WEEK 13 – What have we done during the Trimester? Topic 1 – Money Simple Interest Formula Finding P, R, T, I, M Compound interest formula Finding M, P Multiple cash flows F

business test / quiz prep

Pratice E–x-a-m Time: 24/01/2023 09:00 AM (Adelaide Time)
Duration : 2hrs, 30min
Question structure:
The questions will be comparable to those covered in class; therefore, the best form of revision will be to work through lecture and tutorial questions, exercises and problems.
Topic covered: Similar to that screenshot attached
Requirements: 2 hours | .doc file
WEEK 13 – What have we done during the Trimester?
Topic 1 – Money
Simple Interest Formula
Finding P, R, T, I, M
Compound interest formula
Finding M, P
Multiple cash flows
Fv annuity
Find Fv, R
Finding Interest
Pv Annuity
Find Pv, R
Finding Interest
Outstanding balance
Combination of any above
Topic 2 ? Break even
Graphing BE
Finding BE point
Finding s, v, f, x
Breakeven from a table
Topic 3 ? Linear Programming
Drawing a 2D diagram
Answer Report
Optimal solution
Profit or loss
Slack values
Sensitivity Report
Changing the item price
Changing the ingredient
Shadow Price
Topic 4 ? Centre and Spread
Drawing a histogram
Interpreting a histogram
Shape
Outlier
Central Tendency
Dispersion
Standard deviation table
Topic 5 – Regression
Dependent & Independent
Interpret a scatterplot
Types of correlation
Line of best fit
Equation
Find x or y
Explain slope, intercept, r, r^2
Interpolation or extrapolation
Excel report
Residual plot
Topic 6 – Probability
Basic
And / or / not
Given / Suppose
Dependent / Independent
Venn Diagram
Topic 7 – Normal Curve
Draw or sketch a normal curve
68-95-99.7 rule
Z score and tables
Finding a probability
Finding a value
Normal curve and sampling
Topic 8 ? Confidence Interval
Finding the interval
Assessing the claim
Changing a variable
Finding the sample size
Topic 9 ? Hypothesis testing
The 5 steps
Z-test
Assessing the claim
Type 1 and 2 error
Topic 1 ? Money (How to choose)
Simple Interest
See ?simple or
?not compounding
Formula: I = PRT
Compound Interest
One payment at the start/beginning
Formula: M = P x [COL A]
Money
Sue is looking to upgrade her small business and decides to apply for a $75,000 loan at 8% p.a. compounding quarterly over 7.5 years.
Find the quarterly repayment
Loan ? Quarterly repayment ? Column D
A = 75000 I = 8/4 = 2 n = 7.5×4 = 30 r = ?
75000 = R [22.39645555]
75000 / 22.39645555 = R
R = $3349.70
Calculate the total interest
Interest = $3349.70 x 30 ? 75000 = $25491
Find the outstanding balance after 6 years
There 1.5 years or 6q left so n=6
OB = 3349.70 x 5.60143089
OB = $18763

Pauline is trying to calculate her savings goal based on the following conditions over 5 years
Choice 1:
At the start she deposits $16,800, then deposits $2,500 one year later and lastly deposits $13,600 one year later. The interest rate is 4% p.a. compounding yearly.
Choice 2:
A regular deposit of $5,200 every 6 months at 6% p.a. compound biannually.
Choice 1: Single payments at different stages ? that continuously build the amount
M = P (1 + i ) ^ n
M = 16800 [ 1.04 ] i = 4% n = 1 ColA M = 17472
M = (17472 + 2500) [ 1.04 ] i = 4% n = 1 ColA M = 20770.88
M = (20770.88 + 13600) [1.12486400] i = 4% n = 3 ColA M = $38662.56
Note: Build the total and add the single payment while decreasing the number of years
Choice 2: A regular annuity
S = 5200 x [11.46387931] i = 6/2 = 3 n = 10 Col C S = $59612.12
Tim is planning to have $125,000 over the next 18 months to start expanding his business. He agrees to a savings account that will earn 9% per year (p.a.) compounded monthly. What amount should he deposit into this account at the end of each month, in order to achieve its savings goal?
Regular savings annuity
S = 125000 r = ? i = 9/12 = 3/4 n = 18 Col C
125000 = R x [19.19471849]
R = $651.22 / m
Break even
A manufacturing company is planning a new sponge pack. Each pack has 5 sponges and each one has a soft thick foam side and a thin hard scrubbing side. The processing cost of the thick soft side is $0.22 and $0.62 for the thin harder side. The adhesive glue to join both parts is $0.08 each. The package labelling is $2.40 with a plastic wrapping of $0.15. Other costs during the production cycle are Labour $380, utility $240 and remaining is $60. They think a reasonable selling price is about $9.60 per pack.
Calculate the break-even number and amount ($).
Calculate the profit or loss if 150 are sold.
Contribution margin ratio % (cmr)
Complete on the diagram below
a. S = $9.60 V = 5*(0.22+0.62+0.08) + 2.40 + 0.15=7.15 F = $380 + 240 + 60 =680
BE = 680 / (9.60 ? 7.15) = 277.55 pack BE$ = 9.60 * 277.55 = $2664.48
b. TR = 9.60*150= 1440 TC = 680 + 7.15*150 = 1752.50 ?$x loss
c. Help Contribution = S ? V = 9.6 ? 7.15 = $2.45 / pack
CMR% = ( S ? V ) / S = 2.45 / 9.60 * 100 = 25.52%
Linear Programming
A factory makes gas meters and water meters. Gas meters need 4 gears, 1 dial and 8 minutes of assembly time for a profit. Water meters need 12 gears, 1 dial and 4 minutes of assembly time to yield a profit. There are 60 gears, 9 dials and 64 minutes of assembly time available for use in this production.
Answer Report
Sensitivity Report
What are your decision variables?
Gas meters (x) and Water meters (y) [Table 2]
What is the recommended solution and total value?
$213 6 gas and 3 water [Table 1]
Were all materials fully used?
No ? 4 minutes were left over [Table 3]
Can you find the new profit is each price is increased by $8?
OLD Price equation = $20x + $31y
NEW Price equation = $28x +$39y = Allowed = $28(6) +$39(3) = $285
How much will the profit change (if any) if you accidently break 3 of each ingredient?
Currently $213 – 3 gears ? 3 dials ? 3 minutes. All allowed so use the Shadow price for each one
$213 ? 3*$1.375 ? 3*$14.5 ? 3*(0) = $165.37
How much will my profit increase if I buy 23 more gears
Allowed? 23 < 48 YES, so allowed
How much will profit change = 213 + $1.375*23 = $244.62
What does the 1E+30 mean?
Infinity = a very, very big number
Why are you allowed to increase your minutes as high as you want?
Shadow price = $0 so it will not get you any more money? Why?
Because you already have 4 minutes left over [Table 3] slack

