A randomized clinical trial was performed to compare two dif

A randomized clinical trial was performed to compare two different treatments. Some weeks after treatment the effect X was measured. Assume that X is normally distributed (mean, standard deviation) in both groups. Group 1: n= 62, x mean 83,6, sample standard deviation 22,9. Group 2: n=65, x mean 70.2, sample st.dev: 25.4.

a) What test can be used to test wether treatment effect is equal? Perform the test. Show the calculations. Use significance level 5%. b) Find a 95% confidence interval for mu1 - mu2. How do you interpret this interval?

Solution

a)
Set Up Hypothesis
Null , There Is No-Significance between them Ho: u1 = u2
Alternate , There Is Significance between them - H1: u1 != u2
Test Statistic
X(Mean)=83.6
Standard Deviation(s.d1)=22.9 ; Number(n1)=62
Y(Mean)=70.2
Standard Deviation(s.d2)=25.4; Number(n2)=65
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =83.6-70.2/Sqrt((524.41/62)+(645.16/65))
to =3.13
| to | =3.13
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 61 d.f is 2
We got |to| = 3.12527 & | t | = 2
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != 3.1253 ) = 0.003
Hence Value of P0.05 > 0.003,Here we Reject Ho

b)
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=83.6
Standard deviation( sd1 )=22.9
Sample Size(n1)=62
Mean(x2)=70.2
Standard deviation( sd2 )=25.4
Sample Size(n1)=65
CI = [ ( 83.6-70.2) ±t a/2 * Sqrt( 524.41/62+645.16/65)]
= [ (13.4) ± t a/2 * Sqrt( 18.38) ]
= [ (13.4) ± 2 * Sqrt( 18.38) ]
= [4.82 , 21.98]