Problem 1
The average prices for a product in 12 stores in a city are given below.
$1.99, $1.85, $1.25, $2.55, $2.00, $1.99, $1.76, $2.50, $2.20, $1.85, $2.75, $2.85
Test the hypothesis that the average price is higher than $1.87. Use level of significance a = 0.05.
Problem 2
A store wants to predict net profit as a function of sales for next year. Historical data for 8 years is shown in the table below.
|
Year
|
Sales
(thousands of dollars)
|
Net Profit
|
|
1
|
59
|
5.0
|
|
2
|
50
|
8.4
|
|
3
|
51
|
9.5
|
|
4
|
65
|
8.6
|
|
5
|
80
|
1.5
|
|
6
|
85
|
-2.1
|
|
7
|
95
|
1.2
|
|
8
|
90
|
1.8
|
(a) Make a scatter diagram for the data, using Sales for the Net Profit and independent variable for the dependent variable. Insert the trend line and add the equation and R2 value to the diagram.
(b) Evaluate the correlation coefficient. Comment on the value of the correlation coefficient.
(c) Evaluate the predicted value of Y given X = 75. Give an interpretation of the predicted value in the context of the problem.
(d) Prepare an ANOVA table and attach the summary output.
Problem 3
Last week's sales of a product at a retail store are shown in the subsequent table:
|
Day
|
Sales (Dollars)
|
|
1
|
200
|
|
2
|
250
|
|
3
|
180
|
|
4
|
190
|
|
5
|
175
|
|
6
|
170
|
|
7
|
180
|
(a) Use the 3-day moving average method for forecasting days 4-7.
(b) Use the 3-day weighted moving average method for forecasting days 4-7. Use Weight 1 day ago = 4, Weight 2 days ago = 3, and Weight 3 days ago = 2.
(c) Compare the methods using the Mean Absolute Deviation.
Problem 4
The revenue and cost functions for producing and selling quantity x for a certain company are provided below.
R(x) = 12x - x2
C(x) = 21 + 2x
(a) Evaluate the profit function P(x).
(b) Compute the break-even quantities.
(c) Evaluate the average cost at the break-even quantities.
(d) Evaluate the marginal revenue R'(x).
(e) Evaluate the marginal cost C'(x).
(f) At what quantity is the profit maximized?
Problem 5
A company manufactures two products, Product A and Product B. The manufacturing cost and wholesale price of each product are shown below.
|
Item
|
Price
|
Cost
|
Assembly Times (hr)
|
|
A
|
$20
|
$10
|
2
|
|
B
|
$45
|
$20
|
4
|
The company may a minimum of 10,000 of each item. Given the number of hours, the company can sell no more than 18,000 of Item A and 20,000 of Item B. Consider the company has 100,000 hours of assembly time available, how many of each item should it produce in order to maximize profits while meeting all required constraints? Give the LP Model and use both the graphical method and Excel to find the optimal solution.
Problem 6
A company purchases its merchandise for $10 and sells for each item for $20. The pay-off table for the problem is provided below.
|
|
Demand for Item
|
|
Alternative
|
Low
|
Medium
|
High
|
|
Do nothing
|
0
|
0
|
0
|
|
Order Low
|
30,000
|
30,000
|
30,000
|
|
Order Medium
|
10,000
|
50,000
|
40,000
|
|
Order High
|
-10,000
|
40,000
|
80,000
|
|
Probability
|
0.3
|
0.3
|
0.4
|
What is the decision based on each of the subsequent criteria? Show work in making the decision for each criterion.
a) EMV approach
b) EOL approach
Problem 7
During the dinner hour, the distribution of the inter-arrival time of customers at a restaurant is evaluated to be as shown below. The mode of payment and the service times of the credit card and cash customers are shown in the subsequent tables. Complete the tables and simulate the system for 20 customer arrivals and evaluate the average time a cash and credit card customer must wait in line before paying the cashier.
Use Column A of the given random number table to evaluate the customer inter-arrival time. For the arrival time, please start at 0 seconds, and increments in seconds from the results you get from decoding of the random numbers. As an example, the first random number is 6320, which equates to 60 seconds of inter-arrival time. So, your first customer arrival time is 60 seconds. For your second customer, the random number is 4630, which also equates to 60 seconds of inter-arrival time. As a result, your second customer arrival time may be 120 seconds, and so on. Please keep your results in seconds for all customers.
Use Column B to evaluate whether the customer pays with cash or credit, and Column C to determine the service time.
Inter-arrival time
|
Inter-arrival Time
|
Probability
|
Cumulative Probability
|
Random Number Interval
|
|
30 seconds
|
0.45
|
|
|
|
60 seconds
|
0.25
|
|
|
|
90 seconds
|
0.15
|
|
|
|
120 seconds
|
0.10
|
|
|
|
150 seconds
|
0.05
|
|
|
Mode of Payment
|
Payment Mode
|
Probability
|
Cumulative Probability
|
Random Number Interval
|
|
Cash
|
0.6
|
|
|
|
Credit Card
|
0.4
|
|
|
Cash Service Time
|
Service Time
|
Probability
|
Cumulative Probability
|
Random Number Interval
|
|
20 seconds
|
0.35
|
|
|
|
40 seconds
|
0.30
|
|
|
|
60 seconds
|
0.25
|
|
|
|
80 seconds
|
0.10
|
|
|
Credit Card Service Time
|
Service Time
|
Probability
|
Cumulative Probability
|
Random Number Interval
|
|
30 seconds
|
0.20
|
|
|
|
60 seconds
|
0.45
|
|
|
|
90 seconds
|
0.25
|
|
|
|
120 seconds
|
0.10
|
|
|
|
Random Numbers
|
|
(A)
|
(B)
|
(C)
|
|
6320
|
1094
|
1995
|
|
4630
|
7371
|
7971
|
|
8657
|
2809
|
3554
|
|
0030
|
5148
|
6300
|
|
5624
|
9115
|
5495
|
|
6728
|
1469
|
5165
|
|
5925
|
6480
|
9339
|
|
2829
|
2447
|
6997
|
|
7939
|
7031
|
1443
|
|
6476
|
8442
|
3574
|
|
3319
|
7387
|
0150
|
|
8134
|
1788
|
0933
|
|
1712
|
4891
|
7082
|
|
6317
|
1149
|
5025
|
|
6605
|
8822
|
4081
|
|
2734
|
9451
|
4100
|
|
0432
|
2990
|
7190
|
|
3441
|
8314
|
6822
|
|
0726
|
7176
|
5053
|
|
6969
|
2766
|
8284
|