Lecture 9. ANOVA: Random-effects model, sample size
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1 Lecture 9. ANOVA: Random-effects model, sample size Jesper Rydén Matematiska institutionen, Uppsala universitet Regressions and Analysis of Variance fall 2015
2 Fixed or random? Is it reasonable to assume that the levels of the factor come from a probability distribution? NO YES Treat factor as fixed Treat factor as random Searle, Casella, McCulloch (1992). Variance components.
3 Random-effects model Linear statistical model: y ij = µ + τ i + ɛ ij, { i = 1, 2,..., a j = 1, 2,..., n where τ i N(0, στ 2 ) and ɛ ij N(0, σ 2 ) (independence). For an observation y ij, V[Y ij ] = στ 2 + σ 2 where στ 2 and σ 2 are called variance components.
4 Covariance issues Unlike the fixed-effects model, the observations y ij are independent only if they come from different factor levels. Example: a = 3 treatments and n = 2 replicates, thus N = 6 observations. Covariance matrix on black board.
5 ANOVA table: Random-effects model Typical statistical problem, test of hypothesis: H 0 : στ 2 = 0 H 1 : στ 2 > 0 F -test is possible; furthermore, Test: E[MS Treat ] = E[SS Treat /(a 1)] = σ 2 + nστ 2 E[MS E ] = E[SS E /(a(n 1))] = σ 2 F 0 = MS Treat MS E F (a 1, a(n 1)) when H 0 is true. Reject H 0 if f 0 > F α (a 1, a(n 1)).
6 Estimating parameters Estimation of variance components: σ 2 = MS E and σ 2 τ = 1 n (MS Treat MS E ) For unequal sample sizes, replace n by [ a n 0 = 1 ] a n i i=1 n2 i a 1 a i=1 n i Estimation of overall mean µ: i=1 µ = ȳ
7 Negative estimates of variance It might happen that a negative value is obtained. Accept, believe that variance is small/zero Re-estimation with dedicated methods Consider negative estimate as evidence that the assumed linear model is incorrect
8 Confidence intervals Confidence interval for the variance component σ 2 can found (based on χ 2 distribution). Confidence interval for the variance component σ 2 τ : harder. No closed-form expression (linear combination of χ 2 distributed random variables). Confidence interval for the ratio σ 2 τ /(σ 2 τ + σ 2 ), the so-called intraclass correlation coefficient. Blackboard
9 Example. Protein content Agricultural researchers study the protein content in factory feed for animals. From each of 8 randomly selected bags, two were selected for analysis. At the bags, a protein content of 16% are printed when delivered from factory. State a model, estimate parameters. Could one believe in the information from the factory? Sum Blackboard
10 Decisions in hypothesis testing Decision H 0 is true H 0 is false Fail to reject H 0 No error Type II error Reject H 0 Type I error No error Probabilities of error: α = P(Type I error) = P(Reject H 0 when H 0 is true) β = P(Type II error) = P(Fail to reject H 0 when H 0 is false) The power of a statistical test is the probability of rejecting the null hypothesis H 0 when the alternative hypothesis is true. Power is calcuated as 1 β. If power is judged to be too low, an analyst can increase either α or the sample size.
11 Example. One-sample t-test (σ unknown) Observations x 1,..., x n (iid) from N(µ, σ 2 ). The usual test statistic is H 0 : µ = µ 0 H 1 : µ µ 0 T 0 = X µ 0 S/ n and under H 0, T 0 t(n 1). However, for power computations, the distribution of T 0 under H 1 is needed, which belongs to a non-central t distribution.
12 Non-central t distribution When the true value of the mean µ = µ 0 + δ, T t(n 1, δ n/σ). Assume µ 0 = 0.82, δ = 0.02, σ = Is a sample of n = 15 adequate for a power of at least 0.80? R computation: power.t.test(n=15,delta=0.02,sd=0.025,sig.level=0.05, type="one.sample",alternative="one.sided",strict = TRUE) One-sample t test power calculation n = 15 delta = 0.02 sd = sig.level= 0.05 power= alternative = one.sided
13 ANOVA: Power analysis for determining sample size When H 0 is false, the statistic F 0 = MS Treat /MS E is distributed as a non-central F random variable with a 1 and N a degrees of freedom and the non-centrality parameter λ. (With λ = 0, the usual, central, F distribution is obtained.) Power analysis can be carried out by various means: Numerical computation in statistical software packages, e.g. R Operating Characteristic (OC) curves, where β (the probability of type II error) is plotted against a parameter Φ: Φ 2 = n a i=1 τ 2 i aσ 2
14 Example. Plasma-etching experiment Consider the experiment (a = 4, n i = 5). Specifications by the engineer: A power of at least 0.90, α = The four treatment means: µ 1 = 575, µ 2 = 600, µ 3 = 650, µ 4 = 675 Assumption: σ = 25 Å/min (prior experience, preliminary tests, judgment estimate). OC curves will imply that at least n = 4 replicates are needed. Blackboard
15 OC curves
16 Etching example: R code We calculate the power directly in R, by using non-central F distributions. The non-centrality parameter is given as λ = n σ 2 In our example, with a = 4 treatment groups, we have τ 2 i = 6250 and σ 2 = 25 2 = 62 (from specification in the text: σ = 25). Hence, λ = 10n. R-code: a i=1 n = 3; a = 4; fcrit = qf(1-0.01, a-1, a*(n-1)); pow = 1-pf(fc, a-1, a*(n-1), 10*n) We obtain as results (cf. Montgomery, Example 3-10): τ 2 i n = 3 n = 4 n =
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