ACC 544: Rules for Decision Making under Uncertainty Express Different Degree of Decision Maker´s Optimism - Statistics Assessment Answers

August 22, 2017
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Question 1.

(a).

Most of the rules for decision making under uncertainty express a different degree of decision maker´s optimism.

We will describe the most used rules for the choice of the best alternative supposing that the decision criterion expresses the requirement of maximization

Themaximax ruleis appropriate for extreme optimists that expect the most favorable situation (they choose the alternative that could result in the maximum payoff). Under this rule, the decision maker will find the largest payoff in the decision matrix and select the alternative associated with it (the largest payoff is determined for each alternative and then the largest payoff from these values is selected; therefore “maximax”).

Themaximin rule(Wald criterion) represents a pessimistic approach when the worst decision results are expected. The decision maker determines the smallest payoff for each alternative and then chooses the alternative that has the best (maximum) of the worst (minimum) payoffs (therefore “maximin”).

TheHurwicz-criterionrepresents a compromise between the optimistic and the pessimistic approach to decision making under uncertainty. The measure of optimism and pessimism is expressed by anoptimism - pessimism index,<0,1> . The more this index is near to 1, the more the decision maker is optimist. By means of the index, a weighted average of the best payoff (its weight =)and the worst payoff (its weight =1-) is computed for each alternative and the alternative with the largest weighted average should be chosen

Theprinciple of insufficient reason(Laplace criterion) assumes that all states of nature are equally likely. Under this assumption, the decision maker can compute the average payoff for each row (the sum of the possible consequences of each alternative is divided by the number of states of nature) and, then, select the alternative that has the highest row average. (Orms.pef.czu.cz, 2015)

(b).

(i) If Joe is an optimist he will buy equipment A, because he will follow MAXIMAX criteria.

(ii) If Joe is a pessimist he will buy equipment C, because he will follow MAXIMIN criteria.

(iii) Joe would select Equipment A if he follows the criterion of regret.

(iv) If Joe believes that there is a 60% chance of a good economy which type of investment should he select?

For Equipment A = (80000*0.6) + (-20000*0.4) = 40000

Equipment B = (30000*0.6) + (20000*0.4) = 26000

Equipment C = (23000*0.6) + (23000*0.4) = 23000

Joe will buy Equipment A.

(C).

Marginal Profit= $20,000

Marginal Loss= $10,000

Minimum Probability using marginal analysis= ML/MP+ML

= 10,000/10,000+20,000

= 10,000/30,000

= 0.333%

Question 2.

Probability of S1=0.7 & S2=0.3

S1=0.7 S2=0.3
A1 $6,000 $10,000
A2 -1,000 25,000

 

(a).Expected value of information a1= 6,000*0.7+10,000*0.3= 7,200*

Expected value of information a2= -1,000*0.7+ 25,000* 0.3= 6,800

(b). Expected value of perfect information

EVWPI= 6,000(0.7) + 25,000(0.3) = 11,700

STEP1:

P (Y1/S1) =0.8 P (Y1/S2) =0.7

P (Y2/S1) =0.2* P (Y2/S2) =0.3*

1 1

STEP 2: Revised Probabilities

P(Y/S1) = Probability that demand will be poor as predicted.

P(Y/S2) = Probability that demand will good as predicted.

Si P(Si) P(Y1/Si) P(Si,Y1) P(Si/Y1)
S1 0.7 0.8 0.560 0.727
S2 0.3 0.7 0.210 0.273
1 0.770 1.000

If Y2 has been predicted

Si P(Si) P(Y2/Si) P(Si,Y2) P(Si/Y2)
S1 0.7 0.2 0.140 0.609
S2 0.3 0.3 0.090 0.391
1 0.23 1.000

770[P (y1)] + 0.230 [P (y2)] =1.000

STEP 3: Now we should calculate using new probabilities:

S1=0.727 S2=0.273
A1 $6,000 $10,000
A2 -1,000 25,000

E(U/a1,y1,n)= 6000(0.727)+ 10,000(0.273)= 7092*

E(U/a2,y1,n)= -1000(0.727)+ 25,000(0.273)=6098

  • A1 is optimal upon receiving signal Y1.
  • If signal Y2 is received.

