Drug Information Practical Revision PDF

Drug Information Practical Revision In the following table Smokers Non-Smokers Total Cancer Patients 50 40 85 Non-cancer Patients 35 60 100 Total 90...

Drug Information Practical Revision In the following table Smokers Non-Smokers Total Cancer Patients 50 40 85 Non-cancer Patients 35 60 100 Total 90 95 Calculate Odds Ratio: a. (50/35)/(40/60). b. (35/40)/(60/50). c. (35/60)/(50/40). d. (50/40)/(35/60) A study was made on 10,000 drug abusers 1000 of them had Hepatitis C and 200 of them had Hepatitis B. During the follow- up 200 of those who had Hepatitis C and 50 of those who had Hepatitis B died. After the duration of the follow-up which extended for 2 years, 200 new patients were infected with Hepatitis C. The incidence of Hepatitis C in One year was: a. 1000/10,000. b. 200/10,000. c. 100/10,000. d. 100/9000. A study that investigates the prevalence of S.aureus and E.coli in two groups (Breastfed subjects and artificially fed subjects). The most appropriate statistical test to use here is. a. Friedman’s test. b. Unpaired t-test. c. Chi square test. d. None of the above. A study that investigates the LDL levels between two groups, one takes Drug A and the other one takes drug B. The most appropriate statistical test to use here is. a. Paired t-test. b. Unpaired t-test. c. Kruskal-Wallis test. d. Multifactorial ANOVA. If Drug B was offered as a replacement of an existing Drug A and Drug B was more expensive and less effective then it will be: a. Refused because Drug B is dominated. b. Accepted because Drug B is dominant. c. There is no difference between the two drugs. d. None of the above. The following table describes the frequencies of the subjects who were & were not affected with the avian flu, the frequencies of those who were & were not dealing with any type of avian. Avian Flu (Cases) No Avian Flu (Control) Dealing with any avian 40 70 Not dealing with any avian 35 130 a. Calculate Odds Ratio. b. Calculate Relative Risk. c. If you were told that there is a new laboratory test that can detect avian flu and the above results were assessed using this new indicator. Calculate the sensitivity and specificity of this indicator. The following table describes the frequencies of the subjects who were & were not affected with the avian flu, the frequencies of those who were & were not dealing with any type of avian. Avian Flu (Cases) No Avian Flu (Control) Dealing with any avian 40 70 Not dealing with any avian 35 130 a. Calculate Odds Ratio. Odds Ratio= A/B / C/D= 40/35 / 70/130= 2.122. The following table describes the frequencies of the subjects who were & were not affected with the avian flu, the frequencies of those who were & were not dealing with any type of avian. Avian Flu (Cases) No Avian Flu (Control) Dealing with any avian 40 70 Not dealing with any avian 35 130 b. Calculate Relative Risk. Relative Risk= A/A+B / C/C+D= 40/(40+70) / 35/(35+130)= 1.714. The following table describes the frequencies of the subjects who were & were not affected with the avian flu, the frequencies of those who were & were not dealing with any type of avian. Avian Flu (Cases) No Avian Flu (Control) Dealing with any avian 40 70 Not dealing with any avian 35 130 c. If you were told that there is a new laboratory test that can detect avian flu and the above results were assessed using this new indicator. Calculate the sensitivity and specificity of this indicator. Sensitivity= a/(a+b)= 40/(40+35)=0.533. Specificity= d/(c+d)= 130/(70+130)= 0.65. Write down the name of the suitable statistical test for analysis: 1. “Gender Differences in Labetalol Kinetics: Importance of Determining Stereoisomer Kinetics for Racemic Drugs” compared the dose corrected AUC (AUC/dose) in women (n=5) and men (n=14). Null Hypothesis: “There is no difference in labetalol AUC/dose in men and women.” Unpaired t-test Write down the name of the suitable statistical test for analysis: 2. Ondansetron tablets and suppositories were administered to eight men and eight women. Cmax was determined in both groups. Null Hypothesis: There is no difference between Cmax of ondansetron preparations in men and women. Multifactorial ANOVA Write down the name of the suitable statistical test for analysis: 3. St. John’s Wort (250 mg hypericum extract twice daily) was compared to imipramine (75 mg twice daily)in a randomized, multi-center, double-blind, parallel group trial.(zh Woelk: Comparison of St. John’s Wort and imipramine for treating depression: Randomized controlled trial, BMJ; 321: 536-539, 2000) Adverse events per treatment group were counted for the duration of the study. Null Hypothesis: “Adverse events are the same after hypericum as after imipramine”. Chi square test IV. What is the expected value for antibiotic A? Cure 100 No adverse 0.9 reaction 0.8 Failure 200 0.1 Antibiotic A Cure 0.8 150 Adverse reaction Antibiotic B (+) 0.2 Failure 0.2 250 No adverse reaction= (100×0.9) + (200×0.1)= 110. Adverse reaction= (150×0.8) + (250×0.2)= 170. For Antibiotic A= (110×0.8) + (170×0.2)= 122. It has been estimated that approximately 182,000 new cases of breast cancer will be diagnosed in the United States in 2000. Estimate the incidence of the disease expressed as number of new cases per million citizens given that the US population in 2000 was 282 million citizens. › Incidence= Number of new patients with disease/Total population at risk. › Prevalence= Number of patients with disease/Total population. Incidence= 182,000 ÷ 282= 645.39 cases/million citizens. Given an approximately normal distribution, what percentage of all values are: a) above the mean? b) below the mean? c) within one standard deviation from the mean? d) within two standard deviations from the mean? e) within three standard deviations from the mean? a) Above the mean: 50%. b) Below the mean: 50%. c) Within 1 SD: 68%. d) Within 2 SD: 95%. e) Within 3 SD: 99% The time, in hours, that each student spent sleeping on a school night was recorded for 550 secondary school students. The distribution of these times was found to be approximately normal with a mean of 7.6 hours and a standard deviation of 0.5 hours. a) Draw the normal distribution curve to show this data. b) What amount of time did 95% of these students spend sleeping? c) What percentage of students spent between 7.1 and 8.1 hours sleeping? The time, in hours, that each student spent sleeping on a school night was recorded for 550 secondary school students. The distribution of these times was found to be approximately normal with a mean of 7.6 hours and a standard deviation of 0.5 hours. a) Draw the normal distribution curve to show this data. The time, in hours, that each student spent sleeping on a school night was recorded for 550 secondary school students. The distribution of these times was found to be approximately normal with a mean of 7.6 hours and a standard deviation of 0.5 hours. b) What amount of time did 95% of these students spend sleeping? 95% of the students are between μ − 2σ → μ + 2σ = 6.6 → 8.6 hrs. The time, in hours, that each student spent sleeping on a school night was recorded for 550 secondary school students. The distribution of these times was found to be approximately normal with a mean of 7.6 hours and a standard deviation of 0.5 hours. c) What percentage of students spent between 7.1 and 8.1 hours sleeping? 7.1 → 8.1 hrs= μ − σ → μ + σ= 68%