Properties of t distributionConfidence IntervalsExample: if n = 12, degrees of freedom = 11For  known, useBinomial DistributionVar(X) = npqInterval Estimate--Use Normal TheoryInterval EstimationUnderlying distribution is Normal—Central Limit theoremExample: Suppose we draw a sample of children and measure Diastolic Blood Pressure (n=10)x= 58 mm HgNot certain that  = 58 exactly—second sample estimate might be x = 61 mm HgAssume DBP ~ N(, 2)Therefore, if , 2 were known, and because n)σ~N(μx,/295% of all sample means would fall within)nσ.,μnσ.(μ 961961 Similarly, if we convert x to z wherenσ/μxz , then z ~ N(0, 1)Therefore, 95% of the z values are between (-1.96, 1.96)t distributionCalculate ns/μxtShape depends on n—not a unique distribution but a family indexed by a parametercalled degrees of freedomIf X1, X2, … Xn ~ N(, 2) are independent,1n~tns/μXtNotation: td,uWhere d = degrees of freedom u refers to some percentile, like 95thPr(td < td,u)  uFor example: t9, .95Properties of t distribution1. Symmetric about 02. More spread out than N(0, 1)3. For any percentile, ,  > 0.5,td, 1- is always larger than corresponding percentile for N(0, 1)—z1-4. As d increases, t approaches N(0, 1)5. ns/μxt has more variability than nσ/μxzUsing the t-table (Page 823)t10, 0.975 =t20, 0.975 =t40, 0.975 =t120, 0.975 =z0.975 =t20, 0.025 =t20, 0.05 =Confidence Intervalst statistic follows a t distribution with n-1 degrees of freedomPr(tn-1, .025 < t < tn-1, .975) = 0.95orαtttα,nα,n1)Pr(21121that is,21121α,nα,ntns/μxns/μxt andtherefore,α)n*s/txμn*s/txα,nα,n1211211Pr(therefore, CI)( α)%(n*s/tx,n*s/txα,nα,n1100211211orCIα)%()n*s/tx(α,n1100211Example: if n = 12, degrees of freedom = 11)*s/.x,*s/.x(%CI 12201212201295 Example: Estimate the average level of a certain enzyme in a human population. Take a sample of 10 individuals and determine the level of the enzyme in each.45222sx).,.().,.).*.,.*.()*.,*.(%CI798262021779842279842212132262222121322622221045262222104526222295For  known, use)n/*σzx,n/*σzx(%CIαα2121100For large n (n > 200), use)n*s/zx,n*s/zx(%CIαα2121100Interpretation: 95% of such intervals constructed from repeated samples of size n contain . 99% Confidence Interval).,.().()*.()n*s/t(.,8942810615894622104525032222995990% Confidence Interval).,.().()*.()n*s/t(.,8882511218888322104583312222959Length of Confidence Interval determined by1. n2. s3. Binomial DistributionExample: In a study of childhood abuse in psychiatric patients, we found that 166 in a sample of 947 patients reported histories of physical and/or sexual abuse.Construct a point estimate and a 95% Confidence Interval estimate for the population proportion.Let Xi = 1 if the child was abused Xi = 0 if notniiXX1i = 1, 2, …, 947p = prevalence of child abuse in this populationE(X) = npVar(X) = npqPoint EstimationLet pˆ = sample proportion of eventsnXXnnii11E(pˆ) = E(Xi) = pVar(pˆ)=pq/nSE(pˆ)= pq/n--estimated by nqp /ˆˆFor the example,pˆ= 166/947 = 0.175qˆ= 0.825SE = 0123094782501750 ./.*. Interval Estimate--Use Normal Theorypˆ  N(p, pq/n)orX  N(np, npq))/nqpzp(α)CI%()/nqp.p(%CIαˆˆˆˆˆˆ21110096195Use if 5qpnˆˆFor our