1 Chapter 2: Simple Comparative Experiments (SCE) Simple comparative experiments: experiments that compare two conditions (treatments) –The hypothesis testing framework –The two-sample t-test –Checking assumptions, validity

2 Portland Cement Formulation (page 23) Average tension bond sterngths (ABS) differ by what seems nontrivial amount. Not obvois that this difference is large enough imply that the two formulations really are diff. Diff may be due to sampling fluctuation and the two formulations are really identical. Possibly another two samples would give opposite results with strength of MM exceeding that of UM. Hypothesis testing can be used to assist in comparing these formulations. Hypothesis testing allows the comparison to be made on objective terms, with knowledge of risks associated with searching the wrong conclusion

3 Graphical View of the Data Dot Diagram, Fig. 2-1, pp. 24 Response variable is a random variable Random variable: 1.Discrete 2.continuous

4 Box Plots, Fig. 2-3, pp. 26 Displays min, max, lower and upper quartile, and the median Histogram

Probability Distributions 5 Probability structure of a Random variable, y, is described by its probability distribution. y is discrete: p(y) is the probability function of y (F2- 4a) y is continuous: f(y) is the probability density function of y (F2-4b)

Probability Distributions Properties of probability distributions 6 y-discrete: y-continuous:

Probability Distributions mean, variance, and expected values 7

Probability Distributions Basic Properties 1.E(c) = c 2.E(y) =  3.E(cy) = c E(y) = c  4.V(c) = 0 5.V(y) =  2 6.V(cy) = c 2 V(y) = c 2  2 8

Probability Distributions Basic Properties E(y 1 +y 2 ) = E(y 1 )+E(y 2 ) =  1 +  2 Cov(y 1,y 2 ) = E[(y 1 -  1 )(y 2 -  2 )] Covariance: measure of the linear association between y 1 and y 2. E(y 1.y 2 ) = E(y 1 ).E(y 2 ) =  1.  2 (y 1 and y 2 are indep) 9

Sampling and Sampling Distributions The objective of statistical inference is to draw conclusions about a population using a sample from that population. Random Sampling: each of N!/(N-n)!n! samples has an equal probability of being chosen. Statistic: any function of observations in a sample that does not contain unknown parameters. Sample mean and sample variance are both statistics. 10

Properties of sample mean and variance Sample mean is a point estimator of population mean  Sample variance is a point estimator of population variance   The point estimator should be unbiased. Long run average should be the parameter that is being estimated. An unbiased estimator should have min variance. Min variance point estimator has a variance that is smaller than the variance of any other estimator of that parameter. 11

Degrees of freedom (n-1) in the previous eq is called the NDOF of the sum of squares. NDOF of a sum of squares is equal to the no. of indep elements in the sum of squares Because, only (n-1) of the n elements are indep, implying that SS has (n- 1) DOF 12

The normal and other sampling distributions Normal Distribution y is distributed normally with mean  and variance  2 Standard normal distribution:  =0 and  2 =1 13

Central Limit Theorem If y 1, y 2, …, y n is a sequence of n independent and identically distributed random variables with E(y i ) =  and V(y i ) =  2 (both finite) and x = y 1 +y 2 +…+y n, then has an approximate N(0,1) distribution. This implies that the distribution of the sample averages follows a normal distribution with This approximation require a relatively large sample size (n≥30) 14

Chi-Square or  2 distribution If z 1, z 2, …, z n are normally and independently distributed random variables with mean 0 and variance 1 NID(0,1), then the random variable follows the chi-square distribution with k DOF. 15

Chi-Square or  2 distribution The distribution is asymmetric (skewed)  = k  2 = 2k Appendix III 16

Chi-Square or  2 distribution y 1, y 2, …, y n is a random sample from N( ,  ), then SS/  2 is distributed as chi-square with n-1 DOF 17

Chi-Square or  2 distribution If the observations in the sample are NID(  ), then the distribution of S 2 is and,,,, Thus, the sampling distribution of the sample variance is a constant times the chi- square distribution if the population is normally distributed 18

Chi-Square or  2 distribution Example: The Acme Battery Company has developed a new cell phone battery. On average, the battery lasts 60 minutes on a single charge. The standard deviation is 4.14 minutes. a)Suppose the manufacturing department runs a quality control test. They randomly select 7 batteries. The standard deviation of the selected batteries is 6 minutes. What would be the chi-square statistic represented by this test? b)If another sample of 7 battery was selected, what is the probability that the sample standard deviation is greater than 6? DOX 6E Montgomery19

Chi-Square or  2 distribution Solution a) We know the following: –The standard deviation of the population is 4.14 minutes. –The standard deviation of the sample is 6 minutes. –The number of sample observations is 7. To compute the chi-square statistic, we plug these data in the chi-square equation, as shown below.  2 = [ ( n - 1 ) * s 2 ] / σ 2  2 = [ ( ) * 6 2 ] / = 12.6 DOX 6E Montgomery20

b) To find the probability of having a sample standard deviation S > 6, we refer to the Chi- square distribution tables, we find the value of  corresponding to chi-square = 12.6 and 6 degrees of freedom. This will give  0.05 which is the probability of having S > 6 DOX 6E Montgomery21 Chi-Square or  2 distribution

