FEEDBACK TUTORIAL LETTER 2ND SEMESTER 2020 · 2021. 2. 5. · TUTORIAL LETTER MEMO SEMESTER 2/2020 STATISTICS FOR ECONOMISTS 2B SFE612S 4 yy. 4.6 2.534 Since all values of yy ij

FEEDBACK TUTORIAL LETTER

2ND SEMESTER 2020

ASSIGNMENT 1 & 2

Statistics for Economists 2B SFE612S

TUTORIAL LETTER MEMO

SEMESTER 2/2020

STATISTICS FOR ECONOMISTS 2B

SFE612S

1

COURSE: STATISTICS FOR ECONOMISTS 2B

COURSE CODE: SFE612S

TUTORIAL LETTER: 2020

DATE: 04/02/ 2021

Dear Student

Congratulations on the successful completion of your first and second assignments for

semester 2 2020.

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. The attached memoranda are for you to

see the step by step methods of realising the final calculations and will also prepare you for

the comprehensive test.

Regards,

Dr. Dismas Ntirampeba

Tel. +26461 207 2808

Email: [email protected]

mailto:[email protected]


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Assignment 1

Question 1 [50 marks]

1.1

1.1.1

Ho: 1 2 3

i.e. There is no difference in the three aspects of judgment

H1: , for at leat one pair ( , ),i j i j i j

i.e. At least two aspects of judgment affect the quality of judgement differently.

2

2..

1 1

22 2 2 436.5 = 17 18.5 ... 24.2

21

=317.34

a b

TOTij

i j

SSN

yy

23

2..

.1

22 2 2

1

1 1 1 436.8 = 119 142.8 175

7 7 7 21

=225.68

treati

ii

SSn N

yy

317.34 - 225.68

91.66

T TreatSSE SS SS

225.68112.84

1 2

treattreat

SSMS

k

91.665.0922

18

SSEMSE

N k

Test statistic:


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112.8422.159

5.0922

treatMSFMSE

Critical value:

0.05,2,18 3.55F

Because the test statistic ( 22.159F ) is greater than the critical value

( 0.05,2,18 3.55F ) we reject the null hypothesis.

Conclusion

At 5% significance level, the data provide sufficient evidence to indicate the basis of

judgment affects the quality of judgment.

1.1.2

0 : i jH

1 : i jH , i j ( we compare all possible pairs)

Pair of means i and j would be declared significantly different if

1 3

2

1 1. . , n ni j n k

y y t MSE .

Test statistics: . .i jy y ,

corresponding critical values: 2

1 1, i jn nn k

LSD t MSE

Since 1 2 3 7n n n , LSD is constant.

0.051 3

2

1 1,36

1 17 7 =2.101 5.0922

=2.534

n nLSD t MSE

. .i jy y LSD

1. 2.y y 3.4 2.534

1. 3.y y 8 2.534


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2. 3.y y 4.6 2.534

Since all values of . .i jy y are greater than 2.534LSD , we conclude that, at 5% level

of significance, all aspects of judgement are different and the best one is the direct

experience( i.e lower scores are obtained through this aspect of judgement).

1.2

0 1 2 6: ...H

There is no difference in students’ mean performances.

1 : for at least one pair ( , ) and i jH i j i j

At least two two sudents perform differently from others.

Now, we calculate the test statistics.

2

2..

1 1

22 2 2 18132 = 526 594 ... 420

18

=65798

a b

Tij

i j

SSN

yy

26

2..

.1

22 2 2 2 2 2

1

1 18132 = 1590 1770 1374 1680 1344 1308

3 18

=63250

studentsi

i

SSb N

yy

23

2..

de t .

1

22 2 2

1

1 18132 = 3012 3090 2964

6 18

=1348

Aptitu est partsj

j

SSa N

yy

de t 65798 - (63250 1348)

1200

T students aptitu est partsSSE SS SS SS


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( ) 6325012650

1 6 1

Treat students

students

SSMS

a

1200

120-1 -1 10

EE

SSMS

a b

F-Statistic for treatment (student performance) comparison:

12650105.4167

120

studentsMSFMSE

F-critical:

0.05,5,10 3.33F

Because the test statistic 105.4167F is greater than the critical value 0.05,5,10 3.33F ,

we reject the null hypothesis.

Conclusion:

At 5% significance level, there is enough evidence that students perform differently in

three portions of the SAT

0 1 2 3:H

There is no difference in trouble give to students by the three portions of SAT.

