Transcript Session Slides/Handout

Biostatistics in Practice
Session 6: Case Study
Peter D. Christenson
Biostatistician
http://gcrc.humc.edu/Biostat
Case Study
Hall S et al: A comparative study of Carvedilol,
slow release Nifedipine, and Atenolol in the
management of essential hypertension.
J of Cardiovascular Pharmacology 1991;18(4)S35-38.
Data is available at the class website:
http://gcrc.humc.edu/Biostat
Select Courses > Biostatistics in Practice 2004 >
Session 6 > Download Data
Case Study Outline
Subjects randomized to one of 3 drugs for controlling
hypertension:
A: Carvedilol (new)
B: Nifedipine (standard)
C: Atenolol (standard)
Blood pressure and HR measured at baseline and 5 posttreatment periods.
Primary analysis ?
“The present study compares … A, B, and C for the
management of … hypertension.”
Data Collected for Sitting dbp
Number of Subjects
Visit #
Week
Baseline
1
-1
Acute*
2
0
100
93
95
Post 1
3
2
100
93
94
Post 2
4
4
94
91
94
Post 3
5
6
87
88
93
Post 4
6
12
83
84
91
A
B
C
311 total
* 1 hour after 1st dose. We do not have data for this visit.
Sitting dbp from Figure 2 of the Paper
d BP
103
Baseline
A: Carvedilol
B: Nifedipine
C: Atenolol
102
101
100
99
98
97
96
95
94
A
93
92
91
2 Weeks
B
90
C
89
88
- 1
0
1
2
3
4
5
6
7
8
We e k
t r eat
A
B
C
9
10
11
12
Question #1
Describe dbp at baseline for the study population.
Give an appropriate graphical display, and
summarize dbp with just a few numbers.
Is the mean appropriate? Would the median be
better? Is a transformation necessary?
Answer #1
Histogram of dbp1
100.0
Count
75.0
50.0
25.0
0.0
90.0
97.5
105.0
112.5
120.0
N
= 255
Mean = 102.68
SD = 4.63
SEM = 0.29
Min = 92
Max =117
dbp1
Median = 102. Log-transformation gives geometric
mean = 102.58.
No transformation is necessary. Mean is best.
95% of subjects between ~ 102.68 ± 2(4.63) = 93.42 to
111.94
Question #2
It appears that group B may have had lower dbp at
baseline than group A, on the average.
Is there evidence for this? Is the lower group B
mean dbp lower (relative to A) than expected by
chance?
Write out a formal test for this question, and use
software to perform the test.
Answer #2, Part 1
Box Plot
120.0
Drug Mean ± SD
102.9 ± 4.8
B
102.2 ± 4.3
C
103.0 ± 4.8
110.0
dbp1
A
100.0
90.0
A
B
C
treat
So, the mean for B is low, as in the earlier figure, but the
overall distribution is similar to that for A and C, so this
is entirely due to chance, but we will formally test B vs.
A on the next slide. [Would use ANOVA to include C.]
Answer #2, Part 2
We are formally testing, where μx represents the mean
baseline dbp among those who eventually receive
treatment x:
H0: μA = μB vs. HA: μA ≠ μB
Since μA – μB is estimated by 0.75 with a SE of 0.71 , tc =
0.75/0.71 = 1.05 is not larger (~ >2) than expected by
random fluctuation (p=0.29), so there is not sufficient
evidence that the A and B groups differed in their
baseline dbp.
Note that we do not expect A and B to differ at baseline
due to the randomization in the study design.
Question #3
How much can a patient’s dbp be expected to be
lowered after 2 weeks of therapy with A?
We are 95% sure that this lowering will be between
what two values?
Repeat for drug C.
Do the intervals for A and for C overlap considerably?
Can this overlapping be used to compare A and C
in their dbp lowering ability?
Answer #3
How much can a patient’s dbp be expected to be
lowered after 2 weeks of therapy with A? with C?
We are 95% sure that this lowering will be between
what two values?
Ans:
Drug Estimated Δ
~95% Prediction Interval
A
8.13
8.13 ± 2*9.1 = -10.1 to 26.3
C
11.5
11.5 ± 2*8.7 = - 5.9 to 28.9
The intervals for A and for C do overlap considerably.
However, to compare A and C, we need to examine
not these expected intervals for individuals, but
rather the precision of ΔC – ΔA estimated from this
study, which incorporates the Ns.
Question #4
Is there evidence that A and C differ in their dbp
lowering ability at 2 weeks post-therapy?
Formally test for this.
Give a 95% confidence interval for the C-A difference
in change in dbp after 2 weeks.
Answer #4
Is there evidence that A and C differ in their dbp
lowering ability at 2 weeks post-therapy?
Ans:
Test H0: ΔA-ΔC = 0 vs. HA: ΔA-ΔC ≠ 0 with t-test:
Estimate ΔA-ΔC with 3.39, with SE of 1.36.
Since tc = 3.39/1.36 = 2.50 exceeds ~2, choose HA.
95% CI for ΔA-ΔC is 3.39±2*1.36 = 0.67 to 6.11, which
does not include 0, so choose HA.
Question #5
Is there evidence that B and A differ in their dbp
lowering ability at 2 weeks post-therapy?
We want to examine whether the study was large
enough to detect a difference in 2 week changes in
dbp between B and A. To do so, we need the SD of
these changes among subjects receiving B and
among subjects receiving A. Find these SDs.
Answer #5
Is there evidence that B and C differ in their dbp
lowering ability at 2 weeks post-therapy?
Ans:
Test H0: ΔB-ΔA = 0 vs. HA: ΔB-ΔA ≠ 0 with t-test:
Estimate ΔB-ΔA with 0.96, with SE of 1.35.
Since tc = 0.96/1.35 = 0.71 < ~2, choose H0 (p=0.48).
SD for B is 8.29 and SD for A is 9.08.
Question #6
Estimate the true minimal difference in 2 week
changes in dbp between B and C that this study was
able to detect.
1. Use the conventional risks of making incorrect
conclusions that the FDA typically requires.
2. Set both risks of an incorrect conclusion at ≤5%.
Typical Statistical Power Software
Answer #6
1. Use the conventional risks of making incorrect
conclusions that the FDA typically requires.
Use α=0.05, power=0.80, NA=83, NB=82, SDA=9.08,
SDB=8.29. Find Δ from a power calculation to be
3.8.
1. Set both risks of an incorrect conclusion at ≤5%.
Use α=0.05, power=0.95, NA=83, NB=82, SDA=9.08,
SDB=8.29. Find Δ from a power calculation to be
4.9.
Question #7
Suppose that differences in 2 week changes in dbp
between B and C of <2 mmHg is clinically irrelevant,
but we would like to detect larger differences with
80% certainty. How large should such a study be?
Answer #7
Suppose that differences in 2 week changes in dbp
between B and C of <2 mmHg is clinically irrelevant,
but we would like to detect larger differences with
80% certainty. How large should such a study be?
Ans:
Use α=0.05, power=0.80, SDA=9.08, SDB=8.29, Δ=2.
From a power calculation , NA = NB = 297.