 = 1c 1 = 1Critical Region


O E O-E (O-E)/E8 8.4 -0.4 0.0192 1.6 0.4 0.100

34 33.6 0.4 0.0056 6.4 -0.4 0.025

(O-E)/E 0.149G2 = (O E)2 = 0.149

EResult

Since the value ofG2

falls in the acceptance region, we may accept the null hypothesisand conclude that Marks in SI and background in Math are independent of each other.


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Finance

2 -2.99 3 - 4 Total

Background inMath

Yes 10 32 42No 1 7 8

Total 11 39 50Ho: Marks in finance & background in Math are independent of each other.Ha: Marks in finance & background in Math are dependent on each other.E = 0.05r 1 = 1c 1 = 1Critical Region


O E O-E (O-E)/E

10 9.24 0.76 0.06

1 1.76 -0.76 0.3332 32.76 -0.76 0.02

7 6.24 0.76 0.09

(O-E)/E 0.50G2 = (O E)2 = 0.50

EResult

Since the value ofG2 falls in the acceptance region, we may accept the null hypothesisand conclude that Marks in finance and background in Math are independent of eachother.


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Question No. 11Hypothesis testing that all population means are equal or not through ANOVA tableHo: Q1 = Q2 =Q3 = Q4

Ha: At least two of the means are not equalE = 0.05Critical RegionV1 = k 1 = 3V2 = n k = 196F > FE(V1,V2)F > F0.05 (3,196)

F > 2.65Computation:

T.. 17359

T.j 75350509

Xij 1518131TSS = X.i - (T..

2/n) = 11456.595

BSS = (T.i/r) - (T..2

/n) = 335.775WSS = TSS BSS =11120.82Sb2 = BSS / k 1

= 335.775/(4-1)= 111.92

Sw = WSS / (n-k)

= 11120.82/(200-4)= 56.74

F = Sb/Sw= 111.92/56.74= 1.97


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