4.Please provide solutions to the following Questions and show all necessary wor

Question

4.Please provide solutions to the following Questions and show all necessary work.

4. Suppose that we want to compare the effectiveness of two pertensive drugs A and B. Suppose that n-100 patients take drug A and n2 120 patients take drug B. Patients in the two samples are matched on certain important baseline characteristics. One month after taking the drugs, the mean DBP is 78 mm Hg (with SD=7.07 mm Hg), and 82 mm Hg (with SD=10.1 mm Hg) for patients who took drug A and B respectively. Suppose that in the underlying population DBP is normally distributed. (a) Provide point estimates and compute 95% CIs for the variance of the DBP measurements of each group. Compare the variances between the two groups. What is your conclusion? (b) Provide point estimates and compute 95% CIs for the mean DBP of each group. (c) Based on your results from this study what is your conclusion about the effectiveness of the two drugs?

Explanation / Answer

a.
i.
CONFIDENCE INTERVAL FOR VARIANCE FOR DRUG A
ci = (n-1) s^2 / ^2 right < ^2 < (n-1) s^2 / ^2 left
where,
s^2 = variance
^2 right = (1 - confidence level)/2
^2 left = 1 - ^2 right
n = sample size
since aplha =0.05
^2 right = (1 - confidence level)/2 = (1 - 0.95)/2 = 0.05/2 = 0.025
^2 left = 1 - ^2 right = 1 - 0.025 = 0.975
the two critical values ^2 left, ^2 right at 99 df are 128.422 , 73.361
variacne( s^2 )=8.4
sample size(n)=100
confidence interval = [ 99 * 8.4/128.422 < ^2 < 99 * 8.4/73.361 ]
= [ 831.6/128.422 < ^2 < 831.6/73.3611 ]
[ 6.4755 , 11.3357 ]

ii.
CONFIDENCE INTERVAL FOR VARIANCE FOR DRUG B
^2 right = (1 - confidence level)/2 = (1 - 0.95)/2 = 0.05/2 = 0.025
^2 left = 1 - ^2 right = 1 - 0.025 = 0.975
the two critical values ^2 left, ^2 right at 119 df are 151.0844 , 90.7
variacne( s^2 )=8.4
sample size(n)=120
confidence interval = [ 119 * 8.4/151.0844 < ^2 < 119 * 8.4/90.7 ]
= [ 999.6/151.0844 < ^2 < 999.6/90.6996 ]
[ 6.6162 , 11.021 ]

iii.
Given that,
sample 1
s1^2=102.01, n1 =120
sample 2
s2^2 =49.985, n2 =100
null, Ho: ^2 = ^2
alternate, H1: ^2 != ^2
level of significance, = 0.01
from standard normal table, two tailed f /2 =1.656
since our test is two-tailed
reject Ho, if F o < -1.656 OR if F o > 1.656
we use test statistic fo = s1^1/ s2^2 =102.01/49.985 = 2.04
| fo | =2.04
critical value
the value of |f | at los 0.01 with d.f f(n1-1,n2-1)=f(119,99) is 1.656
we got |fo| =2.041 & | f | =1.656
make decision
hence value of | fo | > | f | and here we reject Ho
ANSWERS
---------------
null, Ho: ^2 = ^2
alternate, H1: ^2 != ^2
test statistic: 2.04
critical value: -1.656 , 1.656
decision: reject Ho

no diffrence in variance b/w two drugs

b.
i.
Mean confidence at 95% for DRUG A
given that,
sample mean, x =78
standard deviation, s =7.07
sample size, n =100
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 99 d.f is 1.984
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 78 ± Z a/2 ( 7.07/ Sqrt ( 100) ]
= [ 78-(1.984 * 0.707) , 78+(1.984 * 0.707) ]
= [ 76.597 , 79.403 ]

ii.
Mean confidence at 95% for DRUG B

given that,
sample mean, x =82
standard deviation, s =10.1
sample size, n =120
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 119 d.f is 1.98
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 82 ± Z a/2 ( 10.1/ Sqrt ( 120) ]
= [ 82-(1.98 * 0.922) , 82+(1.98 * 0.922) ]
= [ 80.174 , 83.826 ]

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