assignment 1 stat 217 shatil ff

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Lab Assignment – 1

STA 217

Sec – 4

Submit to : Basanta Kumar Barmon

Department of Business Economic

East West University

Submit by : Shatil Sharyer

ID no. 2007-1-10-081

Date of Submission: March 27, 2011



Answer to question no. 1

YearNetSales Time(X)

New NetSales(Y)

1997 50600 1 50681

1998 67300 2 67381

1999 80800 3 80881

2000 98100 4 98181

2001 124400 5 124481

2002 156700 6 156781

2003 201400 7 201481

2004 227300 8 227381

2005 256300 9 256381

2006 280900 10 280981

i) Here, by using statistical software MS Excel,

Coefficients Standar d Error t Stat P-value

Lower 95%

Upper 95%

Lower 95.0%

Upper 95.0 %

Intercept 5447.667 8377.1820.65029

80.53372

5 -13870.224765.

5 -13870.2 24765

X-Variable 27093.33 1350.10520.0675

7 3.97E-0823979.9

830206.

723979.9

8 30207

Here we find the least square equation,

Y = a + b(t)

Y = 5447.667 + 27093.33(t)

On the Basis of this information,

Here, for 2010 t will be 14

So, The estimated sales for 2010, Y = 5447.667 + 27093.33(14)

= $ 384754.3



ii) Here is the net sales & trend line,


Year Import Time(X)Newimport(Y) Log(Y)

1990 124 1 205 2.311754

1991 175 2 256 2.408241992 306 3 387 2.587711

1993 524 4 605 2.781755

1994 714 5 795 2.900367

1995 1052 6 1133 3.05423

1996 1638 7 1719 3.235276

1997 2463 8 2544 3.405517

1998 3358 9 3439 3.536432

1999 4781 10 4862 3.686815

2000 5388 11 5469 3.737908

2001 8027 12 8108 3.908914

2002 10587 13 10668 4.028083

2003 13537 14 13618 4.134113

Net Sales & Trend Line

0

200000

400000

1 2 3 4 5 6 7 8 9 1

Time

N e t S a l e s

Y

Predicted Y



i) Here, by using statistical software MS Excel,

Coefficient s

Standard Error t Stat P-value

Lower 95%

Upper 95%

Lower 95.0%

Upper 95.0%

Intercept 2.1834980.02601

283.9428

25.45E-

182.12682

42.24017

32.1268

22.24017

3

X Variable 0.1442680.00305

547.2247

45.31E-

150.13761

20.15092

40.1376

10.15092

4

here we find the logarithmic trend,

Y* = log a + log b(t)

Y* = 2.183498 + 0.144268(t)

ii) Here, the annual rate of increase is,

log b = 0.144268

b = antilog(0.144268) - 1

=1.394017 - 1

=.394017

=39.4017%

So, the imports of Carbon block increase at a rate 39.4017% by increasing year by a

unit.

iii) On the Basis of this information,

For 2006, here t will be 17

So, The estimated sales for 2006, log Y = 2.183498 + 0.144268(17)

log Y = 4.636054

Y= antilog(4.636054)

Y = 43256.76 thousands of ton

So, The estimated sales for 2006 will be 43256.76 thousands of ton.




i) Develop a Seasonalize index for each quarter.

Computation of Seasonal Index:

Year Quarter ProductionNewProduction

4 quarterMovingAvg.

