1

FEEDBACK TUTORIAL LETTER

2nd SEMESTER 2017

ASSIGNMENT 2

STATISTICS FOR ECONOMISTS

SFE612S

2

STATISTICS FOR ECONOMISTS(SFE612S)

FEED BACK TUTORIAL LETTER

ASSIGNMENT 2 SEMESTER 2 2017

Dear student,

Congratulations on the successful completion of your second assignment for semester 2 2017.

I am convinced the study guide gave you enough exposure to applications of Statistics for

Economists skills in daily financial transactions.

I have no doubts that working through the questions must have in no small way improved on

your statistical, analytical and other calculation skills.

Most of you have performed well and I recommend that you keep this up.

However, the attached memorandum is for you to see the step by step methods of realising

the final calculations and will also prepare you towards the end of November examinations.

CONGRATULATIONS !!!.

Your Marker-Tutor

Mr D Ntirampeba.

Department of Mathematics and Statistics

Namibia University of Science and Technology

Windhoek

Tel: +264-61-2072808.

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MEMORANDUM

Question 1[35 marks]

1.1 [10]

Step 1:

The null and alternative hypotheses are

1 2 3 4 5 6: 0.12, 0.29, 0.11, 0.10, 0.14, 0.24OH p p p p p p

1H : At least one of the proportions is different from stated proportion.

Step 2

The significance level is 0.01

Step 3: Computation of the test statistic

In step 3, we have to compute the test statistic using the formula 2

2cal

o

e

fn

f

From this formula, we notice that we must find the expected frequencies ef as they are not

given.

We recall that e if np

Thus, the expected frequency for the first category is given by 1 519 0.12 62.28f ,

the expected frequency for the second category is given by 2 519 0.29 150.51f

the expected frequency for the third category is given by 3 519 0.11 57.09f ,

the expected frequency for the first category is given by 4 519 0.1 51.9f ,

the expected frequency for the fifth category is given by 5 519 0.14 72.66f ,

and the expected frequency sixth category is given by 5 519 0.24 125.6f

22

cal

o

e

fn

f

4

2 2 2 2 2 22 88 135 52 40 76 128

51962.28 150.51 57.09 51.9 72.66 125.6

=15.659

cal

Step 4: Select the critical value: 2

, 1k

There are six age categories. Hence, 6k and the number of degrees of freedom is

1 6 1 5k . The critical value is 2

0.01,5 15.08627 .

Step 5: Formulation of the rejection rule

We reject 0H for a value of statistic greater than 2

0.01,5 15.08627 . In our case, 2 =15.609cal

is greater than 2

0.01,5 15.08627 . Hence, 0H is rejected.

Step 6: conclusion

At 1% significance level, we conclude that the data provide enough evidence to indicate that

the the distribution of customer ages in the Snoop report does not agree with that of the

sample report.

1.2 [15]

H0: The performance and the size of a company are statistically independent.

H1: The performance and the size of a company are associated

Computation of expected values ( iE ) and test statistic ( 2D ):

i

Column total row totalE

Grandtotal

22 i

i

OD n

E

5

Company size

Performance(% price change)

<0% 0- 20

%

20- 40% >40% Total

Small 66(77.0) 90(85) 28(29.2) 39(31.9)

223

Medium 24(29.3) 33(32.4)

17(11.1) 11(12.2)

85

Large 55(38.7) 37(42.7)

10(14.7) 10(16.0)

112

Total 145 160 55 60 420

22

2 2 266 90 10 = .... 420

77 85 16

=19.09

i

i

OD n

E

2(0.05)

6 12.592X

Conclusion:

Because 2 19.09D > 2(0.05)

6 12.592X we reject Ho.

Thus our financial analyst has demonstrated that there is significant relationship between the

performance of firm and its size.

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1.3 [10]

Test Hypotheses

𝐻0: 𝑝𝑖 = 1/6 for all 𝑖 = 1, 2, ⋯ ,6 i. e. The die is balanced(fair)

𝐻𝑎: At least one 𝑝𝑖 ≠1

6 for some 𝑖 = 1, 2, 3, 4 i. e. The die is not balanced (fair)

Since we are testing adequacy of a discrete uniform model, 𝑋~𝑈𝑛𝑖(𝑛 = 300), the expected

counts are simply 𝐸𝑖 = 𝑁𝑃𝑖 = 300(1/6) = 50 for all 𝑖 = 1, 2, ⋯ ,6. Then the observed and

expected counts are as follows.

