EEM332 Design of Experiments En. Mohd Nazri Mahmud MPhil (Cambridge, UK) BEng (Essex, UK) nazriee@eng.usm.my Room 2.14 Ext. 6059 EEM332 Lecture Slides

Agenda Statistical design of experiments Basic Statistical Concepts Simple Comparative Experiments Statistical Hypothesis The t-test EEM332 Lecture Slides

Experiment Is it enough to focus on designing experiments only? How about the analysis of the data collected during the experiments? EEM332 Lecture Slides

Statistical design of experiments The process of planning the experiment so that appropriate data that can be analysed by statistical methods will be collected for the purpose of valid and objective conclusions to be made. Three basic principles Randomisation Replication Blocking Randomisation The allocation of the experimental material and the order in which the individual runs or trials of the experiments are to be performed are randomly determined. Assist in “averaging out” the effect of extraneous factors that may be present Use random number generator EEM332 Lecture Slides

Statistical design of experiments Replication An independent repeat of each factor combination. 2 important properties Allows experimenter to obtain an estimate of the experimental error Allows experimenter to obtain a more precise estimate of the true mean response. Blocking A design technique used to reduce or eliminate the variability due to nuisance factors. EEM332 Lecture Slides

Basic Statistical Concepts Run – one observation of an experiment Noise – fluctuations in the observation of the individual runs Sampling – taking a sample from a population for a study to draw conclusions about that population Random samples – any of the member of the population has an equal probability of being chosen as a sample Sample mean – a measure of the central tendency Suppose y1, y2,….,yn represents a sample, the sample mean, EEM332 Lecture Slides

Basic Statistical Concepts Sample variance – a measure of the dispersion of the sample Sample standard deviation – a measure of the dispersion of the sample EEM332 Lecture Slides

Example. From Montgomery p.24- Portland Cement Formulation Refer Table 2-1 p.24 Two treatments or two levels of factor formulations EEM332 Lecture Slides

Example. From Montgomery p.24 Refer Table 2-1 p.24 Work out the samples means and the variances Sample 1 =16.764 Sample 2 =17.042 Sample 1 = 0.104 Sample 2 = 0.061 EEM332 Lecture Slides

Graphical View of the Data Dot Diagram, Fig. 2-1, pp. 24 Enables the experimenter to see quickly the general location or central tendency of the observations and their spread. In the example, the diagram reveals that the two formulations may differ in mean Strength but that both formulations produce about the same variation in strength. EEM332 Lecture Slides

Box Plots, Fig. 2-3, pp. 26 Displays the minimum, maximum, the lower and upper quartiles and the median. In the example, it shows some difference in mean strength and similar spread. EEM332 Lecture Slides

Simple Comparative Experiments Experiments to compare 2 conditions ( or sometimes called treatments) Example. P. 23 – an experiment to determine whether two different formulations of a product give equivalent results. Section 2.4-p. 34 discuss how the data from the simple comparative experiment can be analysed using Hypothesis testing procedures for comparing two treatment means. EEM332 Lecture Slides

The Hypothesis Testing Framework Statistical hypothesis testing is a useful framework for many experimental situations Origins of the methodology date from the early 1900s We will use a procedure known as the two-sample t-test EEM332 Lecture Slides

The Hypothesis Testing Framework Sampling from a normal distribution Statistical hypotheses: EEM332 Lecture Slides

Statistical Hypothesis A statement either about the parameters of a probability distribution or the parameters of a model that reflects some conjecture about the problem situation Hypothesis Testing (or Significance Testing) A technique of statistical inference to assist experimenter in comparative experiments and allows comparison to be made on objective terms, and with knowledge of the risks associated with reaching the wrong conclusion. (Refer example. p. 24)- it is not obvious that the two samples really are different. One may be faced with the problem of making a definite decision with respect to an uncertain hypothesis which is known only through its observable consequences. EEM332 Lecture Slides

Estimation of Parameters EEM332 Lecture Slides

The t-Test A t test is any statistical hypothesis test for two groups in which the test statistic has a t distribution if the null hypothesis is true. A test of the null hypothesis that the means of two normally distributed populations are equal. Used for investigating the statistical significance of the difference between two sample means, and for confidence intervals for the difference between two population means. Given two data sets, each characterized by its mean, standard deviation and number of data points, we can use some kind of t test to determine whether the means are distinct, provided that the underlying distributions can be assumed to be normal. The t-distribution is a probability distribution that arises in the problem of estimating the mean of a normally distributed population when the sample size is small. (Refer to Montgomery p. 606) EEM332 Lecture Slides

The two sample t-Test Testing for a significant difference between two means The t statistics compares two means to see if they are significantly different from each other (For the table of t-statistics see Table 2-3 page 47 and Table 2-4 page 48) Used for investigating the statistical significance of the difference between two sample means, and for confidence intervals for the difference between two population means. Example of two sample t-test p. 36-7 EEM332 Lecture Slides

Degrees of freedom (df) is a measure of the number of independent pieces of information on which the precision of a parameter estimate is based. The degrees of freedom for an estimate equals the number of observations (values) minus the number of additional parameters estimated for that calculation. Example. P. 36 As we have to estimate more parameters, the degrees of freedom available decreases. It can also be thought of as the number of observations (values) which are freely available to vary given the additional parameters estimated. It can be thought of two ways: in terms of sample size and in terms of dimensions and parameters. For a single sample t-test, degrees of freedom are one less than the number of Observation (n-1). For a two sample t-test, degrees of freedom are one less than the number of observation in the first sample plus one less than the number of observation in the second sample ( n1-1 + n2-1) or (n1 + n2 -2). EEM332 Lecture Slides