Centre & Spread
To test a new Energy Bar people were asked to trial a bar over a week and the average loss in calories was recorded. 87 people participated in the trial.
Type of data is being collected?
Calories: Numerical and continuous (measured)
What is the histogram modality (how many peaks) of histogram?
Unimodal (one peak)
Discuss the histograms shape and the supporting statistic?
Left or Negative skewed (therefore most of the data has been pushed high) also the skewness is -1.8
Discuss the best measure of centre and why it was chosen
Choice: mean median or mode Choose median = 177 loss of calories
Why: The middle number resists skewness (you would have to alter half the data to affect the median)

Measure of dispersion
Choice: st dev iqr range Choose IQR = 197.5 ? 155.5 = 42 calories
Why: IQR = is the middle 50% spread of data – it removes the top/bottom 25% area therefore removes the skewness and a major part of the tail
Are there any lower outliers?
LB < Q1 ? 1.5*IQR < 155.5 ? 1.5*(42) Q3 + 1.5*IQR > 197.5 + 1.5*(42) > 260.5 MAX is 224 so none
Lower boundary added to graph
Can you think of other variables that also need to be tested?
FOOD, Exercise, Age, Work
Regression
A company is tracking its share price since the beginning of the year. The analysis below represents the most of the 200 days since Jan 1st.
What is the dependent/response variable?
(y) Closing share price
Discuss the Correlation of the scatterplot
positive or negative positive ? as one variable increases so does the other
What is the y-intercept and explain it meaning
[Intercept]
It is the starting point on the y-axis = point (x, y) = (0, 0.33)
On average the starting point is (0, 0.33) at the beginning (x=0) the closing share price = $0.33.
The starting point in this example is possible
What is the slope and explain its meaning
X Variable 1
0.018 / 1 = y / x = change of closing share price / change of days
On average, for every ?1 day the ?Closing share price increases by 0.018
Make the linear equation (from the bottom of the table)
y = [X variable 1] + [Intercept] y = 0.0187x + 0.3338
Find the closing share price based on 200 days after the start.
x = 200 days
y = 0.0187(200) + 0.3338 = $4.07 (extrapolated) ? Shown on graph with red arrow
Explain r^2 the coefficient of determination
R Squared = 0.9162
On average 91.62% of the change in (y) closing share price is explain by the number of days (x)
Is this a good model?
Depends on two factors
The Residual plot shows a random distribution of points
The R^2 is at least good (>50%)
OR
(1) Bad model! A residual plot with a pattern (n, D, C, M, W, U) OR a very low R^2
Therefore the model is good!
Normal Curve
Explain the highlighted sections
Show the highlighted information of the diagrams below
Suppose that the lifespan of batteries made are normally distributed with a mean length of 80.56 days and a standard deviation of 4.3 days. Determine ?
What is the probability that a battery will last longer than 84 days?
P(x > 84)
z = (x ? mean) / stdev = (84 ? 80.56) / 4.3 = 0.80
Left Prob 0.7881
Right Prob (last longer) = 1 ? 0.7881 = 0.2119 = 21.19%
the probability that the average battery will last longer than 84 days from a pack of 16 batteries?
n = 16 (sample)
z = ( x ? mean) / (stdev / sqrt(n))
Based on the CLT
Average = 80.56 AND StDev = std / srt(n) = 4.3 / sqrt(16) = 4.3/4 = 1.075
z = (x ? mean) / stdev = (84 ? 80.56) / 1.075 = 3.2
Left Prob = 0.9993
Right Prob = 1 ? 0.9993 = 0.0007 = 0.07%
the battery that will be in the top 10% of all batteries

