statistical analysis - graphical techniques

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Statistical Analysis - Graphical Techniques

Dr. Jerrell T. Stracener, SAE Fellow

Leadership in Engineering

EMIS 7370/5370 STAT 5340 : PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS

Systems Engineering ProgramDepartment of Engineering Management, Information and Systems

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•Time Series Graph or Run Chart

• Box Plot

• Histogram and Relative Frequency Histogram

• Frequency Distribution

• Probability Plotting

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• A plot of the data set x1, x2, …, xn in the order in which the data were obtained

•Used to detect trends or patterns in the dataover time

Time Series Graph or Run Chart

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• A pictorial summary used to describe the most prominent statistical features of the data set, x1, x2, …, xn, including its:

- Center or location- Spread or variability- Extent and nature of any deviation from symmetry- Identification of ‘outliers’

Box Plot

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• Shows only certain statistics rather than all thedata, namely

- median- quartiles- smallest and greatest values in the sample

• Immediate visuals of a box plot are the center,the spread, and the overall range of the data

Box Plot

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Given the following random sample of size 25:

38, 10, 60, 90, 88, 96, 1, 41, 86, 14, 25, 5, 16,22, 29, 34, 55, 36, 37, 36, 91, 47, 43, 30, 98

Arranged in order from least to greatest:

1, 5, 10, 14, 16, 22, 25, 29, 30, 34, 36, 36, 37, 38, 41, 43, 47, 55, 60, 86, 88, 90, 91, 96, 98

Box Plot

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•First, find the median, the value exactly in themiddle of an ordered set of numbers.

The median is 37

• Next, we consider only the values to the left ofthe median:

1, 5, 10, 14, 16, 22, 25, 29, 30, 34, 36, 36

We now find the median of this set of numbers. The median for this group is (22 + 25)/2 = 23.5,which is the lower quartile.

Box Plot

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• Now consider the values to the right of themedian.

38, 41, 43, 47, 55, 60, 86, 88, 90, 91, 96, 98

The median for this set is (60 + 86)/2 = 73, whichis the upper quartile.

We are now ready to find the interquartile range (IQR), which is the difference between the upperand lower quartiles, 73 - 23.5 = 49.5

49.5 is the interquartile range

Box Plot

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The lower quartile 23.5The median is 37The upper quartile 73 The interquartile range is 49.5The mean is 45.1

upperquartile

0 10 20 30 40 50 60 70 80 90 100

lowerextreme

upperextreme

lowerquartile

medianmean

Box Plot

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A graph of the observed frequencies in the dataset, x1, x2, …, xn versus data magnitude tovisually indicate its statistical properties, including

- shape- location or central tendency- scatter or variability

Histogram

Guidelines for Constructing Histograms – Discrete Data

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• If the data x1, x2, …, xn are from a discrete random variable with possible values y1, y2, …, yk

count the number of occurrences of each valueof y and associate the frequency fi with yi, for i = 1, …, k,

Note that

k

ii nf

1

Guidelines for Constructing Histograms – Discrete Data

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• If the data x1, x2, …, xn are from a continuousrandom variable

- select the number of intervals or cells, r,to be a number between 3 and 20, as an initial value use r = (n)1/2, where n is the number of observations- establish r intervals of equal width, startingjust below the smallest value of x- count the number of values of x withineach interval to obtain the frequency associated with each interval- construct graph by plotting (fi, i) for i = 1, 2, …, k

Guidelines for Constructing Histograms – Continuous Data

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2.2 4.1 3.5 4.5 3.2 3.7 3 2.63.4 1.6 3.1 3.3 3.8 3.1 4.7 3.72.5 4.3 3.4 3.6 2.9 3.3 3.9 3.13.3 3.1 3.7 4.4 3.2 4.1 1.9 3.44.7 3.8 3.2 2.6 3.9 3 4.2 3.5

Car Battery Lives

To illustrate the construction of a relative frequency distribution,consider the following data which represent the lives of 40 carbatteries of a given type recorded to the nearest tenth of a year.The batteries were guaranteed to last 3 years.

