What is in the Exam 6900 Updated

8/10/2019 What is in the Exam 6900 Updated

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What is in the 6900 exam It will help you with the assignment page page 2 and 23 is like assignmentquestion 5a and b) page 20 and 21 help with question 6 pages 18 and 19 help with question 7Students study the wrong things for the exam so ask me if you are not exactly sure about what to study.

My handouts have always been about what is the most important thing in the exam so I have included thehandouts for weeks 7,8,910,11 and 12 in this document.

The later topics are more important than the recent topics so study them first.

It will be much harder than the week 5 and week 8 test because you cannot be provided a similar test to use as aguideMY GUESS OF THE FORMAT OF THE EXAM, It does not have this format

14% picking the correct finance formula

4% subbing in the proper numbers into the correct finance formula

4% harder finance

10% NPV IRR (there is a small chance you will do 2 annuity questions instead)

13% for not confusing the standard answers of regression, This is easy marks You just use past papers as a guideand give the exact answers by using the numbers in the output and giving the exact explanation of the numbers.

14% time series including explaining EXACTLY why i ncreasing the smoothing constant makes the series lesssmooth (refer to past exams) , Also note which type of smoothing smoothes the series the best, In this case youlook for the smoothing with the lowest MAD and SSE.

7% simple index numbers INCLUDING EXACT INTERPRETATION

7% not confusing the Laspeyres, Paasche, average of relatives and simple aggregate index numbers and actuallysubbing in the numbers properly, Many students do not sub in the numbers properly. . also give exactinterpretation.

5% using the correct confidence interval formula (there are 3 confidence interval formulas) the correct one is

PROBABLY going to be a 95% CI withn

st x where you use the t- table and it probably wont be

n x

2 because using 2 is to easy

5% hypothesis testing of the mean, use the formula test statistic =n s

x

/

you have to compare this to a t-table

value2% subbing in the numbers into the confidence interval formula.

7% Using the z table to find a probability, once with a single value so divide by standard deviation and once witha mean so divide by standard error.

5% From anything in the course.


2/24

Week 12 handout

1) Confidence interval harder version And hypothesis testing of the mean question.Someone doubts that a certain dice is a fair so they throw the dice 16 times and find that the mean is 3.6 andstandard deviation 1.6

a) Find the standard error b) Find a 95% confidence interval for the meanc) Test claim mean is above 3.5 using a 5% level of significanced) What are the assumptions of the test.

Solution n=16, 6.3 x , s=1.6

a) Standard error =166.1

=0.4

n less than 30 and population standard deviation is not given so we must use the t-table instead of the number 2Use t-table row= df=n-1=15, column =0.025 and get the value 2.1315Between 3.6-2.13150.4 and 3.6+2.13150.4So we are 95% confident the mean is between 2.75 and 4.45

c) H0: =3.5, H1: >3.5

Use row= df=n-1=15, column =0.05 and get the value 1.753

Test statistic = 25.04.0

5.36.3

The test statistic is not above the critical value so we do not reject H0.So we do not have strong evidence the mean is above 3.5

d) We need to assume the data is normally distributed.


3/24

2) 6900 only exam question NPV viability and IRR

Project 1 has initial outlay $100,000 and yearly returns of $60,000 and $70,000Project 2 has an initial outlay of $200,000 and yearly returns of $100,000 and $150,000

(a) Find NPV of each project if interest is 10% and comment(b) Find NPV of each project if interest is 20% and comment(c) Find the IRR of each project and comment(d) What are the disadvantages of only using NPV to compare projects or only using IRR to compare projects.

Solution(a) Project 1 NPV if interest is 10%-100,000+60,000(1.1) -1+70,000(1.1) -2= $12396.69Project 2 NPV if interest is 10%-200,000+100,000(1.1) -1+150,000(1.1) -2=14876.03Comment: Both projects are viable because they both have positive NPV,If you only consider NPV Project 2 is better because it has the higher NPV(b) Project 1 NPV if interest is 20%-100,000+60,000(1.2) -1+70,000(1.2) -2= -$1388.89Project 2 NPV if interest is 20%-200,000+100,000(1.2) -1+150,000(1.2) -2=-$12500

Comment: Both projects are not viable because they both have a negative NPV,If you only consider NPV Project 1 is not as bad as project 2 because you lose less money with project 1(c ) Project 1 IRR

IRR =21

1221

N N I N I N

= 1899.0)89.1388(69.12396

1.0)89.1388(2.069.12396

Project 2 IRR

IRR =21

1221

N N I N I N

= 1543.0)12500(03.14876

1.0)12500(2.003.14876

Comment project 1 has a higher IRR, so if you are only considering IRR project 1 is better.(d) NPV and IRR do not consider the initial outlay and how long you have to wait for profit. (see lecture notes forother comments)


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Sinking fund example You do not need to know what a sinking fund is but you do have understand the phrase interest only is paid every quarter

1) A sinking fund is set up to pay off the principal of a $20000 loan in ten years time.Suppose interest on the loan is 12% compounded quarterly. The sinking fund has interest of 10% where the

payments are quarterly.The term of the loan is interest only is paid every quarter.

