Md. Imrul Kaes - Six Sigma 2013-4-20
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Transcript of Md. Imrul Kaes - Six Sigma 2013-4-20
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SIX SIGMA MANAGEMENT
TYPES OF STATISTICS – DESCRIPTIVE STATISTICS AND INFERENTIAL STATISTICS
Descriptive Statistics - methods of organizing, summarizing, and presenting data in an informative way.
Inferential Statistics: A decision, estimate, prediction, or generalization about a population, based on a sample.
Note: In statistics the word population and sample have a broader meaning. A population or sample may consist of individuals or objects
1-2
POPULATION VERSUS SAMPLEA population is a collection of all possible individuals, objects, or measurements of interest.
A sample is a portion, or part, of the population of interest
1-3
Price, $20000
21000
20000
22000
23000
19000
21000
23000
24000
20000
No. of
Cars
3
2
1
19000
20000
21000
22000
23000
24000
Quantitative variable - information is reported numerically. EXAMPLES: balance in your checking account, minutes remaining in class, or number of children in a family.
THE MEDIAN
PROPERTIES OF THE MEDIAN
1. There is a unique median for each data set.
2. It is not affected by extremely large or small values and is therefore a valuable measure of central tendency when such values occur.
3. It can be computed for ratio-level, interval-level, and ordinal-level data.
4. It can be computed for an open-ended frequency distribution if the median does not lie in an open-ended class.
EXAMPLES:
MEDIAN The midpoint of the values after they have been ordered from the smallest to the largest, or the largest to the smallest.
The ages for a sample of five college students are:
21, 25, 19, 20, 22
Arranging the data in ascending order gives:
19, 20, 21, 22, 25.
Thus the median is 21.
The heights of four basketball players, in inches, are:
76, 73, 80, 75
Arranging the data in ascending order gives:
73, 75, 76, 80.
Thus the median is 75.5
3-17
Population mean Sample mean
Limitation:
If one or two of the values are either extremely very high or extremely very low compared to the majority of the data, mean might not be an appropriate average to represent the data
POPULATION MEAN
3-19
SKEWNESS
4-20
Your height 5’-8’’
3’-5’ Average 4’
1’-7’ Average 4’
MEASURES OF DISPERSION A measure of location, such as the mean or the median, only describes the center of the data. It is valuable from that
standpoint, but it does not tell us anything about the spread of the data.
For example, if your nature guide told you that the river ahead averaged 3 feet in depth, would you want to wade across on foot without additional information? Probably not. You would want to know something about the variation in the depth.
A second reason for studying the dispersion in a set of data is to compare the spread in two or more distributions.
RANGE
MEAN DEVIATION
VARIANCE AND STANDARD DEVIATION
3-22
EXAMPLE – MEAN DEVIATION
EXAMPLE:
The number of cappuccinos sold at the Starbucks location in the Orange Country Airport between 4 and 7 p.m. for a sample of 5 days last year were 20, 40, 50, 60, and 80. Determine the mean deviation for the number of cappuccinos sold.
Step 1: Compute the mean
Step 2: Subtract the mean (50) from each of the observations, convert to positive if difference is negative
Step 3: Sum the absolute differences found in step 2 then divide by the number of observations
505
8060504020
n
xx
LO7
3-23
VARIANCE AND STANDARD DEVIATION
The variance and standard deviations are nonnegative and are zero only if all observations are the same.
For populations whose values are near the mean, the variance and standard deviation will be small.
For populations whose values are dispersed from the mean, the population variance and standard deviation will be large.
The variance overcomes the weakness of the range by using all the values in the population
VARIANCE: The arithmetic mean of the squared deviations from the mean.
STANDARD DEVIATION: The positive square root of the variance.
LO7
3-24
EXAMPLE – POPULATION VARIANCE AND POPULATION STANDARD DEVIATION
2912
348
12
1034...1719
N
x
12412
488,1)( 22
N
X
3-25
SAMPLE VARIANCE AND STANDARD DEVIATION
sample the in nsobservatio of number the is
sample the of mean the is
sample the in nobservatio each of value the is
variance sample the is
:Where2
n
X
X
s
EXAMPLE
The hourly wages for a sample of part-time employees at Home Depot are: $12, $20, $16, $18, and $19.
What is the sample variance?
3-26
CHEBYSHEV’S THEOREM AND EMPIRICAL RULE
The arithmetic mean biweekly amount contributed by the Dupree Paint employees to the company’s profit-sharing plan is $51.54, and the standard deviation is $7.51. At least what percent of the contributions lie within plus 3.5 standard deviations and minus 3.5 standard deviations of the mean?
3-27
CHEBYSHEV’S THEOREM AND EMPIRICAL RULE
3-28
Chebyshev’s Theorem
Range Amount
± 2 Ϭ 75%
± 3 Ϭ 88.90%
± 5 Ϭ 96%
Empirical Rule
Range Amount
± 1 Ϭ 68%
± 2 Ϭ 95.44%
± 3 Ϭ 99.73%
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SIX SIGMA
±2Ϭ means 45,600 defects per 1000,000 product
±3Ϭ means 2,700 defects per 1000,000 product
±6Ϭ means 2 defects per 1000,000,000 product
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±2Ϭ MEANS 45,600 DEFECTS PER 1000,000 PRODUCT
± 2 Ϭ = 95.44% (From empirical rule)
So (100- 95.44) % = 4.56% defective100 products – 4.56 defective1 product – (4.56/100) 1000,000 products – (4.56/100) x 1000,000
defective =45,600 defective
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±3Ϭ MEANS 2,700 DEFECTS PER 1000,000 PRODUCT
± 3Ϭ = 99.73% (From empirical rule)
So (100- 99.73) % = 0.27% defective100 products – 0.27 defective1 product – (0.27/100) 1000,000 products – (0.27/100) x 1000,000
defective =2,700 defective
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±6Ϭ MEANS 45,600 DEFECTS PER 1000,000,000 PRODUCT
± 6Ϭ = 99.9999998% (From empirical rule)
So (100- 99.9999998) % = 0.0000002% defective
100 products – 0.0000002 defective1 product – (0.0000002 /100) 1000,000,000 products – (4.56/100) x
1000,000,000 defective =2 defective
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±3Ϭ WITH 1.5 Ϭ MEAN SHIFT IN ANY DIRECTION
Suppose mean shift 1.5 Ϭ to the right
Left side 3Ϭ + 1.5 Ϭ = 4.5 Ϭ = 0.49997
Right side 3Ϭ - 1.5 Ϭ = 1.5 Ϭ = 0.43319
Total area = 0.49997 + 0.43319 = 0.93316
That means 93.316% data are within specification limit
That means 6.684% data are outside specification limit
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100 products – 6.684 defective1 product – 6.684/100 defective1000,000 products – (6.684/100) x 1000,000
defectives= 66,840 defectives
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±6Ϭ WITH 1.5 Ϭ MEAN SHIFT IN ANY DIRECTION
Suppose mean shift 1.5 Ϭ to the right
Left side 6Ϭ + 1.5 Ϭ = 7.5 Ϭ
Area outside (left) = 0.00000000000003
Right side 6Ϭ - 1.5 Ϭ = 4.5 Ϭ
Area outside (right) = 0.00000339767313
Total area outside = 0.00000339
That means 0.000339% data are outside specification limit
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100 products – 0.000339 defective1 product – 0.000339/100 defective1000,000 products – (0.000339 /100) x 1000,000
defectives= 3.39 or 3.4 defectives
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Six Sigma is short cut of saying six standard deviations from the mean (both direction), which specifies a tolerable range.