ECON 705 Assignment 3: Enrollment, Fees, and Elasticity Analysis

This economics assignment, focusing on ECON 705, analyzes the relationship between undergraduate enrollment, credit hour production, and tuition fees using provided data. The assignment begins with calculating the average elasticity of undergraduate enrollment and credit hours with respect to tuition fees, revealing a negative correlation. It then presents a demand curve analysis, calculating price elasticity at various points and determining optimal pricing strategies for revenue maximization. Furthermore, the assignment incorporates regression analysis, exploring the correlation between seasonal and unseasonal variables, constructing and evaluating several regression models to identify the best fit for the data. Finally, the assignment concludes by examining the impact of price, income, and related goods' price changes on the quantity demanded, calculating price, income, and cross-price elasticities, and assessing how changes in these variables affect quantity demanded.

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Question 1
The requisite computation is highlighted below.
The appropriate symbols for the above computation are highlighted below.
LN denotes natural log
UE indicates undergrad enrolment
CH indicates total LSUS credit hour production
Fee indicates undergrad tuition and fees
The average elasticity of UE variable with regards to fee based on the given computations
comes out as -0.203. Also, the average elasticity of CH variable with regards to fee based on
the given computations comes out as -0.039. This clearly highlights that there is a negative
relationship between fees and undergrad enrolment along with credit hour production. It can
be estimated that an increase in the tuition fee by 1% would decrease the undergrad
enrolment by 0.203% while reducing the credit hour production by 0.039%.

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Question 2
a) The requisite regression output of the given data from excel is shown below.
The requisite equation is shown below.
P= 100.148 – 0.043Q
The associated confidence level with the above regression model is 95% considering that the
coefficients have been estimated with the same level of confidence.
The above equation can be rewritten as follows.
Q = (100.148-P)/0.043
b) From the above, we get
Q = (1/0.043)*(100.148-P)
Hence, dQ/dP = (-1/0.043) = -23.31
Elasticity can be expressed using the following formula.
E= (dQ/dP)*(P/Q)

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Using the above formula the elasticity for different price quantity combinations is
summarised below.
c) When price = $49.8, then point price elasticity is -1.00. When price = $51, then point price
elasticity would be -1.04 in accordance with the approach highlighted above.
d) The company is charging $ 49.15. It is apparent that the point price elasticity magnitude is
lesser than 1 which implies that the demand is inelastic. Hence, the price should be
increased to $49.8 per unit where the elasticity becomes equal to 1. If the price is further
increased, then the decrease in demand would exceed the impact of price increase and
hence lower the overall revenue.
The company is now charging $ 51. At this price, the point elasticity had a magnitude which
exceeds 1 which implies that the demand is elastic. As a result, the price should be decreased
to $49.8 per unit where the elasticity becomes equal to 1. If the price is further decreased,
then the decrease in price would exceed the impact of demand increase and hence lower the
overall revenue.
e) TR or Total Revenue = Price * Quantity = (100.148 – 0.043Q)*Q = 100.148Q – 0.043Q2
MR or Marginal Revenue = dTR/dQ = 100.148 – 0.086Q
The requisite graph is indicated below.

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f) In order to determine the price and quantity for which TR would be maximum, MR needs
to be equated to zero.
100.148 – 0.086Q = 0
Solving the above, Q = 1164.5
Also, we are aware that P= 100.148 – 0.043Q
Hence, revenue maximising price = 100.148 – 0.043*1164.5 = $ 50.074
Thus, maximum revenue = 50.074*1164.5 = $58,311.8
Quantity when price is $52.5 = (100.148-52.5)/0.043 = 1108.1
Total Revenue at P of $ 52.5 = 52.5*1108.1 = $58.174.9
Quantity when price is $51 = (100.148-51)/0.043 = 1142.98
Total Revenue at P of $ 51 = 51*1142.98 = $60006.28
Question 3
The correlation coefficient between the unseasonal and seasonal variable has come out to be
0.973 which is quite high considering that the maximum possible value is 1. The relevant
scatter plot also highlights a strong positive relationship between the two.

