bs2506 tutorial3 corrected - wordpress.com · 1. house price (again) predictor (variable)...
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![Page 1: bs2506 tutorial3 corrected - WordPress.com · 1. House price (Again) Predictor (Variable) Coefficient (B) SE (B) Constan -2.5 41.4 X1 1.62 0.21 X2 0.257 1.88 X4 -0.027 0.008 Source](https://reader036.vdocuments.net/reader036/viewer/2022070804/5f0344107e708231d4085c11/html5/thumbnails/1.jpg)
Tutorial 3
Inferential Statistics, Statistical Modelling & Survey Methods
(BS2506)
Pairach Piboonrungroj (Champ)
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1. House price (Again) Predictor (Variable)
Coefficient (B) SE (B)
Constant -2.5 41.4 X1 1.62 0.21 X2 0.257 1.88 X4 -0.027 0.008
Source of variation Sum of Squares Degree of Freedom Mean Squares
Regression 277,895
Residual 34,727
Analysis of Variance (ANOVA)
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1 (a) (i) Write out the estimated regression equation
421 027.0257.062.15.2ˆ XXXY −++−=
Predictor (Variable)
Coefficient (B) SE (B)
Constant -2.5 41.4 X1 1.62 0.21 X2 0.257 1.88 X4 -0.027 0.008
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1 (a) (ii) Test for the significance of regression equation
1058.311,005.0415,201.0,2
===−
tttdfα
01.0=αAt 1% Step1: Critical Value
Step2: t-Statistic i
i SEt i
ββ
β=
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1 (a) (ii) Test for the significance of regression equation
1058.311,005.0 =tAt 1% Step1: Critical Value
Step2: t-Statistic
i
i SEt i
ββ
β=
01.0=α
71.721.062.1
1 ==t
137.088.1257.0
2 ==t
375.3008.0027.0
4 −=−
=t
Reject H0
Do NOT Reject H0
Reject H0
> 3.1058
< 3.1058
< -3.1058
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1. a). (iii) What are DF for SSR & SSE? Predictor (Variable)
Coefficient (B) SE (B)
Constant -2.5 41.4 X1 1.62 0.21 X2 0.257 1.88 X4 -0.027 0.008
Source of variation Sum of Squares Degree of Freedom Mean Squares
Regression 277,895 3 (p)
Residual 34,727 11 (n-p-1)
Analysis of Variance (ANOVA)
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1. a). (iv) Test for Significant relationship X&Y?
Source of variation
Sum of Squares
Degree of Freedom
Mean Squares F Statistic
Regression 277,895 3 92,631 29.341
Residual 34,727 11 3157
Analysis of Variance (ANOVA)
0421 === βββH0:
H1: At least one of the coefficients does not equal 0
217.6)11,3(01.0 =FAt Critical Value 01.0=α
Then we can reject Null hypothesis, there is a relationship between Xs & Y
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1. a). (v) Compute the coefficient of determination and explain its meaning
Source of variation
Sum of Squares
Degree of Freedom
Mean Squares F Statistic
Regression 277,895 3 92,631 29.341
Residual 34,727 11 3157
TOTAL 312,622
Analysis of Variance (ANOVA) R2
R2 = 1 – (34,727/312,622) R2 = 1 – 0.111 R2 = 0.889 = 88.9%
Total Squares SumError Square Sum1−=
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1(b)
41 026.0601.18.1ˆ xxy −+=
880.02 =R
Model 1
6541 371.65794.63026.023.105.64ˆ xxxxy −+−+=
Model 2
935.02 =R
65421 447.65447.63026.0067.022.12.65ˆ xxxxxy −+−−+=Model 3
936.02 =R
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1(b) (i) Compute Adjusted Coefficient of determination for three models
)11)(1(1 222
−−
−−−==
pnnRRRadj
86.0)1215115)(880.01(12
1 =−−
−−−=R
909.0)1415115)(935.01(12
2 =−−
−−−=R
900.0)1515115)(936.01(12
3 =−−
−−−=R
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1(b) (ii) Interpret the coefficients on the house type, Beta5 and Beta6
Prices for Detached houses increase by £63,794
Prices for Terrace Houses decreased by £65,371
