Problem 1
Determine if each of the following satements is true or false.
(a) False, we can not conclude that the rate of cardiovascular problems for those on pioglitazone treatment is higher than those on rosiglitazone based on the count, we must compare proportions. Additionally, this observed difference may be due to chance, thus we can not make any conclusions.
(b) True, the data does suggest that this might be the case.
(c) False, correlation/association in an observation based study gives no ground for causal claims!
(d) True!
Problem 2
(a) 88.23% of patients in the Control group and 65.22% of patients in the Treatment group died.
(b.i.)
Null Hypothesis: The variables Group and Outcome are independent. They have no relationship, and the observed difference between the proportion of those in the control group and those in the treatment group that died, 23.01%, was due to chance.
Alternative Hypothesis: The variables Group and Outcome are not independent. The difference in deate rates of 23.01% was not due to chance, and those in the control group are more likely to die than those in the treatment group.
(b.ii.) 28, 75, 69, 34, 0/zero, as or more extreme.
(b.iii.) The distribution of simulated differences in proportions suggests that the observed difference between control and treatment groups of 23.01% is rare and unlikely to occur just by chance as getting a difference of 23.01% is likely to happen less than 5% of the time.
Problem 3
(a) A RCT (Randomized Controlled Trial)
(b) Yes. The participants are blind to which condition they are in, however it is not a double blind as the description does not tell whether the researchers were aware of who was receiving what treatment.
(c) 77.65% - 80.25% = -2.6%
(d) At first glance neither treatment appears to be more effective for the treatment of sinusitis. If the Null Hypothesis is that there is no difference in improvment between treatment conditions, phat antibiotic - phat placobo = 0, then an observed difference in proportions of -2.6% is too close to 0/zero to reject the Null that there is no difference.
(e)
Null Hypothesis: There is no difference in the improvement in symptoms for those receiving the antibiotic or the placebo treatments.
Alternative Hypotheis Those who receive the antibiotic improve more in their symptoms than those who receive the placebo.
(f) It is very likely and usual that 66 patients taking the antibiotic would self-report an improvment in symptoms
Problem 4
(a)
Null Hypothesis: Having someone yawn near by another person makes no difference in the number of people who yawn.
Alternative Hypothesis: Having someone yawn near by another person makes that person yawn more than if no one near them was yawning.
(b) Treatment 29.41% - Control 25% = Observed difference 4.41%
(c) One might estimate the p-value to be about 0.45, which is greater than 0.05 and would lead us to fail to reject the Null Hypothesis that there is no difference.
Problem 5
(a)
Null Hypothesis: The proportion of high school students that who followed the news about Egypt = 0.69.
Alternative Hypothesis: The proportion of high school students that who followed the news about Egypt is NOT = to 0.69.
(b) phat = 17/30 = 0.567 or 56.7%
(c) To perform a simulation in this context you might take 100 playing cards, where you have 69 black suited cards that will represent individuals that follow the news about Egypt and then 31 red suited cards that will represent the individuals who do not. We will practice sampling with replacment, where we will shuffle the 100 cards and randomly choose 30 cards at a time to represent the high school students, and then return those cards to the deck. We would then calculate the proportion of black cards in the sample of 30 cards to get a sample proportion of those who follow the news about Egypt. You might repeat this about 1,000 times or more times and plot each sample proportion on a distribution of proportions. Because this is a two-tailed hypothesis test, to estimate the p-value you look at the percentage of proportions equal to or greater than 0.57 and multiply that by two.
(d) 0.075 + 0.035 + 0.02 + 0.005 = 0.135*2 = 0.27.
Because the estimated p-value is greater than 0.05, we fail to reject the Null Hpothesis that there is no difference between the proportions of high school students and adults following the news in Egypt.
(e)
nsim<-10000
nullDist<-rbinom(nsim, 30, .69)
hist(nullDist)
abline(v=17, col="red", lwd=2, lty=2)
head(nullDist)
## [1] 25 19 21 19 22 20
head(nullDist<=17)
## [1] FALSE FALSE FALSE FALSE FALSE FALSE
p<-mean(nullDist<=17)
# Our P-Value:
p*2
## [1] 0.2106
Our P-Value is greater than 0.05 so we fail to reject the null that there is no difference between proportions of high school students and adults following the news in Egypt.