E (U/a1, y2, n) = 6,000(0.609) +10,000(0.391) = 7564

E(U/a2,y2,n)= -1,000(0.609) + 25,000(0.391)= 9166*

  • If signal Y2 is received, then A2 is optimal.

Step 4: Expected Utility

E (U/n) = E (U/a*,yk,n) P(yk/n)

= (7092) p(y1) + (9166) p(y2)

= 7092(0.770) + 9166 (0.230)

=$7569.02

Step 5: The benefit of information

= Expected Utility- Expected value

= 7569.02- 7200

=369.02

(c). It is not worth paying consultant $3,000 as information has not increased profit much. The cost of opinion will be more, compare to profit we are garnering.

Question 3.

(a).

For the number demanded the previous match.

Normal view:

Formula view:

For 2500 units-

Normal view-

Formula view-

(b).

Production Average profits Increase in average profits
2300 6900 0
2400 7085 185
2500 7155 70
2600 7080 -75
2700 6980 -100

Simulations were run on games by marginally adding 100 production units and the profits were calculated per game. The marginal calculation shows that the profit increases up to 2500 and then substantially started decreasing, thus 2500 is the optimum number of production.

(c)

Business report:

Introduction: Profit can occur only when there is an appropriate strategy is used. Manager will find the analysis and can decide what strategies are viable for the company and the number of programs they should print.

Recommendation: After considering these excel sheets we can figure it out that both the strategies are only giving the profits at every stage, but the key point is that the profit at each stage in 2500 programs strategy is more than the 2400 programs printed strategy. So Firstly I would recommend the manager for 2500 programs strategy for each game for getting more profits and then 2400 programs strategy because this strategy also giving profits and there is no chance of loss in both the strategies.

Conclusion: 2500 programs strategy is best for every game to get more advantages without suffering loss. Also, 2400 programs strategy is also good so far. As a Manager, we can apply this approach in every game without fear of loss.

Question 4.

(b).

(c).

Question 5.

(a).

Fixed cost = $40000

SP = $10

Cost = $5

CM = $10 - $5 = $5

CM ratio = ($10 - $5)/$10 = 0.5

(i) Breakeven point in units = fixed cost / Contribution margin = $40000 / $5 = 8000 units

(ii) Breakeven point in sales dollar = Fixed cost / CM ratio = $40000 / 0.5 = $80000

(iii) Let x be the number of pizzas to be sold.

Profit = sales – variable cost – fixed cost

$60000 = 10x – 5x - $40000

$100000 = 5x

X = $100000/5 = 20000 units

(b).

(i)

X Y
Selling price 19 15
Variable cost 10 9
Contribution/unit 9 6

Fixed cost = 11000

Average contribution margin = 1/3 (9) + 2/3 (6) = 3+4 = 7

Required volume = 11000 + 10000/7 = 3000 units

(ii)Ratio = 2:2

Average contribution margin = 2/4 (9) + 2/4 (6) = 7.5

Required volume = Fixed cost + ?/ contribution margin

12000 = 11000 + ?/7.5

? = $79000

(C).

(i) Expected Annual profit

? = (p – b) x-a

E ? = E (p – b) x-a

= (p – b) E(x) - a

= (4000 – 3500) 12000 – 2000000

= 6000000 – 2000000

= 40,00,000

(ii) -?X =(p - b) * ?X

= (4000 – 3500) * 4000

= 500 * 4000

= 20,00,000

(iii)Selling price = 4000

Variable cost = 3500

Fixed cost = 2000,000

Mean = 12000 units

Standard Deviation = 4000 units

Breakeven point = Fixed cost

Selling price – Variable Cost

= 2000000

4000 – 3500

= 2000000/500

= 4000 units

This area is under the curve and where breakeven point is less than mean.

