Statistical Project Assignment for ECON7300

Statistical Project Assignment
ECON7300
STUDENT NAME/ID
[Pick the date]

STUDENT NAME:
STUDENT NUMBER:
(a) Regression model
Independent variable: Grade (X1) and hours (X2)
Dependent variable: Wage (Y)
Multiple regression equation
Y =−4.180+ ( 0.699∗X 1 ) +(0.077∗X 2)
Wage=−4.180+ ( 0.699∗Grade )+(0.077∗Hours)
(b) The grade slope implies that the wage would increase by $ 0.699 if there is an increase in the
grade by 1 unit. Also, the hours slope implies that the wage increases by $ 0.077 if there is an
increase in hour by 1 unit.
(c) The wage (Y) when grade (X1) =12 and hours (X2) =34
Y =−4.180+ ( 0.699∗X 1 ) +(0.077∗X 2)
Y =−4.180+ ( 0.699∗12 ) + ( 0.077∗34 )=6.8131
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STUDENT NAME:
STUDENT NUMBER:
(d) 95% confidence interval of the mean Y or employed women when X1 =12, X2 =34
Lower value of theinterval=−5.748+ ( 0.598 X 1 ) + ( 0.053 X 2 )
Upper value of theinterval=−2.612+ ( 0.800 X 1 ) + ( 0.101 X 2 )
Now, X1 =12, X2 =34
Lower limit of 95% confidence interval = −5.748+ ( 0.598∗12 ) + ( 0.053∗34 ) =3.216
Upper limit of 95% confidence interval = −2.612+ ( 0.800∗12 ) + ( 0.101∗34 ) =10.411
Therefore, 95% confidence interval [3.216 10.411]
(e) 95% prediction interval of the mean Y or employed women when X1 =12, X2 =34
Predicted y=−4.180+ ( 0.699∗12 ) + ( 0.077∗34 )=6.8131
Margin of error (M.E.) = t value * Standard error of prediction
Upper limit of 95% prediction interval = Predicted y value + M.E.
Lower limit of 95% prediction interval = Predicted y value - M.E.
Hence, the 95% prediction interval = [-3.9872 17.6136]
(f) Residual plots
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