TEVA Pharmaceuticals USA, INC. of North Wales, PA has published the results of v
ID: 3364821 • Letter: T
Question
TEVA Pharmaceuticals USA, INC. of North Wales, PA has published the results of various clinical trials of their drug Azithromycin (also called Z-Pack) against other drugs for various age groups and for various bacterial agents. Here are a few of the test results from these trials
Case A For sinus infection in adults:
Azithromycin, 500 mg daily for 3 days, versus
Amoxicillin/Clavulante, 500/125 mg 3 times daily for 10 days.
At day 28, subsequent to the medicines start, the following results were observed: For patients on (i) 213 out of 298 people were “cured” of their infection; For patients on (ii) 206 out of 288 people were “cured” of their infection.
Based on a computed confidence interval (use 1-a = 97.5%) could you conclude that there is a statistical difference in the performance of these two treatments? Yes, or no, and why?
Please show detailed step by step work! Thank you!
Explanation / Answer
TRADITIONAL METHOD
given that,
sample one, x1 =213, n1 =298, p1= x1/n1=0.7148
sample two, x2 =206, n2 =288, p2= x2/n2=0.7153
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.7148*0.2852/298) +(0.7153 * 0.2847/288))
=0.0373
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.025
from standard normal table, two tailed z /2 =2.24
margin of error = 2.24 * 0.0373
=0.0836
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.7148-0.7153) ±0.0836]
= [ -0.0841 , 0.083]
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DIRECT METHOD
given that,
sample one, x1 =213, n1 =298, p1= x1/n1=0.7148
sample two, x2 =206, n2 =288, p2= x2/n2=0.7153
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.7148-0.7153) ± 2.24 * 0.0373]
= [ -0.0841 , 0.083 ]
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interpretations:
1) we are 97.5% sure that the interval [ -0.0841 , 0.083] contains the difference between
true population proportion P1-P2
2) if a large number of samples are collected, and a confidence interval is created
for each sample, 97.5% of these intervals will contains the difference between
true population mean P1-P2
since value of zero lies in the confidence calculated, we assume that there is no statistical difference
in the performance of these two treatments
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