The One-Factor Model

1. The One-Factor Model • Statistical model is used to describe data. It is an equation that shows the dependence of the response variable upon the levels of the treatment factors. • Let Yij be a random variable that represents the response obtained on the j-th observation of the i-th treatment. • Let μ denote the overall expected response. • The expected response for an experimental unit in the i-th treatment group is μi = μ + τi • τi is deviation of i-th mean from overall mean; it is referred to as the effect of treatment i. STA305 week2

2. The model is where is the deviation of the individual’s response from the treatment group mean. • is known as the random or experimental error. STA305 week2

3. Fixed Effects versus Random Effects • In some cases the treatments are specifically chosen by the experimenter from all possible treatments. • The conclusions drawn from such an experiment apply only to these treatments and cannot be generalized to other treatments not included in experiment. • This is called a fixed effects model • In other cases, the treatments included in the experiment can be regarded as a random selection from the set of all possible treatments. • In this situation, conclusions based on the experiment can be generalized to other treatments. • When the treatments are random sample, treatment effects, τi are random variables. • This model is called a random effects model or a components of variance model. • The random effects model will be studied after the fixed effects model STA305 week2

4. More about the Fixed Effects Model • As specified in slide (2) the model is Where are i.i.d. with distribution N(0, σ2) • It follows that response of experimental unit j in treatment group i, Yij, is normally distributed with • In other words STA305 week2

5. Treatment Effects • Recall that treatment effects have been defined as deviations from overall mean, and so the model can be parameterized so that: • In the special case where r1 = r2 = · · · = ra = r this condition reduces to • The hypothesis that there is no treatment effect can be expressed mathematically as: H0 : μ1 = μ2 = · · · = μa Ha : not all μi are equal • This can be expressed equivalently in terms of the τi: H0 : τ1 = τ2 = · · · = τa = 0 Ha : not all τi are equal to 0 STA305 week2

6. ’Dot’ Notation • “Dot” notation will be used to denote treatment and overall totals, as well as treatment and overall means. • The sum of all observations in the i-th treatment group will be denoted as • Similarly, the sum of all responses in all treatment groups is denoted: • The treatment and overall means are: STA305 week2

7. Rationale for Analysis of Variance • Consider all of the data from the a treatment groups as a whole. • The variability in the data may come from two sources: 1) treatment means differ from overall mean, this is called between group variability. 2) within a given treatment group individual observations differ from group mean, this is called within group variability. STA305 week2

8. Total Sum of Squares • Total variation in data set as a whole is measured by the total sum of squares. It is given by • Each deviation from the overall sample mean can be expressed as the sum of 2 parts: 1) deviation of the observation from the group mean. 2) deviation of the group mean from the overall mean • In other words… • The SST can then be written as… STA305 week2

9. Expected Sums of Squares • Finding the expected value of the sums of squares for error and treatment will lead us to a test of the hypothesis of no treatment effect, i.e., H0 : τ1 = τ2 = · · · = τa = 0 • We start by finding the expected value of SSE…. • We continue with the expected value of SSTreat STA305 week2

10. Mean Squares • As we have seen in the calculation above, the MSE = SSE/(n − a) is an unbiased estimator of σ2. • The MSE is called the mean square for error. • The degrees of freedom associated with SSE are n − a and it follows that E(MSE) = σ2. • The mean square for treatment is defined to be: MSTreat = SSTreat / (a-1). • The expected value of MSTreat is STA305 week2

11. Hypothesis Testing • Recall that our goal is to test whether there is a treatment effect. • The hypothesis of interest is H0 : τ1 = τ2 = · · · = τa = 0 Ha : not all τi are equal to 0 • Notice that if H0 is true, then • On the other hand, if H0 is false, then at least one τa ≠ 0, in which case and so E (MSTreat) > E (MSE) • On average, then, the ratio MSTreat/MSE should be small if H0 is true, and large otherwise. • We use this to develop formal test. STA305 week2