(a) What is the interest paid(b) How much has to paid into the sinking fund each quarter

(c) What is the total amount paid each quarter.Solution

(a) i.=0.12/4=0.03, So amount paid every quarter is 200000.03=$600

(b) i.=0.10/4=0.025, n=104=40, So 72.296$1)025.1(

025.02000040

R

(c) 296.72+600=$896.72


5/24

WEEK 9 FINANCEYOU CAN BRING IN 2 A4 PAGES (BOTH SIDES) OF NOTES INTO EXAM, I GIVE NO MARKS FORONLY GIVING THE FORMULA SUCH AS OR

Sample finance questions1a) Sue Invests $1000 for 5 years at 12% p.a. compounded monthly, what is the final amount?

b) John Invests $20,000 for 10 years at 12.4% p.a. compounded quarterly, what is the final amount?c)Calculate and interpret Sues effective inter est rate for investing for 1 year.

d)Calculate Johns effective interest rate for investing for 1 year and explain it in the context of the problem.e)Use the effective interest rates to work out who is getting a better deal (WARNING IN THE EXAM THEYWILL JUST ASK PART (E) THEY WILL NOT HELP YOU BY MAKING YOU DO PARTS c) and d)

SOLUTIONa) interest per month i = 0.12/12=0.01, number of months in 5 years n=512=60

Final amount =1000(1+0.01) 60=$1816.70

b) interest per quarter i = 0.124/4=0.031, number of quarters in 10 years n=410=40Final amount =20000(1+0.031) 40 =$67822.94

c) interest per month i = 0.12/12=0.01number of months in 1 year = 12Effective rate= (1+0.01) 12-1=0.1268=12.68.%

If you invest at 12% p.a. compounded monthly your money will grow by 12.68% every year

d) interest per quarter i = 0.124/4=0.031number of quarters in 1 year = 4Effective rate= (1+0.031) 4-1= 0.1299=12.99.%If you invest at 12.4% p.a. compounded quarterly your money will grow by 12.99% every quarter

e) Investing at 12.4% compounded quarterly is better than investing at 12% compounded monthly because it has ahigher interest rate

YOU NEED TO UNDERSTAND QUESTION 2 AND 3 ARE VERY DIFFERENT TO QUESTION 1 THEYARE USING A TOTALLY DIFFERENT FORMULA

2) A small business borrows $60000 which is to be repaid monthly over a five year period. Assumeinterest is constant at 24% per annum, payable monthly. Find the monthly repayment requiredassuming payments are made at the end of each month.

Solution Number of payments =512=60, interest between each payment =0.24/12=0.02,Amount borrowed = 60000

Size of repayments =

3) Mortgage/Home loan questionI Borrow $300,000 to buy a house. What is the size of the monthly repayments if the interest is 6%

p.a. compounded monthly and the term of the loan is 25 years?Solution

Number of payments =2512=300, interest between each payment =0.06/12=0.005,Amount borrowed = 300000

Size of repayments =

07.1726$

])02.01(1[

6000002.060

90.1932$])005.01(1[

300000005.0300

ni P S )1( ])1(1[ ni

iP R


6/24

Week 101) Find the net present value of an investment where I have to initially pay $10,000 but I will get a return of$6,000 next year and $8,000 in two years time if interest is 10% compounded yearly2)a) Fred has agreed to repay a $74,000 reducing balance loan with quarterly installmentsover 20 years at 8% compounded quarterly. Find the installment value.

b) If Sue has borrowed a total $100,000 what are the weekly repayments in interest is 7.3% compounded dailyand the term is 10 yearsc) Find and interpret the effective rate for Fred and Sue, Which person is getting a better deal in terms of effectiveinterest rate?

