ploportih of students taking the drug who break out in hives. Compute a 95% conf
ID: 3364290 • Letter: P
Question
ploportih of students taking the drug who break out in hives. Compute a 95% confidence interval for the proportion of students on the placebo who break out in hives. e) Do the intervals overlap? What, if anything, can you conclude about the effectiveness of the drug? (d) ) Compute a 95% confidence interval for the difference in proportions of students who break out in hives by using or not using this drug and give a sentence interpreting this interval 2. An article in the March 2003 New England Journal of Medicine describes a study to see if aspirin is effective in reducing the incidence of colorectal adenomas, a precursor to most colorectal cancers (Sandler et al. (2003)). Of 517 patients in the study, 259 were randomly assigned to receive aspirin and the remaining 258 received a placebo. One or more adenomas were found in 44 of the aspirin group and 70 in the placebo group. Find a 95% one-sided upper bound for the difference in proportions (pA -Pp) and interpret your interval. 25·The data set Bangladesh has measurements on water quality from 271 wells in Bangladesh (Example 5.3).Explanation / Answer
TRADITIONAL METHOD
given that,
sample one, x1 =44, n1 =259, p1= x1/n1=0.17
sample two, x2 =70, n2 =258, p2= x2/n2=0.271
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( (0.17*0.83/259) +(0.271 * 0.729/258))
=0.036
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.64
margin of error = 1.64 * 0.036
=0.059
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.17-0.271) ±0.059]
= [ -0.161 , -0.042]
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DIRECT METHOD
given that,
sample one, x1 =44, n1 =259, p1= x1/n1=0.17
sample two, x2 =70, n2 =258, p2= x2/n2=0.271
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 = [ (0.17-0.271) ± 1.64 * 0.036]
= [ -0.161 , -0.042 ]
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interpretations:
1) we are 95% sure that the interval [ -0.161 , -0.042] 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, 95% of these intervals will contains the difference between
true population mean P1-P2
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