Homework1Key

Homework 1 Key BIOSTAT 522 1-A. If a 99% confidence interval for  1 -  2 is (2.5, 3.5), which of the following conclusion(s) can be drawn based on this interval? a. Reject H 0 :  1 =  2 at = 0.01 if the alternative is H α A :  1 ≠  2. b. Reject H 0 :  1 =  2 at = 0.01 if the alternative is H α A :  1 <  2. c. Do not reject H 0 :  1 =  2 at = 0.05 if the alternative is H α A :  1 ≠  2. d. Do not reject H 0 :  1 =  2 at = 0.01 if the alternative is H α A :  1 ≠  2. 1-B Suppose that ¯ X = 125.2 and ¯ X 2 = 125.4 are the mean systolic blood pressure for two samples of workers from different plants in the same industry. Suppose, further, that a test of H 0 :  1 =  2 using these samples is rejected at = 0.01. Which of the following α conclusions is most reasonable? a. There is a (clinically) meaningful difference in population means, but not a statistically significant difference. b. The difference in population mean s is both statistically and clinically meaningfully significant. c. There is a statistically significant difference, but not a clinically meaningful difference in population means. d. The sample sizes used must have been quite small. e. There is neither a statistically significant nor a clinically meaningful difference in population means. 1-C. Suppose that Y i is the random variable representing weight, measured in pounds , and that Y i ~ N( 150 , 10 ). Note that the notation is such that N( μ Y , σ Y ). (a) What is the probability that an individual weighs at most 120 pounds? P(Y i ≤ 120) = P(z ≤ -3) = 0.0013 (b) What is the median weight, in kg ? Note that 1 pound equals 0.45 kg. 150 pounds= 67.5 kg (c) What is the probability that an individual weighs more than 80 kg? P(Y i ≥ 80kg) = P(Y i ≥ 177.78) = P(z ≥ 2.78) = 0.0028 1

2-A. Obtain a 95% and 98% confidence interval for mean age of the women who smoke during pregnancy? 2-B. Carry out a hypothesis test for mean Age = 40 . Write the null and alternative hypotheses, find the appropriate test and the test statistic, degrees of freedom and p-value, and draw a conclusion. H 0 : Age = 40 H a : Age ≠ 40 t-statistic = -10.74 df = 19 p-value < 0.0001 Conclusion: We reject the null 2

3-B . [Bivariate Analysis: Graphical] The goal here is to assess if anxiety is predictive of pain . Assess graphically the relationship by first selecting the appropriate Y and X variable. Describe in one sentence what you saw from the graph and submit the graph. 3-C . [Bivariate Analysis: Model] Fit a regression model corresponding to the study goal. Write the model found and interpret estimates for the intercept ( β 0 ) and slope ( β 1 ). Model: ^ Pain i = 2.696 + 0.485 × Anxiet y i Interpretations: Intercept : The estimated average pain score is 2.696 when the anxiety score is 0. Based on the following scatter plot, there seems to be a weak/moderate, positive, linear relationship between pain and anxiety. 4

Slope : For every one-unit increase in anxiety score, the average pain score will increase by 0.485. 3-D . Find the N and % for the following questions. How many non-missing pain observations do you have in this dataset? How about those with non-missing anxiety data? And how many patients provided data for the plot described in problem # 3-B? Lastly, how many observations are used in the regression model in problem #3-C? Non-missing PAIN data: N = 206 (78.9%) Non-missing ANXIETY data: N = 202 (77.4%) Observations used in SCATTER PLOT and REGRESSION MODEL: N = 199 (76.2%) SAS Code proc import datafile = "C:\Users\amwallpa\Downloads\smoke.csv" 5

RUN ; **3c); PROC REG data = tmp; model pain = anxiety / clb ; title ; RUN ; R Code # reading in the data smoke = read.csv("~/Downloads/smoke.csv", header=T) #95 % CI for mean age mean(smoke$A) - qt(1-0.05/2, 19)*sd(smoke$A)/sqrt(20) mean(smoke$A) + qt(1-0.05/2, 19)*sd(smoke$A)/sqrt(20) # 98% CI for mean age mean(smoke$A) - qt(1-0.02/2, 19)*sd(smoke$A)/sqrt(20) mean(smoke$A) + qt(1-0.02/2, 19)*sd(smoke$A)/sqrt(20) # t-test t.test(smoke$A, mu=40) # reading in the data surgery = read.csv(‘~/Downloads/surgery.txt", header=T) # Question 3a ## PAIN ## sum(complete.cases(surgery$pain)) # N mean(surgery$pain, na.rm=TRUE) # mean sd(surgery$pain, na.rm=TRUE) # sd # Histogram with normal curve x=surgery$pain[complete.cases(surgery$pain)] h=hist(x, col = "gray", right=FALSE, xlab="res", main="Histogram for Pain") xfit = seq(min(x),max(x),length=40) yfit = dnorm(xfit, mean=mean(x) ,sd=sd(x)) yfit = yfit*diff(h$mids[1:2])*length(x) lines(xfit, yfit, col="blue", lwd=2) 7

## ANXIETY ## sum(complete.cases(surgery$anxeity)) # N mean(surgery$anxiety, na.rm=TRUE) # mean sd(surgery$anxiety, na.rm=TRUE) # sd # Histogram with normal curve x=surgery$anxiety[complete.cases(surgery$anxiety)] h=hist(x, col = "gray", right=FALSE, xlab="res", main="Histogram for Anxiety") xfit = seq(min(x),max(x),length=40) yfit = dnorm(xfit, mean=mean(x) ,sd=sd(x)) yfit = yfit*diff(h$mids[1:2])*length(x) lines(xfit, yfit, col="blue", lwd=2) # Question 3b # Scatterplot of Pain vs. Anxiety plot(x=surgery$anxiety, y=surgery$pain, main = 'Scatterplot of Pain vs. Anxiety', xlab='Anxiety', ylab='Pain') # Question 3c # Model Y=pain and X=anxiety summary(lm(pain ~ anxiety, data=surgery)) # Question 3d # Number non-missing pain data sum(complete.cases(surgery$pain)) # Number non-missing anxiety data sum(complete.cases(surgery$anxiety)) # Obs used in scatter plot and regression model sum(complete.cases(cbind(surgery$pain, surgery$anxiety))) 8