M350 Wk3 Stats 2 Student t

8/12/2019 M350 Wk3 Stats 2 Student t

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MECH 350 Instrumentation and Design ofExperiments

Assessing and presenting experimental dataMain topics

3 s principle and its applicationThe bound of uncertainty with confidence

level Assessing Small Sample Sizes

Department of Mechanical Engineering andTechnology

Wentworth Institute of Technology


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Gaussian Distributions3 sigma ( s) principle

If the random variable complies with thenormal distribution, the probability of the


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Physical meaning of 3 principle inmeasurement

If the measurement can be described by normaldistribution:

(1) Any measured value in the range ( m-3 s , m+3 s)has the probability of 99.74%. In other word,any real measured value should or must be inthis range because the probability is 99.74%.

%74.999974.0

21

exp2

1)33(

3

3-

s

m

s s m s m

s m

s m

x x P


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(2) The probability of a measured value which islarger than ( m+3 s) or less than ( m-3s) is 0.0036(0.36%). That is, the chance of a realmeasured value out of range ( m-3s , m+3 s) isonly 0.36%.

(3) If a real measured value out of range ( m-3 s ,m+3 s) does happen, this measured value willbe rejected because of:

a) Something is wrong with the instrument, ormeasurement system, or measurement setting, orthere was some mistake.

b) The measured parameter had significant changedue to machine setting.In one word, There was some mistake duringmeasurement or the measured parameter wasnot the same.


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Application of the 3 principle inmeasurement

The 3 s principle is a tool to pre-analyze yourmeasured data to check whether themeasurement system and the experimentalsetting is in normal status. If some data isout the value range defined by 3 s principle,you should stop experiment andmeasurement and find the possible causes

before you resume your experiment andmeasurement. Otherwise, you will wasteyour time because the measured data mightnot be the data you intend to obtain.


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Example: Followings were 20 measured sound levels (dBA) ofa type of machine:101.03; 100.98; 101.30; 100.98; 101.10;101.25; 100.68; 100.75; 101.36; 101.67; 100.56; 100.85;100.39; 101.25; 101.56; 101.23; 101.40; 100.95; 100.67;

100.56. The mean and standard deviation of these dataare 101.03 dBA and 0.35 dBA. Now, another 3 test data:98.10, 101.45 and 103.89 were obtained. Are these threemeasured data acceptable?

Solution:Based on 6 s principle, the measured data should be in therange (101.03-3X0.35=99.98, 101.03+3X0.35=102.08) withthe probability 99.74%. The test data 98.10 was too low and should be rejected

(the cause for this was to use the wrong type of filter) the test data 103.89 was too high and should berejected (the cause for this was to wrong RPM setting).

The test data 101.45 was normal and acceptable.


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Application of probability theory inmeasurement

Population vs. samplePopulation: The set ofdata (numerical or

otherwise)corresponding to theentire collection of unitsabout which information

is soughtSample: A subset of thepopulation data that areactually collected in the

course of a study.


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In any real-life situation, we always deal withsamples from a population and not the populationitself.

Typically, our objectives for measurement are touse measurement of sample to estimate :a) the mean and standard deviation of the

population.b) the bound of measured parameter with a

confidence level.Since we normally use the average of the

measurement data to represent the measuredparameter, it is good to determine the bound of themean with a confidence level.

c) the bound of the mean with a confidence level.


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The confidence intervals for measurement

If the measured parameter complies with normaldistribution and the sample number is big enough (>30),the actual value of the measured parameter inside therange ( m z a/2 s, m z a/2 s ) will have the confidence level a

as listed in following table. m-mean. s -standard deviationTable 1 The confidence intervals

Confidencelevel

85% 90% 95% 99%

Za/2 1.439 1.645 1.960 2.575

Bounds (m-1.439 s ,m+1.439 s )

(m-1.645 s ,m+1.645 s )

(m-1.960 s ,m+1.960 s )

(m-2.575 s ,m+2.575 s )


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Za/2 can be determined by use of table 3.2 in page 50.For example, probability level is 95%. Half of 0.95 is0.475. From table 3.2, the z value for 0.475 is 1.960.