Score 𝑂𝑖 𝐸𝑖 = 𝑁𝑃𝑖

1 62 50

2 45 50

3 63 50

4 32 50

5 47 50

6 51 50

Total 𝑁 = 300 𝑁 = 300 (Note: This table is optional and it has no mark at all. However, if a candidate only gives this

table without explanation just above then will still be awarded the same 2 marks above.)

Observed 𝜒2 − value

𝜒𝑜𝑏𝑠2 = ∑

𝑂𝑖2

𝐸𝑖− 𝑁𝑘

𝑖=1 = ∑𝑂𝑖

2

𝐸𝑖− 806

𝑖=1 = [622

50+ ⋯ +

512

50] − 300

= [313.44] − 300 = 13.44

Critical 𝜒2 − value

𝜒𝑐𝑟𝑖𝑡2 = 𝜒𝛼, 𝑘−𝑝−1

2 = 𝜒0.010,52 = 15.08627

Decision Rule

Reject H0 if 𝜒𝑜𝑏𝑠2 > 𝜒𝑐𝑟𝑖𝑡

2 = 15.08627

Decision Making

Since 𝜒𝑜𝑏𝑠2 = 13.44 < 𝜒𝑐𝑟𝑖𝑡

2 = 15.08627, we do not reject H0

Conclusion

7

At 1% significance level, we conclude that the data do not provide sufficient evidence to infer

that the die is not balanced (fair).

Question 2 [44 marks]

2.1 [12]

Year Quarter rate 4_MA(8marks) Seasonal

ratio (4

marks)

2006 1 0.561

2 0.702

3 0.8 0.6595 121.304

4 0.568 0.66575 85.31731

2007 1 0.575 0.67875 84.71455

2 0.738 0.691875 106.6667

3 0.868 0.698875 124.1996

4 0.605 0.70125 86.27451

2008 1 0.594 0.683875 86.85798

2 0.738 0.665875 110.8316

3 0.729 0.66875 109.0093

4 0.6 0.6685 89.75318

2009 1 0.622 0.674375 92.23355

2 0.708 0.688 102.907

3 0.806 0.697375 115.5763

4 0.632 0.718625 87.94573

2010 1 0.665 0.742875 89.51708

2 0.835 0.756 110.4497

3 0.873

4 0.67

8

2.2 [10]

Q1 Q2 Q3 Q4 Total

2006 121.304 85.31731

2007 84.71455 106.6667 124.1996 86.27451

2008 86.85798 110.8316 109.0093 89.75318

2009 92.23355 102.907 115.5763 87.94573

2010 89.51708 110.4497

Seasonal

median 88.18753 108.5582 118.4401 87.11012 402.296

100

400 = 0.994

402.296

kAdjustment factor

Median seasonal index

Quarter

Seasonal

median(SM)

Seasonal index=𝑆𝑀 ×

𝐴𝑑𝑗. 𝑓𝑎𝑐𝑡𝑜𝑟

Q1 88.18753 87.68422

Q2 108.5582 107.9386

Q3 118.4401 117.7642

Q4 87.11012 86.61296

2.3 [6]

Year Quarter rate SI De-seasonalised index

2006 1 0.561 87.68422 0.639796

2 0.702 107.9386 0.65037

3 0.8 117.7642 0.679324

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4 0.568 86.61296 0.655791

2007 1 0.575 87.68422 0.655762

2 0.738 107.9386 0.683722

3 0.868 117.7642 0.737066

4 0.605 86.61296 0.69851

2008 1 0.594 87.68422 0.677431

2 0.738 107.9386 0.683722

3 0.729 117.7642 0.619034

4 0.6 86.61296 0.692737

2009 1 0.622 87.68422 0.709364

2 0.708 107.9386 0.655928

3 0.806 117.7642 0.684419

4 0.632 86.61296 0.729683

2010 1 0.665 87.68422 0.758403

2 0.835 107.9386 0.773588

3 0.873 117.7642 0.741312

4 0.67 86.61296 0.773556

Interpretation of the first de-seasonilised index (0.6398):

2006 quarter 1 occupancy rate would have been higher at 0.6398, instead of the actual sales of

0.561, had seasonal influences not been present.