P(r >0) = P (Z > 4000 – 12000/4000)

= P (Z > -2) = 1- F (-2)

= P (Z < 2)

After checking the Z table, we get the value of Z= 0.97725

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Solution:

QUESTION 1 Probability

  1. Classical Method:

Classical probability technique deals with relative frequency of each and every event in the sample space. Let us consider the experiment of rolling a die twice and the possible outcomes are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. The frequency of occurrence is given in the table below

Outcome 2 3 4 5 6 7 8 9 10 11 12
Frequency 1 2 3 4 5 6 5 4 3 2 1
Relative Frequency 0.028 0.056 0.083 0.111 0.139 0.167 0.139 0.111 0.083 0.056 0.028

Going through the above relative frequency distribution, we can find the most frequent occurrence of outcomes

  1. Discrete Probability Distribution

When the random variable takes a finite number of values, then the distribution involved will be discrete probability distribution. For example, tossing two coins and counting the number of heads turned up is an example of discrete probability distribution

When the random variable falls between two specified values, then the distribution involved will be continuous probability distribution. For example, the weight of the students studying in class X will fall between 30 kgs and 36 kgs

  1. The probability distribution for the sales unit is given below

Sales Unit Number of Days Probability
8 10 10/50 = 0.2
9 12 12/50 = 0.24
10 14 14/50 = 0.28
11 8 8/50 = 0.16
12 6 6/50 = 0.12
Total 50

 

  1. P (selling 9 or 10 units on any one day) = 0.24 + 0.28 = 0.52
  2. Average daily sales = ?x * P (x) = 8 * 0.2 + 9 * 0.24 + 10 * 0.28 + 11 * 0.16 + 12 * 0.12 = 9.76
  3. P (selling 10 or more) = 0.28 + 0.16 + 0.12 = 0.56
  4. P (Selling 11 or less) = 1 – P (selling 12) = 1 – 0.12 = 0.88

d.The information in the given problem is represented with the following notations

Sample Space = n (s) = 40

Number of females and Australian citizens = n (F ? A) = 12

Number of males and Australian citizens = n (M ? A) = 14

Number of female and overseas citizens = n (F ? OS) = 10

Number of male and overseas citizens = n (M ? OS) = 4

  1. The cross tabulation showing the relationship between gender and citizenships is given below

American Overseas Total
Male 14 10 24
Female 12 4 16
Total 26 14 40

 

2.(i) P (Male) = 24/40 = 0.6

(ii) P (Overseas student) = 14/40 = 0.35

(iii) P(Australian given a female) = 12/16 = 0.75

(iv) P (Overseas student given a male) = 10/24 = 0.417

e.The information in the given problem is represented with the following notations

Population Mean = µ = 50 Weeks

Population Standard deviation = ? = 10 Weeks

1.PX?54=Px-?<54-5010

=PZ<54-5010, where Z=x-?~N(0,1)

= P (Z < 0.40)

= 0.6554 (by referring normal distribution table)

Therefore, the probability that the project will be completed in 54 weeks or less is 0.6554

2.PX?62=Px-?<62-5010

=PZ<62-5010, where Z=x-?~N(0,1)

= P (Z < 1.2)

= 0.8849 (by referring normal distribution table)

Therefore, the probability that the project will be completed in 62 weeks or less is 0.8849

3.PX>70=1-Px-?<70-5010

=1-PZ<70-5010, where Z=x-?~N(0,1)

= 1-P (Z < 2)

= 1 – 0.97725 (by referring normal distribution table)

0.02275

Therefore, the probability that the project will take longer than 70 weeks is 0.02275

QUESTION 2 Research Question, Constructing data table and calculating probabilities

1.The latest figures of age and sex of the Australian population is taken form the link https://populationpyramid.net/australia/2016/

2.

Male Female Total
Age in Years 0-14 2333664 2212119 4545783
15-24 1628703 1531467 3160170
25-54 5056272 5007654 10063926
55-64 1385613 1409922 2795535
65 and over 1750248 1993338 3743586
Total 12154500 12154500 24309000

3.

  • The probability that any person selected at random from the population is a male is 0.5 (12154500/24309000 = 0.5)
  • The probability that any person selected at random from the population is aged between 25 and 54 is 0.414 (10063926/24309000 = 0.414)
  • The joint probability that any person selected at random from the population is a female and aged between 15 and 24 is 0.063 (1531467/24309000 = 0.063)
  • The conditional probability that any person selected at random from the population is 25 or over given that the person is a male is 0.674 (8192133/12154500 = 0.674)

QUESTION 3 Statistical Decision Making and Quality Control

The information in the given problem can be represented with the following notations

Population mean = µ = 20

Population Standard deviation = ? = 9

Sample size = 36

  1. The 95% confidence interval for the true mean labor time to produce a product is calculated by using the formula given below

x-Z1-2*n,x+Z1-2*n

=20-1.96*936,20+1.96*936

= (20 – 2.94, 20 + 2.94)

