Upload
mirzahussain10
View
219
Download
0
Embed Size (px)
Citation preview
8/7/2019 ARM Project(Sample)
1/67
ARM - Final Project
Page 1
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
8/7/2019 ARM Project(Sample)
2/67
ARM - Final Project
Page 2
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
8/7/2019 ARM Project(Sample)
3/67
ARM - Final Project
Page 3
= 88.90 (1.96 x 6.82/50)= 88.90 1.891
= 88.90 1.891 & = 88.90 + 1.891
= 87.009 = 90.791Confidence Interval(87.009 < < 90.791)
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
8/7/2019 ARM Project(Sample)
4/67
ARM - Final Project
Page 4
= 85.36 (1.96 x 6.99/50)= 85.36 1.94
= 85.36 1.94 & = 85.36 + 1.94
= 83.42 = 87.3Confidence Interval(83.42 < < 87.3)
8/7/2019 ARM Project(Sample)
5/67
ARM - Final Project
Page 5
b) Using first 25 as sample size (t-test)Note: is unknown and n
8/7/2019 ARM Project(Sample)
6/67
ARM - Final Project
Page 6
= 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)
8/7/2019 ARM Project(Sample)
7/67
ARM - Final Project
Page 7
= 82.64 2.94= 82.64 2.94 & = 82.64 + 2.94= 79.7 = 85.58
Confidence Interval(79.7 < < 85.58)
8/7/2019 ARM Project(Sample)
8/67
ARM - Final Project
Page 8
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
8/7/2019 ARM Project(Sample)
9/67
ARM - Final Project
Page 9
(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
8/7/2019 ARM Project(Sample)
10/67
ARM - Final Project
Page 10
= 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
8/7/2019 ARM Project(Sample)
11/67
ARM - Final Project
Page 11
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
8/7/2019 ARM Project(Sample)
12/67
ARM - Final Project
Page 12
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
8/7/2019 ARM Project(Sample)
13/67
ARM - Final Project
Page 13
= 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
8/7/2019 ARM Project(Sample)
14/67
ARM - Final Project
Page 14
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
8/7/2019 ARM Project(Sample)
15/67
ARM - Final Project
Page 15
= 4.28 - 3.87 & = 4.28 + 3.87= 0.41 =8.15
Confidence Interval (0.41
8/7/2019 ARM Project(Sample)
16/67
ARM - Final Project
Page 16
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
8/7/2019 ARM Project(Sample)
17/67
ARM - Final Project
Page 17
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
Confidence Interval(84.383 < < 88.497)
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.
8/7/2019 ARM Project(Sample)
18/67
ARM - Final Project
Page 18
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.
8/7/2019 ARM Project(Sample)
19/67
ARM - Final Project
Page 19
C.STATISTICAL INFERENCE
Dx = 86.48S = 8.31n = 50
Confidence Interval(84.18 < < 88.78)Ho: Q = 84.18Ha: Q > 84.18E = 0.05Critical Regionz > zE
z > z0.05z > 1.65Computationz = Dx - Qo
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.
Ho: Q = 88.78Ha: Q < 88.78E = 0.05Critical Regionz < -zE
z < -z0.05z < -1.65Computationz = Dx - Qo
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.
8/7/2019 ARM Project(Sample)
20/67
ARM - Final Project
Page 20
D.FINANCEDx = 85.36
S = 6.99n = 50
Confidence Interval(83.42 < < 87.3)Ho: Q = 83.42Ha: Q > 83.42E = 0.05Critical Regionz > zE
z > z0.05z > 1.65Computationz = Dx - Qo
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.
Ho: Q = 87.3Ha: Q < 87.3E = 0.05Critical Regionz < -zE
z < -z0.05z < -1.65Computationz = Dx - Qo
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.
8/7/2019 ARM Project(Sample)
21/67
ARM - Final Project
Page 21
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.
8/7/2019 ARM Project(Sample)
22/67
ARM - Final Project
Page 22
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.
8/7/2019 ARM Project(Sample)
23/67
ARM - Final Project
Page 23
C. STATISTICAL INFERENCE
Dx = 84.24S = 7.22n = 25
Confidence Interval(81.26 < < 87.22)
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.
8/7/2019 ARM Project(Sample)
24/67
ARM - Final Project
Page 24
D. FINANCEDx = 82.64
S = 7.13n = 25
Confidence Interval(79.7 < < 85.58)
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.
8/7/2019 ARM Project(Sample)
25/67
ARM - Final Project
Page 25
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.
8/7/2019 ARM Project(Sample)
26/67
ARM - Final Project
Page 26
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
8/7/2019 ARM Project(Sample)
27/67
ARM - Final Project
Page 27
Critical 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 = 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
8/7/2019 ARM Project(Sample)
28/67
ARM - Final Project
Page 28
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.
8/7/2019 ARM Project(Sample)
29/67
ARM - Final Project
Page 29
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
Computationz = (Dx1 - Dx2 ) (Q1 - Q2 )
( 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.
8/7/2019 ARM Project(Sample)
30/67
ARM - Final Project
Page 30
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.
8/7/2019 ARM Project(Sample)
31/67
ARM - Final Project
Page 31
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
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
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.
8/7/2019 ARM Project(Sample)
32/67
ARM - Final Project
Page 32
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
Computationt = (Dx1 - Dx2 ) (Q1 - Q2 )
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.
