2.2 Confidence intervals for # Example 3.Darwin's data:gains in height of plants from cross- fertilisation For a given a E(0,1).since we will not reject the null hypothesis X=height(Cross-F)-height(Self-F) H0:μ=0 iff 2logT<x1,1-a:where P(xi<x1,1-a}1-a.For a 0.05, 15 observations: X11-a=3.84. 6.1,-8.4,1.0,2.0,0.7,2.9.3.5,5.1,1.8,3.6,7.0.3.0, Hence a 100(1-a)%confidence interval for u is 9.3,7.5,-6.0 u-2log(L()n")<xi.1-a} =ulogp:()>-0.5x3.1-a-nlogn} The sample meanX=2.61,the standard error s=4.71. =1 =cim)>-05a-小 Is the gain significant? Intuitively:YES,if no two negative observations -8.4 and -6.0. Let u=EXi. (a)Normal plot H0:4=0D8H1:μ>0 QQ-plot: Quantile of (i)Standard approach:Assume {X1,...,X15}is a random sample N(0,1)vs Quantile of the from N(u,2) empirical distribution. MLE:立=X=2.61 -1 Nomal quantiles The t-test statistic: T=√求/s=2.14 (b)Lk likelihood.k=1,2,4.8 Since T~t(14)under Ho.the p-value is 0.06-significant but The profile likelihood l(u)is not overwhelming. plotted against u for k=1 Is N(u,o2)an appropriate assumption?as the data do not (solid),2 (dashed),4 (dot- appear to be normal(with a heavy left tail);see Fig(a). ted),and 8(dot-dashed).2.2 Confidence intervals for µ. For a given α ∈ (0, 1), since we will not reject the null hypothesis H0 : µ = µ0 iff 2 log T < χ2 1,1−α , where P{χ 2 1 ≤ χ 2 1,1−α } = 1 − α. For α = 0.05, χ 2 1,1−α = 3.84. Hence a 100(1 − α)% confidence interval for µ is n µ    − 2 log{L(µ)n n } < χ2 1,1−α o = n µ    Xn i=1 log pi(µ) > −0.5χ 2 1,1−α − n log n o = n µ    Xn i=1 log{npi(µ)} > −0.5χ 2 1,1−α o . Example 3. Darwin’s data: gains in height of plants from cross￾fertilisation X = height(Cross-F) - height(Self-F) 15 observations: 6.1, -8.4, 1.0, 2.0, 0.7, 2.9, 3.5, 5.1, 1.8, 3.6, 7.0, 3.0, 9.3, 7.5, -6.0 The sample mean X¯ = 2.61, the standard error s = 4.71. Is the gain significant? Intuitively: YES, if no two negative observations -8.4 and -6.0. Let µ = EXi . H0 : µ = 0 vs H1 : µ > 0 (i) Standard approach: Assume {X1, · · · , X15} is a random sample from N(µ, σ2) MLE: µb = X¯ = 2.61 The t-test statistic: T = √ nX/s ¯ = 2.14 Since T ∼ t(14) under H0, the p-value is 0.06 — significant but not overwhelming. Is N(µ, σ2) an appropriate assumption? as the data do not appear to be normal (with a heavy left tail); see Fig(a). −1 0 1 −5 0 5 10 Normal quantiles Observed quantiles (a) Normal plot −5 0 5 10 0.0 0.4 0.8 µ Profile likelihood (b) Lk likelihood, k=1,2,4,8 QQ-plot: Quantile of N(0, 1) vs Quantile of the empirical distribution. The profile likelihood lk(µ) is plotted against µ for k = 1 (solid), 2 (dashed), 4 (dot￾ted), and 8 (dot-dashed)