Question

(21.28) Healthy men aged 21 to 35 were randomly assigned to one of two groups: half received 0.82 grams of alcohol per kilogram of body weight; half received a placebo. Participants were then given 30 minutes to read up to 34 pages of Tolstoy's War and Peace (beginning at chapter 1, with each page containing approximately 22 lines of text). Every two to four minutes participants were prompted to indicate whether they were "zoning out." The proportion of times participants indicated they were zoning out was recorded for each subject. The table below summarizes data on the proportion of episodes of zoning out.

(The study report gave the standard error of the mean s/n−−√s/n, abbreviated as SEM, rather than the standard deviation s.)

Group	n	x¯¯¯x¯	SEM
Alcohol	30	0.26	0.05
Placebo	30	0.09	0.02

What are the standard deviations (±±0.001) for the two groups?

For "Alcohol" group s1=s1=

For "Placebo" group s2=s2=

What degrees of freedom does the conservative Option 2 use for two-sample t procedures for these samples?

Using Option 2, a 95% confidence interval (±±0.01) for the mean difference (Alcohol - Placebo) between the two groups is  from to .

Answer 1

TRADITIONAL METHOD
given that,
mean(x)=0.26
standard deviation , s.d1=0.273
number(n1)=30
y(mean)=0.09
standard deviation, s.d2 =0.109
number(n2)=30
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((0.075/30)+(0.012/30))
= 0.054
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 29 d.f is 2.045
margin of error = 2.045 * 0.054
= 0.11
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (0.26-0.09) ± 0.11 ]
= [0.06 , 0.28]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=0.26
standard deviation , s.d1=0.273
sample size, n1=30
y(mean)=0.09
standard deviation, s.d2 =0.109
sample size,n2 =30
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 = [( 0.26-0.09) ± t a/2 * sqrt((0.075/30)+(0.012/30)]
= [ (0.17) ± t a/2 * 0.054]
= [0.06 , 0.28]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [0.06 , 0.28] 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
95% confidence interval for the mean difference (Alcohol - Placebo) between the two groups is [0.06 , 0.28]