A) using OLS, the estimated inverse demand function (P=f(Q)) is? B) using algebr
ID: 1105123 • Letter: A
Question
A) using OLS, the estimated inverse demand function (P=f(Q)) is? B) using algebra to transform the indirect demand function, the direct demand function (Q=f(p)) is? C) What is point price elasticity of demand when P=$3.98? What is point price elasticity of demand when P=$3.81? D) to maximize total revenue, what would you recommend if the company was currently charging P=$4.53? If it was charging $3.81? E) what is the total revenue maximizing price and quantity , and how much revenue is earned there? into Excel: 2. Copy and paste the following data $4.80 1170 $4.53 1235 $3.98 1337 $3.72 1442 $3.49 1548 a. Run OLS to determine the inverse demand function (P f(Q); how much confidence do you b. Using calculus to determine dQ/dP, construct a column which calculates the point-price elasticity c. What is the point price elasticity of demand when P-$3.98? What is the point price elasticity of have in this estimated equation? Use algebra to then find the direct demand function (Q- f(P)). for each (P,Q) combination. demand when P=$3.81? P-$4.53? If it was charging P=$3.81? Use your indirect demand function to determine an equation for TR and MR as a function of Q, and create a graph of P and MR on the vertical and Q on the horizontal axis. What is the total-revenue maximizing price and quantity, and how much revenue is earned there? Compare that to the TR when P = $4.80 and P = $3.81. e. f.Explanation / Answer
a) P = 8.85-0.0035Q
Regression Statistics
Multiple R
0.982068
R Square
0.964458
Adjusted R Square
0.952611
Standard Error
0.119518
Observations
5
ANOVA
df
SS
MS
F
Significance F
Regression
1
1.162867
1.162867
81.40758
0.002875
Residual
3
0.042853
0.014284
Total
4
1.20572
Coefficients
Standard Error
t Stat
P-value
Lower 95%
Upper 95%
Lower 95.0%
Upper 95.0%
Intercept
8.854859
0.529256
16.73077
0.000465
7.17053
10.53919
7.17053
10.53919
X Variable 1
-0.00353
0.000391
-9.02261
0.002875
-0.00477
-0.00228
-0.00477
-0.00228
Confident at95% level of confidence since the value lies in the CI.
b)
P = 8.85-0.0035Q
dP/dQ = -0.0035
dQ/dP = -1/0.0035
dQ/dP = 285.71
Elasticity:
e= dQ/dP*(P/Q)
c) P = $3.98
TR = P*Q = 3.98*1337
=$5321.26
P = $3.81
At P = 3.81
3.81 = 8.85-0.0035Q
0.0035Q = 8.85-3.81
Q = 1440
TR = 3.81*1440
=$ 5486.4
d)
P = $4.53
Q = 1235
TR = 4.53*1235
=$5594.55
TR is maximized at P = $4.53
AT $3.81 Revenue can be increased by charging $4.53 instead.
Regression Statistics
Multiple R
0.982068
R Square
0.964458
Adjusted R Square
0.952611
Standard Error
0.119518
Observations
5
ANOVA
df
SS
MS
F
Significance F
Regression
1
1.162867
1.162867
81.40758
0.002875
Residual
3
0.042853
0.014284
Total
4
1.20572
Coefficients
Standard Error
t Stat
P-value
Lower 95%
Upper 95%
Lower 95.0%
Upper 95.0%
Intercept
8.854859
0.529256
16.73077
0.000465
7.17053
10.53919
7.17053
10.53919
X Variable 1
-0.00353
0.000391
-9.02261
0.002875
-0.00477
-0.00228
-0.00477
-0.00228
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