Assignment 1

19
1. Listed below is the net sales in $ million for home depot, Inc and its subsidiaries from 1997 to 2006. i) Estimate sale of 2010 Year Code Net Sales ($) 1997 1 50,704 1998 2 67,404 1999 3 80,904 2000 4 98,204 2001 5 1,24,5 04 2002 6 1,56,8 04 2003 7 2,01,5 04 2004 8 2,27,4 04 2005 9 2,56,4 04 2006 10 2,81,0 04 2007 11 2008 12 2009 13 2010 14 384772 .7 ii) Plotting of net sales and trend line

Transcript of Assignment 1

Page 1: Assignment 1

1. Listed below is the net sales in $ million for home depot, Inc and its subsidiaries from 1997 to 2006.

i) Estimate sale of 2010

Year Code Net Sales ($)

1997 1 50,7041998 2 67,4041999 3 80,9042000 4 98,2042001 5 1,24,5042002 6 1,56,8042003 7 2,01,5042004 8 2,27,4042005 9 2,56,4042006 10 2,81,0042007 112008 122009 132010 14 384772.7

ii) Plotting of net sales and trend line

Page 2: Assignment 1

2. It appears that the imports of carbon black have been increasing but about 10 percent annually.

i) The logarithmic trend

ii) To find the annual rate of increase:

Log b = 0.1406

b = Anti Log (0.1406)

= 1.38

So, the annual rate of increase: GM = b – 1

= 1.38 – 1

= 0.38

= 38%

So, the imports of carbon black increase 38% annually.

Page 3: Assignment 1

iii) Estimate of imports in 2006

Year Code Import log (imports)1990 1 228 2.3579348471991 2 279 2.4456042031992 3 410 2.6127838571993 4 628 2.7979596441994 5 818 2.9127533041995 6 1156 3.0629578341996 7 1742 3.2410481511997 8 2567 3.4094258691998 9 3462 3.5393270641999 10 4285 3.6319508262000 11 5492 3.7397305292001 12 8131 3.9101439612002 13 10691 4.029018332003 14 13641 4.1348462092004 152005 162006 17 4.6088

Where,

Intercept, a = 2.2186

X Variable 1, b = 0.1406

So, the log trend equation: is: Log Y'= 2.2186+ 0.1406 (t)

Here, 2.2186 is the point through which the trend line crosses the y-axis. If the time is increased by 1 year, then the imports of carbon black will be increased by an amount of $ 0.14 thousands of tons.

Page 4: Assignment 1

3. The quarterly production of pine lumber, in million of board feet by Northwest lumber since 1998 is:

i) Seasonal index for each quarter

Year Quarter Production4QTR Total 4QTR avg Centrd Avg

Seasonality

1998 Winter 194

Spring 189749 187.25

Summer 160 190.375 0.8404774 193.5

Fall 206 194 1.0619778 194.5

1999 Winter 219 195.125 1.1224783 195.75

Spring 193 196.75 0.9809791 197.75

Summer 165 204 0.8088841 210.25

Fall 214 212.875 1.0053862 215.5

2000 Winter 269 220.125 1.2220899 224.75

Spring 214 242 0.88431037 259.25

Summer 202 263.75 0.76591073 268.25

Fall 352 272.25 1.29291105 276.25

2001 Winter 305 277.75 1.09811117 279.25

Spring 246 282.5 0.87081143 285.75

Summer 214 292 0.73291193 298.25

Fall 378 301.125 1.25531216 304

2002 Winter 355 305.875 1.16061231 307.75

Page 5: Assignment 1

Spring 269 311.625 0.86321262 315.5

Summer 229 314.25 0.72871252 313

Fall 409 312.125 1.31041245 311.25

2003 Winter 345 312.125 1.10531252 313

Spring 262 312.25 0.83911246 311.5

Summer 236 314.5 0.75041270 317.5

Fall 403 320.875 1.25591297 324.25

2004 Winter 369 325.5 1.13361307 326.75

Spring 289 331 0.87311341 335.25

Summer 246 337.375 0.72921358 339.5

Fall 437 338.25 1.29191348 337

2005 Winter 386 338.875 1.13911363 340.75

Spring 279 342.875 0.81371380 345

Summer 261 346 0.75431388 347

Fall 454 350.25 1.29621414 353.5

2006 Winter 394 357.25 1.10291444 361

Spring 305 367.25 0.83051494 373.5

Summer 291

Fall 504

Table - 4

Page 6: Assignment 1

Year Winter Spring Summer Fall1998 0.8404 1.06191999 1.1224 0.9809 0.8088 1.00532000 1.2220 0.8843 0.7659 1.29292001 1.0981 0.8708 0.7329 1.25532002 1.1606 0.8632 0.7287 1.31042003 1.1053 0.8391 0.7504 1.2559 CF=0.9810140172004 1.1336 0.8731 0.7292 1.29192005 1.1391 0.8137 0.7543 1.29622006 1.1029 0.8305 0 0Total 9.0840 6.9556 6.1106 9.7698Mean 1.223 0.869 0.764 1.221 4.077Adjusted Mean 1.199682 0.852947 0.749327 1.198043 4

ii) Plotting of the original data and Deseasonalized data-

Production Index CodesDeseasonalized Index

194 1.199682 1 161.71189 0.852947 2 221.58160 0.749327 3 213.52206 1.198043 4 171.95219 1.199682 5 182.55193 0.852947 6 226.27165 0.749327 7 220.20214 1.198043 8 178.62269 1.199682 9 224.23214 0.852947 10 250.89202 0.749327 11 269.58352 1.198043 12 293.81305 1.199682 13 254.23246 0.852947 14 288.41214 0.749327 15 285.59378 1.198043 16 315.51355 1.199682 17 295.91269 0.852947 18 315.38229 0.749327 19 305.61409 1.198043 20 341.39345 1.199682 21 287.58262 0.852947 22 307.17236 0.749327 23 314.95