= (17.06, 22.94)

  1. The 95% confidence interval for the true mean labor time to produce a product is calculated by using the formula given below

x-t1-2*n,x+t1-2*n

=20-2.306*99,20+2.306*99

= (20 – 6.918, 20 + 6.918)

= (13.082, 26.918)

3.Sampling more frequently using 9 observations and setting confidence intervals of 80

The 80% confidence interval for the true mean labor time to produce a product is calculated by using the formula given below

x-t1-2*n,x+t1-2*n

=20-1.397*99,20+1.397*99

= (20 – 4.19, 20 + 4.19)

= (15.81, 24.19)

Maintaining 95% confidence intervals and increasing sample size to 64 observations

The 95% confidence interval for the true mean labor time to produce a product is calculated by using the formula given below

x-z1-2*n,x+z1-2*n

=20-1.96*964,20+1.96*964

= (20 – 2.205, 20 + 2.205)

= (17.795, 22.205)

Setting 95% confidence intervals and using sample sizes of 100 observations.

The 95% confidence interval for the true mean labor time to produce a product is calculated by using the formula given below

x-z1-2*n,x+z1-2*n

=20-1.96*9100,20+1.96*9100

= (20 – 1.764, 20 + 1.764)

= (18.236, 21.764)

On comparing the three confidence intervals, we see that the width of the 95% confidence intervals and using sample sizes of 100 observations is less and the confidence interval is narrower. This is mainly due to larger sample size. Increase in sample size will decrease the width of the confidence interval and hence the interval becomes narrower

  1. Hypothesis Testing

In order to determine whether the mean time until delivery differ significantly from 120 days, we perform single mean z test. The null and alternate hypotheses are given below

Null Hypothesis: H0: µ = 120

That is, the mean time until deliver do not differ significantly from 120 days

Alternate Hypothesis: Ha: µ ? 120

That is, the mean time until deliver differ significantly from 120 days

Level of Significance

Let the level of significance be ? = 0.05

Test Statistic

The z test statistic is

Z=x-??/n=118.5-12012/100=-1.25

The p – value of z test statistic is

P (Z > |-1.25|) = 0.2113 (by referring normal distribution table)

Since the p – value of z test statistic is greater than 0.05, there is no sufficient evidence to reject the null hypothesis at 5% level of significance. Therefore, there is no sufficient evidence to support the claim that the mean time is not 120 days

Group Research project

Here, we wish to start building the first child care center in Australia. Currently, the Australian population is 24309000 people. Out of which 10063926 people aged between 25 years and 54 years live in Australia. The Australian Pyramid map showing the population shape in Australia is given below

It is also seen that, overall in Australia, there are 802197 male children and 753579 female children aged less than 5 years

In addition, we found that the states Victoria (Population = 5,938,100) and New South Wales (Population = 7,618,200) records the second and third largest states in Australia in terms of population size. Therefore, we can select these two states to build a new first child care center in Australia

Assessment item 3

QUESTION 1 Decision Analysis

a.Marginal analysis is considered as one of the most important decision-making tool and now days, it is widely used in many business situations. It allows business operators to calculate the additional benefits on comparing its production activity and cost function. When we are planning to decide whether the project is moving on the right track, we need to use marginal analysis to break down all available options. The main purpose of marginal analysis is to allocating their rare resources to increase their benefits of their products. The formula used to compute the marginal analysis is given below

Marginal Benefit = Increase in Total Benefits per unit of control variable

?TR / ?Qcv = MR

For example, let us consider the situation of doing exercise daily for five days. Then before performing the exercise on day 6, we must do marginal analysis on how much calories burnt in doing exercises and also measure the effort ratio and output ratio to determine whether we are doing the exercise correctly and the work is going in right track or not

Advantages

  • Marginal cost ignores average costs, fixed costs, and sunk costs and thus it aims to reach optimal activity solution
  • It helps the company to maximize their profits\
  • It is used to analyze how complex system is affected by marginal manipulation

Disadvantages

  • The major setback is the range of values that the choice variables can take
  • It assumes single maximization objective and therefore it is not suitable in situations where we have two or more maximization objectives

b.(i)Maximum (Stock Market) = $160000

Maximum (Bonds) = $60000

Maximum (Term Deposit) = $46000

From the above table, it can be clearly seen that the maximum of the maximum payoff = $160000, which is corresponding to the Alternative Stock Market. Thus, an optimist would make a decision to choose Stock Market as their decision alternative