3) I invest the sum of $2500 for 10 years. What is the final amount if interest is 3.65% compounded dailySolution1) Net present value =-10,000+6,000(1.1) -1+8,000(1.1) -2=2066.122a) P=74000, n = 204=80, i= 0.08/4=0.02R=0.0274000/(1-(1.02) -80)=$1861.89

b) P=100,000, n =payments= 5210=520Interest between each payment .i= (1+0.073/365) 7-1=0.0014 **NOTE THE EXAM MARKER KNOWS THISIS TRICKY SO THERE IS A SMALL MARK PENALY FOR GETTING IT WRONGR=0.0014100,000/(1-(1.0014) -520)=270.86c) Fred is 8% compounded quarterlyQuarters in 1 year = 4, Interest per quarter = 0.08/4=0.02

Effective yearly rate = (1+0.02)4

-1=0.0824=8.24%So Freds debt grows by 8.24% every yearSue is 7.3% compounded dailyDays in 1 year = 365, Interest per day = 0.073/365=0.0002Effective yearly rate = (1+0.0002) 365-1=0.07572=7.522%So Sues debt grows by 7.522% every yearIf it is a loan You want the lowest effective rate, If you are making an investment you want the highest effectiverate Since Sue and Fred have a loan Sue has the better deal.

3) P=2500, n = 10365=3650, i= 0.0365/365=0.0001S=2500(1+0.0001) 3650=3601.22

ON RESOURCE DRIVE , LECTURE FOLDER WEEK 10 handout.A)Dave invests 1000 for 2 years at 5.2%p.a compound weekly what is the final amountP=1000, n = 252=104 , i= 0.052/52=0.001S=1000(1+0.001) 104=1107.33B)Fred invests 75 for 3 years at 5%p.a compound quarterly what is the final amountP=75, n = 34=12 , i= 0.05/4=0.0125

S=75(1+0.0125) 12=87.06C) Find and interpret Dave and Freds effective interest rates, which is the bestDave is 5.2 % compounded weeklyWeeks in 1 year =52 , Interest per week = 0.052/52=0.001


7/24

Effective yearly rate = (1+0.001) 52-1=0.0533=5.33% (2 marks)So Daves money grows by 5.33% every year (1 mark)Fred is 5% compounded quarterlyQuarters in 1 year = 4, Interest per quarter = 0.05/4=0.0125Effective yearly rate = (1+0.0125) 4-1=0.05095=5.095% (2 marks)Fred money grows by 5.095% every year (1 mark)Dave has best deal because he has the higher effective rate (2 marks)


8/24

Week 11 handoutEvery student clearly showed that they were not approaching the course in the way we want you to, AND WILLLOSE ABOUT 10% IN THE FINAL EXAM for using the wrong approach. (

WARNING FOR FINAL EXAM YOUR METHOD FOR CONFIDENCE INTERVALS FOR THE MEANIS WRON G (IT IS TO SIMPLE YOU CAN ONLY USE Z=2 FOR a 95% CI for proportions of if isgiven).

For regression and confidence intervals for the mean with small sample size and unkown standard deviation youuse t-table So you have to check the normality assumption (for regression there are other checks as well).Sample finance questions

1) Suppose Interest is 12% compounded quarterly. If I need to borrow $10,000 to buy a car and I make a $5,000dollar payment in 1 years time and a final payment in two years time(a) Find the value of the final payment(b) Find and interpret the effective rate of interest

2) I borrow $200,000 to buy a water jetpack.(a) What are the monthly installments if interest is 12% compounded monthly and the term is 20 years(b) After 15 years the interest rate changes from 12% compounded monthly to 6% compounded monthly. What isthe value of the new repayments.

3) I invest the sum of $1000 how much do I have after 10 years if interest is 10.4% compounded weekly

4) I invest $1000 every week, What is the of the sum of the future value of my investments after 10 years.Interest is 10.4% compounded weekly.

1a) Let the final payment be X10,000=5,000(1.03) -4 +X(1.03) -8

10,000= 4442.44+X(0.7894)So X = 7040.23

b) 4Quarters in 1 year, interest is 0.12/4=0.03 per quarter so effective rate is (1.03) 4-1=0.1255=12.55%

2) a) number of payments n=1220=240, interest between each payment i= 0.12/12=0.01

R= )01.11( 000,20001.0 240 = $2202.17

b) Payment left after 15 years = 512=60, .i=0.01,

amount owing after 15 years =01.0

)01.11(17.2202 60=98,998.64

For the final 5 years number of payments n=512=60, interest between each payment i= 0.06/12=0.005

R=)005.11(64.98998005.0

60 = $1913.92

3) number of compoundings n=5210=520, interest rate .i= 0.104/52=0.002

S=1000(1.002)520

= 2826.28

4) number of payments n=5210=520, interest between each payment i= 0.104/52=0.002

