{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ch3_measures_of_variability.studentview

# And often quoted quantity in variance statistics

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: owever, 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 2 xσn (2) EXE 5 4.803748536 Interpretation: Interpretation The deviation is x1 − µ = 15.2 − 12.8 = 2.4 because Small 2.4<4.8037 Small 3.6<4.8037 x2 − µ = 9.2 − 12.8 = −3.6 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 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 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) 1 SHIFT S-VAR 2 SHIFT S-VAR 2 x EXE (1) EXE xσn (2) EXE DT 55.0 DT 6 20 16.41117505 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 ∑ x = Nµ i ∴ ∑ ( xi − µ ) = ∑ x − 2 µ ( Nµ ) + Nµ = ∑ x − Nµ 2 ∴σ = 2 i ( xi − µ ) 2 ∑ N 2 = xi2 ∑ N 2 i − µ 2 →(2) 2 Example 3: Use formula (2) to compute Use σ for set A. Solution 1: Solution 15.2 xi xi2 9.2 6.7 12.4 20.5 ∑x i 231.04 84.64 44.89 153.76 420.25 ∑ x 2 i N =5 ∑x i ∑x 2 i ∑x µ= i = 64 = 934.58 σ= N 64 = 5 = 12.8 = ∑x 2 i N = 64 = 934.58 − µ2 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 SHIFT S-VAR SHIFT S-VAR x xσn (1) EXE (2) EXE 5 64 934.58 12.8 4.803748536...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online