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Chapter 15 Analysis of Variance The articleto“Could Mean be a In order compare thePlatelet means, Volume the Predictive Marker foruse Acute Myocardial researchers must a procedure called a Infarction?” (Medical Science Monitor, single factor analysis of variance or 2005) described ANOVA.an experiment in which four groups of patients seeking treatment for chest pain were compared with respect of mean platelet volume experiment, the factor is the (MPV). In Thethis purpose of the study was to diagnosis. four groups determineclinical is mean MPV was The different for the were noncardiac pain,attack (2) four groups, in (1) particular for chest the heart angina unstable group. Ifstable so, then MPVpectoris, could be (3) used as an (4) myocardial indicator ofangina heart pectoris, attack risk. infarction (heartor attack). Researchers need to compare the means from When two or more populations treatments fourcompared, treatmentthe groups to determine arethe being characteristic thatif m1 = m2 = m3 = m if at least or onetreatments of the means distinguishes the populations 4 or differ from the rest. from one another is called factor. Whether the null (ofthat equal In hypothesis Group A, notice themeans) three In Group B, notice that the comes three samples The phrase “analysis of variance” from thehave should be rejected depends how substantially samples seem toon have very different the same means as Group A. However, due to idea of analyzing variability in the data to see means and very little variability in each the samples from the different populations or the large amount of variability in each sample how much differ can be attributed to differences in the sample. This one would lead us toConsider doubt treatments from another. andhow the much fact that the samples overlap, it is m’s and isclaim due to variability in the that m = m = m . 1 2 come 3 from the following example. plausible that the samples could individual populations. populations with equal means. Mean of Sample 1 Mean of Sample 2 Mean of Sample 3 Mean of Sample 1 Mean of Sample 2 Mean of Sample 3 Graph A Graph B ANOVA Notation k = the number of populations or treatments being compared N n1 n2 ... nk The total number of observations T n1x1 n2x2 ... nk xk T x N Grand Mean Grand Total ANOVA Notation Continued . . . A measure of differences among the sample means is the treatment sum of squares, denoted by SSTr and given by SSTr n1 x1 x 2 n2 x2 x 2 ... nk xk x 2 The number of error degrees of freedom comes from A measure variation the k samples, calledassociated error sum adding of the numberwithin of degrees of freedom of squares andwith denoted is sample variances: eachSSE, of the (n1 – 1)2+ (n2 – 1) + …2(nk – 1) = N - k 2 SSE n1 1s1 n2 1s2 ... n k 1sk Each sum of squares has an associated df: treatment df = k – 1 error df = N – k A mean square is a sum of squares divided by its df. SSTr MSTr k 1 SSE MSE N k The Single Factor ANOVA F test Null hypothesis: H0: m1 = m2 = … = mk Alternative hypothesis: Ha: at least two m’s are different MSTr Test Statistic: F MSE When H0 is true, with df1 = k – 1 and df2 = N – k mMSTr = mMSE P-value: the area under the appropriate F curve When H0 is false, to the right of the calculated F value mMSTr > mMSE The Single Factor ANOVA F Test Continued . . . Assumptions: 1. Each of the k population or treatment response distributions is normal. 2. The k normal distributions have identical standard deviations. (ssizes = … large, = sk) individual boxplots or If sample 1 = s2 are probability plots for each can be one 3. Thenormal observations in the sample fromsample any particular used to or check for normality. of the k populations treatments are independent of the sample are small, then a combined While theresizes is a formal procedure to check for one If another. normal equal probability plotdeviations, should be used to is check standard itskuse not 4. When comparing population means, the random for normality. Firstdue findtothe recommended its deviations sensitivityfrom to any samples are selected mean independently of each other. the respective in each The sample, Then departure from normality. ANOVA F test When comparing treatment are combine deviations to the normal canthe safely be usedmeans, if create the treatments largest sample assigned at random to subjects or objects. probability plot standard deviation is not more than twice the smallest sample standard deviation. Heart Attack Risk Continued . . . Here are the summary statistics for the four groups: Group Description 1 Noncardiac chest pain 2 Stable angina pectoris 3 Unstable angina pectoris 4 Myocardial Infarction Sample Sample Sample standard sizesubjects mean were deviation The randomly selected from groups0.69 of 35 10.89 individuals who had been 35 11.25 0.74 diagnosed with the four 35 11.37 0.91 conditions. 35 11.75 H0: m1 = m2 = m3 = m4 1.07 State theassumptions. hypotheses. Verify H : at least two m’s are different a To verify the equality of the standard deviations, notice that the largest sample The four boxplots are deviation (group 4) isapproximately less than twice that symmetrical with no of the smallest standard deviation (group 1).outliers, so the assumption of normality is plausible. Heart Attack Risk Continued . . . Here is the summary statistics for the four groups: Sample size Group Description Sample Sample standard mean deviation 1 Noncardiac chest pain 35 10.89 0.69 2 Stable angina pectoris 35 11.25 0.74 3 Unstable angina pectoris 35 11.37 0.91 4 Myocardial Infarction 35 11.75 1.07 x 11.315 Calculate the the sum Foftest squares terms. Calculate statistic. SSTr 35(10.89 11.315)2 35(11.25 11.315)2 35(11.37 11.315)2 35(11.75 11.315)2 13.199 SSE 34(.069)2 34(0.74)2 34(0.91)2 34(1.07)2 101.888 13.199 4.400 3 101.88 MSE 0.749 136 MSTr 4.400 F 5.87 0.749 Heart Attack Risk Continued . . . H0: m1 = m2 = m3 = m4 Ha: at least two m’s are different Test Statistic: 4.400 F 5.87 0.749 with df1 = 3 and df2 = 136 P-value < .001 a = .05 Since the P-value < a, we reject H0. There is convincing evidence to conclude that mean MPV is not the same for all four patient populations. Summarizing an ANOVA ANOVA calculations are often summarized in a tabular format called an ANOVA table. To understand such a table, we need one more sum of squares term. Total sum of squares, denoted by SSTo, is given by x x This is the fundamental identity for SSTo with df = N – 1. 