DST Feedback2Indrani shakya

Indrani Shakya (Rajbhandari) [4002166718]USQ Sydney Education CentreL-1 29-35 Bellevue StreetSurry HillsSydney NSW 2010

CMA feedbackProcessed: Fri May 22 14:55:14 EST 2009

Indrani Shakya (Rajbhandari) [4002166718]0050091119STUDENT:

Decision Support ToolsFIN5003COURSE:

CMA ASSIGNMENT86167-2290-1ASSESSMENT:

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CMA test summary

MarkJudgedResponseQuestion

1.0✓(1)1

1.0✓(2)2

1.0✓(1)3

1.0✓(1)4

1.0✓(2)5

1.0✓(1)6

0.0✗(2)7

1.0✓(2)8

1.0✓(4)9

1.0✓(1)10

1.0✓(3)11

0.0✗(1)12

1.0✓(1)13

1.0✓(1)14

1.0✓(4)15

1.0✓5.516

1.0✓0.6117

1.0✓-1.701918

0.0✗(1)19

1.0✓3223.8120

1.0✓59.6921

1.0✓1622

0.0✗0.344623

1.0✓[12.1155, 12.1645]24

1.0✓1.921625

1.0✓27.425426

1.0✓190.764927

1.0✓-0.373128

1.0✓129

1.0✓(1)30

1.0✓(4)31

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1.0✓0.1032

1.0✓0.391333

1.0✓(1)34

1.0✓0.46435

1.0✓68.68336

1.0✓(4)37

1.0✓0.42538

1.0✓0.36639

1.0✓0.60340

Correct = 36Incorrect = 4Not answered = 0Your mark = 36.0 out of a possible 40.0, which is 90%.This test's contribution to your total percentage for this unit is 36% out of a possible 40%.

True or False: Whether the university is private or public is an example of a nominalscaled variable.

Question 1

(1) True(2) False

Correct.

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TABLE 2-5The following are the durations in minutes of a sample of long-distance phone callsmade within the continental United States reported by one long-distance carrier.

Question 2

RelativeFrequency

Time (in minutes)

0.370 but less than 5






0.0230 or more

Referring to Table 2-5, if 100 calls were sampled, __________ of them would havelasted less than 15 minutes.(1) 26(2) 74(3) 10(4) None of the above.

Correct.

According to the empirical rule, if the data form a ‘bell-shaped’ normal distribution,__________ percent of the observations will be contained within 1 standard deviationaround the arithmetic mean.

Question 3

(1) 68.26(2) 75.00(3) 88.89(4) 93.75

Correct.

True or False: As a general rule, an observation is considered an extreme value if itsZ score is less than –3.

Question 4

(1) True(2) False

Correct.

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The owner of a fish market determined that the average weight for a catfish is 3.2pounds with a standard deviation of 0.8 pound. A citation catfish should be one ofthe top 2% in weight. Assuming the weights of catfish are normally distributed, atwhat weight (in pounds) should the citation designation be established?

Question 5

(1) 1.56 pounds(2) 4.84 pounds(3) 5.20 pounds(4) 7.36 pounds

Correct.

True or False: The probability that a standard normal random variable, Z, is between1.50 and 2.10 is the same as the probability Z is between –2.10 and –1.50.

Question 6

(1) True(2) False

Correct.

Why is the Central Limit Theorem so important to the study of sampling distributions?Question 7(1) It allows us to disregard the size of the sample selected when the population is

not normal.(2) It allows us to disregard the shape of the sampling distribution when the size of

the population is large.(3) It allows us to disregard the size of the population we are sampling from.(4) It allows us to disregard the shape of the population when n is large.

The correct answer is:

● It allows us to disregard the shape of the population when n is large.

True or False: As the sample size increases, the standard error of the mean increases.Question 8(1) True(2) False

Correct.

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It is desired to estimate the average total compensation of CEOs in the Serviceindustry. Data were randomly collected from 18 CEOs and the 97% confidenceinterval was calculated to be ($2,181,260, $5,836,180). Based on the interval above,do you believe the average total compensation of CEOs in the Service industry ismore than $3,000,000?

Question 9

(1) Yes, and I am 97% confident of it.(2) Yes, and I am 78% confident of it.(3) I am 97% confident that the average compensation is $3,000,000.(4) I cannot conclude that the average exceeds $3,000,000 at the 97% confidence

level.

Correct.

True or False: The t distribution is used to construct confidence intervals for thepopulation mean when the population standard deviation is unknown.