1 : for at least one pair ( , ) and i jH i j i j

At least two two portions of SAT give different trouble to students.

( ) 1348674

1 3 1

block aptitude

aptitude

SSMS

b

F-Statistic for block (three portions of the SAT) comparison:

( ) 6745.617

120

block aptitudeMSF

MSE


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F-critical:

0.05,2,10 4.10F

Because the test statistic 5.617F is greater than the critical value 0.05,2,10 4.10F , we

reject the null hypothesis.

Conclusion:

At 5% significance level, there is enough evidence to indicate that the three portions

of the SAT do not give the students the same trouble.


2.1

Temperature (oC) 22 17 24 26 28 29

Number of passengers 173 149 175 188 186 198

2.1.1 Fit an ordinary least square (OLS) simple linear regression model for predicting number

of passengers throughout a particular hour given the temperature at the beginning of the hour.

[7]

Let

Model is

2.1.2 Without using Pearson (product moment) correlation formula, compute coefficient of

determination and interpret it. [5]

Meaning: About 86.14% of total variation in number of passengers ( is explained (accounted

for) by temperature ( .

2.1.3 Compute the standard error of the estimate. [3]


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[OR using ]

Then

2.1.4 Construct the 90% prediction interval (P.I.) for number of passengers in an hour with a

beginning temperature of 25oC. [5]

Note: [Rounding appears incorrect but it is correct since these are number of passengers; we

round LPL down and UPL up regardless of value of the tenth digit]

2.2

Yes, a linear equation seems to a be good fit for the transformed data


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log log logy x

Let0log , log , ,and logY y b b X x . Then the above transformed equation can

be rewritten as

Y bo bX

We use the same computational procedure as in 2.1.2 to obtain the coefficient of the regression model.

0

0.3047 , 0.1498 , 0.727, 0.481

0.14980.492

0.3047

0.481 0.492 0.727 0.128

0.1283 0.492

X XY

XY

X

SS SS X Y

SSb

SS

b Y bX

Y X

Thus,

0.1283 log and 0.492b

0.12830.1283 log 10 1.344

2.3

2.3.1

1 2 3 4 5 6ˆ 18.859 16.202 0.175 11.526 13.580 5.311x x x x x x

2.3.2


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2 1

1

6541.41121 =1

95980.35226

=0.992

SSEn p

AdjRSST

n

Interpretation: Accounting for degrees of freedom, about 99.2% of the total variation in

1x̂ is mutually explained by 2x , 3x ,…,and 6x ).

2.3.3

Test for overall adequacy of the fitted model at 5% level?

Hypotheses

against

Observed F-value

Critical F-value

Decision Rule

Conclusion

Since , we highly reject and conclude that overall fitted

model is significantly adequate at 5% level.

2.3.4

against

Since the p-values associated with the regression coefficients are all less than 0.05,

we reject the null hypotheses and concluded that all regression coefficient are

statistically different from zero.

Therefore, we would not consider eliminating any variable.


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Total: 90

Assignment 2

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.


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We recall that e if np

Thus, the expected frequency for the first category is given by1 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

2 2 2 2 2 22 88 135 52 40 76 128 519

62.28 150.51 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 than2

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

cal

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 ):


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i

Column total row totalE

Grandtotal

22 i

i

OD n

E

Company size

Performance(% price change)

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


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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.

1.3 [10]

Test Hypotheses

)

Since we are testing adequacy of a discrete uniform model, , the expected

counts are simply for all . Then the observed and

expected counts are as follows.

Score

1 62 50

2 45 50

3 63 50

4 32 50

5 47 50


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6 51 50

Total

(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

Critical

Decision Rule

Decision Making

Conclusion

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

that the die is not balanced (fair).


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


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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

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=

Q1 88.18753 87.68422

Q2 108.5582 107.9386

Q3 118.4401 117.7642

Q4 87.11012 86.61296


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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

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


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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]


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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]

Quarter

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

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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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

2.6 [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 .


Year Quarter Sales

Exponentially smoothed sales

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.


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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

1982

1981

1982

Food 30 25 40 45 1350 1200 1125

Accommodation

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

5 1505


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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

Total:100

Statistics for Economists 2B SFE612SSFE612S_2020_FTL MEMO_Final (003)

Documents

FEEDBACK TUTORIAL LETTER 2ND SEMESTER 2020 · 2021. 2. 5. · TUTORIAL LETTER MEMO SEMESTER 2/2020 STATISTICS FOR ECONOMISTS 2B SFE612S 4 yy. 4.6 2.534 Since all values of yy ij