Business Statistics – NMIMS

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NMIMS Global Access

School for Continuing Education (NGA-SCE)

Course:  Business Statistics

Internal Assignment Applicable for September 2020 Examination

Assignment Marks: 30

Instructions:

  • All Questions carry equal marks.
  • All Questions are compulsory
  • All answers to be explained in not more than 1000 words for question 1 and 2 and for question 3 in not more than 500 words for each subsection. Use relevant examples, illustrations as far aspossible.
  • All answers to be written individually. Discussion and group work is not advisable.
  • Students are free to refer to any books/reference material/website/internet for attempting theirassignments, but are not allowed to copy the matter as it is from the source of reference.
  • Students should write the assignment in their own words. Copying of assignments from otherstudents is not allowed.
  • Students should follow the following parameter for answering the assignment questions.
For Theoretical Answer For Numerical Answer
Assessment Parameter  Weightage  Assessment Parameter  Weightage  
Introduction20%Understanding and usage of the formula20%
Concepts and Application related to the question60%Procedure / Steps50%
Conclusion20% Correct Answer & Interpretation30%

Question 1: Data Set:

Sample of 7 different species of fish has been taken and their weight in grams, lengths (vertical, diagonal, cross given as length 1, length 2 and length 3 respectively), height and width in cm is given below:

SpeciesWeightLength1Length2Length3HeightWidth
Bream72531.83540.916.366.0532
Bream720323540.616.36186.09
Bream71432.73641.516.5175.8515
Bream85032.83641.616.88966.1984
Bream100033.53742.618.9576.603
Bream9203538.544.118.03696.3063
Bream9553538.54418.0846.292
Bream92536.239.545.318.75426.7497
Bream97537.44145.918.63546.7473
Bream950384146.517.62356.3705
Roach01920.522.86.47523.3516
Roach11019.120.823.16.16773.3957
Roach12019.42123.76.11463.2943
Roach15020.42224.75.80453.7544
Roach14520.52224.36.63393.5478
Roach16020.522.525.37.03343.8203
Roach1402122.5256.553.325
Roach16021.122.5256.43.8
Roach169222427.27.53443.8352
Roach1612223.426.76.91533.6312
Roach20022.123.526.87.39684.1272
Roach18023.625.227.97.08663.906
Roach290242629.28.87684.4968
Roach272252730.68.5684.7736
Roach39029.531.7359.4855.355
Whitefish27023.62628.78.38044.2476
Whitefish27024.126.529.38.14544.2485
Whitefish30625.62830.88.7784.6816
Whitefish54028.5313410.7446.562
Whitefish80033.736.439.611.76126.5736
Whitefish100037.34043.512.3546.525
Parkki5513.514.716.56.84752.3265
Parkki6014.315.517.46.57722.3142
Parkki9016.317.719.87.40522.673
Parkki12017.51921.38.39222.9181
Parkki15018.42022.48.89283.2928
Parkki1401920.723.28.53763.2944
Parkki1701920.723.29.3963.4104
Parkki14519.821.524.19.73643.1571
Parkki20021.22325.810.34583.6636
Parkki27323252811.0884.144
Parkki30024262911.3684.234
Perch5.97.58.48.82.1121.408
Perch3212.513.714.73.5281.9992
Perch4013.815163.8242.432
Perch51.51516.217.24.59242.6316
Perch7015.717.418.54.5882.9415
Perch10016.21819.25.22243.3216
Perch7816.818.719.45.19923.1234
Perch8017.21920.25.63583.0502
Perch8517.819.620.85.13763.0368
Perch8518.220215.0822.772
Perch110192122.55.69253.555
Pike43035.53840.57.294.5765
Pike3453638.5416.3963.977
Pike4564042.545.57.284.3225
Pike5104042.545.56.8254.459
Pike54040.14345.87.7865.1296
Pike5004245486.964.896
Pike56743.24648.77.7924.87
Pike77044.84851.27.685.376
Pike95048.351.755.18.92626.1712
Pike1250525659.710.68636.9849
Smelt6.79.39.810.81.73881.0476
Smelt7.51010.511.61.9721.16
Smelt710.110.611.61.72841.1484
Smelt9.710.411122.1961.38
Smelt9.810.711.212.42.08321.2772
Smelt8.710.811.312.61.97821.2852
Smelt1011.311.813.12.21391.2838
Smelt9.911.311.813.12.21391.1659
Smelt9.811.41213.22.20441.1484
Smelt12.211.512.213.42.09041.3936

Q1: Find the mean and standard deviation for each type of fish for every variable.