Question 10

(1) True(2) False

Correct.

A manager of the credit department for an oil company would like to determinewhether the average monthly balance of credit card holders is equal to $75. An auditorselects a random sample of 100 accounts and finds that the average owed is $83.40with a sample standard deviation of $23.65. If you wanted to test whether the auditorshould conclude that there is evidence that the average balance is different from $75,which test would you use?

Question 11

(1) Z-test of a population mean(2) Z-test of a population proportion(3) t-test of a population mean(4) t-test of a population proportion

Correct.

True of False: A sample is used to obtain a 95% confidence interval for the mean ofa population. The confidence interval goes from 15 to 19. If the same sample hadbeen used to test the null hypothesis that the mean of the population is equal to 18versus the alternative hypothesis that the mean of the population differs from 18, thenull hypothesis could be rejected at a level of significance of 0.05.

Question 12

(1) True(2) False

The correct answer is:

● False

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The Y-intercept (b0) represents theQuestion 13(1) predicted value of Y when X = 0.(2) change in estimated average Y per unit change in X.(3) predicted value of Y.(4) variation around the sample regression line.

Correct.

True of False: If a time series does not exhibit a long-term trend, the method ofexponential smoothing may be used to obtain short-term predictions about the future.

Question 14

(1) True(2) False

Correct.

Blossom’s Flowers purchases roses for sale for Valentine’s Day. The roses arepurchased for $10 a dozen and are sold for $20 a dozen. Any roses not sold onValentine’s Day can be sold for $5 per dozen. The owner will purchase 1 of 3 amountsof roses for Valentine’s Day: 100, 200, or 400 dozen roses. Given 0.2, 0.4, and 0.4are the probabilities for the sale of 100, 200, or 400 dozen roses, respectively, thenthe optimal EMV for buying roses is

Question 15

(1) $700(2) $900(3) $1,700(4) $1,900

Correct.

What is the interquartile range (midspread) for this data __________?Answer should be between one or two decimal places e.g. 10.1, 10.2 etc.Please do not include units (grams) in your answer.

Question 16

Correct.Placing the data in ascending order (or descending) as follows:

30252424222220201919181810

We can derive Q1 = 18.5 (average of 3rd and 4th observations) and Q3 = 24 (average of 10th and 11thobservation). Therefore IQR = 24 – 18.5 = 5.5. Note that we cannot round this value to 6.0 as that is incorrectand gives the impression that the spread is larger than it actually is. The actual interquartile range is 5.5 andshould be reported as such.

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Given that a randomly selected cellular call is one that has interference, what is theprobability it came from company A?Answer should be between two and four decimal places e.g. 0.12, 0.123, 0.1234 etc.Please do NOT include any units in your answer.

Question 17

Correct.The correct answer is 0.61.Refer to following table which follows the format listed in Appendix 1 of Module 4 in your study book withP(A1) representing company A’s cellular market share and P(A2) representing company B’s share. TheConditional probabilities are listed for cellular calls that have interference.

PosteriorJointConditionalPriorEvent

0.61*0.01400.020.70A1

0.390.00900.030.30A2

0.0230Totals

The posterior probability of 0.61 represents ‘given that the call has interference, the probability it came fromcompany A’.

The value of the test statistic in this problem is approximately equal to __________?Answer should be between two and four decimal places e.g. 1.23, 1.234, 1.2345 etc.Please do NOT include any units in your answer.

Question 18

Correct.t value = –1.70193Mean = 1.6Sample std deviation = 0.91025899Hence t = (1.6 – 2) ÷ (0.9103 divided by square root of 15) = –1.702

What would be your decision if a hypothesis test was conducted on this problem withthe null hypothesis given as H0 : μ ≥ 2 and the alternate hypothesis given as H1 < 2?

Question 19

(1) Reject H0 at the 10%, 5% and 1% level of significance.(2) Reject H0 at the 10% and 5% level of significance but do not reject H0 at the 1%

level of significance.(3) Reject H0 at the 10% level of significance but do not reject H0 at the 5% or 1%

level of significance.(4) Do not reject H0 at either the 10%, 5% or 1% level of significance.

Incorrect.Critical values (using your t tables) for a one tailed text to the left with n – 1 or 15 – 1 = 14 degrees of freedomare –1.3450 at 10% level of significance, –1.7613 at 5% level of significance and –2.6245 at 1% level ofsignificance. Thus, we would reject H0 if our t value is less than these corresponding critical values. Since–1.702 is less than the critical value for 10% but not the 5% or 1% level, our answer is (3).