12. Cochran’s Theorem • Let Z1,Z2, . . . ,Zn be i.i.d. N(μ, 1). • Suppose that where Qj has d.f vj. • A necessary and sufficient condition for the Qj to be independent of one another, and for Qj ~ χ2(vj) is that . • Cochran’s theorem implies that SSE/σ2 and SSTreat/ σ2 have independent χ2 distributions with n – a and a − 1 d.f., respectively. • Recall: If X1 and X2 are two independent random variables, each with a χ2 distribution, then STA305 week2

13. Hypothesis Test for Treatment Effects • Cochran’s theorem and the result just stated provide the tools to construct a formal hypothesis test of no treatment effects. • The Hypothesis again are: H0 : τ1 = τ2 = · · · = τa = 0 Ha : not all τi are equal to 0 • The Test Statistic is: Fobs = MSTreat/MSE • Note that if H0 is true, then Fobs ~ F(a − 1, n − a). • So the P-value = P(F(a − 1, n − a) > Fobs). • We reject H0 in favor of Ha if P−value < α. • Alternatively, reject H0 in favor of Ha if Fobs > Fα(a − 1, n − a), where Fα(a − 1, n − a) is the upper 100 × α%-ile point of the F(a − 1, n − a) distribution. STA305 week2

14. Analysis of Variance Table • The results of the calculations and the hypothesis testing are best summarized in an analysis of variance table • The ANOVA Table is given below STA305 week2

15. Estimable Functions of Parameters •  A function of the model parameters is estimable if and only if it can be written as the expected value of a linear combination of the response variables. • In other words, every estimable function is of the form where the cij are constants • It can be shown that from previous sections, μ, μi, and σ2 are estimable. STA305 week2

16. Example - Effectiveness of Three Methods for Teaching a Programming Language • A study was conducted to determine whether there is any difference in the effectiveness of 3 methods of teaching a particular programming language. • The factor levels (treatments) are the three teaching methods: 1) on-line tutorial 2) personal attention of instructor plus hands-on experience 3) personal attention of instructor, but no hands-on experience • Replication and Randomization: 5 volunteers were randomly allocated to each of the 3 teaching methods, for a total of 15 study participants. • Response Variable: After the programming instruction, a test was administered to determine how well the students had learned the programming language. • Research Question: Do the data provide any evidence that the instruction methods differ with respect to test score. • The data and the solutions are…. STA305 week2

17. Conducting an ANOVA in SAS • There are several procedures in SAS that can be used to do an analysis of variance. • PROC GLM (for generalized linear model) will be used in this course • To do the analysis for the Example on slide 16, start by creating a SAS dataset: data teach ; input method score ; cards ; 1 73 1 77 ..... 3 71 ; run ; STA305 week2

18. Use this dataset to conduct an ANOVA using the following SAS code: proc glm data = teach ; class method ; model score = method / ss3 ; run ; quit ; • The output produced by this procedure is given in the next slide. STA305 week2

19. STA305 week2

20. Estimating Model Parameters • The ANOVA indicates whether there is a treatment effect, however, it doesn’t provide any information about individual treatments or how treatments compare with each other. • To better understand outcome of experiment, estimating mean response for each treatment group is useful. • Also, it is useful to obtain an estimate of how much variability there is within each treatment group. • This involves estimating model parameters. STA305 week2

21. Variability • Recall, on slides (9 and 10) we have showed that the MSE is unbiased estimator of σ2. • Further, Cochran’s Theorem was used to show that SSE/ σ2 ~ χ2(n − a). • We can use this result to calculate a 100 × (1 − α)% confidence interval for σ2. • The CI is give by where and are the upper and lower percentage points of the χ2 distribution with n − a d.f., respectively. STA305 week2

22. Overall Mean • As discussed in the beginning, the overall expected value is μ. • Show that is unbiased estimator of μ… • The variance of is σ2/n. • So the 100 × (1 −α)% confidence interval for μ is: • Further, a 100 × (1 −α)% confidence interval for μi is: • It follows that is an unbiased estimator of the effect of treatment i, τi. STA305 week2

23. Differences between Treatment Groups • Differences between specific treatment groups will be important from researcher’s point of view. • The expected difference in response between treatment groups i and j is: μi − μj = τi – τj. • Since treatment groups are independent of each other, it follows that • Therefore, a 100 × (1 −α)% confidence interval for τi – τj is: STA305 week2