(ii)Minimum (Stock Market) = - $40000

Minimum (Bonds) = $40000

Minimum (Term Deposit) = $46000

From the above table, it can be clearly seen that the maximum of the minimum payoff = $46000, which is corresponding to the Alternative Term Deposit. Thus, a pessimistic would make a decision to choose Term Deposit as their decision alternative

(iii)In this procedure we first find the maximum of each economy structure (states of nature) and subtract it from the other elements. The next step is to find the maximum for each alternative and then we need to find the minimum of the maximum alternatives

Decision Alternative Good Economy Poor Economy
Stock market 0 $86,000
Bonds $100,000 $6,000
Term deposit $114,000 $0

Maximum (Stock Market) = $86000

Maximum (Bonds) = $100000

Maximum (Term Deposit) = $114000

Thus the Minimax Regret criteria is to select the Decision Alternative Stock Market

(iv)EMV (Stock Market) = 0.7 * $160,000 + 0.3 * (-$40,000) = $ 100000

EMV (Bond) = 0.7 * $60,000 + 0.3 * $40,000 = $ 540000

EMV (Term Deposit) = 0.7 * $46,000 + 0.3 * $46,000 = $ 46000

Form the above workings, we need to select the Decision Alternative Stock Market

(v)In general, the expectation value of perfect information = EVPI = EV|PI –EOL

.where:

EMV = Expected monetary value = i = 1, 2, 3

EV|PI = Expected value given perfect information=

EV|PI = 0.7 * 160000 + 0.3 * 46000 = $125800

EVPI = EV|PI – EOL = 125800 – 100000 = $ 25800

QUESTION 2 Value of information

a.EMV (a1) = 0.4 * $24,000 + 0.6 * $15,000 = $ 18600

EMV (a2) = 0.4 * $4,000 + 0.6 * $40,000 = $ 22400

Form the above workings, we need to select the a2

b.Expected value of perfect information

EVWPI= 24000 (0.4) + 40000 (0.6) = 33600

It is given that

P (Y1/S1) =0.8 and P (Y2/S1) =0.2

P (Y1/S2) =0.7 and P (Y2/S2) =0.3

Then, we have

Si P(Si) P(Y1/Si) P(Si,Y1) P(Si/Y1)
S1 0.7 0.8 0.560 0.727
S2 0.3 0.7 0.210 0.273

Suppose, if we predict Y2

Si P(Si) P(Y2/Si) P(Si,Y2) P(Si/Y2)
S1 0.7 0.2 0.140 0.609
S2 0.3 0.3 0.090 0.391

c.Now, the posterior probability is

E(U/a1,y1,n)= 24000(0.727)+ 15000(0.273)= 21543

E(U/a2,y1,n)= -4000 (0.727)+ 40,000(0.273)=8012

d.Here, the alternative a1 is the optimal when we use first strategy Y1

E(U/a1,y1,n)= 24000(0.609)+ 15000(0.391)= 18519

E(U/a2,y1,n)= -4000 (0.609)+ 40,000(0.391)=22796

Here, the alternative a2 is the optimal when we use first strategy Y2

EV|PI = 0.77 * 21543 + 0.23 * 22796 = 21831.9

EVPI = 21831.9 – 22400 = -568.1

  1. Since the EVPI value turns negative, it is not advisable to engage the consultant since it do not add any value added information for our situation

QUESTION 3 Simulation

a.The simulation is performed to calculate the profit and is given below

Arrival Time Probability Service Time Probability Arrival Time Service Time Sales Revenue
0.338333 0.526782 4 5 25
0.759587 0.750612 6 7 25
0.842755 0.121937 6 7 25
0.229497 0.326476 4 6 25
0.489786 0.290529 5 8 25
0.309932 0.309861 4 6 25
0.319198 0.335065 4 7 25
0.575895 0.109284 6 7 25
0.150147 0.010189 3 6 25
0.264109 0.153145 4 6 25
0.638102 0.993949 6 6 25
0.896226 0.779099 6 6 25
0.835057 0.526628 6 5 25
0.601559 0.393617 6 6 25
0.06525 0.417294 3 6 25
0.097007 0.862628 3 7 25
0.763174 0.139116 6 6 25
0.124239 0.997282 3 7 25
0.92512 0.940369 7 8 25
0.402377 0.016425 5 6 25
Total 97 128 500

