Question

Scenario 3:

To determine the effectiveness of a new diuretic drug, urine output was measured in 20 men and compared to 20 men that received placebo control. The data are summarized in Table 2 below. The null hypothesis is: H₀: μ _Placebo= μ  _drug.

Table2.  Effectiveness of New Diuretic Drug

	Mean Urine Output (mg/kg/hr)	SD
Placebo	1.58	0.66
Drug	2.78	0.77

What is the null hypothesis equation for this experiment?

What statistical test would you use?

t-test

ANOVA

Relative Risk

POWER

What is the 95% confidence intervals for this 2 sample t-test?

4.                          Based on the statistical test that you chose, do you accept or reject the null hypothesis at P=0.05? Please paste your data output file in the area below.

5.                          What is the Power of this test?Please paste your data output file in the area below.

How many samples are needed to achieve a power of 0.80?  Please paste your data output file in the area below.   

Answer 1

3.
TRADITIONAL METHOD
given that,
mean(x)=1.58
standard deviation , s.d1=0.66
number(n1)=20
y(mean)=2.78
standard deviation, s.d2 =0.77
number(n2)=20
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((0.44/20)+(0.59/20))
= 0.23
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 19 d.f is 2.09
margin of error = 2.093 * 0.23
= 0.47
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (1.58-2.78) ± 0.47 ]
= [-1.67 , -0.73]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=1.58
standard deviation , s.d1=0.66
sample size, n1=20
y(mean)=2.78
standard deviation, s.d2 =0.77
sample size,n2 =20
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 = [( 1.58-2.78) ± t a/2 * sqrt((0.44/20)+(0.59/20)]
= [ (-1.2) ± t a/2 * 0.23]
= [-1.67 , -0.73]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-1.67 , -0.73] 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
4.
Given that,
mean(x)=1.58
standard deviation , s.d1=0.66
number(n1)=20
y(mean)=2.78
standard deviation, s.d2 =0.77
number(n2)=20
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.093
since our test is two-tailed
reject Ho, if to < -2.093 OR if to > 2.093
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =1.58-2.78/sqrt((0.4356/20)+(0.5929/20))
to =-5.292
| to | =5.292
critical value
the value of |t α| with min (n1-1, n2-1) i.e 19 d.f is 2.093
we got |to| = 5.29169 & | t α | = 2.093
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -5.2917 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -5.292
critical value: -2.093 , 2.093
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that compare between placebo and drugs

5.
power of the test = 0.80
power of the test = 1- type2 error
p(type2 error) = 1-0.80 =0.20
Zalpha at 0.05 =2.093
Zbeta at 0.20 = 0.8416
n= ((σ(Zalpha +Zbeta))/(U-Uo))^2
n =((( 2.093+0.8416))/(2.78-1.58))^2
n = 5.98 =6