24. Example - Methods for Teaching Programming Language Cont’d • Back to the example of three teaching methods and their effect on programming test score. • Based on the ANOVA developed earlier, we found significant difference between the three methods. • Which method had the highest average? • What is a 95% CI for mean difference in test scores for the 2 instructor-based methods? STA305 week2

25. Comparisons Among Treatment Means • As mentioned above, ANOVA will indicate whether there is significant effect of treatments overall it doesn’t indicate which treatments are significantly different from each other. • There are a number of methods available for making pairwise comparisons of treatment means. STA305 week2

26. Least Significant Difference (LSD) • This method tests the hypothesis that all treatment pairs have the same mean against the alternative that at least one pair differs, that is the hypothesis are: H0 : μi − μj = 0 for all i, j Ha : μi − μj ≠ 0 for at least one pair i, j • In testing difference between any two specific means, reject the null hypothesis if: • In the case where the design is balanced and ri = r for all i, the condition above becomes: STA305 week2

27. In other words, the smallest difference between the means that would be considered statistically significant is: • This quantity, LSD, is called the least significant difference. • LSD method requires that the difference between each pair of means be compared to the LSD. • In cases where difference is greater than LSD, we conclude that treatment means differ. STA305 week2

28. Important Notes • As in any situation where large number of significance tests conducted, the possibility of finding large difference due to chance alone increases. • Therefore, in case where the number of treatment groups is large, the probability of making this type of error is relatively large. • In other words, probability of committing a Type I error will be increased above α. • Further, although the ANOVA F-test might find a significant treatment effect, LSD method might conclude that there are no 2 treatment means that are significantly different from each other. • This is because ANOVA F-test considers overall trend of effect of treatment on outcome, and is not restricted to pairwise comparisons. STA305 week2

29. Other Methods for Pairwise Comparisons • Other methods for conducting pairwise comparisons are available. • The methods that are implemented in PROC GLM in SAS include: – Bonferonni – Duncan’s Multiple Range Test – Dunnett’s procedure – Scheffe’s method – Tukey’s test – several otheres • Chapter 4 of Dean & Voss discusses some of these methods. STA305 week2

30. Pairwise Comparisons in SAS • Pairwise comparisons can be requested by including a means statement. • The code below requests means with LSD comparison: proc glm data = teach ; class method ; model score = method / ss3 ; means method / lsd cldiff ; run ; • The part of the output containing the pairwise comparisons is shown in the next slide. STA305 week2

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33. Contrasts • ANOVA test indicates only whether there is an overall trend for the treatment means to differ, and does not indicate specifically which treatments are the same, which are different, etc. • In the last few slides looked at pairwise comparisons between treatment means. • However, comparisons that are of interest to researcher may include more then just two group. They can be linear combination of means. STA305 week2

34. Example - Does Food Decrease Effectiveness of Pain Killers? • Researchers at pain clinic want to know whether effectiveness of two leading pain killers is same when taken on empty stomach as when taken with food. • A study with four treatment groups was designed: 1. aspirin with no food 2. aspirin with food 3. tylenol with no food 4. tylenol with food • In addition to determining whether there is a difference between the four treatment groups, researchers want to determine whether there is a difference between taking medication with food and taking it without. • This second hypothesis can be expressed statistically as: H0 : μ1 + μ3 = μ2 + μ4 Ha : μ1 + μ­ ≠ μ2 + μ4 STA305 week2

35. The point estimate of difference between fed and not fed conditions is based on sample means: STA305 week2

36. Hypothesis Tests Using Contrasts • As in the example on the previous slide, the comparison of treatment means that is of interest might be a linear combination of means. That is, the hypothesis of interest would be of the form H0 : c1μ1 + c2μ2 + · · · + caμa = 0 Ha : c1μ1 + c2μ2 + · · · + caμa ≠ 0 • The ci are constants subject to the constraints: (i) ci > 0 for all i, and (ii) • Test of this hypothesis can be constructed using sample means for each treatment group. • The linear combination c1μ1 + c2μ2 + · · · + caμa is called a contrast. STA305 week2

