Student Name : Zainab Kashani

Student ID : 63075

Quantitative Techniques for Management Sciences

Assignment 1

Date of Submission : Sep 16 , 2020

1. Consider the following regression results

Where, Y = labour force participation rate of women in 1972 and X = labour

force participation rate of women in 1968. The regression results were obtained
from a sample of 19 cities in the United States.
a. Interpret the regression coefficients.

Solution: It has a positive relationship between x axis and y axis because when
the x-axis increases the Y-axis also increases positively

b. Are the regression coefficients statistically significant at the 5% level? Show

necessary calculations. (Hint: Test H0:B1=0 and H0: B2=0 through the t-test
approach).
Cofficient Estimated−Null
Solution: For the t approach=T 1=
Standard error

0.2033−0
For the t approach=T 1= =2.089
0.0976

0.6560−0
For the t approach=T 2= =3.346
0.9161

Df =N−K
Df =19−2=17

According to the T table at 17 with it’s two tail so at 0.05 significant CV at 17 will be
+2.110 and -2.110
The T1 value lies 2.089 as it is laying intside the CV so we accept the null.
The T2 value lies 3.346 as it is laying outside the CV so we reject the null.

c. Explain the value of r-squared.

Solution: TSS=0.0544+ 0.0358=0.0902
ESS 1−RSS
R 2= =
TSS TSS
0.0358 1−0.0544
R 2= =
0.0902 0.0902

0.0358 1−0.0544
R 2= =
0.0902 0.0902

0.9456
R2=0.397=
0.0902

R2=0.397=10.49

R2=0.0398

R2=0.397=10.49

R2=0.0398 Poor lift

d. Test the null hypothesis H0: B2=1 using the t-test at the 1% level of significance.
Solution:

Cofficient Estimated−HO
For the t approach=T 1=
Standard error
 0.6560−1
T 1=
0.1961

−0.344
T 1=
0.1961

T 1=−1.754

Df =N−K
Df =19−2=17

+2.567
The T1 value lies -1.756 as it is laying outside the CV so we reject the HO

2. Table 5.5 gives data on the average public teacher pay (annual salary in dollars) and
spending on public schools per pupil (dollars) for several states in the US.

a. Prepare a scatter plot with salary on the Y axis and spending on the X axis. Also
fit a regression line through the scatter plot.
40000
35000
30000
25000
20000

2000 4000 6000 8000

SPENDING

SALARY Fitted values

 b. Generate and discuss the descriptive statistics of the variables.

. summarize salary spending

Variable Obs Mean Std. Dev. Min Max

salary 51 24356.22 4179.426 18095 41480

spending 51 3696.608 1054.761 2297 8349

c. Generate and discuss the correlations between the variables.

Solution

. p wc or sa la ry s pe nd in g

sa la ry s pe nd in g

s al ar y 1. 0 0
s pe nd in g 0. 83 47 1 .0 0

d. Estimate the regression of salary on spending and interpret the results.

Solution
. regress salary spending

Source SS df MS Number of obs = 51

F( 1, 49) = 112.60
Model 608555015 1 608555015 Prob > F = 0.0000
Residual 264825250 49 5404596.94 R-squared = 0.6968
Adj R-squared = 0.6906
Total 873380265 50 17467605.3 Root MSE = 2324.8

salary Coef. Std. Err. t P>|t| [95% Conf. Interval]

spending 3.307585 .3117043 10.61 0.000 2.681192 3.933978

_cons 12129.37 1197.351 10.13 0.000 9723.204 14535.54
Interpretation of constant
The intercept is 12129.37 it means that spending is positive. It makes a logical sense
by showing a positive relation among them

Interpretation of slope coefficient:

The intercept of slope coefficient is 3.307585.
This indicates that salary is increased by 1 rupee and spending increase by
3.307585. there is a positive relationship among them and it makes a logical sense as
well as among them.

e. Generate the ANOVA table and explain how it can be used.

ANOVA table is use to identify how different groups respond with a null
hypotheses. When there are statistically significant results it means that the two
groups are different

Solution
. regress salary spending

Source SS df MS Number of obs = 51

F( 1, 49) = 112.60
Model 608555015 1 608555015 Prob > F = 0.0000
Residual 264825250 49 5404596.94 R-squared = 0.6968
Adj R-squared = 0.6906
Total 873380265 50 17467605.3 Root MSE = 2324.8

f. Use the 95% confidence interval to test the null hypothesis that the slope
coefficient is equal to zero.
 Solution :We reject the null hypothesis because the p value less than 0.05 at the 5%
level of significant.

g. Use the 99% confidence interval to test the null hypothesis that the slope
coefficient is equal to one.

