Please show step by step by hand calculation Thank you 7. In a clinica l study o

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Please show step by step by hand calculation

Thank you

7. In a clinica l study of a medication for hypertension, 36 patients were randomly selected. Each patient had two blood pressure reading months after receiving the medication daily. The mean blood pressure of the reading before the medication was 140 mmHg, and the mean reading after the medication was 135 mmHg Assuming the blood pressure follows a normal distribution. The sample standard deviation of the reading before medication was 20 mmHg, and after medication was 10 mmHg. The difference between the two readings of each individual had a standard deviation of 15 mmHg. s, one before taking the medication and one 2 a) Conduct a test on the effect of the medication (ie. whether the medication lowers the blood pressure) at the significance level 0.05. b) Construct a 95% confidence interval of the blood pressure change after taking the medication. c) Test on the equality of the standard deviations of the before medication and after medication.

Explanation / Answer

7.
a)
Given that,
mean(x)=140
standard deviation , s.d1=20
number(n1)=36
y(mean)=135
standard deviation, s.d2 =10
number(n2)=36
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.69
since our test is left-tailed
reject Ho, if to < -1.69
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =140-135/sqrt((400/36)+(100/36))
to =1.342
| to | =1.342
critical value
the value of |t | with min (n1-1, n2-1) i.e 35 d.f is 1.69
we got |to| = 1.34164 & | t | = 1.69
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:left tail - Ha : ( p < 1.3416 ) = 0.90582
hence value of p0.05 < 0.90582,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 < u2
test statistic: 1.342
critical value: -1.69
decision: do not reject Ho
p-value: 0.90582
we do not have enough evidence to support the claim
b)
TRADITIONAL METHOD
given that,
mean(x)=140
standard deviation , s.d1=20
number(n1)=36
y(mean)=135
standard deviation, s.d2 =10
number(n2)=36
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((400/36)+(100/36))
= 3.727
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table,right tailedand
value of |t | with min (n1-1, n2-1) i.e 35 d.f is 1.69
margin of error = 1.69 * 3.727
= 6.298
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (140-135) ± 6.298 ]
= [-1.298 , 11.298]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=140
standard deviation , s.d1=20
sample size, n1=36
y(mean)=135
standard deviation, s.d2 =10
sample size,n2 =36
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 140-135) ± t a/2 * sqrt((400/36)+(100/36)]
= [ (5) ± t a/2 * 3.727]
= [-1.298 , 11.298]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-1.298 , 11.298] contains the true population proportion
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 true population proportion
c.
Given that,
sample 1
s1^2=20, n1 =36
sample 2
s2^2 =10, n2 =36
null, Ho: ^2 = ^2
alternate, H1: ^2 != ^2
level of significance, = 0.05
from standard normal table, two tailed f /2 =1.961
since our test is two-tailed
reject Ho, if F o < -1.961 OR if F o > 1.961
we use test statistic fo = s1^1/ s2^2 =20/10 = 2
| fo | =2
critical value
the value of |f | at los 0.05 with d.f f(n1-1,n2-1)=f(35,35) is 1.961
we got |fo| =2 & | f | =1.961
make decision
hence value of | fo | > | f | and here we reject Ho
ANSWERS
---------------
null, Ho: ^2 = ^2
alternate, H1: ^2 != ^2
test statistic: 2
critical value: -1.961 , 1.961
decision: reject Ho

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