8/7/2019 ARM Project(Sample)
33/67
ARM - Final Project
Page 33
D. BUSINESS MATHS & STATISTICS and STATISTICAL INFERENCEDx1 = 88.52
S1 = 6.36n1 = 25
Dx2 = 84.24
S2 = 7.22n2 = 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
Computationt = (Dx1 - Dx2 ) (Q1 - Q2 )
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.
8/7/2019 ARM Project(Sample)
34/67
ARM - Final Project
Page 34
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
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
Computationt = (Dx1 - Dx2 ) (Q1 - Q2 )
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.
8/7/2019 ARM Project(Sample)
35/67
ARM - Final Project
Page 35
F. STATISTICAL INFERENCE and FINANCEDx1 = 84.24
S1 = 7.22n1 = 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
Computationt = (Dx1 - Dx2 ) (Q1 - Q2 )
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.
8/7/2019 ARM Project(Sample)
36/67
8/7/2019 ARM Project(Sample)
37/67
ARM - Final Project
Page 37
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
8/7/2019 ARM Project(Sample)
38/67
ARM - Final Project
Page 38
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.
8/7/2019 ARM Project(Sample)
39/67
ARM - Final Project
Page 39
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
8/7/2019 ARM Project(Sample)
40/67
ARM - Final Project
Page 40
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.
8/7/2019 ARM Project(Sample)
41/67
ARM - Final Project
Page 41
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
8/7/2019 ARM Project(Sample)
42/67
ARM - Final Project
Page 42
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
8/7/2019 ARM Project(Sample)
43/67
ARM - Final Project
Page 43
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
8/7/2019 ARM Project(Sample)
44/67
ARM - Final Project
Page 44
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
8/7/2019 ARM Project(Sample)
45/67
ARM - Final Project
Page 45
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
8/7/2019 ARM Project(Sample)
46/67
ARM - Final Project
Page 46
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
8/7/2019 ARM Project(Sample)
47/67
8/7/2019 ARM Project(Sample)
48/67
ARM - Final Project
Page 48
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
8/7/2019 ARM Project(Sample)
49/67
ARM - Final Project
Page 49
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)
At 5% significance level value forH2 lies between 24.66 and 78.28.
2.99 % confidence interval
n = 25S = 6.36E = 0.01E/2 = 0.005
8/7/2019 ARM Project(Sample)
50/67
ARM - Final Project
Page 50
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
C. STATISTICAL INFERENCE
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)
At 5% significance level value forH2 lies between 31.78 and 100.89.
8/7/2019 ARM Project(Sample)
51/67
ARM - Final Project
Page 51
2.99 % confidence interval
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)
At 1% significance level value forH2 lies between 27.46 and 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
8/7/2019 ARM Project(Sample)
52/67
ARM - Final Project
Page 52
(n 1)S2 < H2 < (n 1)S2 .G2E/2(n-1) G
2(1 - E/2)(n-1)
(30.99 < H2 < 98.39)
At 5% significance level value forH2 lies between 30.99 and 98.39.
2.99 % confidence interval
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)
At 1% significance level value forH2 lies between 26.78 and 123.42.
8/7/2019 ARM Project(Sample)
53/67
ARM - Final Project
Page 53
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
E = 0.05 ,1 - E = 0.95n - 1 = 24Critical Region
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.
8/7/2019 ARM Project(Sample)
54/67
ARM - Final Project
Page 54
B. BUSINESS MATHS & STATISTICSn = 25S = 6.36
Confidence Interval
(24.66 < H2
< 78.28)Ho: H2 = 24.66Ha: H2 > 24.66
E = 0.05 ,1 - E = 0.95n - 1 = 24Critical Region
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
E = 0.05 ,1 - E = 0.95n - 1 = 24Critical Region
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.
8/7/2019 ARM Project(Sample)
55/67
ARM - Final Project
Page 55
C. STATISTICAL INFERENCE
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.
8/7/2019 ARM Project(Sample)
56/67
ARM - Final Project
Page 56
D. FINANCE
n = 25S = 7.13
Confidence Interval
(30.99 < H2
< 98.39)Ho: H2 = 30.99Ha: H2 > 30.99
E = 0.05 ,1 - E = 0.95n - 1 = 24Critical Region
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
E = 0.05 ,1 - E = 0.95n - 1 = 24Critical Region
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.
8/7/2019 ARM Project(Sample)
57/67
ARM - Final Project
Page 57
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.
8/7/2019 ARM Project(Sample)
58/67
ARM - Final Project
Page 58
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.
8/7/2019 ARM Project(Sample)
59/67
ARM - Final Project
Page 59
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.
8/7/2019 ARM Project(Sample)
60/67
ARM - Final Project
Page 60
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.
8/7/2019 ARM Project(Sample)
61/67
8/7/2019 ARM Project(Sample)
62/67
ARM - Final Project
Page 62
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.
8/7/2019 ARM Project(Sample)
63/67
ARM - Final Project
Page 63
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
8/7/2019 ARM Project(Sample)
64/67
ARM - Final Project
Page 64
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
G2 > G2E(r 1)(c 1)G2 > G20.05(1)G2 > 3.841Computation
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.
8/7/2019 ARM Project(Sample)
65/67
ARM - Final Project
Page 65
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
G2 > G2E(r 1)(c 1)G2 > G20.05(1)G2 > 3.841Computation
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.
8/7/2019 ARM Project(Sample)
66/67
ARM - Final Project
Page 66
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
8/7/2019 ARM Project(Sample)
67/67