Steps for the two sample t-test State the null hypothesis and the alternative hypothesis 2. Set the alpha level. Example= .05, we have 5 chances in 100 of making a type I error. 3. Select an appropriate test statistics from Table 2-3p 47 or 2-4 p.48 4. Calculate the value of the appropriate statistic. Also indicate the degrees of freedom for the statistical test. (n1+ n2 -2) 5. Write the decision rule for rejecting the null hypothesis. Refer Table 2-3 / 2-4 Example : Reject H0 if t0 is >= talpha or if t0 <= taplha 6. Write a summary statement based on the decision. Example: Reject H0 7. Write a statement of results in standard English. Example: There is a significant difference in the sample mean between the two groups EEM332 Lecture Slides

Exercise 1 A design engineer would like to compare the mean burning times of chemical flare of two different formulations. The burning times (in minutes) are tabulated below Formulation A Formulation B 65 82 64 56 81 67 71 69 57 59 83 74 66 75 70 79 State the hypothesis to be tested and using alpha = 0.05, test the hypotheses? EEM332 Lecture Slides

Answer Test statistic to be used ; Degrees of freedom = n1 + n2 -2 To find Sp use formula for the estimate of the common variance ( ie Formula 2-25.p36) y1 mean = 70.4 Y2 mean = 70.2 Sp = 9 t0=0.050 So, do not reject H0 T0.025,18=2.101 EEM332 Lecture Slides

Exercise 2 A new filtering device is designed for installation in a chemical system. Before its installation, a random sample yielded the following information about the percentage of impurity: After installation, a random sample yielded Has the filtering device changed the percentage of impurity significantly? Use alpha= 0.05. EEM332 Lecture Slides

Answer Test statistic to be used ; Degrees of freedom = n1 + n2 -2 To find Sp use formula for the estimate of the common variance ( ie Formula 2-25.p36) Sp = 9.89 t0=0.479 Do not reject H0 T0.025,15=2.131 So, There is no evidence to indicate that the new filtering device has affected the mean EEM332 Lecture Slides

The case of different variances If we are testing and cannot reasonably assume that the variances of the two samples are equal, We must use the test statistic given by equation 2-31 p 45 and degrees of freedom given by equation 2-32 p 45. The case when the variances for both populations are known If we are testing and the variances for both populations are known, use the test statistics given by equation 2-33 p. 45. EEM332 Lecture Slides

Comparing a single mean to a specified value Some experiments involve comparing only one population mean to a specified value The hypotheses are Use the test statistics given by equation 2-35 p. 46. Example: Ex 2-1 p46 If the variance of the population is unknown use the test statistic given by Equation 2-37 p.47. EEM332 Lecture Slides

Exercise: The case when the variances for both populations are known Two devices are used for filling containers with a net volume of 16.0 litre. The filling processes can be assumed to be normal, with standard deviation of = 0.015 and = 0.018. The quality engineering department suspects that both devices fill to the same net volume, whether or not this volume is 16.0 litre. An experiment is performed by taking a random sample from the output of each device. Device 1 Device 2 16.03 16.01 16.02 16.04 15.96 15.97 16.05 15.98 15.99 16.00 State the hypotheses that should be tested in this experiment and test these hypotheses using =0.05. What are your conclusions? EEM332 Lecture Slides

Answer Test statistic to be used = equation 2-33 z0=1.35 z0.025 =1.96 since z0=1.35 is not greater than Therefore, do not reject H0 EEM332 Lecture Slides

Confidence Interval An interval within which the value of the parameter or parameters in question would be expected to lie Why ? Hypothesis testing is useful but sometimes does not tell the entire story It is often preferable to provide an interval within which the value of the parameter or parameters in question would be expected to lie. In many engineering and industrial experiments, the experimenter already knows That the means differ, so they are more interested in a confidence interval on the difference in means. EEM332 Lecture Slides

Confidence Interval: Example. p.44 In the portland cement problem, the estimate for the difference in mean Tension bond strength for the formulations can be expressed in the form of its 95 percent confidence interval. General form of a confidence interval The 100(1- α)% confidence interval on the difference in two means: EEM332 Lecture Slides

Exercise: Confidence Interval Two devices are used for filling containers with a net volume of 16.0 litre. The filling processes can be assumed to be normal, with standard deviation of = 0.015 and = 0.018. The quality engineering department suspects that both devices fill to the same net volume, whether or not this volume is 16.0 litre. An experiment is performed by taking a random sample from the output of each device. Device 1 Device 2 16.03 16.01 16.02 16.04 15.96 15.97 16.05 15.98 15.99 16.00 Find a 95 percent confidence interval on the difference in the mean fill volume for the two machines. EEM332 Lecture Slides

Answer Test statistic to be used = equation 2-33 z0=1.35 z0.025 =1.96 since z0=1.35 is not greater than Therefore, do not reject H0 For 95% confidence interval use equation 2-34 EEM332 Lecture Slides