Quantitative Analysis for Emperical Research

Jointly organized by JNARDDC & IAPQR, Kolkata A Condensed Review By Amit Kamble

Bridging Training and Research for Industry and the Wider Community An approach for listening to the data with an open mind, using descriptive and graphical tools . Planning of Experiments

Planning is the first step for human activity – for undertaking any scientific, technological or industrial experiment. We are told: If you fail to plan, you are only planning to fail !

An experiment is a means of getting answer to the question experimenter has in mind. Broadly experiments are of two types: Experiments for determining the properties of defined sets of things, e.g. assess proof stress, thermal conductivity, yield of product etc. Experiments comparative in nature, e.g. to assess effect of different % of an alloying element on tensile strength of aluminum alloy

The experimenter has to have a clear idea about the objective , as many of the facets of planning depend on this. Objective Includes: Response Variable Treatment Experimental Unit Experimental Error

Treatment The different procedures , or objects, or levels of factor under comparison in an experiment are different treatments . e.g. different alloying conditions . Response Variable The variable(s) on which measurements are to be taken for analysis – is the response variable(s) . e.g. effect of chemical composition on quality of alloy, is to be decided whether alloy addition, temp., thermal conductivity, or subset of these would be response variable(s)

Experimental Unit An experimental unit is the material to which the treatment is applied and on which the response variable(s) is (are) measured. e.g. a specimen of an alloy under defined conditions would be the experimental unit, in an experiment to develop a new alloy. Experimental Error The unexplained random part of the variation termed as the experimental error . e.g. variation in the measurement of mechanical properties of alloy under different sets of load, etc.

Experimental Error This is technical term, includes all types of uncontrollable extraneous variations due to (i) inherent variability in experimental units (ii) errors associated with measurements (iii) lack of representativeness of sample to the population under study This part of variation cannot be totally eliminated – we have to live with this !

Basic Principles of experimental design (by R.A. Fisher) Fisher’s Diagram *Replication * Randomization * Local Control (desirable) (vital)

Planning of experiments falls into two almost distinct parts, dealing with the principles that should govern (a) The choice of treatments to be compared i.e. observations to be made of, and experimental units to be used (b) The method of assigning treatments to the experimental units and the decision about how many units to be used Techniques of local control Use of homogenous blocks Use of supporting variables Confounding in factorial experiments

The requirements are: The treatment comparisons should be as far as possible free from systematic error The treatment comparisons should be made sufficiently precisely The conclusion should have a wide range of validity The experimental design should be as simple as possible The uncertainty in the conclusions should be assessable

The precision of an experiment is measured by the reciprocal of the variance of a mean 1/  2 x = n/  2 The standard error of the estimate of the difference between two treatments is inversely proportional to the square root of the no. of units for each treatment. Standard error is: standard deviation x  (2/No. of treatments per unit) 1 1 standard deviation x  No. of No. of units of A units of B

Data Modeling Analysis of Dependence Categorical Data Analysis Testing of Statistical Hypothesis

Describing Variation No two units of product produced by a manufacturing process are identical. e.g. the net content of can of soft drink varies slightly from can to can A solved example Data for Forged piston-ring inside diameter (mm) Sample no. Observations 1 74.030 74.002 74.019 73.992 74.008 2 73.995 73.992 74.001 74.011 74.004 3 74.009 73.994 73.997 73.987 73.993 … … … … … … 25 73.982 73.984 73.995 74.017 74.013

A frequency distribution of piston ring data and a histogram of frequencies vs. the ring diameter is as shown: The histogram presents a visual display of the data in which one may see: 1. Shape 2. Location or central tendency 3. Scatter or spread Ring Diameter Frequency 73.965-73.970 1 73.970-73.975 0 73.975-73.980 0 73.980-73.985 8 73.985-73.990 10 73.990-73.995 19 73.995-74.000 23 74.000-74.005 22 74.005-74.010 22 74.010-74.015 13 74.015-74.020 4 74.020-74.025 2 74.025-74.030 1

Numerical summery of data histogram is helpful o use numerical measure of tendency and scatter suppose that x 1 ,x 2 ,x 3 ,..x n are the observations in sample. The central tendency (average) x = x 1 +x 2 +x 3 +...+x n =  x n n The scatter or spread in sample data is measure by sample variance S 2 =  (x i – x) 2 n-1 The sample average of piston ring data = 9250.125/125 = 74.001 mm For piston ring we find S 2 = 0.000102 mm 2 & S = 0.010 mm (S.D.)

Probability Distributions A probability distribution is a mathematical model that relates the value of the variable with the probability of occurrence of that value in the population. Some discrete distributions The Hypergeometric Distribution suppose a finite population of N items. Say D (D  N) of these items falls into a class of interest. A random sample of n items is selected from the population without replacement, and no. of items in sample that fall in class of interest, say x, is observed. Then x is hypergeometric random variable with probability distribution p(x) = D C x N-D C n-x N C n x=0,1,2,3…,n

The mean and variance of hypergeometric distribution are  = nD  2 = nD (1 – D) (N – n) N N N N – 1 Binomial Distribution Consider a process that consists of a sequence of n independent trials where outcome of each trial is either “success” or “failure” (Bernoulli trials), say p, is constant, then no. of “successes” ‘x’, in Bernoulli has binomial distribution as p(x) = n C x p x (1 – p) n-x x= 0,1,2,…,n The mean and variance of binomial distribution  = np  2 = np (1 – p )

Poisson Distribution Another important discrete distribution is Poisson distribution , p (x) = e –m m x x! Where, m>0 & x = 0,1,… The mean and variance of Poisson Distribution  = m  2 = m Ex: Suppose that the no. of wire bonding defects per unit that occur in a semiconductor device is Poisson distributed with parameter m =4. Then the probability that randomly selected semi conductor device will contain two or fewer wire bounding defects is p(x  2) =  e –4 4 x x = 0,1,2 x! = 0.0183+0.0733+0.1464=0.2380

Correlation and Regression Many situation arise in which we may have to study two variables simultaneously, say x and y. and we may be interested to measure numerically the strength of this association between variables. This is problem of Correlation . secondly if one variable is of interest and other variable is auxiliary, in such case we are interested in using mathematical equation for making estimates regarding principle variable. This is known as Regression .