DOX 6E Montgomery22

t distribution with k DOF If z and are indpendent normal and chi- square random variables, the random variable Follow t distribution with k DOF as follows: 23

t distribution with k DOF  = 0 and  2 = k/(k-2) for k>2 If k=infinity, t becomes standard normal If y 1, y 2, …, y n is a random sample from N( ,  ), then is distributed as t with n-1 DOF 24

t distribution - example Example: Acme Corporation manufactures light bulbs. The CEO claims that an average Acme light bulb lasts 300 days. A researcher randomly selects 15 bulbs for testing. The sampled bulbs last an average of 290 days, with a standard deviation of 56 days. If the CEO's claim were true, what is the probability that 15 randomly selected bulbs would have an average life of no more than 290 days? DOX 6E Montgomery25

t distribution – Example Solution To find P(x-bar<290), The first thing we need to do is compute the t score, based on the following equation: –t = [ x - μ ] / [ s / sqrt( n ) ] t = ( ) / [ 56 / sqrt( 15) ] = P(t 0.692) DOX 6E Montgomery26

t distribution – Example From the t-distribution tables, we have  = 25% which correspond to the probability of having the sample average less than 290 DOX 6E Montgomery27

DOX 6E Montgomery28

F distribution If and are two independent chi- square random variables with u and v DOF, then the ratio Follows the F dist with u numerator DOF and v denominator DOF 29

F distribution Two independent normal populations with common variance  2. If y 11, y 12, …, y 1n1 is a random sample of n 1 observations from 1 st population and y 21, y 22, …, y 2n2 is a random sample of n 2 observations from 2 nd population, then 30

31 The Hypothesis Testing Framework Statistical hypothesis testing is a useful framework for many experimental situations We will use a procedure known as the two- sample t-test

32 The Hypothesis Testing Framework Sampling from a normal distribution Statistical hypotheses:

33 Estimation of Parameters

34 Summary Statistics (pg. 36) Formulation 1 “New recipe” Formulation 2 “Original recipe”

35 How the Two-Sample t-Test Works:

36 How the Two-Sample t-Test Works:

37 How the Two-Sample t-Test Works: Values of t 0 that are near zero are consistent with the null hypothesis Values of t 0 that are very different from zero are consistent with the alternative hypothesis t 0 is a “distance” measure-how far apart the averages are expressed in standard deviation units Notice the interpretation of t 0 as a signal-to-noise ratio

38 The Two-Sample (Pooled) t-Test

39 The Two-Sample (Pooled) t-Test So far, we haven’t really done any “statistics” We need an objective basis for deciding how large the test statistic t 0 really is. t 0 = -2.20

40 The Two-Sample (Pooled) t-Test A value of t 0 between –2.101 and is consistent with equality of means It is possible for the means to be equal and t 0 to exceed either or –2.101, but it would be a “rare event” … leads to the conclusion that the means are different Could also use the P-value approach t 0 = -2.20

Use of P-value in Hypothesis testing P-value: smallest level of significance that would lead to rejection of the null hypothesis H o It is customary to call the test statistic significant when H o is rejected. Therefore, the P-value is the smallest level  at which the data are significant. 41

42 Minitab Two-Sample t-Test Results

Checking Assumptions – The Normal Probability Plot Assumptions 1.Equal variance 2.Normality Procedure : 1.Rank the observations in the sample in an ascending order. 2.Plot ordered observations vs. observed cumulative frequency (j- 0.5)/n 3.If the plotted points deviate significantly from straight line, the hypothesized model in not appropriate. 43

44 Checking Assumptions – The Normal Probability Plot

The mean is estimated as the 50 th percentile on the probability plot. The standard deviation is estimated as the differnce between the 84 th and 50 th percentiles. The assumption of equal population variances is simply verified by comparing the slopes of the two straight lines in F

46 Importance of the t-Test Provides an objective framework for simple comparative experiments Could be used to test all relevant hypotheses in a two-level factorial design.

47 Confidence Intervals (See pg. 43) Hypothesis testing gives an objective statement concerning the difference in means, but it doesn’t specify “how different” they are General form of a confidence interval The 100(1- α)% confidence interval on the difference in two means:

Hypothesis testing The test statitic becomes This statistic is not distributed exactly as t. The distribution of t o is well approximated by t if we use as the DOF 48

Hypothesis testing The test statitic becomes If both populations are normal, or if the sample sizes are large enough, the distribution of z o is N(0,1) if the null hypothesis is true. Thus, the critical region would be found using the normal distribution rather than the t. We would reject H o, if where z  is the upper  2 percentage point of the standard normal distribution 49

Hypothesis testing The 100(1-  ) percent confidence interval: 50

Hypothesis testing Comparing a single mean to a specified value The hypothesises are: H o :  =  o and H 1 :  ≠  o If the population is normal with known variance, or if the population is non-normal but the sample size is large enough, then the hypothesis may be tested by direct application of the normal distribution. Test statistic If Ho is true, then the distribution of zo is N(0,1). Therefore, H o is rejected if 51

Hypothesis testing Comparing a single mean to a specified value The value of  o is usually determined in one of three ways: 1.From past evidence, knowledge, or experimentation 2.The result of some theory or model describing the situation under study 3.The result of contractual specifications 52

Hypothesis testing Comparing a single mean to a specified value If the variance of the population is known, we assume that the population is normally distributed. Test statistic H o is rejected if The 100(1-  ) percent confidence interval 53

The paired comparison problem The two tip hardness experiment Statistical model j th paired difference Expected value of paired difference Testing hypothesis: H o :  d =0 and H 1 :  d ≠0 Test statistic: 54

The paired comparison problem The two tip hardness experiment Randomized block design Block: homogenous experimental unit The block represents a restriction on complete randomization because the treatment combinations are only randomized within the block 55 Randomized BlockComplete Randomization DOFn-1 = 92n-2 = 18 Standard deviationS d = 1.2S p = 2.32 Confidence interval on  1 -  ± ±2.18

Inferences about the variablity of normal distributions H o :  2 =  o and H 1 :  2 ≠  o Test statistic Ho is rejected if or The 100(1-a) percent confidence interval 56

Inferences about the variablity of normal distributions Test statistic H o is rejected if or 57