S=002.0

)1)002.1((1000 520= $913,140.04


9/24

Week 11 handout harder finance questions

Harder questionIf I borrow $3,000 to buy a new computer and I make monthly repayments of $50 how long does itTake to pay of the loan if interest is 12% compounded monthlySolution

R = 50, P=3000, .i=0.12/12=0.01

So 93 payments so 93 months, this is 93/12=7.75 years so 7 years and 9 months

Another harder question A $200,000 conventional variable-rate mortgage has monthly payments over 30 years and the interestrate is 4.4% p.a. payable monthly

(a) What is the size of the repayment ?(b ) What is the outstanding balance at the end of two years with this interest rate (assuming allmonthly payments have been paid)?(c) After two years, the interest rate for this mortgage is changed to 6.5%.What is the new monthly payment amount (payable for 28 years) at thenew interest rate of 6.5%?

Solution a) number of payments n=1230=360, interest i= 0.044/12=0.00367

R=)00367.11(

000,20000367.0360 = $1001.99

b) Payment left after 2years = 2812=336, .i=0.00367,

amount owing after 2 years =00367.0

)00367.11(99.1001 336 =193,288.37

c) number of payments n=2812=336, interest i= 0.065/12=0.00542

R=)00542.11(37.288,19300542.0

336 = $1251.11

.

08.9201.01log

50)300001.050(log1log

)(logi

RiP Rn


10/24

WEEK 8 HANDOUT.INDEX NUMBERS SAMPLE QUESTIONS1) Find and interpret the simple index number for bread using the year 2000 as the base.Food item Year 2000 $Price Year 2010 $PriceBread 2.00 3.00Solution: simple index = 3/2=1.5 So the bread price has increased by 50%2) Using the information belowa)Find and interpret the Laspeyres index of Prices uses the year 2000 as the base.

b) Find and interpret the Paasche index of Prices using the year 2000 as the base.

c) State the disadvantages of each indexFood item Year 2000

$PriceYear 2000Quantity

Year 2010$Price

Year 2010Quantity

Bread 2.00 10 3.00 12Chocolate 4.00 1 5.00 3

Solution

a) Laspeyres index= 46.11410215103

so the price of food items has increased by 46%

b) Paasche index= 42.13412235123

so the price of food items has increased by 42%

c) Laspeyres index overweights quantities that increase in price, The Passche under weights quantities thatdecrease in price (NOTE FOR STUDENTS This is a direct quote from lectures this is acceptable because you arecommenting on the disadvantages of a method)

3) Find the simple price index numbers using the year 2004 as the base and explain it in the context of the problemYear USB Price2004 1002006 502010 10Solution:for 2006 the simple index = 50/100=0.5 So the USB price has decreased by 50% when you compare 2010 to 2004for 2010 the simple index = 10/100=0.1 So the USB price has decreased by 90% when you compare 2010 to

2004

4) Using the information belowa)Find and interpret the Laspeyres index of Prices uses the year 2004 as the base.

b) State the differences between the Laspeyres Index and the Paasche Index.Food item Year 2004 $Price Year 2004 Quantity Year 2010 $PriceUSB 100 2 10Blank CD 1.20 10 1.18

Solution

a) Laspeyres index= 15.0102.121001018.1210

so prices have decreased by 85% when you compare 2004 to 2010

b) Laspeyres Uses the base year quantities as the weights and Paasche uses the current year quantities as theweights


11/24

WEEK 8 HANDOUT3)A hospital buys four products with the following characteristics:

Price paid per unit ($)

Product Year 2001 Year 2002

Bandages 10 11

Saline solution 23 25

Sheets 17 17

Thermometers 19 20a) Find the average of relatives index number for 2002 using 2001 as the base

b) Find the simple aggregate index number for 2002 using 2001 as the base year

solutiona) *** Note you do not need this much setting out in the exam****

Price paid per unit ($)

Items Year 2001 Year 2002 Index for 2002 using 2001 as the baseBandages 10 11

11/10=1.1Thermometers 23 25

25/23=1.09Sheets 17 17

17/17=1Bedpans 19 20

20/19=1.05

Average or relatives=average of indexes =(1.1+1.09+1+.05) /4=1.06 b)The simple Aggregate index is the sum of the current year prices divided by the sum of the base year prices

Simple aggregate Index 058.16973

1917231020172511

1) The advertising expenditure by a supermarket was recorded over a five year period(a) Find the simple index numbers using year 2001 as the base

Year 2001 2002 2003 2004 2005

Advertising ($) 3000 5880 6600 8250 9240

(a) Find the simple index numbers using year 2001 as the base(b) Find the simple index numbers 2003 as the base year

