Statstcal tables are provded Two Hours UNIVERSITY OF MNCHESTER Medcal Statstcs Date: Wednesday 4 th June 008 Tme: 1400 to 1600 MT3807 Electronc calculators may be used provded that they conform to Unversty Regulatons nswer LL fve questons n SECTION (40 mars) nswer TWO of the three questons n SECTION ( 0 mars each) The total number of mars on the paper s 80 Page 1 of 7
SECTION nswer LL fve questons 1. refly explan what s meant by loc randomsaton. Illustrate ths by showng how you mght prepare a randomsaton lst for frst twenty patents n a tral wth two treatments. How mght you use bloc randomsaton to mprove balance between two treatment groups n a dchotomous prognostc factor? [6 Mars]. clncal tral compared an analgescs gel wth a placebo gel wth no actve ngredent for the treatment of jont pan. Usng randomsaton, 30 patents were allocated to the new gel and 30 to the placebo. Patents were assessed at the end of the two-wee treatment perod. The swellng was eradcated for 1 patents n the new treatment group and 15 patents n the placebo group. n absolute dfference n the success rate of the two treatments of 10% was consdered to be a clncally mportant effect. () State the hypotheses you mght use to compare the treatments. () Carry out a statstcal test to compare the treatments specfyng the assumptons that you mae. () Comment on the results of the tral. [10 Mars] Page of 7
3. You are ased to advse on the analyss of a double-blnd randomsed controlled tral of treatments for depresson comparng fluoxetne wth placebo. The prmary outcome measure s the ec Depresson Inventory (DI-FU), whch s a contnuous outcome measure that s consdered to be approxmately normally dstrbuted. Ths same measure has also been recorded at baselne (DI- SE). pparently, DI-FU s expected to be strongly correlated wth DI-SE. The followng three statstcal analyses are beng consdered: Carry out a t-test comparng DI-FU at follow-up. Calculate the change from baselne, that s DI-CHG = DI-FU DI-SE, and then apply a t-test. Ft a lnear model wth DI-FU as the dependent varable wth co-varates DI-SE and treatment group. What advce would you gve regardng the choce of statstcal analyss to be ncluded n statstcal analyss plan justfyng that choce? [5 Mars] 4. Consder a randomzed controlled tral. Suppose the patent populaton can be dvded nto three latent sub-groups as follows: Complers: patents who wll comply wth the allocated treatment, lways control treatment: patents who wll receve control treatment regardless of allocaton, lways new treatment: patents who wll receve the new treatment regardless of allocaton. Ths assumes that there are no defers, that s patent who wll always receve the opposte of the treatment to whch they are randomzed. ssumng that the proporton and characterstcs of complers, always control treatment, always new treatment s the same n both arms and that randomzaton can only affect the outcome through the recept of treatment, show that: () n ntenton-to-treat estmate of the treatment effect s based towards the null hypothess of no treatment effect. () per-protocol estmate of the treatment effect may be based ether towards or away from the null hypothess of no treatment effect. [10 Mars] Page 3 of 7
5. The Mntab prnt-out below gves the results of analyss of a randomsed controlled -perod - crossover tral on 45 patents comparng the effect of eatng butter and margarne on total blood fats. Patents are randomsed to receve utter then Margarne or Margarne then utter. nalyss of Perod 1 Two-Sample T-Test and CI Sample N Mean StDev SE Mean utter then Marg 3 6.0 0.870 0.18 Marg then utter 5.910 0.780 0.17 Dfference = mu (utter then Marg) - mu (Marg then utter) Estmate for dfference: 0.310 95% CI for dfference: (-0.188, 0.808) T-Test of dfference = 0 (vs not =): T-Value = 1.6 P-Value = 0.16 DF = 43 oth use Pooled StDev = 0.873 nalyss of Perod - Perod 1 Two-Sample T-Test and CI Sample N Mean StDev SE Mean utter then Marg 3-0.70 0.560 0.1 Marg then utter 0.30 0.590 0.13 Dfference = mu (utter then Marg) - mu (Marg then utter) Estmate for dfference: -0.500 95% CI for dfference: (-0.846, -0.154) T-Test of dfference = 0 (vs not =): T-Value = -.9 P-Value = 0.006 DF = 43 oth use Pooled StDev = 0.5748 () () () (v) Usng the analyss for Perod 1 output gve the estmate and 95% confdence nterval of the treatment effect for Margarne as compared to utter. Usng the analyss for Perod Perod 1 gve the estmate and 95% confdence nterval of the treatment effect for Margarne as compared to utter. What s the advantage of a crossover desgn as compared to a parallel group desgn? Gve one lmtaton of a crossover desgn as compared to a parallel group desgn. [9 Mars] Page 4 of 7
