please provide citaions for this including in text Here, Variable cost (TVC) = m
ID: 2948452 • Letter: P
Question
please provide citaions for this including in text
Here, Variable cost (TVC) = materials + energy cost + wage
Fixed cost, TFC = Business license + Insurance + Business lease rentals
AVC = TVC / Q
AFC = TFC / Q
A)
Q
TFC
TVC
AFC
AVC
875
4515
29170
5.16
33.34
670
4515
20810
6.74
31.06
1675
12715
52285
7.59
31.21
1155
4515
35750
3.91
30.95
1845
4515
55300
2.45
29.97
1650
4515
50530
2.74
30.62
1995
4515
62750
2.26
31.45
2845
4515
90000
1.59
31.63
2265
6715
68995
2.96
30.46
3470
4400
108500
1.27
31.27
3665
4400
115540
1.20
31.53
3750
4400
129050
1.17
34.41
4595
4400
151900
0.96
33.06
4060
4400
132090
1.08
32.53
3575
14680
112000
4.11
31.33
4380
5030
137580
1.15
31.41
5575
5030
218500
0.90
39.19
7870
5030
325500
0.64
41.36
6750
5030
269350
0.75
39.90
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.825528097
R Square
0.681496639
Adjusted R Square
0.662761148
Standard Error
139.6143834
Observations
19
ANOVA
df
SS
MS
F
Significance F
Regression
1
709020.7512
709020.75
36.3746331
1.34871E-05
Residual
17
331366.9928
19492.176
Total
18
1040387.744
Coefficients
Standard Error
t Stat
P-value
Lower 95%
Upper 95%
Lower 95.0%
Upper 95.0%
Intercept
766.0997098
63.78614199
12.010441
9.9293E-10
631.522715
900.6767046
631.522715
900.6767046
Q
0.10087006
0.016724878
6.0311386
1.3487E-05
0.065583652
0.136156469
0.065583652
0.136156469
D)
The quadratic expression for AVC is (regression output below):
AVC = 766.099 + 0.1009Q
The parameters are both positive, as expected. When quantity produced is 0, AVC is 766.099. With each unit increase in output, AVC increases by 0.1009 units.
The regression was performed at 95% level of confidence, signifying that above regression relationship holds with 95% likelihood.
R2 is 0.6814, or 68.14%, indicating that the model explains 68.14% of the variability of response data around its mean. It signifies a good fit of the model.
Q
TFC
TVC
AFC
AVC
875
4515
29170
5.16
33.34
670
4515
20810
6.74
31.06
1675
12715
52285
7.59
31.21
1155
4515
35750
3.91
30.95
1845
4515
55300
2.45
29.97
1650
4515
50530
2.74
30.62
1995
4515
62750
2.26
31.45
2845
4515
90000
1.59
31.63
2265
6715
68995
2.96
30.46
3470
4400
108500
1.27
31.27
3665
4400
115540
1.20
31.53
3750
4400
129050
1.17
34.41
4595
4400
151900
0.96
33.06
4060
4400
132090
1.08
32.53
3575
14680
112000
4.11
31.33
4380
5030
137580
1.15
31.41
5575
5030
218500
0.90
39.19
7870
5030
325500
0.64
41.36
6750
5030
269350
0.75
39.90
Explanation / Answer
Firstly, we need to understand that what are our variables. Our 'Y' variable is AVC or we can call it as dependent variable. and 'X' variable is 'Q' or 'Quantity' and we can call it as an independent variable. After fitting regression, our regression line looks like AVC = 766.099 + 0.1009Q It means that, When quantity produced is 0, AVC is 766.099. With each unit increase in quantity, AVC increases by 0.1009 units. Also, it concludes that Quantity and AVC are directly proportional since the parameter of Quantity is positive. Coming onto the summary table: Multiple R = 0.82 ~ 82% and can be interpreted as the correlation between observed values and fitted values of AVC. that is they are 82% related with each other. R Square = 0.68 ~ 68% also known as coefficient of determination and it can be interpreted that 68% of the variability in AVC can be explained by its independent variable which is Quantity. Adjusted R Square = 0.66 ~ 66% The adjusted R-squared is a modified version of R-squared that has been adjusted for the number of predictors in the model. It is always recommended to use Adjusted R Square in place of R Square. Standard error = 139.61, it is the estimate of measure of the accuracy of predictions, smaller values of standard error implies more accurate results. Coming onto the Hypothesis Testing: For parameters our Null hypothesis remains: Ho : Bj = 0 and alternative hypothesis remains: H1 : Bj ? 0 For intercept, intercept = 766.09 and p-value = 9.9293E-10 < 0.05 which implies that we reject Ho and conclude that intercept is not equal to 0. For Q, Q = 0.1008 and p-value = 1.3487E-05 < 0.05 which implies that we reject Ho and conclude that coefficient of Q is not equal to 0. Coming onto ANOVA Table : df SS MS F Significance F Regression 1 709020.751 709020.8 36.3746331 1.35E-05 Residual 17 331366.993 19492.18 Total 18 1040387.74 Total SS = ?(Yi – mean of Y) ^ 2 tells us how much variation is there in the dependent variable. Regression SS = ?(Estimated(Yi) – mean of Y) ^ 2 the regression sum of squares measures how much variation there is in the modelled values and this is compared to the Total SS Residual SS = ?(Yi – Estimated(Yi)) ^ 2 It is a measure of the discrepancy between the data and an estimation model. All values of respective sum of squares is shown above. Mean sum of squares can be estimated by dividing respective SS by their df. And, finally F statistic can be calculated by dividing both MS obtained above. Now, concluding the Significance F column : the P value for the F-test of overall significance test is less than your significance level that is 1.35E-05 < 0.05 , you can reject the null-hypothesis and conclude that your model provides a better fit than the intercept-only model.Related Questions
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