Solution, NOTE THAT MULTIPLYING BY 100 IS AN OPTIONAL STEP YOU DONT HAVE TO DO IT

Year 2001 2002 2003 2004 2005

(a) Simple Index 2001 as base year 3000/3000100= 100 196 220 275 308

(b) Simple Index Using 2003 as a base year 6600/3000100= 45.5 89.1 100 125 140


12/24

6900 handout for the tutorial week 91) Quartely sales for 3 years are given in the plot below, A linear trend line was fitted to the data. and was foundto be Y = 310-10 X

The seasonal indices areQuarter 1 2 3 4Index 0.8 1.01 0.99 1.2

a) Identify the Independent variable and dependent variable in the regression line. b) Interpret seasonal index for quarter 1c) Predict sales for quarter 1 2011 The regression lined) Predict sales for quarter 1 2011using regression line and the index numbere) Predict sales for quarter 2 2011The regression linef) Predict sales for quarter 2 2011using regression line and the index numberg) Comment on the time seriesh) Did we use a Multiplicative model or an additive model

0100200300400

Q12008

Q22008

Q32008

Q42008

Q12009

Q22009

Q32009

Q42009

Q12010

Q22010

Q32010

Q42010

S a

l e s

Quarter

0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

0 2 4 6 8 10 12 14

D e

t r e n

d e

d S a

l e s

Time


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6900 tutorial handoutSolutiona) The independent variable is Time and the dependent variable is Sales

b) Sales in quarter 1 are 20% lower than the regression line on averagec) Sales = 310-1013=180d) 1800.8=144e) Sales = 310-1014=170f) 1701.01=171.7g)

*there is a Strong seasonal component, Quarter 4 has high sales and quarter 1 has low sales*No apparent cyclic component*Irregular component, there are minor variations each year*There is a decreasing trend component since the prices are decreasing as time increases.h) Since the points are scattered about 1 in the Detrended plot, We are using the multiplicative model.

2) Using the information belowa)Find and interpret the Laspeyres index of Prices uses the year 2000 as the base.

b) Find and interpret the Paasche index of Prices using the year 2000 as the base.c) What is the difference between The Laspeyres index and the Paasche index

Solution

a) Laspeyres index= 17.10100603540506006041406

so the price of has increased by 17%

b) Paasche index= 48.12100833515526008341156

so the price of shares has increased by 48%

c) Laspeyres index uses the old quantities as weights whereasPaasche uses current quantities as the weights

3) Using the information belowa)Find and interpret the simple index number for Bread Using the year 2000 as the base year

b) Find and interpret the simple index number for Bread Using the year 2000 as the base yearc) Find the average of relatives index for Food items and explain the value in the context of the problem.Food item Year 2000

$PriceYear 2010$Price

Bread 2.00 3.00Chocolate 4.00 5.00

Solutiona) 3/2=1.5 so the price of bread has increased by 50%

b) 5/4=1.25 so the price of chocolate has increased by 25%

c) Find the average of all the index numbers, the average is 375.12

25.15.1 so prices have increased by

37.5%

Share Price 2004 SharePrice

2004quantity

2010 share price

2010 quantity

Toyota 5 40 6 15Coco Cola 35 60 41 83Google 100 0 600 2


14/24

WEEK 8 HANDOUT 1 page 1Sample time series question,

1) Smooth the following series with a 3 point moving average, Exponential smoothing with smoothingconstant w=0.2 and exponential smoothing with w=0.7 and plot the original series with the smoothedseries.

WeekAttendance in

'000's1 35

2 233 454 325 386 737 598 61

Solution NOTE THAT I HAVE ONLY GIVEN YOU SOME OF THE VALUES BECAUSE THAT IS WHATHAPPENS IN THE EXAM.

WeekAttendance

in '000's Moving AverageExponentially smoothed

= 0.2 Exponentially

Smoothed, = 0.7 1 35 35 352 23 34.33 32.60 26.603 45 33.33 35.08 39.484 32 38.33 34.46 34.24

5 38 (32+38+73)/3=47.670.2*38+(1-0.2)34.46

=35.17 0.7*38+(1-0.7)34.24 =36.87

6 73 56.67 42.74 62.167 59 64.33 45.99 59.958 61 48.99 60.68

01020304050

607080

0 5 10

A t t e n

d a n c e

Week

Weekly Football Attendance

0.00

10.00

20.00

30.00

40.00

50.00

60.00

70.00

80.00

1 2 3 4 5 6 7 8

Week

A t t e n

d a n

c e A

B

C

D

D is the least smooth so it isThe original series

C has missing values so itMust be the moving average

Series A has the most smoothing soit is so it is exponential smoothingwith the lowest value so this isexponential smoothing with w=0.2