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Using a three period moving average (i.e. MA(3)) as a forecasting method, what isthe MSE for this forecasting model?Answer should be between two and four decimal places e.g. 1.23, 1.234, 1.2345 etc.Please do NOT include any units in your answer.

Question 20

Correct.The correct answer is 3223.81Refer following table with MSE = 22566.67 ÷ 7 = 3223.81

Month SalesSmoothed

value Forecasts Errors Errors^2Absolute

error MAPD

1 4402 480 503.333 590 490.004 400 496.67 503.33 -103.33 10677.78 103.33 0.2583335 500 483.33 490.00 10.00 100.00 10.00 0.0200006 550 506.67 496.67 53.33 2844.44 53.33 0.0969707 470 506.67 483.33 -13.33 177.78 13.33 0.0283698 500 523.33 506.67 -6.67 44.44 6.67 0.013333

600 540.00 506.67 93.33 8711.11 93.33 0.155556520 523.33 -3.33 11.11 3.33 0.006410

540.00

Sum 30.00 22566.67 283.33 0.578971

MAPE 0.082710MAD 40.48MSE 3223.81

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Using simple exponential smoothing (with a smoothing constant of 0.2) as aforecasting method, what is the MAD for this forecast model?Answer should be between two and four decimal places e.g. 1.23, 1.234, 1.2345 etc.Please do NOT include any units in your answer.

Question 21

Correct.The correct answer is 59.69Refer following table with MAD = 537.19 ÷ 9 = 59.69

Month Sales Smoothedvalue Forecasts Errors Errors^2 Absolute

error MAPD

1 440 440.002 480 448.00 440.00 40.00 1600.00 40.00 0.0833333 590 476.40 448.00 142.00 20164.00 142.00 0.2406784 400 461.12 476.40 -76.40 5836.96 76.40 0.1910005 500 468.90 461.12 38.88 1511.65 38.88 0.0777606 550 485.12 468.90 81.10 6577.86 81.10 0.1474627 470 482.09 485.12 -15.12 228.52 15.12 0.0321638 500 485.67 482.09 17.91 320.64 17.91 0.035813

600 508.54 485.67 114.33 13070.26 114.33 0.190542520 510.83 508.54 11.46 131.34 11.46 0.022039

510.83

Sum 354.16 49441.23 537.19 1.020791

MAPE 0.113421MAD 59.69MSE 5493.47

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If a frequency distribution for the defects data is constructed, using ‘0 but less than5’ as the first class, what would be the relative frequency of the ‘10 but less than 15’class __________%?Answer should be a percentage value to whole numbers e.g. 12, 23, 34 etc.Please do NOT include units (%) in your answer.

Question 22

Correct.The correct answer is any one of the following:

● 16

● 0.16

Refer the following table with answer listed by asterisk. Note that the question asked for RELATIVE frequencynot frequency, hence the answer 4 is not what the question asked (nor is this value a percentage)

Cumulative relativefrequency

Relative frequencyFrequencyClass

0.160.1640 but less than 5

0.360.2055 but less than 10

0.520.16 *410 but less than 15

0.560.04115 but less than 20

0.760.20520 but less than 25

1.000.24625 but less than 30

What is the probability that the sample mean will be more than 48 minutes__________?Answer should be to four decimal places which is consistent with the number ofdecimal places listed in your appendix tables e.g. 0.1234 etc.Please do NOT include any units in your answer.

Question 23

Incorrect.The correct answer is 0.6554Note that this problem has a sample size of 64 which must be incorporated into our Z calculations (refer

equation 7.4 of Levine et al. 5th edn on p. 266). Thus Z = 48 – 50 ÷ 40 √64 = –0.40. Therefore P(Z > –0.40)= 1 – 0.3446 = 0.6554.If you ignored the sample size, you would have derived a Z value of –0.05 leading to an area of 0.5199 forthis question.

What are the two limits of the confidence interval __________ and __________?Answer should be to four decimal places e.g. 1.2345 and 2.3456.Please do NOT include any units in your answer.

Question 24

Correct.The confidence interval ranges from 12.14 ± 1.96 × 0.15 ÷ √144 i.e. the range is from a minimum of 12.1155to a maximum of 12.1645.

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What is the least squares estimate of the slope?Answer should be to four decimal places e.g. 1.2345.Please do NOT include any units in your answer.