Centeredmovingaverage

SpecificSeasonal

Winter 90 171

Spring 85 166

1998 164.25

Summer 56 137 167.375 0.818521

170.5

Fall 102 183 171 1.070175

171.5

Winter 115 196 172.125 1.138707172.75

Spring 89 170 173.75 0.978417

1999 174.75

Summer 61 142 181 0.78453

187.25

Fall 110 191 189.875 1.005925

192.5

Winter 165 246 197.125 1.247939

201.75

Spring 110 191 219 0.872146

2000 236.25

Summer 98 179 240.75 0.74351245.25

Fall 248 329 249.25 1.31996

253.25

Winter 201 282 254.75 1.106968

256.25

Spring 142 223 259.5 0.859345

2001 262.75

Summer 110 191 269 0.710037

275.25

Fall 274 355 278.125 1.276404

281

Winter 251 332 282.875 1.173663

284.75

Spring 165 246 288.625 0.852317

2002 292.5

Summer 125 206 291.25 0.707296

290

Fall 305 386 289.125 1.335063288.25



Winter 241 322 289.125 1.113705

290

Spring 158 239 289.25 0.826275

2003 288.5

Summer 132 213 291.5 0.730703

294.5

Fall 299 380 297.875 1.275703

301.25

Winter 265 346 302.5 1.143802

303.75

Spring 185 266 308 0.863636

2004 312.25

Summer 142 223 314.375 0.709344

316.5

Fall 333 414 315.25 1.313243

314

Winter 282 363 315.875 1.149189

317.75Spring 175 256 319.875 0.800313

2005 322

Summer 157 238 323 0.736842

324

Fall 350 431 327.25 1.317036

330.5

Winter 290 371 334.25 1.109948

338

Spring 201 282 344.25 0.819172

2006 350.5

Summer 187 268

Fall 400 481

Calculation of Seasonal Index :

Year Winter Spring Summer Fall

1998 0.8185213 1.070175

1999 1.138707 0.978417 0.7845304 1.005925

2000 1.247939 0.872146 0.7435099 1.31996

2001 1.106968 0.859345 0.7100372 1.276404

2002 1.173663 0.852317 0.7072961 1.335063

2003 1.113705 0.826275 0.7307033 1.275703

2004 1.143802 0.863636 0.7093439 1.3132432005 1.149189 0.800313 0.7368421 1.317036

2006 1.109948 0.819172

Total 9.183921 6.871621 5.9407841 9.91351 Total

Mean 1.14799 0.858953 0.742598 1.2391893.98872946

Adjusted Mean 1.151234 0.86138 0.7446963 1.24269 4

Index 115.1234 86.13797 74.46963 124.269



Correction factor for adjusting quarterly means:

Correction factor =

=

= 1.002826

Adjusted Mean = Mean * Correction Factor

Mean Adjusted Mean Index

Winter 1.14799007 1.151233827 115.123383

Spring 0.858952656 0.86137971 86.137971

Summer 0.742598019 0.744696301 74.4696301

Fall 1.239188714 1.242690162 124.269016

Interpretation : The production of Winter quarter are 15.12% above the typical quarter, The production

of Spring quarter are 13.87% below the typical quarter, The production of Summer quarter are 25.54%

below the typical quarter, The production of Fall quarter are 24.26% above the typical quarter.

ii) Production for 2007 for different quarters:

Estimated Sale:

Intercept = 5.5629X variable (a) = 163.71

Y = a + b(t)

Y = 163.71 + 5.5629(t)

Quarterly forecast production = Estimated sale(Y) * Seasonal Index

Quarter Time(t)Estimatedsale(Y)

SeasonalIndex

Quarterlyforecastproduction

Winter 37 369.5373 1.151234 425.4238

Spring 38 375.1002 0.86138 323.1037

Summer 39 380.6631 0.744696 283.4784

Fall 40 386.226 1.24269 479.9593



iii) Plot of original data and deseasonalize data :

Deseanolizes Production =

Year Quarter Production

New

Production Code(t)