B has the less smoothing relative toA so it is so it is exponentialsmoothing with the higher value so ithas smoothing w=0.7


15/24

WEEK 8 HANDOUT 1 page 22) Time series question, Consider the time series plot and corresponding detrended plot for quarter sales dataStarting at quarter 1 2002 and finishing in quarter 4 2006 (So 5 years of data)Original data Detrended sales plot

The regression line to model the trend component is Y = 480+20XQuarter 1 2 3 4Index 0.99 1.2 0.98 0.83

i) Identify the Independent variable and dependent variable in the regression line. j) Interpret seasonal index for quarter 1k) Predict sales for quarter 1 2007 The regression linel) Predict sales for quarter 1 2007 using regression line and the index numberm) Comment on the time seriesn) Did we use a Multiplicative model or an additive model

Solutiona) The independent variable is Time and the dependent variable is Sales

b) Sales in quarter 1 are 1% lower than the regression line on averagec) Sales = 480+2021=900d) 9000.99=891e) Strong seasonal component, Quarter 2 has high sales and quarter 4 has low sales

No apparent cyclic componentIrregular component, there are minor variations each yearThere is an increasing trend componentf) Since the points are scattered about 1 in the Detrended plot, We are using the multiplicative model.

4205206207208209201020

112012201320

0 10 20 30

S a

l e s

$ ' 0 0 0 s

Time (quarters)0

1

2

0 5 10 15 20

Time


16/24

Week 8 handout 2, The sort of time series question you USUALLY get in the exam.Time series question: A key performance indicator for the tourism industry is the quarterly room night occupancy figures

for accommodation. The table below shows the number of room nights occupied in Australia for each quarter from2001 to 2003. The graph shows the quarterly data from 2001 to 2006. Also shown are two smoothed series of theoriginal data, including 4-point moving average and a centred moving average.

a) Determine the respective smoothed values for the missing data points. b) Explain why it was appropriate to use a centred moving average in this case.c) Which of the two smoothed series has been plotted with the original data below?d) What is the purpose of smoothing data such as this?

e) Identify the time series components that are evident in the plot below.f) Which Smoothed time series fits the data best.g) What is the effect of making the smoothing constant w smaller?

Qtr Occupancy 4-Pt M.A4pt centered

MA

Exp smoothw=0.4

1 2001 10088 10088

2 10000 10052.8

3 10449

10270

10282.63 10211.3

4 10543

------------

10279.38 10344

1 2002 10189

10263.5

10284.75 10282

2 9873

10306

10300.13 --------

3 10619

10294.25

10324.13 10318.6

4 10496

10354

------------ 10389.6

1 2003 10428

10341.75

10361.63 10404.9

. . MAD=5SSE=36

MAD=9SSE=92

a) 4-Pt M.A = (10000+10449+10543+10189)/4 = 10295.25

CMA (4 point centered moving average)= (10354.0+10341.75)/2 = 10347.88

Exponential smoothing (using w=0.4) = 0.4 9873+0.6 10282= 10118.4

b) When calculated a 4-Pt MA is placed between the 2 nd and 3 rd quarters. This becomes a problem whenplotting. Centring these 4- Pt MAs results in mapping them against 3 rd quarter, 4 th quarter, etc. This resultsin more regular plots, easy to interpret, etc

c) The Centred Moving Averaged) Smoothing data such as this smoothes out seasonal and irregular components in a time series and tends to

highlight the trend.

e) There is a general positive trend over the 5-year period.The seasonal patterns are very clear with major peaks occurring in the 3 rd and 4 th quarters and distinct trough in the2nd quarter.There is no clear evidence of any cyclical pattern.The differing sizes of the peaks and troughs represent the irregular component of the time series.

f) the center moving average has the lower MAD and SSE so it is a better fitg) If w get smaller then there is less weight placed on the original values so the series will become more smooth

Measures of accuarcy4pt centered MA Exp smooth w=0.4

MAD=5 SSE=36 MAD=9 SSE=92


17/24

Week 7 handout

The sample slope 1

=b1 estimates the population slope 1

The sample intercept 0

=b0 estimates the population intercept 0 The standard error of the line x y 10

will always be given it measures how close the pointsare to the line.The sample slope and sample intercept are called the coefficients, you are given a standard errornext to the coefficient this tells you an idea of how close the sample coefficient is to the

population coefficient

1) Using the data below(a) State the regression line(b) Predict sales in January 2011 (this is time 5)(c) Test the claim that time is a useful predictor(d) Do you think the model is a good model(e) Interpret the coefficient of determination (f) Comment on the relationship(g) Interpret the slope

