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BMED2803Brani VidakovicThursday, 3/15/2007.
1. Alzheimer’s. A medical research team wished to evaluate a proposed screening test forAlzheimer’s disease. The test was given to a random sample of 450 patients with Alzheimer’sdisease and to an independent sample of 500 subjects without symptoms of the disease.
Figure 1: The Reverend Thomas Bayes. He did not have Alzheimer’s but can help you withthe solution.
The two samples were drawn from the population of subjects who are 65 years old or diagnosed Alzheimer, D no Alzheimer’s symptoms, Dc (a) Using the numbers from the table estimate P (T |D) and P (T c|Dc). Interpret these probabilities in terms of the problem, one sentence for each.
Probability of D (prevalence) is the rate of the disease in the relevant population (≥ 65 Find P (D|T ) (positive predicted value) using Bayes formula. You cannot find P (D|T ) using information from the table only – you need external info, such as prevalence.
Solution. P (T |D) = 436/450 = 0.9689 and P (T c|Dc) = 495/500 = 0.99. The first is theprobability that a patient who shows symptoms of Alzheimer’s disease would test positive(sensitivity) and the second is the probability that a subject who does not have symptomsof Alzheimer would test negative (specificity). Note that P (T c|D) = 1 − P (T |D) = 0.0311and P (T |Dc) = 1 − P (T c|Dc) = 0.01.
P (D|T ) = P (T |D)P (D)/P (T ) P (T |D)P (D) P (T |D)P (D) + P (T |Dc)P (Dc) 0.9689 × 0.113 + 0.01 × 0.887 1Evans, D. A. et al. (1990). Estimated Prevalence of Alzheimer’s Disease in the United States. Milbank 2. Random Variables as Models. Tubert-Bitter et al (1996) 2 found that number ofserious gastrointestinal reactions reported to the British Committee on Safety Medicine was538 for 9,160,000 prescriptions of anti-inflammatory drug Piroxicam. (a) What is the rate of gastrointestinal reactions per 10,000 prescriptions?(b) Using the Poisson model with rate λ as in (a), find the probability of exactly two gastrointestinal reactions per 10,000 prescriptions.
(c) Find the probability of at least two gastrointestinal reactions per 10,000 prescriptions.
Solution: (a) The rate of gastrointestinal reactions per prescription is 538/9160000, and per 10000 prescriptions is 538/9160000 × 10000 = 0.5873. (b) If λ = 0.5873, and X is the number of gastrointestinal reactions per 10000 prescrip- tions, then suggested model is X ∼ Poi(0.5873). Further, (c) The probability to be found is P (X ≥ 2) which is equal to 1−P (X < 2) = 1−P (X ≤ 1), since the Poisson model is discrete and P (X < 2) = P (X ≤ 1). Then, P (X ≥ 2) = 1 − P (X ≤ 1) = 1 − P (X = 0) − P (X = 1) = 1 − i.e., about 11.8%. Recall, 0! = 1, by definition.
>> lambda = 538/9160000 * 10000 %answer for (a)lambda = >> p2 = lambda^2/2 * exp(-lambda) %answer for (b)p2 = >> p2plus = 1 - lambda^0/1 * exp(-lambda).
- lambda^1/1 * exp(-lambda) %answer for (c) Parts (b) and (c) can be found via poisspdf and poisscdf as 2Tubert-Bitter et al (1996). Comparing the Toxicity of Two Drugs in the Framework of Spontaneous Reporting: A Confidence Interval Approach. Journal of Clinical Epidemiology, 49, 121–123.
3. Psoriasis. Woo and McKenna (2003) 3 investigated the effect of broadband ultravioletB (UVB) therapy and topical calcipotriol cream used together on areas of psoriasis. One ofthe outcome variables is the Psoriasis Area and Severity Index (PASI) where lower score isbetter. The following table gives PASI scores for 20 subjects measured at baseline and aftereight treatments. Do these data provide sufficient evidence, at 0.05 level of significance, toindicate that the combination therapy reduces PASI scores? The data set is available online at http://zoe.bme.gatech.edu/~bv20/bmestat/exams, (a) Import the data to MATLAB and test the hypothesis that PASI index significantly decreased after the treatment. Use α = 0.05. Be careful, this is a 2-sample problem but theobservations are subject-blocked (dependent). Perform the test by any method you choose,either rejection region or p-value (or both if you have time).
(b) Find the 99% confidence interval for the difference between the population means ∆ = µ1 − µ2. You may want to consult the equation 8.6 (page 303).
baseline = [5.9 7.6 12.8 16.5 6.1 14.4 6.6 5.4 .
9.6 11.6 11.1 15.6 9.6 15.2 21 5.9 10 12.2 20.2 6.2]; after = [5.2 12.2 4.6 4 0.4 3.8 1.2 3.1 3.5 4.9 .
3Woo, W. K. and McKenna, K. E. (2003). Combination TL01 Ultraviolet B Phototherapy and Topical Calcipotriol for Psoriasis: A Prospective Randomized Placebo-Controlled Clinical Trial, Brithish Journal ofDermatology, 149, 146–150.