Scatter diagram showing different types of degree of Correlation Correlation Coefficient (r):= Covariance (x,y)  var(x)  var(y) x y x y r= +1 x y r= -1 x y x y r = 0

A categorical variable or attribute is one for which the measurement scale consists of set of categories. variables that do not have natural ordering is ‘nominal’ Categorical variables having ordered status is called ‘ordinal’ Nominal : public school, private school Ordinal : primary school, secondary school, college, university Interval : years of schooling

Types of data Distribution by single attribute Distribution by several populations by single attribute Distribution by two attributes Distribution by more than two attributes Purpose of study Estimation of incidence of levels Measurement of association between attributes Testing homogeneity of several populations in respect of single attribute Testing significance of association between two or more attributes Testing goodness of fit

2 x 2 Contingency Table A has two forms: A (presence or higher level) &  (absence or lower level) B has two forms: B (presence or higher level) &  (absence or lower level) Example: Smoking & Lung Cancer Smoker Lung Cancer Patent Total Yes (A) No (  ) Yes (B) 183 (AB) 645 (  B) 828 (B) No (  ) 59 (A  ) 2113 (  ) 2172 (  ) Total 242 (A) 2578 (  ) 3000 (n)

Relative risk = (AB/B)/(A  /  ) = (183/828)/(59/2172) = 8.125 Odds B = (AB/B)/(1 – (AB/B)) = 0.2210/0.7790 = 0.2837 Odds  = (A  /  )/(1 – (A  /  )) = 0.272/0.9728 = 0.0280 Odds Ratio = Odds B /Odds  = (AB *  )/(A  *  B) = 10.1611 Independence Implies AB = (A*B) / n Positive Association implies AB > (A*B) / n Negative Association implies AB < (A*B) / n Here (A*B) / n = (242*828)/3000 = 66.792 Which implies positive association

An introduction through examples (Single mean) Ex. 40 samples of an specimen of an aluminum alloy (Sn 6.1%,Cu 1.2%,Ni0.9% rest Al) were tested for density (g/cc). The result obtained were mean x = 2.61and  variance = std. deviation = 0.605. Do the data support the conjecture that the mean density of alloy is less than 2.84? Here: H 0 :  =2.84 against H 1:  <2.84 The test statistics is T = (x -  0 )  n = (2.61-2.84)  40 S 0.605 = - 2.404 since -  .05 = -1.645 and -  . 01 = -2.326 the observed value of T is less than both these values, we conclude that the mean density of the alloy under reference is significantly lower than 2.84

Example of mean from two samples Ex. 32 samples of an specimen of an aluminum alloy (Sn 20.3%,Cu 1.1%,rest Al) were tested for 0.2% compression strength (MN/m 2 ). The result obtained were mean x 1 = 102.8 and S 1 = 7.9. A set of 35 samples of an specimen of an aluminum alloy (Pb 20.6%,Cu 1.1%,rest Al) were tested for same property, result obtained were mean x 2 = 102.8 and S 2 = 8.4 do the data support the conjecture that two alloys have identical status in respect of property? Here: H0:  1=  2 against H1:  1   2 The test statistics is T = x 1 – x 2 S 2 S 1 n1 n2 2 2 +

Thus T = 102.8 – 106.5 = - 3.7/  (3.9693) = - 1.8578  . 025 = 1.960 and  . 005 = 2.576 Since |T| is less than both these values, we may conclude that in light of given sample, the alloys may be taken to have identical mean 0.2% compression strength 32 8.4 7.9 2 35 2 +

Ex: 40 sample data (double mould) is taken for the study of variation of Mg% in FeSiMg alloy the observed results were  1 =7.494 and S 1 =0.18004. A second set of 45 sample data (single mould) gave values  2 =7.5949 and S 2 =0.19082. Do the data support the hypothesis that the two alloys have identical mean value of Mg% in the population? Here H 0 :  1 =  2 H 1 :  1  2 By following the test statistics from the previous problem; we have, T = 7.494 – 7.5979/  ((0.18004) 2 /45+(0.19082) 2 /40) T = – 1.82245  . 025 = 2.014 and  . 005 = 2.968 now, |T| = 1.82245 is less than both these values, we may conclude that in the light of given data the two alloys may be taken to have identical mean of Mg%

The Greatest value of a picture is when it forces us to notice what we never expected to see. – John W. Tukey

Temperature Variation Vs. Slag

Plot of comparison of outer temperature of ladle, ladle lining namely silica & High Al 2 O 3

Lastly Some Facts The no. of human beings killed by an Hippopotamus annually is more than a yearly plane crash. No paper of any size can be folded in half for more than 8 times. Approximately a human being spend nearly 2 weeks of his life waiting at Red Traffic Signal.

Quantitative Analysis for Emperical Research

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Quantitative Analysis for Emperical Research