P (x > k) = 0.10 OR
P (x < k) = 0.90
From table: z = 1.28
z = (x ? mean) / stdev
1.28 = (x ? 80.56) / 4.3
1.28 x 4.3 + 80.56 = x
x = 86.064

Probability
P(male) = 150 / 400
P(not liking) = (100 + 50) / 400
P(a neutral male) = P(neutral and male) = 50 / 400
P(female or like) = (30 + 30 + 0 + 220) / 400
Suppose you like it what is the probability you are a female? Not out of 400 = CONDITIONAL
P(female| like) = 220 / 250 = 0.88 = 88%
What is the probability of being a male IF you are neutral? P(male | neutral) = 50 / 50= 1 = 100%
Prove if: males and neutral events are independent?
P(males) * p(neutral) = p(males and neutral)
LHS 150/400 * 50/400 = 0.046 RHS = 50/400 = 0.125 different = both events dependent
Complete the Table
60% of the people chose the Morning time
12% of people said they would choose Take-away in the afternoon
Of the 55% of people that chooses to Eat-in, 27% of people said they would do so in the morning
Is the choice of a Morning time independent of choosing Take-Away
P(morning) * p(take-away) = p(morning and TA)
LHS 0.60 x 0.45 = 0.27 RHS = 0.33 Events are dependent (different)
Interval
The battery company wanted to test their competitors product and see how reliable their claim is. The competitor claim states that their batteries are all better than the average life span of at least 80.5 days. To test this they select a sample of 32 batteries that produced an average life span of 85 days with a population standard deviation of 22.5 days.
Construct a 95% confidence interval
CI = 0.95 Sample mean = 85 n = 32 st dev = 22.5 z = +/- 1.96
85 ? 1.96 x ( 22.5 / sqrt (32) ) < pop mean < 85 + 1.96 x ( 22.5 / sqrt (32) )
77.204 < pop mean < 92.796
Explain the confidence interval
Based on a 95% confidence interval (95 out of 100 samples) the population mean has a range from
77.3 days to 92.7 days
Assess the claim
the claim is inconclusive as the range only represents part of the interval ? some of it is witihin and some is outside
Assuming a 95% confidence interval what sample size would be needed for the claim to be true?
Choose the lower end
85 ? 1.96 x ( 22.5 / sqrt(n) ) = 80.5
? 1.96 x ( 22.5 / sqrt(n) ) = 80.5 – 85
? 1.96 x ( 22.5 / sqrt(n) ) = – 4.5
1.96 x ( 22.5 / sqrt(n) ) = 4.5
( 22.5 / sqrt(n) ) = 4.5 / 1.96
22.5 / sqrt(n) = 2.2959
22.5 / 2.2959 = sqrt n
(22.5 / 2.2959)^2 = n n = 96.04 = 95 batteries
Hypothesis Testing
A repair shop states that all their complex repairs average about 6.2 days to fix. To test this a hypothesis was done on 40 complex repairs which produced a sample average of 6 days. Suppose that the population standard deviation is 0.7 days at the 5% level of significance does their claim hold true?
Step #1
Step #2
Step #3
Step #4
Step #5
At the 5% level of significance there was in-sufficient evidence that the sample result was different than the claim.
The claim is correct that the average of all repairs take approximately 6.2 days to fix
Page 1 of 2 MATH 1053 Quantitative Methods for Business Primary Examination ? Trimester 1, 2022 Time Allowed Normal Exam time: 150 min Extra time: +15 min Total time: 165 min Exam Format Hand-written Y Allowed Materials ? All materials summarised in Task 1 on the Student Lounge ? Additional materials include: Pages (no.) Blank: 10 Lined: 0 Graph: 1 Calculator None Basic Y Scientific Y Graphical Y Paper Dictionary English only Y Dual Language Hand-written Notes None Y Unlimited Single Side (1pg) Single Side (2pg) Double Sided (1pg) Double Sided (2pg) Text book Moodle Page Other 3 formulae pages may be removed from the exam Forbidden Materials Scrap paper, Paper Booklets, Printed Materials, Online Dictionary, Online Translator Important Information All students must be ready and adhere to Task 1 before the exam can begin. The Invigilator will then open the Moodle examination link and you can refresh the page, open the exam and beginThere are (enter) parts to the examination. Part A (enter) marks Part B (enter) marks At the end of the exam, submit your work as per Task 2 instructions on the Student Lounge
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