ch3measuresofvariability-3.studentview

ch3measuresofvariability-3.studentview - Chapter 3: Chapter...

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Unformatted text preview: Chapter 3: Chapter Measures of Variability Measures Variability, dispersion and volatility are Variability, treated as synonyms. treated A: The Range Example 1: Find the range for each of the Find sets of data for Company A and Company B from Chapter 2 Ex1 and Ex2. from A : 6.7, 9.2, 12.4, 15.2, 20.5 6.7, B : 7.3, 8.7, 10.7, 16.5, 21.8, 55.0 7.3, Solution: Solution For Company A, For R = Max – Min = 20.5 - 6.7 = 13.8=$13,800 For Company B, R = Max - Min = 55 - 7.3 = 47.7=$47,700 Comment: Data set B is much more dispersed than set Data A, as B’s range is much larger than A’s. A, Note: The range is simple to understand. However, it The only looks at the 2 extreme values, ignoring the information inside. Therefore, the range is not much used, especially when N is large. used, Computer steps: Computer ON MODE COMP (1) MODE SD (4) 15.2 M + 9.2 DT 6.7 DT 12.4 DT 20.5 DT 15.2 DT SHIFT S-SUM n (3) SHIFT EXE (1) EXE 1 SHIFT S-VAR 5 12.8 x 2 SHIFT S-VAR minX (1) EXE 6.7 B: Mean Absolute Deviation B: Consider the data of Company A: Consider 15.2, 9.2, 6.7, 12.4, 20.5 15.2, We know that the mean is µ = 12.8 We Had there been no variation in the data, each data value would have been 12.8, that is, identical to µ The actual situation is not so. Some data are smaller than 12.8, some are larger. Some are above the average, some are below. ____|________|_______|___|_______|__________|____ ____|________|_______|___|_______|__________|____ 6.7 9.2 12.4 12.8 15.2 20.5 6.7 µ The deviations are as follows: The ____|________|_______|___|_______|__________|____ 6.7 9.2 12.4 12.8 15.2 20.5 6.7 µ x 6.7 9.2 12.4 15.2 20.5 total d i = ( xi − µ ) = ( xi − 12.8) -6.1 -3.6 -0.4 +2.4 +7.7 d ∑ i= 1 i N = 0 Note: The sum of the deviations is ALWAYS zero. These deviations may be large, may be small, may be positive, may be negative. We are naturally interested in the “average.” However, Average deviation = Average ∑d N i =1 i N = 2.4 − 3.6 − 6.1 − 0.4 + 7.7 0 = =0 5 5 The source of trouble is the signs. To get rid of the sign The trouble, we could use the absolute values: trouble, Mean Absolute Deviation= ∑ |x − µ | = 2.4 + 3.6 + 6.1 + 0.4 + 7.7 = 20.2 MAD = i N 5 5 Thus, on the average, each datum is 4.04 units from Thus, the mean Note: The idea of MAD is simple. But, in mathematics, Note The absolute values are not easy to handle. So, MAD is not popularly used. not = 4.04 C: Standard Deviation C: To get rid of the “sign effect”, and to avoid using absolute values, we find “squaring” a useful tool. After squaring, After (i) All values become positive (ii) A number originally small in magnitude remains small (ii) and a number originally big remains big. and So the average squared deviation , called the average variance, can describe the variability of the variance can data. data. (2.4) 2 + (−3.6) 2 + (−6.1) 2 + (−0.4) 2 + (7.7) 2 115.38 Variance = = = 23.076 5 5 Variance is a useful and often quoted quantity in Variance statistics. However, for the beginner, this quantity has certain difficulty in interpretation, as the unit of measurement (here, it is “squared thousand $”) is unusual. To go back to our familiar unit ($), we take