b.Formula

  1. The initial procedure had a total service time of 128 minutes which is approximately 2.13 hours, mean that, it takes an average of 6.4 minutes to serve an customers which is too long. In order to reduce service time, manager suggest to reorganization of the tool shop could reduce the time taken to serve the toolmakers by 2 minutes. By reducing the service time to 2 minutes and service time ranges between 3 minutes and 7 minutes, it is found that average of 4.77 minutes to serve a customer and therefore, it is of acceptable time and the total time taken to serve the customers in a 2 hour time period is 105 minutes which is pretty less when compared to old technique. Therefore, we can recommend the manager to suggest to reorganization of the tool shop could reduce the time taken to serve the toolmakers by 2 minutes

QUESTION 4 Regression Analysis and Cost Estimation

  1. High-Low method

To determine the slope using high value method, we first need to find the minimum and maximum value of the variables taken into consideration

For variable Administrative cost, the minimum value is y1 = 4100 and the maximum value y2 = 16100

For variable patient load, the minimum value is x1 = 300 and the maximum value x2 = 1500

Variable Cost per Unit=y2-y1x2-x1=16100-41001500-300=10

Total Fixed Cost = y2 ? bx2 = 16100 – 10 * 1500 = 1100

Therefore, the regression model is

Administrative Cost = 1100 + 10 * Patient Load

b.Here, we perform three regression models with dependent variable as Administrative Costs

Model 1 (Dependent Variable – Administrative Costs and Independent Variable – Patient Load)

SUMMARY OUTPUT
Regression Statistics
Multiple R 0.957909
R Square 0.91759
Adjusted R Square 0.909349
Standard Error 995.7963
Observations 12
ANOVA
df SS MS F Significance F
Regression 1 1.1E+08 1.1E+08 111.3447 9.69E-07
Residual 10 9916103 991610.3
Total 11 1.2E+08
Coefficients Standard Error t Stat P-value Lower 95% Upper 95%
Intercept 2670.705 731.3395 3.651799 0.004449 1041.179 4300.231
Patient Load (Number) 7.812068 0.74034 10.552 9.69E-07 6.162488 9.461649

Administrative Cost = 2670.705 + 7.812 * Patient Load

Going through the ANOVA table, we see that the value of F test statistic is 111.3447 and its corresponding p – value is 0.0000 < 0.05, indicating that the estimated regression model is good fit in predicting the dependent variable Administrative Cost

Model 2 (Dependent Variable – Administrative Costs and Independent Variable – Emergency Procedure)

SUMMARY OUTPUT
Regression Statistics
Multiple R 0.685904
R Square 0.470464
Adjusted R Square 0.417511
Standard Error 2524.228
Observations 12
ANOVA
df SS MS F Significance F
Regression 1 56609415 56609415 8.884472 0.013792
Residual 10 63717252 6371725
Total 11 1.2E+08
Coefficients Standard Error t Stat P-value Lower 95% Upper 95%
Intercept 2115.091 2668.472 0.792623 0.446394 -3830.64 8060.818
Emergency Procedures (Number) 734.5512 246.4372 2.980683 0.013792 185.4549 1283.648

Administrative Cost = 2115.091 + 734.5512 * Emergency Procedure

Going through the ANOVA table, we see that the value of F test statistic is 8.884472 and its corresponding p – value is 0.0138 < 0.05, indicating that the estimated regression model is good fit in predicting the dependent variable Administrative Cost

Model 3 (Dependent Variable – Administrative Costs and Independent Variables – Patient Load and Emergency Procedure)

SUMMARY OUTPUT
Regression Statistics
Multiple R 0.963963
R Square 0.929225
Adjusted R Square 0.913497
Standard Error 972.7451
Observations 12
ANOVA
df SS MS F Significance F
Regression 2 1.12E+08 55905285 59.08194 6.68E-06
Residual 9 8516098 946233.1
Total 11 1.2E+08
Coefficients Standard Error t Stat P-value Lower 95% Upper 95%
Intercept 1769.331 1029.328 1.718919 0.119748 -559.17 4097.832
Patient Load (Number) 7.101339 0.929748 7.637918 3.2E-05 4.998103 9.204576
Emergency Procedures (Number) 148.5074 122.0906 1.21637 0.254781 -127.681 424.6956