(relative to Semi- detached)
6541 371.65794.63026.023.105.64ˆ xxxxy −+−+=(model 2)
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1(b) (iii) At 0.05 level of significance, determine whether model 2 is superior to model1
6541 371.65794.63026.023.105.64ˆ xxxxy −+−+=Model 2 41 026.0601.18.1ˆ xxy −+=Model 1
qppn
RRR
FComplete
strictedComplete
−
−−×
−
−=
11 2
2Re
2
231.4241415
935.01880.0935.0
=−
−−×
−
−=F
231.4103.410,2,05.0)1415,24(,05.0)1,(, <=== −−−−−− FFF pnqpα
Significant i.e., Model 2 is better than Model 1
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1(b) (iv) At 0.05 level of significance, determine whether model 3 is superior to model 2
6541 371.65794.63026.023.105.64ˆ xxxxy −+−+=Model 2
qppn
RRR
FComplete
strictedComplete
−
−−×
−
−=
11 2
2Re
2
141.0451515
936.01935.0936.0
=−
−−×
−
−=F
141.0117.59,1,05.0)1515,45(,05.0)1,(, >=== −−−−−− FFF pnqpα
NOT Significant i.e., Model 3 is NOT better than Model 2
65421 447.65447.63026.0067.022.12.65ˆ xxxxxy −+−−+=Model 3
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6541 371.65794.63026.023.105.64ˆ xxxxy −+−+=
0*371.651*794.63)5*250(026.0250*23.105.64ˆ −+−+=y
844,402£ˆ =y
1(b) (v) From model2, estimate the price of 5 years old detached house with 250 square meters
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2. Advertising expenditure X, Advertising
(£000) Y, Sales
(£000) 5.5 90
2.0 40
3.2 55
6.0 95
3.8 70
4.4 80
6.0 5.0 6.5 7.0
88 85 92 91
R square 0.97 Adjusted R Square 0.96 Standard error of regression 3.37
DF Sum Square Mean Square Regression 2,904 Residual 80.0
Analysis of variance
Variables in the Equation Variable B SE B
Advert 31.79 4.48 Advert-square -2.30 0.485 (constant) -17.22 9.65
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2.(a) State the regression equation for the curvilinear model.
230.279.3122.17ˆ XXYt −+−=
Variables in the Equation Variable B SE B
Advert 31.79 4.48 Advert-square -2.30 0.485 (constant) -17.22 9.65
2210
ˆ XXYt βββ −+=
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2.(b) Predict the monthly sales (in pounds) for a month with total advertising
expenditure of £6,000 230.279.3122.17ˆ XXYt −+−=
X = 6
Yt = −17.22+31.79(6)− 2.30(6)2 = 90.720
720,90£000,1*720.90 ==Sales
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2.(c) Determine there is significant relationship between the sales and advertising expenditure at
the 0.01 level of significance
DF Sum Square Mean Square F
Regression 2 2,904 1,452 127.05
Residual 7 80.0 11.428
Analysis of variance
547.5)7,2(01.0 =FAt Critical Value 01.0=α
Then we can reject Null hypothesis, there is a curvilinear relationship between sales and advertising expenditure
021 == ββH0:
H1: At least one of the coefficients does not equal 0
2210
ˆ XXYt βββ −+=
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2 (d) Fit a linear model to the data and calculate SSE for this model
∑∑
−
−= 221
ˆxnxyxnxy
β
xy 10ˆˆ ββ −=
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2 (d) Fit a linear model to the data and calculate SSE for this model
ID X
Advertising Y
Sales
1 5.5 90
2 2 40
3 3.2 55
4 6 95
5 3.8 70
6 4.4 80
7 6 88
8 5 85
9 6.5 92
10 7 91
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2 (d) Fit a linear model to the data and calculate SSE for this model
ID X
Advertising Y
Sales xy x^2 y^2
1 5.5 90 495 30.25 8100 2 2 40 80 4 1600 3 3.2 55 176 10.24 3025 4 6 95 570 36 9025 5 3.8 70 266 14.44 4900 6 4.4 80 352 19.36 6400 7 6 88 528 36 7744 8 5 85 425 25 7225 9 6.5 92 598 42.25 8464
10 7 91 637 49 8281
Sum 49.4 786 4127 266.54 64764