the square root (to off-set the previous step of “squaring”), and call it the σ standard deviation, which is denoted by standard which σ = Variance = 23.076 = 4.8037 [ σ = sigma, the Greek “s”. ] sigma, Formula: σ = Variance = ( xi − µ ) 2 ∑ N →(1) Computer steps: Computer For Company A data ON MODE COMP (1) MODE SD (4) 15.2 M + 9.2 DT 6.7 DT 12.4 DT 20.5 DT 15.2 DT SHIFT S-SUM n (3) SHIFT EXE 1 SHIFT S-VAR 5 xσn (2) EXE 2 4.803748536 Interpretation: Interpretation Thus, the standard deviation is a “not-large”, “not-small” deviation. The standard deviation is a deviation standard, used to tell the largeness or smallness of any other individual deviation. individual It can be seen that the variance is the average of the squared deviations. The s.d. is the square root of the variance. the The deviation is x1 − µ = 15.2 − 12.8 = 2.4 because Small 2.4<4.8037 x2 − µ = 9.2 − 12.8 = −3.6 Small 3.6<4.8037 x3 − µ = 6.7 − 12.8 = −6.1 Large 6.1>4.8037 x4 − µ = 12.4 − 12.8 = − 0.4 Small 0.4<4.8037 x5 − µ = 20.5 − 12.8 = 7.7 Large 7.7>4.8037 Example 2: Example Find the standard deviation for each of the data sets for Company A and Company B. Company A : 6.7, 9.2, 12.4, 15.2, 20.5 µ = 20 B : 7.3, 8.7, 10.7, 16.5, 21.8, 55.0 Solution: For Company A, σ A = 4.8037 , found above. Solution For Company B, For (8.7 − 20) 2 + (10.7 − 20) 2 + ......... + (55 − 20) 2 ∴σ B = 6 1615.96 6 = 16.4112 = Comment: Comment Data set B is much more variable than Set A since the standard deviation for data company B is larger than for company A. for Computer steps : For Company B data Computer ON MODE COMP (1) MODE SD (4) 8.7 DT 10.7 DT 7.3 DT 16.5 DT 21.8 SHIFT S-SUM n (3) EXE DT 55.0 DT 1 SHIFT S-VAR 6 20 16.41117505 x 2 2 (1) EXE (2) EXE SHIFT S-VAR xσn For data, x1 , x2 ,.............., xN , the standard deviation is: is: σ = Variance = ( xi − µ ) 2 ∑ N Often the data x1 , x2 ,.............., xN are “neat” numbers such as integers, but µ involves a lot of decimal places. The results of ( xi − µ ) 2 will also be ugly, i.e. involving many decimal places, so that, at the end, the trouble of rounding will appear. It would be useful to derive another formula that does not call upon µ until a very late stage. We shall derive a practical formula for σ . practical 2 ( x1 − µ) 2 = x1 − 2 µx1 + µ2 2 ( x N − µ) 2 = x N − 2 µx N + µ2 _____________________________ ( xi − µ) 2 = ∑xi2 − 2 µ∑xi + Nµ2 ∑ Now, ∑ x , so that µ= i N 2 ∑ x = Nµ i 2 2 i 2 ∴ ∑ ( xi − µ ) = ∑ x − 2 µ ( Nµ ) + Nµ = ∑ x − Nµ 2 i ∴σ = ( xi − µ ) 2 ∑ N = xi2 ∑ N − µ 2 →(2) Example 3: Use formula (2) to compute Use Solution 1: Solution xi xi2 σ for set A. 15.2 9.2 6.7 12.4 20.5 ∑x 2 i i = 64 231.04 84.64 44.89 153.76 420.25 ∑ x = 934.58 N =5 ∑x ∑x µ= N i σ= = ∑x N 2 i − µ2 i 2 i = 64 = 934.58 ∑x 64 = 5 = 12.8 934.58 −12.8 2 5 = 186.916 −163.84 = 4.8037 Calculator steps: Calculator For Company A data ON MODE COMP (1) MODE SD (4) 15.2 DT 9.2 DT 6.7 DT 12.4 DT 20.5 DT 15.2 SHIFT SHIFT SHIFT SHIFT S-SUM n (3) EXE S-SUM ∑ x (2) EXE S-SUM ∑ x 2 (1) EXE 5 64 934.58 12.8 4.803748536 SHIFT S-VAR SHIFT S-VAR x xσn (1) EXE (2) EXE ...
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This note was uploaded on 04/22/2011 for the course STAT 301 taught by Professor Gabrille during the Spring '11 term at HKU.

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