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Considering the high degree of correlation between the two, it is evident that there is no need
of any seasonal adjusted data.
Regression Model 1
The relevant screenshot is given below.
Regression equation is given below.
US NSA = 112.86 -1.867t + 0.0099t2

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Coefficient of Determination = 0.852
The slope of all the independent variables are significant considering the p value for each is 0
and hence lesser than assumed significance value of 0.01.
Regression Model 2
The relevant screenshot is given below.
Regression equation is given below.
US NSA = 108.257-1.86t + 0.0099t2+ 8.73D
Coefficient of Determination = 0.8839
The slope of all the independent variables are significant considering the p value for each is 0
and hence lesser than assumed significance value of 0.01.
Regression Model 3
The relevant screenshot is given below.

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Regression equation is given below.
US SA = 1338.81 -21.98t + 0.117t2
Coefficient of Determination = 0.8924
The slope of all the independent variables are significant considering the p value for each is 0
and hence lesser than assumed significance value of 0.01.
Regression Model 4
The relevant screenshot is given below.

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Regression equation is given below.
US SA = 1339.01 -21.98t + 0.117t2 -0.43D
Coefficient of Determination = 0.8924
The slope of all the independent variables are significant considering the p value for each is 0
and hence lesser than assumed significance value of 0.01.
Best MODEL
The best model is model 4 i.e. which taken the dependent variable as US SA with
independent variables as t, t2 and D. This model is considered to be the best model as the R2
and adjusted R2 are the highest for this particular model and hence the predictive power is the
maximum.
Question 4
a) The adjusted R2 is 0.7711. Hence, the coefficient of determination would be greater than
0.7711. This implies that the given regression model is capable of explaining atleast
77.11% movements in the dependent variable based on corresponding movements of the
independent variable. As a result, it would be apt to conclude that the regression model is

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significant and would generate good estimates. This is also validated from the significant
slope coefficients of variable P and M at 5% level of confidence.
b) The applicable regression equation is highlighted below.
Q = -25.995 -4.582P + 0.0063M + 3.09R
The requisite input values are P = $15, M = $35,000, R = $ 17
Substituting the above inputs, we get the following.
Q = -25.995 -4.582*15 + 0.0063*35000 + 3.09*17 = 178
C) Values of own price, income and cross price elasticities.
 Own price
Own price=
dQ
Q
dP
P =
dQ
dP ∗P
Q =−4.582∗( 15
178.305 )=−0.385
 Income
Income=
dQ
Q
dM
M =
dQ
dM ∗M
Q =0.0063∗( 35000
178.305 )=1.236
 Cross price
Income=
dQ
Q
dR
R =
dQ
dR ∗R
Q =3.09∗( 17
178.305 )=0.295
d) If P has increased by 4% then the quantity demanded would decrease. Therefore, the new
price would be calculated as shown below:
P New = 15 * 1.04 = $15.6

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QNew=−25.995−4.582QNew+0.0063 M+ 3.09 R
QNew=−25.995− ( 4.582∗15.6 ) + ( 0.0063∗35000 ) + ( 3.09∗17 ) =175.56
Change∈Q= ( 175.56−178.305 )∗100
178.305 =−1.541 %
e) If M has increased by 3% then the quantity demanded would decrease. Therefore, the new
price would be calculated as shown below:
P New = 35000 * 1.03 = $36050
QNew=−25.995−4.582 P+0.0063 M +3.09 R
QNew=−25.995− ( 4.582∗15 ) + ( 0.0063∗36050 ) + ( 3.09∗17 )=184.92
Change∈Q= ( 184.92−178.305 )∗100
178.305 =3.710 %
f) If R has decreased by 5% then the quantity demanded would decrease. Therefore, the new
price would be calculated as shown below:
P New = 17 * 0.95 = $16.15
QNew=−25.995−4.582QNew+0.0063 M+ 3.09 R
QNew=−25.995− ( 4.582∗15 ) + ( 0.0063∗35000 ) + ( 3.09∗16.15 ) =175.678
Change∈Q= ( 175.678−178.305 ) ∗100
178.305 =−1.473 %