Family and statement (individual) confidence coefficientsBonferroni joint confidence intervalsBonferroni procedureBonferroni procedure4.1 Joint estimation of 0 and 1 Suppose separate 95% confidence intervals are constructed for 0 and 1. What is the confidence level that BOTH confidence interval inferences are correct? NOT 95%! If these inferences were independent, then the probability of both being correct is 0.952=0.9025. However, they are not independent, which makes the actual confidence level difficult to calculate. Family and statement (individual) confidence coefficientsA set of estimates or hypothesis tests of interest are called a “family”. If we are interested in inferences for 0 and 1 simultaneously, then they form a family. Each individual estimate or hypothesis test has a confidence level associated with it. For example, a 95% confidence interval for 1. The family confidence level corresponds to the level of confidence that all inferences are correct for the entire family.  2012 Christopher R. Bilder4.1For example, a C% confidence level that both 0and 1 are within their confidence interval limits. Bonferroni joint confidence intervalsFind an expression for the probability that both confidence intervals contain their corresponding parameters; i.e., find the family confidence level. Let A1 = the event that 0 is not in the C.I.Let A2 = the event that 1 is not in the C.I.The probability that at least one confidence interval doesnot contain i is P(A1 or A2) = P(A1  A2). At least one means that A1 is true or A2 is true or A1 and A2 are true.The probability that BOTH confidence intervals contain their corresponding parameters is 1-P(A1  A2) (use the complement). From the addition theorem of probability (p. 662 of KNN),P(A1  A2) = P(A1) + P(A2) – P(A1  A2) where P(A1  A2) = P(A1 and A2) = probability both confidence intervals do NOT contain their correspondingparameters.  2012 Christopher R. Bilder4.2Thus, 1 - P(A1  A2) = 1 - P(A1) - P(A2) + P(A1  A2)  1 - P(A1  A2)  1 - P(A1) - P(A2) In words: The probability that both confidence intervals contain their corresponding parameters is greater than or equal to 1 minus the probability 0 is not in its confidence interval minus the probability 1 is not in its confidence interval.For a (1-)100% confidence interval, P(Ai)= for i=1,2.If 95% confidence intervals are formed for 0 and 1, then the probability that both intervals contain their corresponding parameters is  1 – 0.05 – 0.05 = 0.90. Suppose a family confidence level of at least 95% is desired. Then using 97.5% confidence intervals for eachgives the desired level. In general for a “Bonferroni” family confidence level of at least (1-)100%, the confidence levels for each of 2 C.I. are 1-/2. This is because 1-P(A1  A2)  1 - P(A1) - P(A2) = 1 - /2 - /2 = 1-. The C.I.s for 0 and 1 are: For 0: 0 0b t(1 / 4,n 2) * Var(b )�� - a -For 1: 1 1b t(1 / 4,n 2) * Var(b )�� - a - 2012 Christopher R. Bilder4.3Example: HS and College GPA (HS_college_GPA_ch4.R)Find C.I.s for 0 and 1 using a family confidence level of95%. > #Read in the data> gpa<-read.table(file = "C:\\chris\\UNL\\STAT870\\Chapter1\\gpa.txt", header=TRUE, sep = "")> head(gpa) HS.GPA College.GPA1 3.04 3.12 2.35 2.33 2.70 3.04 2.05 1.95 2.83 2.56 4.32 3.7 > mod.fit<-lm(formula = College.GPA ~ HS.GPA, data = gpa)> sum.fit<-summary(mod.fit)> sum.fit$coefficients Estimate Std. Error t value Pr(>|t|)(Intercept) 0.7075776 0.19941429 3.548279 2.296706e-03HS.GPA 0.6996584 0.07319166 9.559264 1.778933e-08 > alpha<-0.05> g<-2> qt(p = 1-alpha/(2*g), df = mod.fit$df.residual)[1] 2.445006For 0: 0 0b t(1 / 4,n 2) * Var(b )�� - a -0.70758 t(1 0.05 / 4,20 2) 0.199410.70758 t(0.9875,18) 0.199410.70758 2.4450 0.19941(0.2200, 1.1951)      2012 Christopher R. Bilder4.4For 1: 1 1b t(1 / 4,n 2) * Var(b )�� - a -0.6997 t(0.9875,18) * 0.0732(0.5207, 0.8787)We are at least 95% confident that 0 is between 0.2200 and 1.1951 and 1 is between 0.5207 and 0.8787. More R code and output:> mod.fit$coefficients[1]-qt(p = c(1-alpha/(2*g), alpha/(2*g)), df = mod.fit$df.residual) *sum.fit$coefficients[1,2][1] 0.2200086 1.1951467> mod.fit$coefficients[2]-qt(p = c(1-alpha/(2*g), alpha/(2*g)), df = mod.fit$df.residual) *sum.fit$coefficients[2,2][1] 0.5207044 0.8786124> #Another way> confint(object = mod.fit, level = 1 - alpha/g) 1.25 % 98.75 %(Intercept) 0.2200086 1.1951467HS.GPA 0.5207044 0.8786124Notes:1.The Bonferroni (1-)100% family confidence level is a lower bound. The actual family confidence level may be higher. 2.This procedure of finding a lower bound can be extended to g simultaneous confidence intervals. Instead of using 1- for the individual confidence levels, 1-/g is used to assure at least a (1-)100% family  2012 Christopher R. Bilder4.5confidence level. Thus, 1-/(2g) needs to be used whenfinding the t-distribution quantile because a two-sided confidence interval is being found. 3.It is best to use this procedure when g is small. If g is large, the individual confidence levels become close to 1which causes the confidence intervals to be very wide.  2012 Christopher R. Bilder4.64.2 Simultaneous estimation of mean responsesRemember that E(Yh) is the mean value of Yh at a specific Xh. The (1-)100% C.I. for E(Yh) at Xh is (from Section 2.4): h hˆ ˆY t(1 / 2,n 2) Var(Y )�� - a - where 2hh2i1 (X X)ˆVar(Y ) MSEn(X X)    Suppose we wanted to find C.I.s for E(Yh) at many different Xh values with a family confidence level of 1-. Bonferroni procedureTo have a family confidence level of at least 1-, the Bonferroni confidence limits are: h hˆ ˆY t 1 ,n 2 Var(Y )2g�� �a� - -� �� �where g is the number of C.I.s in the family. Example: College and HS GPA (HS_college_GPA_ch4.R)Find C.I.s for E(Yh) at Xh=2.0, 2.5, 3.0, 3.5, 4.0 with at least a 95% family confidence level.  2012 Christopher R. Bilder4.7> more.gpa<-data.frame(HS.GPA = c(2, 2.5, 3, 3.5, 4))> g<-nrow(more.gpa)> round(predict(object = mod.fit, newdata = more.gpa, interval = "confidence", level = 1-alpha/g), 2) fit