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Question No 1. (Marks 10) i. Find the correlation coefficients between wheat and all other independent variables and test the


Wheat Rainfall Pesticide Fertiliser 163.97632 147.69494 160.01483 115.34989 173.29319 156.14844 156.94424 155.24754 160.95077
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Answer #1

Excel > Data > Data Analysis > Regression

SUMMARY OUTPUT
Regression Statistics
Multiple R 0.281179532
R Square 0.079061929
Adjusted R Square 0.011676217
Standard Error 23.46976611
Observations 45
ANOVA
df SS MS F Significance F
Regression 3 1938.82388 646.2746268 1.173274367 0.331641182
Residual 41 22584.02676 550.829921
Total 44 24522.85064
Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0%
Intercept 156.9759072 57.47053394 2.731415501 0.009257536 40.91180936 273.0400051 40.91180936 273.0400051
Rainfall 0.10225057 0.196587703 0.52012699 0.605772383 -0.29476635 0.499267489 -0.29476635 0.499267489
Pesticide -0.232545938 0.190739308 -1.21918204 0.229743027 -0.617751785 0.152659908 -0.617751785 0.152659908
Fertilizer 0.089801399 0.169024693 0.531291596 0.598083496 -0.251550894 0.431153692 -0.251550894 0.431153692

a)

Correlation coefficient (r) = 0.2812

Hypothesis:

H0​: ρ = 0

HA​: ρ not = 0

df = n−2 = 45−2 = 43, α = 0.02

Test:

t = r*SQRT((n-2)/(1-r^2)) = 0.2812*SQRT((45-2)/(1-0.2812^2)) = 1.9215

P value = 0.0613

P value > 0.02, Do not reject H0

There is not enough evidence to conclude that significant relationship between dependent and independent variables

Correlation coefficient between wheat and rainfall is 0.1925

Hypothesis:

H0​: ρ = 0

HA​: ρ not = 0

df = n−2 = 45−2 = 43, α = 0.02

Test:

t = r*SQRT((n-2)/(1-r^2)) = 0.1925*SQRT((45-2)/(1-0.1925^2)) = 1.2864

P = 0.2052 ≥ 0.02

P value > 0.02, Do not reject H0

There is not enough evidence to conclude that significant relationship between wheat and rainfall

b)

Y = 156.9759+0.1023*Rainfall-0.2325*Pesticide+0.0898*Fertilizer

c)

IV Coefficients Standard Error t Stat P-value alpha Significance
Rainfall 0.10225057 0.196587703 0.52012699 0.605772383 > 0.01 No
Pesticide -0.23254594 0.190739308 -1.21918204 0.229743027 > 0.01 No
Fertilicer 0.089801399 0.169024693 0.531291596 0.598083496 > 0.01 No

Overall F test

F stat = 1.1733

P value = 0.3316

P value > 0.01, Overall model is not significant

R^2 = 0.0791 and it close to 0, which means model does not fit

d)

Water = 111, Fertilizer = 113 and pesticide = 12

Y = 156.9759+0.1023*Rainfall-0.2325*Pesticide+0.0898*Fertilizer

Y = 156.9759+0.1023*111-0.2325*12+0.0898*113 = 175.6886

Observation Predicted Wheat Residuals
1 145.2255817 18.75073825
2 150.386875 22.90631504
3 155.6244229 5.326347117
4 143.0967454 -15.26429536
5 143.3028411 -20.87583111
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