Clinical Trials

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❘❙❚■ Chapter 17 | Types of Data and Normal DistributionThe one-sided area above 130 + 15 mm Hg is 15.87% (calculated using the procedureset out in Question 1). Therefore, the two-sided area above 130 + 15 mm Hg orbelow 130 – 15 mm Hg is twice the one-sided area, ie, 2 × 15.87% = 31.74%.This result suggests that, for any normal distribution, about 68% of allobservations are bounded within one standard deviation distance either side ofthe mean (eg, for SBP data, 130 ± 15 mm Hg). It can be similarly calculated thatabout 95% of all observations fall within the mean ± 1.96 × standard deviationseither side of the mean, which is often called the 95% reference range.How do we assess normality?As mentioned, the normality assumption is a prerequisite for many statisticalmethods and models, such as t-tests, analyses of variance, and regression analysis.This assumption should be checked on a given dataset when conducting statisticalanalysis, especially with smaller sample sizes.There are two ways of doing this:• graphical inspection of the data to visualize differences betweendata distributions and theoretical normal distributions• formal numeric statistical testsHistograms are sometimes used for visual inspection, but are unreliable for smallsample sizes (as demonstrated in Figure 6). The most common graphical approachis the inverse normal plot, or the quantile–quantile plot (Q-Q plot), which comparesordered values of a variable with corresponding quantiles of a specific theoreticalnormal distribution. If the data and the theoretical distributions match, the pointson the plot form a linear pattern that passes through the origin and has a unit slope.Figure 7 shows the inverse normal plots for the data represented in Figure 6.These demonstrate that the plots are almost linear and pass through the origin,suggesting that these samples represent a normal distribution. Figure 8 displaysa histogram and inverse normal plot for creatine kinase data for 2,668 patientswith chronic heart failure from a clinical trial. The histogram indicates that thedistribution of creatine kinase is not normal – it is positively skewed. The inversenormal plot shows marked departure from a linear pattern, especially at the lowerand upper range of the data, indicating that a distribution other than the normaldistribution would better fit these data.182
<strong>Clinical</strong> <strong>Trials</strong>: A Practical Guide ■❚❙❘Figure 7. Visual inspection of normality by inverse normal plot using data from Figure 6.210–1–2210–1–2210–1–2210–1–2n = 20 4n = 20 20n = 2021000–2–10–4–20–2 –1 0 1 2–4 –2 0 2 4–20 –10 0 10 20Inverse normalInverse normalInverse normaln = 40 4n = 40 20n = 4021000–2–10–4–20–2 –1 0 1 2–4 –2 0 2 4–20 –10 0 10 20Inverse normalInverse normalInverse normaln = 100 4n = 100 20n = 10021000–2–10–4–20–2 –1 0 1 2–4 –2 0 2 4–20 –10 0 10 20Inverse normalInverse normalInverse normaln = 400 42n = 400 2010n = 40000–2–10–4–20–2 –1 0 1 2–4 –2 0 2 4–20 –10 0 10 20Inverse normalInverse normalInverse normalFigure 8. Histogram (left) and inverse normal plot (right) for creatine kinase data for 2,668 patients withchronic heart failure.0.018000.008600Density0.0060.0040.002Creatine kinase400200000 200 400 600 800–200–200 0 200 400 600 800Creatine kinase (U/L)Inverse normal183
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❘❙❚■ ContributorsStephen L
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Clinical Trials: A Practical Guide
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■■❚❙❘Chapter 1RandomizedC
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❘❙❚■ Chapter 2 | Uncontroll
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■■❚❙❘Chapter 3Protocol De
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■■❚❙❘Chapter 4EndpointsAm
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■■❚❙❘Chapter 5Patient Sel
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■■❚❙❘Part IIAlternativeTr
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❘❙❚■ Chapter 10 | Crossover
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❘❙❚■ Chapter 11 | Factorial
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■■❚❙❘Chapter 21Analysis o
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■■❚❙❘Part IVSpecial Trial
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■■❚❙❘Chapter 22Intention-
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■■❚❙❘Chapter 23Subgroup A
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■■❚❙❘Chapter 24Regression
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■■❚❙❘Chapter 25Adjustment
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■■❚❙❘Chapter 26Confoundin
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■■❚❙❘Chapter 29Multiplici
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■■❚❙❘Chapter 32Overview o
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■■❚❙❘Chapter 33Trial Prof
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■■❚❙❘Chapter 34Presenting
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■■❚❙❘Chapter 35Use of Tab
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■■❚❙❘GlossaryGlossary453
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■■❚❙❘IndexIndex467
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