Histogram and Relative Frequency Example

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For this example, using the guidelines for constructing a histogram,the number of classes selected is 7 with a class width of 0.5. Thefrequency and relative frequency distribution for the data are shownin the following table.

Class Class Frequency Relativeinterval midpoint f frequency1.5-1.9 1.7 2 0.0502.0-2.4 2.2 1 0.0252.5-2.9 2.7 4 0.1003.0-3.4 3.2 15 0.3753.5-3.9 3.7 10 0.2504.0-4.4 4.2 5 0.1254.5-4.9 4.7 3 0.075

Total 40 1.000

Relative Frequency Distribution ofBattery Lives

Histogram and Relative Frequency Example

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The following diagram is a relative frequency histogram of the batterylives with an approximate estimate of the probability density functionsuperimposed.

Relative frequency histogram

0.000

0.050

0.100

0.150

0.200

0.250

0.300

0.350

0.400

1.7 2.2 2.7 3.2 3.7 4.2 4.7

Battery Lives (years)

Rel

ativ

e F

req

uen

cy

Histogram and Relative Frequency

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• Data are plotted on special graph paper designed for a particular distribution

- Normal - Weibull- Lognormal - Exponential

• If the assumed model is adequate, the plotted points will tend to fall in a straight line

• If the model is inadequate, the plot will not be linear and the type & extent of departures can be seen

• Once a model appears to fit the data reasonably well, percentiles and parameters can be estimated from the plot

Probability Plotting

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We need value estimates corresponding to each of the sample values in order to plot the data on the probability paper. These estimates are accomplished with what are called median ranks.

Median ranks represent the 50% confidence level (“best guess”) estimate for the true value of F(t), based on the total sample size and the order number (first, second, etc.) of the data.

Probability Plotting General Procedure

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There is an approximation that can be used to estimate median ranks, called Benard’s approximation. It has the form:

where n is the sample size and i is the sample order number. Tables of median ranks can be found in many statistics and reliability texts.

%)100(0.4n

0.3iMRxF̂ ii

Benard’s Approximation

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• Step 1: Obtain special graph paper, known asprobability paper, designed for the distribution underexamination. Weibull, Lognormal and Normal paper are available at:http://www.weibull.com/GPaper/index.htm

• Step 2: Rank the sample values from smallest to largest in magnitude i.e., X1 X2 ..., Xn.

Probability Plotting Procedure

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• Step 3: Plot the Xi’s on the paper versus or ,

depending on whether the marked axis on the paper refers to the % or the proportion of observations. The axis of the graph paper on which the Xi’s are plotted is referred to as the observational scale, and the axis for as the cumulative scale.

%100*4.0

3.0)(

n

ixF i 40

30

.n

.i)F(x

^

i

%100*4.0

3.0)(

^

n

ixF i


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• Step 4: If a straight line appears to fit the data, draw a line on the graph, ‘by eye’.

• Step 5: Estimate the model parameters from the graph.

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If

the cumulative probability distribution function is

We now need to linearize this function into the form y = ax +b

t

et 1)(F

θβ,W~T

Weibull Probability Plotting Paper

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Then

which is the equation of a straight line of the form y = ax +b

lnln)T(F1

1lnln

ln)T(F1lnln

)T(F1ln

ln)T(F1ln

x

x

x

ex


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where

and

)t(F1

1lnlny

a

tx ln

i.e., ,ln b


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which is a linear equation with a slope of b and an intercept of . Now the x- and y-axes of the Weibull probability plotting paper can be constructed. The x-axis is simply logarithmic, since x = ln(T) and

ln xy

ln

)t(F1

1lnlny


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cumulativeprobability

(in %)

x


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To illustrate the process let 10, 20, 30, 40, 50, and 80 be a random sample of size n = 6.

Probability Plotting - Example

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Based on Benard’s approximation,

we can now calculate F(t) for each observed value of X.