Seasonal

Index

Deseanolizes

ProductionWinter 90 171 1 1.151233827 148.5362886

1998 Spring 85 166 2 0.86137971 192.7140819

Summer 56 137 3 0.744696301 183.9676117

Fall 102 183 4 1.242690162 147.2611642

Winter 115 196 5 1.151233827 170.2521203

1999 Spring 89 170 6 0.86137971 197.3577948

Summer 61 142 7 0.744696301 190.6817581

Fall 110 191 8 1.242690162 153.6988108

Winter 165 246 9 1.151233827 213.6837836

2000 Spring 110 191 10 0.86137971 221.737287

Summer 98 179 11 0.744696301 240.3664416

Fall 248 329 12 1.242690162 264.7482133

Winter 201 282 13 1.151233827 244.9545813

2001 Spring 142 223 14 0.86137971 258.8869896

Summer 110 191 15 0.744696301 256.4803929

Fall 274 355 16 1.242690162 285.6705645

Winter 251 332 17 1.151233827 288.3862446

2002 Spring 165 246 18 0.86137971 285.5883383

Summer 125 206 19 0.744696301 276.6228322

Fall 305 386 20 1.242690162 310.6164448

Winter 241 322 21 1.151233827 279.6999119

2003 Spring 158 239 22 0.86137971 277.4618409

Summer 132 213 23 0.744696301 286.0226372Fall 299 380 24 1.242690162 305.7882099

Winter 265 346 25 1.151233827 300.5471103

2004 Spring 185 266 26 0.86137971 308.8069024

Summer 142 223 27 0.744696301 299.45093

Fall 333 414 28 1.242690162 333.1482076

Winter 282 363 29 1.151233827 315.3138759

2005 Spring 175 256 30 0.86137971 297.1976203

Summer 157 238 31 0.744696301 319.5933692

Fall 350 431 32 1.242690162 346.8282065

Winter 290 371 33 1.151233827 322.262942

2006 Spring 201 282 34 0.86137971 327.3817537

Summer 187 268 35 0.744696301 359.8782477

Fall 400 481 36 1.242690162 387.0634972

Interpretation : We get the deseanolize data in by dividing the New production by seasonalize index. As

a result we remove the new seasonality from the production.



Using of Deseasonalize Data to Forecast y = 5.5629x + 163.71

R 2

= 0.8864

0

100

200

300

400

500

0 10 20 30 40

Deseasonalized

Index

Linear (Deseasonalized

Index)




i) Here, we select a random sample of 50 cases from 105 cases.

When k = 6,

> 50 or, 64> 50

Class interval range =

=

= 343.23 = 400

confidance Interval

3 6.0 6.0 6.0

22 44.0 44.0 50.0

12 24.0 24.0 74.0

7 14.0 14.0 88.0

5 10.0 10.0 98.0

1 2.0 2.0 100.0

50 100.0 100.0

1300-1700

1700-2100

2100-2500

2500-2900

2900-3300

3300-3700

Total

ValidFrequency Percent Valid Percent

Cumulative

Percent

Confidance Interval

50 0

1.1843

5.00

1.00

6.00

Valid Missing

N

Std. Deviation

Range

Minimum

Maximum



ii)

iii) Bar Diagram for Variable “Confidance Interval” :

confidance Interval

confidance Interval

3300-37002900-33002500-29002100-25001700-21001300-1700

30

20

10

0

confidance Interval

50

0

Valid

Missing

N

Selling Price

50

0

2224.4520

2110.6500

2090.30 a

467.7359

218776.9

2059.40

1710.1400

1875.4000

2110.6500

2525.3000

2938.3100

Valid

Missing

N

Mean

Median

Mode

Std. Deviation

Variance

Range

10

25

50

75

90

Percentiles

Multiple modes exist. The smallest value is shown a.



Comment: From the Bar diagram, we find, the the majority number of selling price (22) in 1700-2100

class interval. and fewer number of selling price (01) is in 3300-3700 class interval.

iv)

Bar Diagram for Variable “Township” :

Comment: From the Bar diagram, we find, the the majority number of apartment situated in Dhanmondi

(14) and fewer apartment is situated in Banani (6).

Pie chart for Variable “Township” :

Comment: From the Pie chart, we find, the the majority portion of apartment situated in Dhanmondi

(28.0%) and fewer apartment is situated in Banani (12.0%).

Statistics

Tow nship

50

0

Valid

Missing

N

Township

Township

BananiDhanmondiDOHSUttaraGulshan

16

14

12

10

8

6

4

2

0

Township

12.0%

28.0%

22.0%

18.0%

20.0%

Banani

Dhanmondi

DOHS

Uttara

Gulshan



v) Develop a box plot for variable “Distance”

Comment : a) The distribution is negatively skewed.b) Mean < Median

c) There is no outlier.


i) Determining the coefficient of skewness for variable “ Selling Price”

Comment : A positive skewness indicates a greater number of smaller values, and a negative value

indicates a greater number of larger values. The coefficient of skewness is .696 .So the distribution

is positively skewed. And mean > median .

Case Process ing Summ ary

50 100.0% 0 .0% 50 100.0%Distance from the

centre of the city

N Percent N Percent N Percent

Valid Missing Total

Cases

Distance from the ce

3020100

Selling Price

50 0

.696

.337

Valid

Missing

N

Skewness

Std. Error of Skewness



ii)

Comment : From the box plot, we find, there are no Outlier. And the distribution is positivelyskewed. And mean > median .

Case Process ing Summ ary

50 100.0% 0 .0% 50 100.0%Selling Price

N Percent N Percent N Percent

Valid Missing Total

Cases

Selling Price

4000300020001000

assignment 1 stat 217 shatil ff

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