Administrative Cost = 1769.331 + 7.10134 * Patient Load + 148.51 * Emergency Procedures

Going through the ANOVA table, we see that the value of F test statistic is 59.08194 and its corresponding p – value is 0.00000686 < 0.05, indicating that the estimated regression model is good fit in predicting the dependent variable Administrative Cost

On comparing the coefficient of determination (R2) for three models, we see that the R2 value for Model 3 is high when compared with other two regression models . Therefore, we can say that the regression equation predicting Administrative cost using Patient Load and Emergency Procedures as independent variables as the best model

(c)The regression model chosen is

Administrative Cost = 2670.705 + 7.812 * Patient Load

When there were 1,000 Patients, we have

Administrative Cost = 2670.705 + 7.812 * 1000 = 10482.77292

QUESTION 5 CVP Analysis

a.The unit contribution margin for each product is given below

Unit Contribution Margin= Unit Specific Revenue-Unit Specific CostUnit Specific Revenue

=500-300500=0.40

(b) The break even workings is given below

Road Bikes Sale Price Per Unit Variable Cost Fixed Cost Total Cost Revenue
150 500 300 65000 110000 75000
175 500 300 65000 117500 87500
200 500 300 65000 125000 100000
225 500 300 65000 132500 112500
250 500 300 65000 140000 125000
275 500 300 65000 147500 137500
300 500 300 65000 155000 150000
325 500 300 65000 162500 162500
350 500 300 65000 170000 175000
375 500 300 65000 177500 187500
400 500 300 65000 185000 200000
425 500 300 65000 192500 212500
450 500 300 65000 200000 225000
475 500 300 65000 207500 237500
500 500 300 65000 215000 250000
525 500 300 65000 222500 262500
550 500 300 65000 230000 275000

Sales Price = 500 * 325 (Road bikes) = $162500

Cost = 65000 +300 * 325 (Road bikes) = $162500

Therefore, the manufacturer should sell at least 325 Road Bikes

(c) Sales Price = 500 * 325 (Road bikes) = $162500

The breakeven sales volume for the year in sales dollars is $162,500

(d) The table given below shows the revenue and profit workings

Road Bikes Sale Price Per Unit Variable Cost Fixed Cost Total Cost Revenue Profit
500 500 300 65000 215000 250000 35000
525 500 300 65000 222500 262500 40000
550 500 300 65000 230000 275000 45000
575 500 300 65000 237500 287500 50000
600 500 300 65000 245000 300000 55000
625 500 300 65000 252500 312500 60000
650 500 300 65000 260000 325000 65000
675 500 300 65000 267500 337500 70000
700 500 300 65000 275000 350000 75000
725 500 300 65000 282500 362500 80000
750 500 300 65000 290000 375000 85000
765 500 300 65000 294500 382500 88000
790 500 300 65000 302000 395000 93000
815 500 300 65000 309500 407500 98000
840 500 300 65000 317000 420000 103000
865 500 300 65000 324500 432500 108000
890 500 300 65000 332000 445000 113000

Profit earned = $382500 - $294500 = $88000

(ii) The table given below shows the workings of profit calculated after tax

Road Bikes Total Cost Revenue Profit Profit after tax
700 275000 350000 75000 52500
725 282500 362500 80000 56000
750 290000 375000 85000 59500
775 297500 387500 90000 63000
800 305000 400000 95000 66500
825 312500 412500 100000 70000
850 320000 425000 105000 73500
875 327500 437500 110000 77000
900 335000 450000 115000 80500
925 342500 462500 120000 84000
950 350000 475000 125000 87500
975 357500 487500 130000 91000
1000 365000 500000 135000 94500
1025 372500 512500 140000 98000
1050 380000 525000 145000 101500
1075 387500 537500 150000 105000
1100 395000 550000 155000 108500

Sales Price = 500 * 1075 (Road bikes) = $537500

Cost = 65000 +300 * 1075 (Road bikes) = $387500

Profit earned = $537500 - $387500 = $150000

Tax = 150000 * 0.3 = 45000

Profit after tax = $150000 - $45000 = $105000

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