(a) Expected Selling Price
Price (r) Probability (p) Expected Selling Price 
  r= r x p 
10,000 0.2 2000
15,000 0.3 4500
20,000 0.4 8000
25,000 0.1 2500
  17000
  
  
(b) Standard Deviation
Price (r) Probability (p) (r  r )2 p
  
10,000 0.2 9,800,000
15,000 0.3 1,200,000
20,000 0.4 3,600,000
25,000 0.1 6,400,000
  21,000,000
  

σ=(r  r )2 p
σ=21,000,000
σ=4582.5756
(c) Coefficient of variation
v=σr
v=4582.575617,000
v=0.26956
QUESTION 2
(d) Output level of Maximize profits
MR=MC
Q=8
(e) Total Profit at profit maximizing output level
Total Profit = Total Revenue – Total Cost
= 216 – 124
= 92
QUESTION 3
TC = 20 + 5Q + Q2
Q = 25  P
P = 25 – Q
(a) Total Profit = Total Revenue – Total Cost
= ( P x Q ) – TC
= ( 25 – Q ) (Q) – ( 20 + 5Q + Q2 )
= 25Q – Q2 – 20 5Q – Q2
=  2Q2 + 20Q – 20
(b)
TR = 25Q  Q2MR=dTRdQ=252Q TC = 20 + 5Q + Q2MC=dTCdQ=5+2Q
MR=MC
252Q =5+2Q
2Q2Q=525
4Q= 20
Q=5
(c)
Total Profit =  2Q2 + 20Q – 20=  2(52) + 20(5) – 20 = 50 +10020 = 30  Selling Price = 25 – Q = 25 – 5 = 20  (d)
TP =  2Q2 + 20Q – 25
dTPdQ=4Q+20 dTPdQ=0 4Q+20=0 4Q= 20 Q=5 Total Profit =  2Q2 + 20Q – 25=  2(52) + 20(5) – 25 = 50 +10025 = 25  Even fixed cost change, output still 5.
(e)
TR = 25Q  Q2MR=dTRdQ=252Q TC = 20 + 5Q + Q2MC=dTCdQ=5+2Q QUESTION 4
(a) Ex=Q2Q1P2P1 ×P2 + P1Q2 + Q1
2.2=10,0008,000P22.98 × P2 +2.98110,000 +8,000
2.2=2,000 (P2 +2.981)18,000 (P22.98 )
2.2=2,000P2 +5,96018,000 P253,640 )
QUESTION 5
(a)
(b) Estimated Regression
b=nƩƩxyƩƩxƩynƩƩx2(ƩƩx)2
=7552.34 50.9(75.6)7374.51 (50.9)2 =3866.333848.042621.572590.81
=18.3430.76
=0.5962
y=ƩƩyn=75.67=10.8 x=ƩƩxn=50.97=7.2714
a=yb x
= 10.8 – 0.5962 (7.2714)
= 10.8 – 4.3352
= 6.4648
Y= 6.4648 + 0.5962X
(c) Hypothesis Testing
H0 : β=0
Ha : β≠0
Se =ƩƩy2aƩƩybƩƩxyn 2
=818.086.464875.6 0.5962(552.34)7 2
=818.08488.74 329.315
=0.08487
Sb =Se2ƩƩx2 –(ƩƩx)2n
=0.084872374.51 –(50.9)27
=0.00720244.3943
=0.04049
t=b βSb
=0.596200.04049
=14.7264
t0.052 , n2=2.571
From the tdistribution, the value is 2.571. since the calculated tvalue (14.7264) is greater than value from the table, we reject null hypothesis that there is no relationship between the variable.
(d) Proportion of total variation
r2=SSRSST= (y  y )2 (y  y )2
= 1.562081.6
=0.9763 @ 97.63%
The regression equation “explains” 97.63% of variation in the company sales.
(e)
F=SSRSST= (y  y )2 (y  y )2 n2
= 1.562080.0378972
= 206.1335
The value of (F0.05 , 1,5) from the Fdistribution 6.61. Since the calculated Fvalue (206.1335) is greater than value from the table, we reject null hypothesis that there is no relationship between the no of construction permits issued and sales.
(f) Y= 6.4648 + 0.5962X
Y= 6.4648 + 0.5962(8.00)
Estimated sales for Phoenix Lumber Company in 1998 would be 11.234 million.
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