Statistics 101. Why statistics ?. To understand studies in clinical journals. To design and analyze clinical research studies. To be better able to explain epidemiologic research to patients. To answer questions on board examinations. Types of Clinical Research Studies.
AST, CK, glucose, etc
OLD vs YOUNG
SD= ((differences from the mean2 )/n-1)
14.5 = 3.8
SD = 3.8
Mean, 140; median 140; range, 135-145 mM; standard deviation2
Standard deviations (SD) from the mean.
95% of values are within 1.96 SD of mean
Ann Int Med 2009: 150: JC6-16
95% Confidence interval (CI): Example 1
H. pylori eradication/NSAID study with outcomeof ulcer or no ulcer (categorical outcome):
5 of 51 (10%, or .10) Hp+ pts. who received antibiotics got ulcers when exposed to NSAID.
… and 15 of 49 (31%, or .31) Hp+ pts. who did not receive antibiotics got ulcers when exposed to NSAID.
What is the chance this difference in outcome occurred due to chance and not the antibiotics?
Lancet 2002; 359:9-13.
The proportions, p1 and p2, of patients who got ulcers in the 2 groups are an estimate of the true rate. However, from this estimate we can be 95% confident that the actual rates ranges from A to B, with p1 and p2 in the center of the interval from A to B. A and B are the 95% confidence intervals.
A→B is t h e 9 5 % c o n f i d e n c e i n t e r v a l
To calculate the 95% CI for p (i.e., A and B), use this formula:
p ± 1.96 [(p)(1-p)/n]
The larger the n, which is in the denominator, the smaller (more precise) the CI
5 of 51 (p1=10%, or .10) of the antibiotic group got ulcers when exposed to NSAID for a fixed time
15 of 49 (p2=31%, or .31) of the placebo- group got ulcers when exposed to NSAID for a fixed time
Note: the two 95% CIs do not overlap, which means that differences
are unlikely to be due to chance. But is the ARR significant?
21% 1.96 (p1)(1-p1)/n1+(p2)(1-p2)/n2)= 21% 15%, or [6%, 36%].
Example : A new protease inhibitor is tested in chronic hepatitis C, genotype 1. The new therapy (added to the standard therapy, interferon alpha/ribavirin) or standard therapy is randomly given to 200 patients for 48 weeks. Sustained viral response rates were as follows:
What is the N needed to treat to achieve 1 additional SVR?
(SVR, NEW / # NEW) – (SVR,CONTROL / # CONTROL)
(83/99) –( 50/101)
Note that the denominator , .343 (34.3%) , is the absolute risk reduction ( ARR).
95% CI of ARR = 0.222 to 0.465.
95% CI of NNT = 2.2 to 4.5.
VTE or no VTE (categorical outcome)
14 of 255 (p1=5.5%, or .055) patients with VTE switched to low-intensity warfarin developed another VTE
… and 37 of 253 (p2=14.6%, or .146) switched to placebo developed another VTE
Is this 9.1% difference in VTE likely to be due to chance?
New Engl. J. Med. 2003; 348: 1425-1434
(used for categorical outcomes)
New Rx a b a+b
Standard Rx cd c+d
a + c b + d a+b+c+d=n=total patients in study
New Rx: 90(a) 30(b) 120(a+b)
Standard Rx: 75(c) 50(d) 125(c+d)
165 80 245(a+b+c+d)=n
2 = n (ad-bc- n/2)2
2 = 6.264 (p=0.0123)
Fisher exact test, p=0.0143
New Rx a=90b=30
Standard Rxc= 75d=50
odds ratio = ad/bc
odds ratio = 4,500/ 2,250= 2
But this odds ratio of 2 could have occurred by chance.
We can calculate the 95% CI of the odds ratio to see if the CI overlaps 1 or not. If not, it favors the new treatment with >95% confidence.