Question 25

Correct.The correct answer is 1.9216 (i.e. 1921.60).The Excel printout for this problem is presented below with Amount of Life Insurance being the dependantvariable (Y) and Income the independent variable (X). For those who undertook this problem by hand, thefollowing will allow you to check your results:

ΣX = 820, ΣY = 1850, ΣXY = 162000, ΣX2 = 72600, ΣY2 = 364500, n = 10,Mean of X = 82, Mean of Y = 185.

P-valuet StatStandard errorCoefficients

0.2156230141.34462484120.396301127.42537313Intercept

4.2596E-058.0276666620.2393773771.921641791Income

What is the least squares estimate of the Y intercept?Answer should be to four decimal places e.g. 1.2345.Please do NOT include any units in your answer.

Question 26

Correct.The correct answer is 27.4254 (i.e. 27,425.40).The Excel printout for this problem is presented below with Amount of Life Insurance being the dependantvariable (Y) and Income the independent variable (X). For those who undertook this problem by hand, thefollowing will allow you to check your results:

ΣX = 820, ΣY = 1850, ΣXY = 162000, ΣX2 = 72600, ΣY2 = 364500, n = 10,Mean of X = 82, Mean of Y = 185.


0.2156230141.34462484120.396301127.42537313Intercept

4.2596E-058.0276666620.2393773771.921641791Income

What is the prediction for the amount of life insurance for a family whose income is$85,000?Answer should be to four decimal places and be consistent with your original dataset, e.g. if your answer was $75,410.90, you would enter 75.4109 as your answer.Please do NOT include any units in your answer.

Question 27

Correct.To predict when X = 85, Y = 27.4254 + 1.9216 ( 85 ) ≈ 190.7614This is $190,761.40.

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What would be the residual (error) term for a family income of $90,000?Answer should be to four decimal places and be consistent with your original dataset e.g. if your answer was $940.90, you would enter 0.9409. If your answer was$9,400.90, you would enter 9.4009 etc.Please do NOT include any units in your answer.

Question 28

Correct.The correct answer is 0.3694 or –0.3694, i.e. 369.40 or –369.40.To calculate the error term, we need to know the actual value and predicted value of Y when X = 90. Lookingat your original table, when X = 90 the actual Y was 200. The predicted Y becomes Y = 27.4254 + 1.9216(90) = 200.3694. Hence the error term becomes 200 – 200.3694 = –0.3694 (0.37). Some students may havenoticed that we can derive a listing of all error terms by selecting the residuals check box from the regressiondialog box in Excel (the answer is the same at 2 decimal places).

What is the opportunity loss for a General Distribution for a Modest level of success?Answer should be to whole numbers only. You do not need to put any units in youranswer.

Question 29

Correct.The opportunity loss for a general distribution release for a modest level of success is 1 as the alternativeaction yields a higher return (9 compared to 8) resulting in an opportunity loss of 1.

What would the optimal action be for International before running the sneak preview?Question 30(1) Run a limited release with an expected payoff of $7.20m(2) Run a limited release with an expected payoff of $6.20m(3) Run a general distribution with an expected payoff of $7.20m(4) Run a general distribution with an expected payoff of $6.20m

Correct.E(Limited release) = 22 × 0.3 + 9 × 0.4 – 10 × 0.3 = $7.20mE(General distribution) = 12 × 0.3 + 8 × 0.4 – 2 × 0.3 = $6.20mThus, optimal action under uncertainty is Limited release with expected payoff of $7.20m

What is the maximum amount of money that International would be prepared to payfor an absolutely reliable forecast of the movies’ level of success?

Question 31

(1) $9.6m(2) $7.2m(3) $6.2m(4) $2.4m

Correct.EVPI = [ 22 × 0.3 + 9 × 0.4 – 2 × 0.3 ] – [ 7.20 ] = $2.40mHence, we would be prepared to pay $2.4m for ‘perfect’ information or for an absolutely reliable forecast.

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What would be the joint probability for a ‘modest success’ and good preview giventhat in the past, it was found that 25% of all modest successes were rated good?Answer should be to two decimal places e.g. 0.12, 0.23, etc.Please do NOT include any units in your answer.