Year month Time Sales

2010 Sep 1 902010 Oct 2 822010 Nov 3 702010 Dec 4 61

SUMMARY OUTPUTR Square 0.995Standard Error 1.162

CoefficientsStandard

error P-valueIntercept 100.5 15 0.0002Time -9.9 2 0.0027Solution

(a) Sales = 100.5-9.9 Time(b) Sales =100.5-9.95=51 (c) H0: 1=0 H1: 10

The pvalue is 0.0027 this is less than 0.05 so we have strong evidence the slope is significantlydifferent to zero so we have strong evidence that Time is a useful predictor

(d) The model is a good model since the slope is significantly different to 0 (pvalue < 0.05)The standard error is smaller than the spread of the y values and the R squared is high(R 2>0.64 so strongRelationship)

(e) R 2=0.995 so 99.5% of variation in sales is explained by the line(f) Strong negative linear relationship


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(g) Increasing time by 1month decreases sales by 9.9units


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2) Using the data below(a) State the regression line(b) Predict sales in January 2011 (this is time 5)(c) Test the claim that time is a useful predictor(d) Do you think the model is a good model(e) Interpret coefficient of determination (f) Comment on the relationship(g) Interpret the slope

Year month Time Sales

2010 Sep 1 812010 Oct 2 1002010 Nov 3 902010 Dec 4 80

SUMMARY OUTPUTR Square 0.018Standard Error 11

Coefficients Standard erro P-valueIntercept 102 20 0.041Time -2 4 0.865Solution

(a) Sales = 102- 2Time(b) Sales =102-25=92 (c) H0: 1=0 H1: 10

The p=value is 0.865 this is not less than 0.05 so the slope is not significantly different to zeroso we do not have evidence that Time is a useful predictor

(d)The model is a not a good model since the slope is not significantly different to 0 (pvalue>0.05)The standard error is not smaller than the spread of y values and the R squared is low(R 2


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5) Suppose we have performed a regression where the Toyota return is the y variable (the dependent variable)and Honda return is the x variable (the independent variable) and we obtained the output below. (We used

percentage units)Regression Statistics

R Square 0.98

Standard Error 0.06Coefficients standard error P-value

Intercept 0.01 0.007 0.31Honda 0.99 0.05 4.50E-89

(a) State the regression equation(b) Interpret the gradient(c) Predict the Toyota return if the Honda return is 0.5 (using percentage units)(d) Test the claim that the slope is different to 0

(a) Toyota = 0.01+ 0.99Honda(b) If Honda increases by 1 percentage unit then Toyota increases by 0.99 percentage units(c) Toyota = 0.01+ 0.990.5=0.505 percentage units(d) H0: 1=0 H1: 10The pvalue is 4.50E-89 this is less than 0.05 so we have strong evidence the slope is significantly different tozero so we have strong evidence that Honda is a useful predictor

These last two questions are index number questions they should help you understand what interpret means.5) Find and interpret the simple index number for bread using the year 2000 as the base.Food item Year 2000 $Price Year 2010 $PriceBread 2.00 3.00Solution: simple index = 3/2=1.5 So the bread price has increased by 50%

6)Find and interpret the simple index numbers for taxis using 2007 as a baseYear Price$2007 2.002008 2.202009 2.40

Solution(a) 2008 Index = 2.2/2=1.1 so prices have increased by 10%

2009 Index =2.4/2=1.2 so prices have increased by 20%


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1. The selling price of a house can depend upon many factors. A study of recent selling prices for a sample of Ballarathouses took into consideration the size of the house, the age of the house and the size of the land upon which the housewas built. Correlations were obtained for each of the variables and are shown in the output below.

Price Size AgeSize 0.849Age -0.589 -0.809Land 0.247 -0.171 0.410

a) Which variable correlates highest (highest strength) with Selling price ?

b) Is the relationship in part (a) positive or negative ?

c) What does the answer to (a) and (B) mean in terms of the two variables?

d) Which two variables show the least correlation?

e) Explain why you think the variables in b have low correlation

answer1. a) Size (r=0.849 has the largest absolute value) ; b) Positive; c) As the size of the house increases so to does the price.;

d) Between size and land; (the absolute value of -0.171 is 0.171 this is the smallest) e) Perhaps people tend to build ahouse they can afford regardless of the size of the land and buy a block of land they can afford, regardless of the sizeof the house they intend to build.