2.4 [10]

Quarter 𝑦 𝑥 𝑥𝑦 𝑥2

1 0.561 -19 -10.659 361

2 0.702 -17 -11.934 289

3 0.8 -15 -12 225

4 0.568 -13 -7.384 169

1 0.575 -11 -6.325 121

10

Trend Line formula: abxy

2

6.9770.00263

2660

13.889int 0.6945

20

xySlope b

x

yy ercept a

n

Trend Line: 0.00263 0.6945y x

2.5 [2]

Estimate the trend value of the time series for Quarter 3 in 2011. .

Quarter 3 in 2011: ˆ 0.00263(25 0.6945 0.76T y

2 0.738 -9 -6.642 81

3 0.868 -7 -6.076 49

4 0.605 -5 -3.025 25

1 0.594 -3 -1.782 9

2 0.738 -1 -0.738 1

3 0.729 1 0.729 1

4 0.6 3 1.8 9

1 0.622 5 3.11 25

2 0.708 7 4.956 49

3 0.806 9 7.254 81

4 0.632 11 6.952 121

1 0.665 13 8.645 169

2 0.835 15 12.525 225

3 0.873 17 14.841 289

4 0.67 19 12.73 361

Tot 13.889y 0x 6.977xy 2 2660x

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2.6 [4]

The seasonally-adjusted trend estimate is calculated usingT S .

where the S is the seasonal index for the specific period (Q3)

Thus, 117.7642

0.76 0.895100

T S .

Question 3 [6 marks]

Year Quarter Sales

Exponentially smoothed

sales(𝑤 = 0.1)

2007 1 18 18.000

2 22 18.400

3 27 19.260

4 31 20.434

2008 1 33 21.691

2 20 21.522

3 38 23.169

4 26 23.452

2009 1 25 23.607

2 36 24.846

3 44 26.762

4 29 26.986

Question 4[15 marks]

4.1.

Price relative for food= 1

0

45100% 100% 112.5%

40

p

p

This shows that since 1981, the price of food has increased by 12. 5%.

Price relative for accommodation = 1

0

40100% 100% 133.33%

30

p

p

This shows that the average price of accommodation has increased by 33.33% since 1981.

12

Price relative for petrol = 1

0

1.40100% 100% 107.69%

1.30

p

p

This shows that the average price of petrol has increased by 7.69% since 1981.

4.2

Quantity relative for food= 1

0

25100% 100% 83.33%

30

p

p

This shows that since 1981, the food consumed by typical family of four has decreased by

16.67%.

Quantity relative for accommodation = 1

0

5100% 100% 62.5%

8

p

p

This shows that the consumption of accommodation units has decreased by 37.5% since 1981.

Quantity relative for petrol = 1

0

100% 75 100% 80.00%p

p

This shows that the number of units of petrol consumed by a typical family of four has

decreased by 20% since 1981.

4.3-4.5

Quantities Prices

Item

1981

(𝑞0)

1982

(𝑞1)

1981

(𝑝0)

1982

(𝑝1)

𝑝1𝑞0 𝑝0𝑞0 𝑝1𝑞1

Food 30 25 40 45 1350 1200 1125

Accommodatio

n 8 5 30 40

320 240 200

Petrol 75 60 1.3 1.4 105 97.5 84

Entertainment 10 8 10 12 120 100 96

Total

∑ 𝑝0

= 81.3

∑ 𝑝0

= 98.4

∑ 𝑝1𝑞0

= 1895

∑ 𝑝0𝑞0 =

1637.55

∑ 𝑝1𝑞1 =

1637.5150

5

13

4.3.

Laspeyres price index = 1 0

0 0

1895100% 100% 115.725%

1637.5

p q

p q

If quantities (number of units consumed) are held constant at 1981(base period) levels, the

price of the number of units consumed by a typical family has increased by, on average

15.725% since 1981. [4]

4.4

Paasche’s quantity index= 1 1

1 0

100% 100% 79.42%p q

p q

If prices are held constant at current period levels (1982), the quantities (the number of units

consumed by a typical family) would have increased by 20.58%. [4]

4.5 [1]

Unweighted composite (aggregate) price index=∑ 𝑝1

∑ 𝑝0× 100% =

98.4

81.3× 100% = 121.03%