Question 32

Correct.The joint probability table is shown below with I1 = sneak preview is good and I2 = sneak preview is excellent:(the required answer is shown with an asterisk)

TotalS3S2S1

0.30.40.3

0.3 × 0.60.4 × 0.250.3 × 0.1I1

0.31= 0.18= 0.1*= 0.03

0.3 × 0.40.4 × 0.750.3 × 0.9I2

0.69= 0.12= 0.3= 0.27

1.000.30.40.3Total

What is the posterior probability of a smash success given the sneak preview indicatesexcellent?Answer should be to four decimal places e.g. 0.1234, 0.2345 etc.Please do NOT include any units in your answer.

Question 33

Correct.The correct answer is 0.3913.The following posterior probabilities are derived from the joint probability table (the required answer is shownwith an asterisk):P( Smash success / sneak preview indicates good ) = 0.03 ÷ 0.31 = 0.0968P( Modest success / sneak preview indicates good ) = 0.1 ÷ 0.31 = 0.3226P( Bomb / sneak preview indicates good ) = 0.18 ÷ 0.31 = 0.5806P( Smash success / sneak preview indicates excellent ) = 0.27 ÷ /0.69 = 0.3913*P( Modest success / sneak preview indicates excellent ) = 0.3 ÷ 0.69 = 0.4348P( Bomb / sneak preview indicates excellent ) = 0.12 ÷ 0.69 = 0.1739The total probability of a good preview is P(I1) = 0.31 and the total probability of an excellent preview is 0.69.

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What is the maximum amount that should be paid for the sneak preview (i.e. whatwould be the expected value of sample information (EVSI))? Select the closest correctanswer.Use four decimal places in your calculations but you can round off the final answerto 2 decimal places.)

Question 34

(1) $1.04 million(2) $2.58 million(3) $7.2 million(4) $8.24 million

Correct.The correct answer is $1.04 million. Refer to the EVSI diagram below.

7.8262

.1739

.4348

.1739

10.7828

.4

.3

.4

.3

7.2

• No Sneak Preview

.4348Sneak Preview

Preview good

Previewexcellent

0.31

0.69

8.24

2.5812

.3913

.3913

.3

.3

10.7828

2.5812

-0.773

6.2

7.2

.0968.3226.5806

.0968.3226.5806

22

9-10

12

8-222

9

-10

12

8

-222

12

9

-10

8

-2

Limited

Limited

Limited

General

General

GeneralEVSI = 8.24- 7.2 = 1.04

EVSI tree diagram

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What is the centred moving average that would correspond to Quarter 1 in 2006?Answer should be consistent with the data provided and be to three decimal placese.g. 0.123, 0.456 etc.Please do NOT include any units in your answer.

Question 35

Correct.Excel printout shown below with answer shown next to Q1 2006.

MA(2×4)Occupancyrate

QuarterTimeYear

0.261112004

0.30222

0.4130.50233

0.4200.56844

0.4310.295152005

0.4450.32426

0.4530.56837

0.4570.61548

0.464 *0.312192006

0.4780.345210

0.4950.598311

0.5060.698412

0.5240.3621132007

0.5410.388214

0.5500.696315

0.5580.738416

0.5730.3961172008

0.5920.418218

0.782319

0.802420

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What is the adjusted seasonal index for Quarter 1 __________%?Answer should be listed to three decimal places in the form 0.123 i.e. 0.123 represents12.3%, 1.234 represents 123.4% etc.Please do NOT include the units (%) in your answer.

Question 36

Correct.Table of unadjusted and adjusted seasonal indices presented below in excel format (answer listed under Quarter1 column as adjusted seasonal index).

TotalsQuarter 4Quarter 3Quarter 2Quarter 1Year

1.3541.2172004

1.3451.2550.7290.6852005

1.3791.2090.7220.6732006

1.3221.2650.7170.6912007

0.0000.0000.7070.6912008

15.9615.4004.9462.8752.741Total

3.9901.3501.2370.7190.685Average

Correction factor 4 ÷ 3.990 = 1.0025

4.0001.3531.2400.7200.687*Adjusted

Sum of Corrected = 4.00Our seasonal indices clearly show that Quarters 3 and 4 have a relative weighting more than the averagewhereas the remaining quarters indicate a weighting less than the average. This would indicate that occupancyrate is obviously much higher in the ‘tourist’ season which comprises the third and fourth quarters. The specificmeaning of each index can be illustrated as follows: The corrected index for quarter 1 represents an occupancyrate which is approx 31.3% below the yearly average whereas the corrected index for Quarter 3 is approx 24%above the yearly average.

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The trend line for this decomposition model was calculated to be Y = 0.36806 +0.01195 T (where T represents time). What would be the coefficient of determination

(R2) for this trend line? (Select the closest correct answer.)

