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Short Course: Biostatistics in Practice Fall 2014
HW #3 (total 46 pts, Due by Oct 21)
Last Name: ____________________
First Name: _____________________
1. CIRCLE T (true) or F (false). If you answer is “F” then explain WHY it is a false statement and provide the
CORRECT statement. Each question is worth 2 points ( total 10pts).
T
F Type I error is also called as the significance level of test
T
F The p-value represent how rare the observed statistics when the H0 is true.
T
F If we have a two-sided Ha then one-sided p-value should be calculated.
T
F If p-value < α, then we reject H0 at α significance level.
T
F From two sample independent test comparing two groups, suppose p-value=0.01.
Then this means that 1 % of experiments like this would yield a difference of at least as large as
you observed.
2. Suppose a mortality of 20% for a heart operation in children less than one year old. Suppose we have 10
babies to be operated. Answer the following questions (16 points)
(a) Suppose the results of a heart operation only take either success or failure, do you think each the
results of a heart operation follows the Bernoulli distribution? Why or Why not. Provide your
reasoning. (4pts).
The result of a heart operation follows the Bernoulli distribution since the outcome can be either
success or failure.
(b) Suppose a random variable, denoted as X, is defined as the number of success for 10 heart surgery, do
you think X follows binomial distribution? Why or Why not. Provide your reasoning.(4pts)
X follows the binomial since X is the total number of success from independent Bernoulli trials
(c) What is the probability of 8 out of 10 will be survived through this heart operation. Provide the solution as
well as the answer. You will need to set up the proper the equation using formula in the lecture note 5 and
solve your equation to get the answer. (4pts)
For each Bernoulli trial, we have the success rate of 0.8
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Thus, if X = the total number of success, then X ~ B( 10, 0.8)
Then,
P(X=8)= n !/ [x!(n-x)!] pX(1-p)n-X
= 10 !/ [8!(10-8)!] 0.8 X (1-0.8)10-8
≈ 0.30
(d) What is the probability of at least 8 out of 10 will be survived through this heart operation. Provide the
solution as well as the answer. You will need to set up the proper the equation using formula in the lecture
note 5 and solve your equation to get the answer. (4pts)
P(at least 8 out of 10 will be survived)=P(X ≥ 8) =P(X=8) + P(X=9) + P(X=10)
Then we already have P(X=8)=0.30 from (c)
P(X=9)= n !/ [x!(n-x)!] pX(1-p)n-X
= 10 !/[9! (10-9)!] (0.8)9 (1-0.8) 10-9
≈ 0.27
P(X=10)= n !/ [x!(n-x)!] pX(1-p)n-X
= 10 !/[10! (10-10)!] (0.8)10 (1-0.8) 10-10
≈ 0.11
Thus, P(at least 8 out of 10 will be survived)= P(X ≥ 8)
= P(X=8) + P(X=9) + P(X=10)
≈ 0.30+ 0.27 + 0.11
= 0.67
3. A randomized controlled trial was conducted to investigate the cost effectiveness of community leg ulcer
clinics (Morrell et al 1998). A total of 233 patients were randomly allocated to either intervention (120
patients, treatment at a leg ulcer clinic) or control (113, usual care at home by district nursing service) . At the
end of 12 months the mean time (in weeks) that each patient was free from ulcers during follow up was 20.1
and 14.2 in the clinic and control groups, respectively. On average, patients in the clinic group had 5.9 more
ulcer-free weeks (95% CI: 1.2 to 10.6 weeks) than the control patients. Mean total costs were ₤ 878 per year
for the clinic group and ₤863 per year for the control group ( p = 0.89). (Total 20 points)
(a) What is the statistical null hypothesis for this study? What is the alternative hypothesis? ( 4 pts)
This study is to evaluate whether or not a new intervention bring a better outcome. The comparison was
conducted between two groups (treatment, control) and the outcome measure was the number of ulcer
free weeks. Thus the alternative hypothesis is the number of ulcer free weeks, on average, of patients in
the treatment group is different from the ones in the control group. Thus, H0 and Ha are as follows.
H0: . H0: µT= µC (or µT -µC =0) vs. Ha: µT ≠ µC (or µT -µC ≠ 0)
where µT=the population mean number of ulcer free weeks in the treatment group
µC = the population mean number of ulcer free weeks in the control group
(b) What can we conclude from the 95% CI? (4pts)
IF we interpret the 95% CI, that means that with 95% confidence, the true mean difference in the
number of ulcer free weeks is at least 1.2 weeks and at most 10.6 weeks, where the positive difference is
higher ulcer free weeks for the intervention group. Therefore, the true difference is at least great than
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zero and that means that there is a significant difference in mean number of ulcer free weeks between
two groups , thus, new intervention increase the number of ulcer free weeks. on average.
(c) Is there a statistically significant difference between the two groups with respect to the number of ulcerfree week? (4pts)
Since the 95% CI does not include zero, we can conclude that there is a significant difference between
two group.
(d) Is there a statistically significant difference between the two groups with respect to the cost of treating
the patients over the 12 months period? Would you expect the confidence interval of this difference to
include the value of ‘no difference’? (4pts)
Mean costs were ₤878 per year for the treatment group and ₤863 for the control group and the pvalue for the difference was 0.89, which is greater than 0.05. Thus our conclusion would be there is no
statistically significant evidence to support that the cost of treatment is different from the cost of control.
(e) What would you conclude from the information above? (4pts)
From the information above, it would be reasonable to conclude that community based leg ulcer
clinics are more effective than traditional home-based treatment, in terms of the number of ulcer-free
weeks. This benefit is achieved at a marginal additional cost. If we divide the mean difference in costs
between the two groups by the mean difference between the groups in ulcer-free weeks we get
something called the incremental cost-effectiveness ratio. This gives ₤2.50 to achieve and extra-ulcer
free week.