D = baseline - after;n = length(D);Dbar = mean(D) %Dbar = 6.3550sdD= sqrt(var(D)) %sdD = 4.9309tstat = Dbar/(sdD/sqrt(n)) %tstat = 5.7637 critpt = tinv(0.95, n-1) %critpt = 1.7291 % Rejection region (1.7291, infinity). Reject H_0 since% tstat=5.7637 falls in the rejection region.
% TEST using the p-value p_value = 1-tcdf(tstat, n-1) %p_value = 7.4398e-006 % Reject H_0 at the level alpha=0.05 since the p_value = 0.00000744 < 0.05.
%(b) alpha = 0.01LB = Dbar - tinv(1-0.01/2, n-1)*(sdD/sqrt(n)) UB = Dbar + tinv(1-0.01/2, n-1)*(sdD/sqrt(n)) % UB = 9.5094% 99% CI is [3.2006, 9.5094] Alternatively if you used D1 = after - baseline, then D1 = after - baseline;n = length(D1);D1bar = mean(D1) %D1bar = -6.3550sdD1= sqrt(var(D1)) %sdD1 = 4.9309tstat = D1bar/(sdD1/sqrt(n)) %tstat = -5.7637 critpt = tinv(0.05, n-1) %critpt = -1.7291 % Rejection region (-infinity, -1.7291). Reject H_0 since% tstat=-5.7637 falls in the rejection region.
% TEST using the p-value p_value = tcdf(tstat, n-1) %p_value = 7.4398e-006 % Reject H_0 at the level alpha=0.05 since the p_value = 0.00000744 < 0.05.
% (b) alpha = 0.01LB1 = D1bar - tinv(1-0.01/2, n-1)*(sdD1/sqrt(n)) UB1 = D1bar + tinv(1-0.01/2, n-1)*(sdD1/sqrt(n)) % UB1 = -3.2006% 99% CI is [-9.5094, -3.2006] 4. Hypertension. Dernellis and Panaretou (2002) 4 examined a small number of subjectswith hypertension and healthy control subjects. One of the variables of interest was the aorticstiffness index. Measures of this variable were calculated from the aortic diameter evaluatedat M-mode echocardiography and blood pressure measured by a sphygmomanometer. Gen-erally, physicians wish to reduce aortic stiffness. From n1 = 15 patients with hypertension(Group 1), the mean aortic stiffness index was ¯ X1 = 16.16 with a standard deviation of s1 = 4.29. In the n2 = 16 control subjects (Group 2), the mean aortic stiffness index was X2 = 10.53 with a standard deviation of s2 = 3.33. (a) Test the hypothesis that the population mean aortic stiffness indices for the two groups differ (use the two sided alternative). Perform the test by both methods rejectionregions and p-value. Take α = 0.05 and assume that the population variances are the same.
(b) What is the (observed, retrospective)5 power of this test against the specific alterna- tive ∆ = |µ2 − µ1| = 3. Use equation 8.28 (page 333) by estimating σ2 and σ2 by s2 and s2.
(c) Pretend that this study is only a pilot-study needed to design more elaborate clinical trial of the same type. What group sample sizes are needed to assure 99% power against thefixed alternative ∆ = |µ2 − µ1| = 3? Use Equation 8.26 (page 332) and estimate σ2 and σ2 from the pilot study. (Do not estimate ∆ as the book does, take ∆ = 3.) % population variances assumed the same, need s_p sp2 = ((n1 -1)* s1^2 + (n2-1) * s2^2)/(n1 + n2 - 2);sp = sqrt(sp2) % sp = 3.8237 tstat = (X2bar - X1bar)/(sp * sqrt(1/n1 + 1/n2)) %TEST two sided H_1 rejection region method alpha = 0.05;tcrit = tinv(1-alpha/2, n1 + n2 - 2) %tcrit = 2.0452 %and the rejection region RR is (-infinity, -tcrit) U (tcrit, infinity)%which is (-infinity, -2.0452) U (2.0452, infinity).
%Reject H_0 since tstat falls in the RR.
pval = 2 * tcdf(-abs(tstat), n1 + n2 - 2) % pval = 3.0733e-004 4Dernellis, J. and Panaretou, M. (2002) Effect of Thyroid Replacement Therapy on Arterial Blood Pres- sure in Patients with Hypertension and Hyperthyroidism. American Heart Journal, 143, 718–724.
5Attributes observed and retrospective stand for power analysis of the test that is already performed.
This is the same as the prospective analysis but the σ’s are replaced by estimators s.
% which is the same as 2*tcdf(tstat, n1+n2-2) since% tstat < 0.
% Reject H_0 since pval = 0.0003 < 0.05 = alpha.
(b) We will use equation 8.28 (text page 333): σ21/n1 + σ22/n2 with ∆ = 3, α = 0.05, and σ2, σ2 replaced by s2 = 4.292 and s2 = 3.332. Delta = 3;pow = normcdf( -norminv(1 - alpha/2) + Delta/( sqrt(s1^2/n1 + s2^2/n2) ) ) (c) We will use equation 8.26 (text page 332): 1−α/2 + z1−β )2 , with ∆ = 3, α = 0.05, β = 0.01 and σ2, σ2 replaced by s2 = 4.292 and s2 = 3.332. beta = 0.01;n = (s1^2 + s2^2)*(norminv(1-alpha/2) + norminv(1-beta))^2 /Delta^2

Source: http://zoe.bme.gatech.edu/~bv20/bmestat/Bank/midtermkey.pdf