Sample final exam hypothesis test question1)In a sample of 235 babies where the father was a dwarf and the mother was not a dwarf the baby was a dwarf108 cases. Test the claim the proportion is different to 50%. Use a 5% level of significance.

Solution : n=235, 46.0235108

p , standard error is =

235

5.015.0 =0.0326

H0: p=0.5, H 1 : p0.5

We have to use the t- table, but for proportion use z which is t with df=

Using the z row (bottom row) of the t-table, we need 0.05/2=0.025 on each side so use column 0.025

test stat z=

235)5.01(5.0

5.046.0)1(

n p p

p p=

0326.05.046.0

=0326.0

04.0=-1.23,

The test stat is not in the rejection region so do not reject H 0There is not strong evidence the proportion is different to 50%

Note that z scores can give probabilities as well so if you were asked what is the probabilityThat the sample proportion of dwarves is less than 46% if you have a sample of 235 babies andthe proportion of dwarves is 0.solution

z=

235)5.01(5.0

5.046.0)1(

n p p

p p=

0326.05.046.0

=0326.0

04.0=-1.23,

so probability P(Z


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If I borrow $6000 how many repayments do I have to makeIf I make a $2000 repayment each monthand interest is 24% pa compounded monthly?

Solution , i = 0.24/12=0.02 , P =6000, R=2000,

12.3)02.1log(

)2000/)600002.02000log(()1log(

)/)log(()1log(

)/)log((i

RiP Ri

RiP Rn

So you have to make 4 payments

Since 3.12 payments is would be 3 payments of 2000 and one smaller payment.

You can think of payments as delicious donuts.

If you see 3.12 delicious donuts there are 4 things not 3 things.

1. Student marks depend on many factors

Marks Study DrugsStudy 0.9Drugs -0.95 -0.809Prayer 0.01 -0.2 0.3

a) Which variable has the highest correlation

b) Is the relationship in part (a) positive or negative ?

c) What does the answer to (a) and (B) mean in terms of the two variables?

d) Which two variables show the least correlation?e) Explain why you think this may be so

Answer

A)Marks and drugs

B) negative

C) Increasing drugs decreases marks

D) prayer and marks

E) Increasing prayer simply wont help you in exams.


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Hypothesis testing question similar to task 5b of the assignment and hypothesis testingquestion in the exam.1) In a sample of 25 people the mean wage $40,000 with standard deviation $5,000

e) Find the standard errorf) Find a 95% confidence interval for the priceg) Test claim mean is less than $39,000 use a 5% level of significanceh) Test claim mean is less than $30,000 use a 5% level of significancei) Discuss the assumptions of the tests in parts c and d.

Solution n=25, 40000 x s=5,000

a) Standard error =25

5000=1000

b) we estimate the population mean is between40000 2.063910 and 40000+2.06391000so between 37936.1 and 42063.9

YOU DO NOT NEED TO WRITE THE FOLLOWINING IN THE EXAM JUST UNDERSTAND ITUse the t-table column n-1=24 column 0.025 and get the value 2.0639We are 95% confident the population mean is within 2.306 standard errors

c) H 0:: =39000, H 1:: >39000

Test statistic =1000

3900040000=1

We are testing that the mean is more than 41000 however the test statistic isnot more than the critical value of 1.7109So we do not reject H 0So there is not strong evidence the mean is more than $39000

** NOTE TO STUDENTS YOU DO NOT NEED TO WRITE THE FOLLOWINING IN THE EXAM JUST

UNDERSTAND ITThe critical value comes from the t table row n-1=24, column 0.05 and get critical value 1.710You are using the 0.05 column because you are using a on sided test,

d) H 0:: =30000, H 1 >30000

Test statistic =1000

3000040000=10

We are testing that the mean is more than 41000 and the test statistics is more than the critical valueof 1.7109So we do reject H 0So there is strong evidence the mean is more than 30000


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What to discuss if the question asks you about assumptions required for a t-test (the comments for regression are similar)

Heights are normally distributed so the assumptions required for hypothesis tests and confidence intervals onsmall samples are are met.

Wages not normally distributed so the assumptions required for hypothesis tests and confidence intervals on small

samples are not met,

For example if you have the sample 70,80,50,60,2000The average is (70+80+50+60+2000)/5=452 which is above most the values, You estimate of the mean andstandard deviation are not reliable so you will have severe problems when using the formula

Or the confidence interval for small n between and where t * is the value from the t-tableThe assumptions required for hypothesis tests and confidence intervals is not met

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