Question 37

(1) 0.2973 (i.e. 29.73%)(2) 0.5452 (i.e. 54.52%)(3) 0.7165 (i.e. 71.65%)(4) 0.9886 (i.e. 98.86%)

Correct.Our trend equation is estimated to be Y = 0.36806 + 0.01195 T. Complete regression printout is presentedbelow:

Regression Statistics

0.994286339Multiple R

0.988605323R Square

0.987791418Adjusted R Square

0.006323483Standard Error

16Observations


5.52273E-2193.591632360.00393260.368058456Intercept

5.2507E-1534.851741540.0003429390.011952022X Variable 1

Note that our trend equation is based on 16 observations as our dependent variable was our centred movingaverages and our time variable was from t = 3 to t = 18. This trend equation has performed exceptionally well

with an R2 of 98.86%. If we examine the plot of the original data over time (presented below), the reason isclear. There is a very clear noticeable trend and our equation is reflecting this fact.

Occupancy rate

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

NOTE* If we used the original data to generate our trend line (i.e. using our original data as our dependentvariable and time from t = 1 to t = 20 as our independent variable), we would be using a series that seasonalinfluences mixed up in them and, as such, our trend line would worse. This trend line will be Y = 0.32442 +

0.01658 T (with a R2 of 29.73%). In this case our coefficient of T is biased by seasonal influences.

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What would be the forecast in Quarter 1, 2009 using the trend line given previously(recall that to three decimal places it was Y = 0.368 + 0.012 T) and the relevantadjusted seasonal index?Answer should be consistent with the data provided and be to three decimal placese.g. 0.123, 0.456 etc.Please do NOT include any units in your answer.

Question 38

Correct.Our forecasts for the next 12 months are listed below.

0.426*=(0.368 + 0.012 × 21) × 0.687Quarter 1

0.455=(0.368 + 0.012 × 22) × 0.720Quarter 2

0.798=(0.368 + 0.012 × 23) × 1.240Quarter 3

0.888=(0.368 + 0.012 × 24) × 1.353Quarter 4

Note on all forecasts derived. While we should calculate some of the measures of forecasting accuracy to seehow well our model has performed (measures such as MSE, MAE, MAPE ) these are of limited value since,by themselves, these measures convey little information. What we require is an alternative forecasting modelto compare the resulting MSE, MAE, MAPE with our decomposition model to see which model has performed

the best in terms of forecasting accuracy. Notwithstanding the above points, the high R2 associated with ourtrend equation would lead us to have some confidence in the final performance of our multiplicativedecomposition model.

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If we exponentially smooth the data in Table 1 with a smoothing constant of 0.2, thesmoothed value for Quarter 4 in 2004 would be?Answer should be consistent with the data provided and be to three decimal placese.g. 0.123, 0.456 etc.Please do NOT include any units in your answer.

Question 39

Correct.Refer to the following Excel table:

W = 0.2Occupancyrate

QuarterTimeYear

0.2610.261112004

0.2690.30222

0.3160.50233

0.366 *0.56844

0.3520.295152005

0.3460.32426

0.3910.56837

0.4360.61548

0.4110.312192006

0.3980.345210

0.4380.598311

0.4900.698412

0.4640.3621132007

0.4490.388214

0.4980.696315

0.5460.738416

0.5160.3961172008

0.4970.418218

0.5540.782319

0.6030.802420

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86167-2290-10050091119

If we exponentially smooth the data in Table 1 with a smoothing constant of 0.2, theforecast for Quarter 1 2009 would be?Answer should be consistent with the data provided and be to three decimal placese.g. 0.123, 0.456 etc.Please do NOT include any units in your answer.

Question 40

Correct.For the exponential smoothing answers please refer to following Excel original table. The forecast for Q1,2009 is the last smoothed value available (i.e. the smoothed value for Q4, 2008).

W = 0.2Occupancyrate

QuarterTimeYear

0.2610.261112004

0.2690.30222

0.3160.50233

0.3660.56844

0.3520.295152005

0.3460.32426

0.3910.56837

0.4360.61548

0.4110.312192006

0.3980.345210

0.4380.598311

0.4900.698412

0.4640.3621132007

0.4490.388214

0.4980.696315

0.5460.738416

0.5160.3961172008

0.4970.418218

0.5540.782319

0.603 *0.802420

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86167-2290-10050091119

DST Feedback2Indrani shakya

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