Solution: We reject the null hypothesis as the p value is less than 0.01.

. lincom spending-1, level(99)

( 1) spending = 1

salary Coef. Std. Err. t P>|t| [99% Conf. Interval]

(1) 2.307585 .3117043 7.40 0.000 1.472233 3.142937

3. Table 7.7 provides data on a number of variables.

Y = number of oil wells drilled

X2= price of the wellhead in the previous period
X3= domestic output
X4= GNP in dollars
X5= trend variable

a. Generate descriptive statistics and the pairwise correlations.

 Solution:
. summarize y x2 x3 x4 x5

Variable Obs Mean Std. Dev. Min Max

y 31 10.64613 2.351515 6.92 16.17

x2 31 4.449677 .8135334 3.39 6.12
x3 31 7.524839 1.126851 5.05 9.18
x4 31 882.789 271.3539 487.67 1385.1
x5 31 16 9.092121 1 31

b. Estimate a multiple regression of Y on the X variables. Also interpret the regression

results in detail.
Solution :
. pwcorr y x2 x3 x4 x5

y x2 x3 x4 x5

y 1.0000
x2 0.1352 1.0000
x3 -0.4266 -0.3054 1.0000
x4 -0.5574 0.1820 0.8271 1.0000
x5 -0.5299 0.1609 0.8481 0.9906 1.0000

. regress y x2 x3 x4 x5

Source SS df MS Number of obs = 31

F( 4, 26) = 8.99
Model 96.277981 4 24.0694953 Prob > F = 0.0001
Residual 69.6107522 26 2.67733662 R-squared = 0.5804
Adj R-squared = 0.5158
Total 165.888733 30 5.52962444 Root MSE = 1.6363

y Coef. Std. Err. t P>|t| [95% Conf. Interval]

x2 2.701012 .6957689 3.88 0.001 1.270839 4.131186

x3 3.059606 .9373141 3.26 0.003 1.132929 4.986282
x4 -.0160601 .0081788 -1.96 0.060 -.032872 .0007517
x5 -.0227016 .2723061 -0.08 0.934 -.5824348 .5370317
_cons -9.854598 8.895195 -1.11 0.278 -28.13893 8.429736
 Interpretation of regression of coefficient B1:
The intercept is -9.854598 it indicates that when the price of the wellhead in the
previous period, domestic output, GNP in Dollars and trend variable are equal to zero
number of oil wells drilled will be -9.854598.
And it doesnot make any logical sense in it .
Interpretation of regression of coefficient B2:
value of regression coefficient is 2.701012
It means when there is $100 increase in number of oil wells drilled, the price of the
wellhead in the previous period also increase by 2.701012
This shows a logical sense among them as when there oil wells drilled the price of
wellhead also increased.
Interpretation of regression of coefficient B3:
value of regression is 3.059606.
This means that when there is increase of $100 in oil wells drilled the domestic
output is also increasing by 3.059606 which makes the independent variable constant
and it shows a logical sense as well
Interpretation of regression of coefficient B4:
Value of regression is -0.0160601.
The GNP is negative this means that when the oil wells drilled is zero then the GNP
will be negative by putting all the independent variables zero and it makes no sense
at all because the gnp can never be zero
Interpretation of regression of coefficient B5:
Value of regression is -0.0227016.
This means that the trend variables must be also negative because oil wells drilled is
zero by putting all the independent variables at constant.
c. Test the overall significance of the regression model using the F-test.
Solution : The overall significance of regression value of F statistic is 8.99 and
the p value is 0.0001.
It means independent variables has no affect on the dependent variable.
 We reject the null hypothesis so the overall model is statistically significant
d. Comment on the goodness of fit of the regression model. How would you explain
adjusted r-squared.
The regression result shows that the value of R square is 0.5804
This value implies that 58.04% variation of dependent variable number of oil wells
drilled is explained by the independent variables.
As we can see that R square value is somehow close to 1 so it seems a good fit.
Adjusted R value is 0.5158.
This value implies that 51.58% which is closer to 1. It’s a good fit
e. Use the residuals of the regression to examine the hypothesis that the error term is
normally distributed. Explain your answer.

We accept the null hypothesis because it p value is greater than 0.05. We use
normally test because this is assumption and in this residuals must be distributed
normally so that it can make the assumption valid.

Shapiro-Wilk W test for normal data

Variable Obs W V z Prob>z

U 31 0.97204 0.91 -0.194 0.57690