Page 7: Assignment 1

403 1.198043 24 336.38369 1.199682 25 307.58289 0.852947 26 338.83246 0.749327 27 328.29437 1.198043 28 364.76386 1.199682 29 321.75279 0.852947 30 327.10261 0.749327 31 348.31454 1.198043 32 378.95394 1.199682 33 328.42305 0.852947 34 357.58291 0.749327 35 388.35504 1.198043 36 420.69

Comments: Here, the plot which shows the original data and the deseasonalized data.

Graph: production and deseasonalize data

Page 8: Assignment 1

Graph: Code & deseasonalize data

Answer to the Question no.4

1. First enter the Wage and wage earners data into SPSS and then select a sample of size, n=50

i) Select an appropriate class interval and organize the selling price into a frequency distribution.

Descriptive Statistics

N Minimum Maximum

Selling price 50 1488.9 3548.30

Valid N (listwise)

50

Maximum (H) = 3548.30

Minimum (L) = 1488.9

Page 9: Assignment 1

Select an appropriate class interval:

2k= rule to decide the way of class

The sample size n= 50

So, 26= 64

Hence, k= 6

Class Interval (i) = H-L

K

= 3548.30 - 1488.9

6

=343.2333

We will take class interval (i) as 500.

Graph: Selling Price frequency distribution

Page 10: Assignment 1

ii) Compute the mean, median, mode, standard deviation, variance, quartiles, 9 th

deciles, 10th percentiles and range of “selling price” from the raw data of your sample.

StatisticsSelling Price

N Valid 50Missing 0

Mean 2422.8520Median 2312.6500

Mode 2292.30Std.

Deviation465.5080

Variance 216697.7319

Skewness .711Std. Error of

Skewness.337

Range 2059.40Percentiles 10 1907.4000

25 2077.400050 2312.650075 2688.0500

iii) Histogram for the variable “class interval”

Page 11: Assignment 1

iv) Pie chart for the variable “Township”

Statistics

Township

N Valid 50Missing 0

Mean 2.94 Median 3.00

Mode 4 Std. Deviation 1.33

Variance 1.77 Range 4

Percentiles 10 1.00

25 2.0050 3.0075 4.00

Township

Frequency Percent Valid Percent

Cumulative Percent

Valid Gulshan 10 20.0 20.0 20.0

Uttara 9 18.0 18.0 38.0

DOHS 11 22.0 22.0 60.0

Dhanmondi 14 28.0 28.0 88.0

Banani 6 12.0 12.0 100.0

Total 50 100.0 100.0

Page 12: Assignment 1

Graph: pie chart for the variable Township

Interpretation: From the above pie chart, Dhanmondi shows the highest township

(28%) and Banani contains the lowest township (6%).

Bar diagram for the variable “Township”

Statistics

Township

N Valid 50Missing 0

Mean 2.94 Median 3.00

Mode 4 Std.

Deviation1.33

Variance 1.77 Range 4

Percentiles

10 1.00

25 2.0050 3.0075 4.00

Page 13: Assignment 1

TownshipFrequen

cyPercent Valid

PercentCumulati

ve Percent

Valid Gulshan 10 20.0 20.0 20.0Uttara 9 18.0 18.0 38.0DOHS 11 22.0 22.0 60.0

Dhanmondi

14 28.0 28.0 88.0

Banani 6 12.0 12.0 100.0Total 50 100.0 100.0

Graph: Bar chart for the variable Township

Interpretation: In this Bar diagram of township by using “question a”, we found that

there are 10 homes in Gulsan, 9 in Uttara, 11 in DOHS, 14 in Dhanmondi and 6 in

Banani for sale.

v) Develop a Box plot for the variable “Distance”. What information can you give from the plot?

Page 14: Assignment 1

Case Processing SummaryCases

Valid Missing TotalN Percent N Percent N Percent

Distance from

the centre of

the city

50 100.0% 0 .0% 50 100.0%

Graph: Box plot for the variable Distance

Answer to the Question no.5

For the variable “Selling Price”

i) Determine the coefficient of skewness. Is the distribution positively or negatively skewed?

Page 15: Assignment 1

StatisticsSelling Price

N Valid 50Missing 0

Skewness .693

Std. Error of

Skewness

.337

Comment: As the value of the coefficient of skewness is .693 which is a positive one so we can

say that it is a distribution which is positively skewed. We can also say it a moderate positive

skewed distribution.

ii) Develop a Box plot. Are there any outliers?

Case Processing SummaryCases

Valid Missing TotalN Percent N Percent N Percent

Selling Price

50 100.0% 0 .0% 50 100.0%

Graph: Box plot for variable Selling price