2 all N obs. single-factor ANOVA. The relationship between the three sum of squares is: SSTo = SSTr + SSE The General Format for a SingleFactor ANOVA Table Source of Variation Sum of Mean df Squares Square F P-value SSTr MSTr Treatment k – 1 SSTr k-1 MSE SSE When the analysis is done by Error N–k SSE N -software, k statistical then the Total N-1 SSToP-value appears here. Heart Attack Risk Continued . . . This is the ANOVA table for this data set. Source Treatment Error df 3 136 SS 13.199 101.888 Total 139 115.087 MS 4.400 0.749 F 5.87 P-value 0.000 Now we know that at least two of the means are different – but which two? To answer the question in this study we need to know if the mean MPV for the heart attack group is the mean that is different. Tukey-Kramer (T-K) Multiple What do we do now that we Comparison Procedure know that at least two of the population or treatment means are different? • This procedure is based on calculating confidence intervals for the difference between each possible pair of m’s. • If the interval contains the value zero, then there is no significant difference between We need to use a multiple theHow means can involved. we tell which of the comparison procedure, which • If, however, the interval does NOT contain mean(s) is/are different? is a method of identifying the value zero, then the two means are differences between m’s. significantly different. Tukey-Kramer (T-K) Multiple Comparison Procedure When there are sample k populations or are treatments If the sizes are thebased same, on webeing can use T-K intervals probability compare, the number of confidence intervals necessary is distributions called studentized range MSE given by q k (k 1) distributions. n 2 For mi – mj: x i xj MSE 1 1 q 2 ni n j where q is the relevant Studentized range critical value The two means are judged to differ significantly if the interval does not contain 0. Heart Attack Risk Revisited . . . Sample size Group Description Sample Sample standard mean deviation 1 Noncardiac chest pain 35 10.89 0.69 2 Stable angina pectoris 35 11.25 0.74 so 11.37 there is not0.91 a Sample sizes are the same in is the Infarction critical value 35 for 95% confidence 4 This Myocardial 11.75 1.07 significant difference in the mean MVP each treatment. when k = 4 and df = 120 (closest df in the between patients with noncardiac chest pain tablewith to 136). 4(4 1) and patients stable angina. 6 3 Unstable angina pectoris This interval contains 35 0, Number of confidence intervals to compute: For m1 – m2: 2 How many confidence intervals will we need to compute? 0.749 (10.89 11.25) 3.68 (.898, .178) 35 Heart Attack Risk Revisited . . . For m1 – m2 m1 – m3 m1 – m4 m2 – m3 m2 – m4 m3 – m4 95% Confidence Interval (-0.898, 0.178) (-1.018, 0.058) (-1.398, -0.322) (-0.658, .0418) (-1.038, 0.038) (-0.918, 0.158) The only interval that does not contain 0 is for the difference in mean MPV between The remaining confidence patients with noncardiac chestin pain intervals are calculated theand patients with heart attacks. same manner. They are . . . Summarizing the Results of the Tukey-Kramer Procedure 1. List the sample means in increasing order, identifying population just above each x Population If the3 sample 2 means 1 for 4 5 populations 3, 2, and 1 are then Sample Mean x3 notx2significantly x1 x4different, x5 draw a line under them. 2. Use the T-K intervals to determine the group of means that do not differ significantly from the first in the list. Draw a horizontal line extending from the smallest mean to the last mean in the group identified, Population Sample Mean 3 x3 2 x2 1 x1 4 x4 5 x5 Summarizing the Results of the Tukey-Kramer Procedure If the sample means for population 2 is not 3. Use the T-K intervals different to determine the group significantly from 1 and 4, of butmeans is that are not significantly from the second different from 5,different then draw a line under 2, smallest in the list. If this 1, entire and 4. group of means is not underscored, draw a horizontal line extending from the smallest mean to the last mean in the new group, Population Sample Mean 3 x3 2 x2 1 x1 4 x4 4. Continue considering the means in the order listed, adding new lines as needed. 5 x5 Heart ... BasedAttack on theseRisk data,Revisited we have evidence that the mean MPV is not the same for the noncardiac chest pain For group and 95% Confidence Interval the heart attack group. But since the mean Should in (-0.898, means is 0.178) small compared toMPV the be m1 – mdifference 2 variability among the individuals in each group, it would used as a 0.058) mstill 1 – mbe 3 difficult to (-1.018, distinguish the two groups based on predictor of an individual MPV value. And we don’t have evidence (-1.398, -0.322) m1 – m4 heart that the mean is different for the heart attack group (-0.658, 0.418) m2 – m3 attacks? and the two angina groups. So, MPV is probably not 0.038) m2 – m4 useful as a(-1.038, predictor of heart attack. (-0.918, 0.158) m3 – m4 Population: 1 2 3 4 SampleLet’s summarize these T-K intervals. 10.89 11.25 11.37 11.75 Mean: