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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

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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

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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

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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

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SKEWNESS

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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

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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

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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

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EXAMPLE – POPULATION VARIANCE AND POPULATION STANDARD DEVIATION

2912

348

12

1034...1719

N

x

12412

488,1)( 22

N

X

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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?

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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?

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CHEBYSHEV’S THEOREM AND EMPIRICAL RULE

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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.