# HOMEWORK #D1

HOMEWORK#D1

Reviewquestions

1. Variance is the average found from the sum of the square of deviations from the mean by each value in a sample. The deviations are squared so as to avoid working with the negative values given by some deviations in the sample. If they are not squared, the negative deviations will cancel out the positive deviations. The standard deviation is the square root of variance which therefore returns the variance to the original value. Hence the relationship between the two is that variance is the square of the standard deviation [ CITATION Owe813 l 1033 ].

2. The two basic formulas for computing variance are

1. Variance = ∑(x – m) 2 / n m = mean, n = number of items in the sample

2. Variance = ∑x2 /n – (∑x /n) 2

Thefirst equation is used when we calculate the deviations using themean while the second one is used when the deviations are calculatedwithout using the value of the mean.

1. The variance of any sample is always an underestimate of the variance of the population. To correct the situation and give a variance that is close to the population variance, the sample variance and also the sample standard deviation is divided by n-1 and not n.

2. In using a spreadsheet program intrinsic function to compute a standard deviation, the items are presented in a column of cells below which a standard deviation formula is inserted. After selecting the data, the spreadsheet will automatically give the standard deviation of the data.

Computationalexercise

1. Find the sample standard deviation for the following data sets, using the basic definitional formula. Show your work

1. 3, 2, 5, 6

Mean= (3 + 2 + 5 + 6)/4 = 4

Variance= (-12+ -22+ 12+ 22)/3= 3.333

Standarddeviation = √3.333 = 1.83

1. 587, 589, 588

Mean= (587 + 589 + 588)/3 = 588

Variance= (-12+ 12+ 02)/2= 1

Standarddeviation = √1 = 1

1. 12, 15, 18, 16, 14

Mean= (12 + 15 + 18 + 16 + 14)/5 = 15

Variance= (-32+ 02+ 32+ 12+ -12)/4= 5

Standarddeviation = √5 = 2.24

1. 6, 1, 8, 5

Mean= (6 + 1 + 8 + 5)/4 = 5

Variance= (12+ -42+ 32+ 02)/3= 8.667

Standarddeviation = √8.667 = 2.94

1. 42, 42, 42, 42, 42, 42

Mean= (42 + 42 + 42 + 42 + 42 + 42)/6 = 42

Variance= (02+ 02+ 02+ 02+ 02+ 02)/5= 0

Standarddeviation = √0 = 0

1. 473875978239874398753425453879654641,

473875978239874398753425453879654642,

473875978239874398753425453879654643

Mean= 473875978239874398753425453879654642

Variance= (-12+ 02+ 12)/2= 1

Standarddeviation = √1 = 1

1. Find the sample standard deviation using alternate computational formula

1. 3, 2, 5, 6

Variance= (32+ 22+ 52+ 62)/4– ((3 + 2 + 5 + 6)/4)2=18.5 – 16 = 2.5

Samplevariance = 2.5 x 4/3 = 3.333

Standarddeviation = √3.333 = 1.83

1. 12, 9, 14, 6, 3, 11

Variance= (122+ 92+ 142+ 62+ 32+ 112)/6– ((12 + 9 + 14 + 6 + 3 + 11)/6)2=97.83 – 84.03 = 13.8

Samplevariance = 13.8 x 6/5 = 16.56

Standarddeviation = √16.56 = 4.07

1. Excel

1. 3.46, 2.19, 8.23, 1.38, 9.24, 8.15, 9.90, 3.27, 8.64, 9.01, 7.42, 8.23

Usingexcel the standard deviation is 3.072791

1. 98347, 27894, 68287, 91684, 47383, 73436

Usingexcel the standard deviation is 26626.68

1. Amazon- average= 148.64

Standarddeviation = 104.91

KO- average= 26.52

Standarddeviation = 8.37

Datasource- [ CITATION Ras14 l 1033 ]

Thenumbers indicate that Amazon has higher average daily returns thanKO. However, the percentage standard deviation in relation to themean in Amazon is higher than KO which means that the risks in Amazonare higher than the risks in KO.

References

Owen, F., &amp Jones, R. (1981). Statistics. London: Polytech.

Rasp, J. D. (2014). Data set for classroom use. Retrieved January 21, 2015, from Dr. John Rasp`s statistics website: http://www2.stetson.edu/…jrasp/data.htm