Case B For sore throat due to strep infection in children: Azithromycin, 12 mg/k

Question

Case B

For sore throat due to strep infection in children:

Azithromycin, 12 mg/kg once daily for 3 days, versus

Penicillin V, 250 mg 3 times per day for 10 days

At day 30, subsequent to the medicines start, the following results were observed:

For patients on (i) 255 out of 330 children were “cured” of their infection;

For patients on (ii) 206 out of 325 children 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?

Explanation / Answer

TRADITIONAL METHOD
given that,
sample one, x1 =255, n1 =330, p1= x1/n1=0.7727
sample two, x2 =206, n2 =325, p2= x2/n2=0.6338
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( (0.7727*0.2273/330) +(0.6338 * 0.3662/325))
=0.0353
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 * 0.0353
=0.0791
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.7727-0.6338) ±0.0791]
= [ 0.0598 , 0.218]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample one, x1 =255, n1 =330, p1= x1/n1=0.7727
sample two, x2 =206, n2 =325, p2= x2/n2=0.6338
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 = [ (0.7727-0.6338) ± 2.24 * 0.0353]
= [ 0.0598 , 0.218 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 97.5% sure that the interval [ 0.0598 , 0.218] 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 not lies in the confidence calculated, we assume that there is statistical difference
in the performance of these two treatments

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