nswer TWO of the three questons n ths secton SECTION 6. randomsed controlled tral s planned to compare a new antbotc treatment () wth the current standard therapy () for patents wth T. t sx months follow-up t s recorded whether the dsease s stll present n the patent. () Why t s mportant to estmate sample sze n a clncal tral. [ Mars] () The two-sample test of proportons wth statstc z gven by z = p p ( p ( 1 p) )( 1 n + 1 n) wll be used to test the null hypothess of no treatment effect, where n, n are the number of subjects allocated to each treatment, p = r n, r n patents n whch T was absent after 6 months for each treatment and Suppose that patents are to be allocated n the rato of :1 wth n =.n. p = wth r, r are the numbers of n. p p = n + n + n. p. ssumng that the test statstc z has a normal dstrbuton under the null and alternatve hypotheses and usng a two-sded α sze test, show that the power ( ) ( ).. 1 zα λ π π τ 1 β α, τ 1 Φ π( 1 π) π( 1 π) + n n, where π, π, and π are the populaton proportons correspondng to p, p, and p, λ = 1 n + 1 n, and Φ s the standard normal cumulatve densty functon. [7 Mars] () Show that the sample sze requred for group to gve a power (1-β ) s approxmately n ( zα π ( 1 π )( 1+ ) + z β π ( 1 π ) + π ( 1 π )) ( π π ) =. [7 Mars] (v) The nvestgators plannng the randomsed controlled tral expect that the proporton of patents that recover n the current standard therapy group () wll be 50%. n mprovement to 65% wth the new medcaton () s consdered to be clncally mportant. Estmate the total sample sze that would be requred usng a to 1 allocaton rato (=), assumng a power of 80% and a two-sded 5% sgnfcance level. [4 Mars] [Total 0 Mars] Page 5 of 7
7. In a parallel group non-nferorty tral a new treatment T s beng compared wth a control treatment C usng a contnuous normally dstrbuted outcome measure Y. Let y T, y C, μ T and μ C be the sample and populaton means of Y for each treatment, n T and n C be the sample szes, and s be the pooled wthn-group sample standard devaton of Y. Defne the treatment effect τ = μt μc. () Explan why a sgnfcance test of the hypothess H : τ 0 vs H : 0 1 τ < would be napproprate n a non-nferorty tral. 0 = () Suppose that the null hypothess H : 0 μt μc τn s rejected f the (1-α) sngle sded [4 Mars] confdence nterval, gven by yt yc z α λs wth λ = 1 nt + 1 nc, s greater than τ N. Show that Pr[Reject H 0 τ] ( τ + τ) N = 1 Φ + z sλ α where Φ s the cumulatve dstrbuton functon of the standard normal dstrbuton. [7 Mars] () Show that Pr[Reject H 0 τ] has a maxmum under H 0 when τ = τ N. Hence show that ths (v) procedure has a type I error α. [6 Mars] randomsed controlled non-nferorty tral s carred out to test whether a new generc drug s as effectve as a current standard drug for controllng pan measured by a 100 mm analogue scale wth hgher scores representng greater pan. Ffty patents are randomsed to the standard treatment and 5 to the new generc treatment. The Mntab output s gven below. Two-Sample T-Test and CI Sample N Mean StDev SE Mean Current standard drug 50 65.5 18.5.6 New generc drug 5 66.1 18.8.6 Dfference = mu (Current standard drug) - mu (New generc drug) Estmate for dfference: -0.60 Standard Error (SE) for dfference: 3.69 95% CI for dfference: (-7.93, 6.73) T-Test of dfference = 0 (vs not =): T-Value = -0.16 P-Value = 0.871 DF = 100 oth use Pooled StDev = 18.6536 dfference of 10 mmhg was consdered by researchers to be the mnmum that was clncally mportant. Usng a 5% sgnfcance level test whether the new medcaton s non-nferor to the current standard drug. [3 Mars] [Total 0 Mars] Page 6 of 7
8. In meta-analyss suppose θˆ s an estmate of the treatment effect for the th study, assumed to be normally dstrbuted, and let Var ˆ θ be ts samplng varance. The fxed effect estmator ˆ =, where w are weghts, wth Var[ ˆ ] θ () w ˆ θ w [ θ ]. ˆ Var w θ =. w Usng the Lagrange multpler method show that the mnmum varance estmator of θ, say ˆ θ, s obtaned when w [ ] MV 1/ Var θˆ to Var ˆ θmv = = 1 and show that the mnmum varance estmate s equal 1 1 Var ˆ θ [1 Mars] The table below summarzes the outcome of three trals comparng detary advce gven by a detcan wth that gven by a doctor for patents for wth hgh blood cholesterol. The treatment effect for each study ( θˆ, = 1,,3 ) s the dfference n mean cholesterol between detcan advce group and doctor advce group. Var ˆ θ s the sample varance estmate the th study. Study Dfference n blood cholesterol, θˆ V ar θˆ Dyson1996-0.34 0.089 Thomson 00-0.18 0.079 Smth 1989-0.7 0.0676 () Compute the mnmum varance estmate of the overall treatment effect,, and determne θˆmv ts 95% confdence nterval statng any assumptons that you mae. [6 Mars] () What do you conclude from the meta-analyss? [ Mars] [Total 0 Mars] End of Examnaton Paper Page 7 of 7