Answer: 1. To compute the mean and standard deviation we will use the below formulas.

The mean  \small \overline{X} is computed as follows,

\small \overline{X}=\frac{1}{n}\sum_{i=1}^{n}X_{i}

Also, the sample variance \small s^{2}is,

\small s^{2}=\frac{1}{(n-1)}
\small \left ( \sum_{i=1}^{n} X_{i}^{2} -\frac{1}{n}\left ( \sum_{i=1}^{n}X_{i} \right )^{2}\right )

Question 2. If you need to choose a fish on the basis of weight, which fish you choose?

Answer   : The mean and standard deviation for Weight variable for all species is below:

SpeciesMeanStandard Deviation
Bream873.4113.1471019
Roach176.466666788.95574073
Whitefish531309.6029716
Parkki154.818181878.75508642

Question.3. Find mean, median, quartiles for the entire data set for each variable. (10 marks)

Answer :  Mean:

 By using the same formula as in part (a), the mean are,

The sample size is n = 73. The provided sample data along with the data required to compute the sample mean \small \overline{X} for each variable of all species is shown in the table below:

Mean

Question 2 : Regress the following: (10 Marks)

  1. Taking weight as dependent variable and height as independent variable. Is variable is found to be significant?

Answer :  Below is the output:

Regression Statistics
Multiple R0.094799718
   
  1. Taking weight as dependent variable and width as independent variable. Is variable is found to be significant?

Answer : 

Regression Statistics
Multiple R0.913804714
R Square0.835039055
  1. Taking weight as dependent variable and length1, length 2 and length 3 as independent variable. Are variables is found to be significant? Which variable is not significant?

Answer : 

Regression Statistics
Multiple R0.938718572
R Square0.881192557
Adjusted R Square0.876027017
  1. Taking weight as dependent variable and height, width, length 1, length 2 and length 3 as independent variable.

Answer : 

Regression Statistics
Multiple R0.940567787
R Square0.884667763
Adjusted R Square0.876060879
Standard Error117.7864532
Observations73
ANOVA
 dfSSMSFSignificance F
Regression57130089.4181426017.884102.78607524.85005E-30
 
  1. On basis of adjusted R square compare the model of part a, b, c and d. which model is best to predict weight?
ModelsAdjusted R Square
Weight vs Height-0.004970943
Weight vs Width0.832715661

Question 3.  The daily COVID 19 cases (in hundred) for Delhi for past 2 week is summarize in the following table:

Day1234567891011121314
Cases2829333137343643413234373932
  1. Using exponential smoothing method forecast the cases for 15 days, taking alpha as 0.3 and Initial forecast is the average of all data. (5 Marks)

Answer : First we need to compute mean of the data.

The mean  \small \overline{X} is computed as follows,

\small \overline{X}=\frac{1}{n}\sum_{i=1}^{n}X_{i}

   Mean = (1/14)* 486 = 34.7142857.

  1. Using linear trend analysis, find the trend line for number of COVID 19 cases in Delhi and forecast for next 3 days. Also compute the Mean Square Error. (5 Marks)

Answer : The independent variable is Time, and the dependent variable is Cases. In order to compute the regression coefficients, the following table needs to be used:

Days (X)Cases (Y)Days*Cases (XY)Days^2 (X^2)Cases^2 (Y^2)
128281784
229584841
3339991089
43112416961
537185251369
634204361156
736252491296
843344641849
941369811681
10323201001024
11343741211156
12374441441369
13395071691521
14324481961024
Sum =1054863756101517120

Hello MBA aspirants,

Get MBA assignments of NMIMS University solved by educational professionals at a nominal charge.

Mail us at: help.mbaassignments@gmail.com

Call us at: 08263069601

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