37. If the assumptions of the model are satisfied, then: • If σ2 was known, a test of H0 could be done using: • Since σ2 is unknown, we use its unbiased estimate, the MSE, and conduct a t-test with n − a d.f.. The test statistics is • Recall, if X is a random variable with t(v) distribution, then X2 has F(1, v) distribution. STA305 week2

38. So an equivalent test statistic is: • At level α , reject H0 in favour of Ha if Fobs > Fα(1, n − a), or equivalently if |tobs| > tα/2 (n − a). • The sum of squares for contrast is: • Each contrast has 1 d.f., so the mean square for contrast is: MScontrast = SScontrast/1 STA305 week2

39. Summary • The hypothesis: H0 : c1μ1 + c2μ2 + · · · + caμa = 0 Ha : c1μ1 + c2μ2 + · · · + caμa≠ 0 • Test Statistic • Decision Rule: reject H0 if Fobs > Fα(1, n − a) STA305 week2

40. Orthogonal Contrasts • Very often more than one contrast will be of interest. Further, it is possible that one research question will require more than one contrast, i.e., H0 : μ1 = μ3 and μ2 = μ4 • Ideally, we want tests about different contrasts to be independent of each other. • Suppose that the two contrasts of interest are: c1μ1 + c2μ2 + · · · + caμa and d1μ1 + d2μ2 + · · · + daμa. • These two contrasts are orthogonal to each other they iff they satisfy: • If there are a treatments then, SSTreat can be decomposed into set of a − 1 orthogonal contrasts, each with 1 d.f. as follows SSTreat = SScontrast1 + SScontrast2 + · · · + SScontrasta−1. • Unless a = 2, there will be more than one set of orthogonal contrasts. STA305 week2

41. Example - Food / Pain Killers Continued • Refer back to the example on slide 31. The study designed with 4 treatment groups. • The treatment sum of squares can be decomposed into 3 orthogonal contrasts. • Since researcher interested in difference between fed & unfed, makes sense to use the following contrasts: STA305 week2

42. Exercise: verify that each is in fact a contrast. • Exercise: verify that contrasts are orthogonal. • Note, there is more than one way to decompose treatment sum of squares into set of orthogonal contrasts. • For example, instead of comparing aspirin and Tylenol, might be interested in comparing food with no food. • In this case, compare (i) aspirin with food and Tylenol with food, (ii) aspirin without food and Tylenol without food, and (iii) the 2 food groups to the 2 no-food groups. STA305 week2

43. ANOVA Table for Orthogonal Contrasts • Contrasts to be used in experiment must be chosen at the beginning of the study. • The hypotheses to be tested should not be selected after viewing the data. • Once the treatment SS has been decomposed using preplanned orthogonal contrasts, the ANOVA table can be expanded to show decomposition as shown in the next slide. STA305 week2

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45. Example - Pressure on a Torsion Spring STA305 week2

46. The figure above shows a diagram of a torsion spring. • Pressure is applied to arms to close the spring. • A study has been designed to examine pressure on torsion spring. • Five different angles between arms of spring will be studied to determined their impact on the pressure: 67º, 71 º, 75 º, 79 º, and 83 º. • Researchers are interested in whether there is an overall difference between different angle settings. • In addition would like to study set of orthogonal contrasts which compares the 2 smallest angles to each other and 2 largest angles to each other. • The data collected are shown in the following slide. STA305 week2

47. Torsion Spring Data STA305 week2

48. Solution STA305 week2

49. Contrasts in SAS • To do the analysis for the last example, start by creating a SAS dataset: data torsion ; input angle pressure; cards ; 67 83 67 85 71 87 71 84 ........... 79 90 83 90 83 92; run ; STA305 week2

50. Here is an additional code that is required to specify the contrasts of interest: proc glm data = torsion ; class angle ; model pressure = angle / ss3 ; contrast ’67-71’ angle 1 -1 0 0 0 ; contrast ’79-83’ angle 0 0 0 1 -1 ; contrast ’sm vs lg’ angle 1 1 0 -1 -1 ; contrast ’mid vs oth’ angle 1 1 -4 1 1 ; run ; quit ; STA305 week2