For example, for x2=20,

%6.26

%100*0.46

0.3220F̂

^


%)100(0.4n

0.3iMRxF̂ ii

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i xi F(xi)

1 10 10.9%2 20 26.6%3 30 42.2%4 40 57.8%5 50 73.4%6 80 89.1%

In summary,


^

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Now that we have y-coordinate values to go with the x-coordinate sample values so we can plot the points on Weibull probability paper. xx F̂,

F(x)(in %)

x

^


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The line represents the estimated relationship between x and F(x):

x

F(x)(in %)

^


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In this example, the points on Weibull probability paper fall in a fairly linear fashion, indicating that the Weibull distribution provides a good fit to the data. If the points did not seem to follow a straight line, we might want to consider using another probability distribution to analyze the data.


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34


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Probability Paper - Normal

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Probability Paper - Lognormal

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Probability Paper - Exponential

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Given the following random sample of size n=8, which probability distribution provides the best fit?

i x i

1 79.409682 88.120933 91.063944 98.730945 104.15366 105.10197 106.50368 112.0354

Example - Probability Plotting

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40 specimens are cut from a plate for tensile tests. The tensile tests were made, resulting in Tensile Strength, x, as follows:

i x i x i x i x1 48.5 11 55.0 21 53.1 31 54.62 54.7 12 55.7 22 49.1 32 49.93 47.8 13 49.9 23 55.6 33 44.54 56.9 14 54.8 24 46.2 34 52.95 54.8 15 49.7 25 52.0 35 54.46 57.9 16 58.9 26 56.6 36 60.27 44.9 17 52.7 27 52.9 37 50.28 53.0 18 57.8 28 52.2 38 57.49 54.7 19 46.8 29 54.1 39 54.8

Perform a statistical analysis of the tensile strength data.

40 Specimens

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Time Series plot:

By visual inspection of the scatter plot, there seems to be no trend.Therefore, sample appears to be a random sample.

40 Specimens

30.0

35.0

40.0

45.0

50.0

55.0

60.0

65.0

0 5 10 15 20 25 30 35 40

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40 Specimens

Descriptive Statistics

Count 40Minimum 42.35Maximum 61.18Range 18.84Sum 2104.82Mean 52.62Median 53.03Sample Variance 19.83Standard Deviation 4.45Kurtosis 2.51Skewness -0.34

Using the descriptive statistics function in Excel, the following were calculated:

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40 Specimens

From looking at the Histogram and the Normal Probability Plot, we see that the tensile strength can be estimated by a normal distribution.

Using the histogram feature of excel the following data was calculated:

and the graph:

Bin Frequency40 045 350 1055 1660 9

More 2

Histogram of Tensile Strengths

0

2

4

6

8

10

12

14

16

18

40 45 50 55 60 More

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40 Specimens

Box Plot

The lower quartile 49.45The median is 53.03The mean 52.6The upper quartile 55.3The interquartile range is 5.86

40 45 50 55 60 65

lowerextreme upper

extreme

lowerquartile

upperquartile

medianmean

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40 Specimens

Normal Probability Plot

0.10%

1%

5%

10%

20%

30%

40%

50%

60%

70%

80%

90%

95%

99%

99.90%

40 45 50 55 60 65

45

40 Specimens

LogNormal Probability Plot

0.10%

1%

5%

10%

20%

30%

40%

50%

60%

70%

80%

90%

95%

99%

99.90%

10 100

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40 Specimens

Weibull Probability Plot

0.10%

0.20%

0.30%

0.50%

1%

2%

3%

5%

10%

20%

30%

40%50%60%70%80%

90%95%

99%

99.90%

41 44 48 52 56 61

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The tensile strength distribution can be estimated by

40 Specimens

45.4ˆ,62.52μ̂N~X

0

0.2

0.4

0.6

0.8

1

49 50 51 52 53 54 55

f(x)

F(x)

^

^

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Solve the Example using Minitab

http://www.minitab.com/en-US/default.aspx

statistical analysis - graphical techniques

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