Question

clear answer, please!

study investigated dose-related effects of methadone in subjects with torsade de pointes (a specific form of polymorphic ventricular tachycardia in patients with a long QT interval), a polymorphic ventricular tachycardia. In the study of 17 subjects, nine were being treated with methadone for opiate dependency and eight for chronic pain. The mean daily dose of methadone in the opiate dependency group was 541 mg/day with a standard deviation of 156, while the chronic pain group received a mean dose of 269 mg/day with a standard deviation of 316. Construct 90, 95, and 99 % confidence intervals for the difference between population means Where appropriate, state the assumptions that make your method valid.

Answer 1

a.
at confidence interval is 90%
TRADITIONAL METHOD
given that,
mean(x)=541
standard deviation , s.d1=156
number(n1)=9
y(mean)=269
standard deviation, s.d2 =316
number(n2)=8
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((24336/9)+(99856/8))
= 123.231
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.1
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 7 d.f is 1.895
margin of error = 1.895 * 123.231
= 233.524
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (541-269) ± 233.524 ]
= [38.476 , 505.524]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=541
standard deviation , s.d1=156
sample size, n1=9
y(mean)=269
standard deviation, s.d2 =316
sample size,n2 =8
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 = [( 541-269) ± t a/2 * sqrt((24336/9)+(99856/8)]
= [ (272) ± t a/2 * 123.231]
= [38.476 , 505.524]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [38.476 , 505.524] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion
b.
at confidence interval is 95%
TRADITIONAL METHOD
given that,
mean(x)=541
standard deviation , s.d1=156
number(n1)=9
y(mean)=269
standard deviation, s.d2 =316
number(n2)=8
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((24336/9)+(99856/8))
= 123.231
II.
margin of error = t a/2 * (standard 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 7 d.f is 2.365
margin of error = 2.365 * 123.231
= 291.442
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (541-269) ± 291.442 ]
= [-19.442 , 563.442]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=541
standard deviation , s.d1=156
sample size, n1=9
y(mean)=269
standard deviation, s.d2 =316
sample size,n2 =8
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 = [( 541-269) ± t a/2 * sqrt((24336/9)+(99856/8)]
= [ (272) ± t a/2 * 123.231]
= [-19.442 , 563.442]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-19.442 , 563.442] 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.
at confidence interval is 99%
TRADITIONAL METHOD
given that,
mean(x)=541
standard deviation , s.d1=156
number(n1)=9
y(mean)=269
standard deviation, s.d2 =316
number(n2)=8
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((24336/9)+(99856/8))
= 123.231
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.01
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 7 d.f is 3.499
margin of error = 3.499 * 123.231
= 431.187
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (541-269) ± 431.187 ]
= [-159.187 , 703.187]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=541
standard deviation , s.d1=156
sample size, n1=9
y(mean)=269
standard deviation, s.d2 =316
sample size,n2 =8
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 = [( 541-269) ± t a/2 * sqrt((24336/9)+(99856/8)]
= [ (272) ± t a/2 * 123.231]
= [-159.187 , 703.187]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [-159.187 , 703.187] contains the true population proportion
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 true population proportion