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Example 1 During a 12-hour test of a steam generator, the inlet

pressure was measured 100 times. The inlet pressure

should be held constant at 4.00 MPA. The results of a 12-hour pressure test is shown in following table. (page 52).The histogram is shown in following.

Pressure, p , in

MPA

Number of results,

m(3.965,3.975)(3.975,3.985)(3.985,3.995)(3.995,4.005)(4.005,4.015)(4.015,4.025)(4.025,4.035)(4.035,4.045)

(4.045,4.055)

13

1225331762

1


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The mean and standard deviation are m=4.008 MPA,s =0.014 MPA. For proper performance, the pressureshould no deviate from setting value 4.00 MPA by morethan about 1%.

(1) Determine the interval containing 95% of pressurereadings.

(2) Based on the test measurement, what is the probabilityof pressure reading whose deviation from setting value

4.00MPA is less than 1%. Solution:(1) The probability level is 99%. From previous table, we

have Z a/2 =2.575. The interval containing 99% of

pressure reading is:(m 2.575 s MPA )=(4.008 2.575X0.014 MPA)=(4.008 0.036 MPA)So, the interval containing 99% of pressure reading is:

(3.972, 4.044) MPA


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The probability level for mean atlarge samples (n>=30)

If the bias errors have been reduced to minimum(proper calibration on the proper instrument withproper measurement procedure), it is reasonable

to assume that the mean of the measurement isvery close to the true mean value in idealcondition.

Now lets assess the uncertainty between themeasured mean and the true mean


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If we do 100 measurement at one time, wecalculate the arithmetic average and will getone value for the mean.

If we do another 100 measurement, we will getanother value of the mean, which will certainlydifferent from previous one.

If we repeat these kind of measurement, we wouldobtain a set of samples for the mean. (this isnot the set of sample of measurement.)

Actually, we dont need to do above in orderto assess the uncertainty between themeasured mean and the true mean. We needuse one of probabilistic theory (Central limittheorem).


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Let us set a new random variable forrepresenting the mean:

x is the arithmetic averagen-is the sample size of the measurements is the standard deviation of the measurement

Here z is the random variable of mean andstill complies with normal distribution.

n

x x z

xs

m

s

m


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Example 2 (cont.)Determine a 99% confidence interval for the mean

pressure calculated in example 1. Solution :For the 100 pressure measurement, the mean and

standard deviation are 4.008 and 0.014 MPA. Son=100, x=4.008 MPA, S x=0.014 MPA

For probability level of 99% from table 1, z c/2 =2.575.

The intervals of mean pressure with probability level99% is:

0036.0008.4

100

014.0575.2008.42/

n

S z x x

c


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Compare and notice the conceptualdifference

(1) The interval containing 99% of pressurereadings is: (expected range of measuredpressures)(m 2.575 s MPA )=(4.008 2.575X0.014 MPA)=(4.008 0.036 MPA)

(2) The interval of mean pressure with probabilitylevel 99% is: (Range of the true mean)

0036.0008.4100

014.0575.2008.42/

n

S z x x

c


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Probability intervals for mean atsmall samples (n


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In figure v is degree of freedom. v=n-1. Theshape of t-distribution is similar to the normaldistribution. When v is big enough, the t-

distribution is the same as normal distribution.


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The two-sided confidence interval for meanx of a small sample (n


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Example 3The 12 sound pressure-level measurement has mean101.35 dBA and standard deviation 0.27 dBA. What is thetwo-sided 95% confidence interval for the true meansound pressure-level?Solution:The required probability level is 95%, so a =1-0.95=0.05.v=n-1=12-1=11. According to table 3-5 in page 58:

t a/2, v =t 0.025, 11 =2.201.Hence the two-sided 95% confidence interval for soundpressure level is:

53.10117.101 11

27.0201.235.101

11

27.0201.235.101

,2/,2/

m

m

m n

S

t xn

S

t x x

vc

x

vc


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