EMBAnswerforstudentPowerCI2011April21

- Epi/Biostats/Medical Informatics Confidence Interval and Power(Answer Key 10:00 – 11:50am Lin Liu PhD [email protected](858 552-8585 ext 2841

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Unformatted text preview: Epi/Biostats/Medical Informatics Confidence Interval and Power (Answer Key) April 21, 2011 10:00 – 11:50am Lin Liu, PhD [email protected] (858) 552-8585 ext 2841 Richard Garfein, PhD, MPH [email protected] (858) 822-3018 Learning Objectives: • Understand the interactions between power, sample size, magnitude of difference, variability of the outcome, and significance level. • Know how to state study questions in terms of null and alternate hypotheses. • Know how to interpret p-values and confidence intervals. • Know the definition of a type I and type II error. I. Hypothesis Testing and Power Analysis Question 1: Becker et al. (2008) compared the lipid-lowering effects of an ‘alternative’ dietary regimen (therapeutic lifestyle change, read yeast rice and fish oil) to a standard dose of a cholesterol-lowering agent (Simvastatin, 40 mg/d) with traditional diet and exercise counseling. The primary end point was the percentage change in low- density lipoprotein cholesterol (LDL-C). What were the null and alternative hypotheses that were tested for the primary end point in the article? Was this a two- sided or one-sided test? Answer: μ 1 : mean percentage change in LDL-C in alternative treatment group μ 2 : mean percentage change in LCL-C in Simvastatin treatment group H : μ 1 = μ 2 H A : μ 1 ≠ μ 2 It was a two-sided test. Question 2: In Table 4 (Becker et. al. 2008), we can see that the percentage changes in LDL-C were -42.4% ( x 1 ) and -39.6% ( x 2 ), and estimated standard deviations were 14.8% ( s 1 ) and 20.2% ( s 2 ) in alternative and Simvastatin treatment group, respectively. There are 37 ( n 1 =n 2 =37) subjects in each treatment group. Assume the percentage changes in LDL-C in both treatment groups are normally distributed with equal variances (or standard deviations), we can use two-sample t-test to study the hypotheses in Question 1. Estimated common standard deviation (pooled standard deviation) is s p = ( n 1- 1) s 1 2 + ( n 2- 1) s 2 2 ( n 1 + n 2- 2) = 36 * (14.8 2 + 20.2 2 ) (37 + 37- 2) = 17.7 a) Write out the test statistic t for t-test. b) The two-sided p-value is twice the probability of t >| t | (| t | is the absolute value of t ). From the above test statistic, we obtain the two-sided p-value of 0.50, what does that mean? What are you going to conclude at a significance level of 0.05? Answer: a) t = ( x 1- x 2 ) s p 1 n 1 + 1 n 2 = (- 42.4- (- 39.6)) 17.7 * 2 37 = - 0.68 b) The p-value of 0.50 means the probability of observing the results that we observed in our study when the null hypothesis (there is no difference in mean percentage change in LDL between two treatment groups) is true is 0.50. We will not reject the null hypothesis at a significance level of 0.05....
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This note was uploaded on 09/14/2011 for the course PHARM ERM taught by Professor Staff during the Spring '11 term at UCSD.

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- Epi/Biostats/Medical Informatics Confidence Interval and Power(Answer Key 10:00 – 11:50am Lin Liu PhD [email protected](858 552-8585 ext 2841

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