Comparing two population proportions Suppose that a medical experiment randomly

Question

Comparing two population proportions

Suppose that a medical experiment randomly assigns patients to either an “old” or a “new” treatment for a certain condition. Suppose that 60% of patients on the old treatment plan recover and that 70% of patients on the new treatment plan recover. These results were each based on samples of 50,000 persons undergoing treatment. (a) Perform a test to statistically determine if the new treatment approach has a higher recovery rate than the old treatment. Use a level of significance of 1%. (b) Construct a 99% confidence interval for the difference in recovery rates. Are you confident that the new treatment is statistically significantly better and should be medically adopted for general use for such patients?

Please show detailed, step by step work! Thank you :)

Explanation / Answer

PART A.

Given that,
sample one, n1 =50000, p1= x1/n1=0.6
sample two, n2 =50000, p2= x2/n2=0.7
null, Ho: p1 > p2
alternate, H1: p1 < p2
level of significance, = 0.01
from standard normal table,left tailed z /2 =2.326
since our test is left-tailed
reject Ho, if zo < -2.326
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.6-0.7)/sqrt((0.65*0.35(1/50000+1/50000))
zo =-33.15
| zo | =33.15
critical value
the value of |z | at los 0.01% is 2.326
we got |zo| =33.15 & | z | =2.326
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: left tail - Ha : ( p < -33.1497 ) = 0
hence value of p0.01 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 < p2
test statistic: -33.15
critical value: -2.326
decision: reject Ho
p-value: 0

we have evidence that new treatment approach has a higher recovery
rate than the old treatment

PART B.

TRADITIONAL METHOD
given that,
sample one, n1 =50000, p1= x1/n1=0.6
sample two, n2 =50000, p2= x2/n2=0.7
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( (0.6*0.4/50000) +(0.7 * 0.3/50000))
=0.003
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.58
margin of error = 2.58 * 0.003
=0.0077
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.6-0.7) ±0.0077]
= [ -0.1077 , -0.0923]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample one, n1 =50000, p1= x1/n1=0.6
sample two, n2 =50000, p2= x2/n2=0.7
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 = [ (0.6-0.7) ± 2.58 * 0.003]
= [ -0.1077 , -0.0923 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ -0.1077 , -0.0923] 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, 99% of these intervals will contains the difference between
true population mean P1-P2

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