Average 4.94 78.6 412.7 26.654 6476.4
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2 (d) Fit a linear model to the data and calculate SSE for this model
∑∑
−
−= 221
ˆxnxyxnxy
β 85.10)94.4(1054.266)6.78)(94.4(104127ˆ
21 =−
−=β
xy 10ˆˆ ββ −= 0.25)94.4(85.106.78ˆ
0 =−=β
xy 85.100.25ˆ +=
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2 (d) Fit a linear model to the data and calculate SSE for this model
ID X
Advertising Y
Sales xy x^2 y^2
1 5.5 90 495 30.25 8100 2 2 40 80 4 1600 3 3.2 55 176 10.24 3025 4 6 95 570 36 9025 5 3.8 70 266 14.44 4900 6 4.4 80 352 19.36 6400 7 6 88 528 36 7744 8 5 85 425 25 7225 9 6.5 92 598 42.25 8464
10 7 91 637 49 8281
Sum 49.4 786 4127 266.54 64764
Average 4.94 78.6 412.7 26.654 6476.4
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2 (d) Fit a linear model to the data and calculate SSE for this model
ID X
Advertising Y
Sales xy x^2 y^2 predicted Y
1 5.5 90 495 30.25 8100 84.68 2 2 40 80 4 1600 46.70 3 3.2 55 176 10.24 3025 59.72 4 6 95 570 36 9025 90.10 5 3.8 70 266 14.44 4900 66.23 6 4.4 80 352 19.36 6400 72.74 7 6 88 528 36 7744 90.10 8 5 85 425 25 7225 79.25 9 6.5 92 598 42.25 8464 95.53
10 7 91 637 49 8281 100.95
Sum 49.4 786 4127 266.54 64764
Average 4.94 78.6 412.7 26.654 6476.4
XYt 85.1025ˆ +=
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2 (d) Fit a linear model to the data and calculate SSE for this model
ID X
Advertising Y
Sales xy x^2 y^2 predicted Y
Square Error
1 5.5 90 495 30.25 8100 84.68 28.35 2 2 40 80 4 1600 46.70 44.92 3 3.2 55 176 10.24 3025 59.72 22.29 4 6 95 570 36 9025 90.10 24.00 5 3.8 70 266 14.44 4900 66.23 14.20 6 4.4 80 352 19.36 6400 72.74 52.69 7 6 88 528 36 7744 90.10 4.41 8 5 85 425 25 7225 79.25 33.05 9 6.5 92 598 42.25 8464 95.53 12.43
10 7 91 637 49 8281 100.95 99.01
Sum 49.4 786 4127 266.54 64764
Average 4.94 78.6 412.7 26.654 6476.4
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2 (d) Fit a linear model to the data and calculate SSE for this model
ID X
Advertising Y
Sales xy x^2 y^2 predicted Y
Square Error
1 5.5 90 495 30.25 8100 84.68 28.35 2 2 40 80 4 1600 46.70 44.92 3 3.2 55 176 10.24 3025 59.72 22.29 4 6 95 570 36 9025 90.10 24.00 5 3.8 70 266 14.44 4900 66.23 14.20 6 4.4 80 352 19.36 6400 72.74 52.69 7 6 88 528 36 7744 90.10 4.41 8 5 85 425 25 7225 79.25 33.05 9 6.5 92 598 42.25 8464 95.53 12.43
10 7 91 637 49 8281 100.95 99.01
Sum 49.4 786 4127 266.54 64764 335.36 Average 4.94 78.6 412.7 26.654 6476.4
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2(e) At 0.01 level of significance, determine whether the curvilinear model is superior to the
linear regression model
Linear Regression Model
Curvilinear Model
qppn
SSESSESSEF
rCurvilinea
rCurvilineaLinear
−
−−×
−=
1
3125.22121210
8080335
=−
−−×
−=F
3.2225.127,1,01.0)1210,12(,01.0)1,(, <=== −−−−−− FFF pnqpα
Significant i.e., Curvilinear effect make significant contribution and should be included in the model.
230.279.3122.17ˆ XXYt −+−=XYt 85.1025ˆ +=
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2 (f) Draw a scatter diagram between the sales& Advertising expenditure.
Sales
0
1020
3040
50
6070
8090
100
0 1 2 3 4 5 6 7 8
Observed
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2 (f) Sketch the Linear regression
Sales
0
1020
3040
50
6070
8090
100
0 1 2 3 4 5 6 7 8
ObservedLinear Regression
XYt 85.1025ˆ +=
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2 (f) Sketch the Quadratic regression
Sales
0
1020
3040
50
6070
8090
100
0 1 2 3 4 5 6 7 8
ObservedLinear Regression
Quadratic Regression
230.279.3122.17ˆ XXYt −+−=
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