Question

A study2examines chocolate’s effects on blood vessel function in healthy people. In the randomized, double-blind, placebo-controlled study, 11 people received 46 grams (1.6 ounces) of dark chocolate (which is naturally flavonoid-rich) every day for two weeks, while a control group of 10 people received a placebo consisting of dark chocolate with low flavonoid content. Participants had their vascular health measured (by means of flow-mediated dilation) before and after the two-week study. The increase over the two-week period was measured, with larger numbers indicating greater vascular health. For the group getting the good dark chocolate, the mean increase was 1.3 with a standard deviation of 2.32, while the control group had a mean change of -0.96 with a standard deviation of 1.58.Find a 95%confidence interval for the difference in means between the two groups CN−, whereCrepresents the mean increase in flow-mediated dilation for people eating dark chocolate every day and Nrepresents the mean increase in flow-mediated dilation for people eating a dark chocolate substitute eachday. .Does the scenario represent a confidence interval or a hypothesis test?b.Does the scenario represent a one population or two populations?c.What is the appropriate formula for this scenario?d.What does each symbol represent in the appropriate formula?

Answer 1

a.
TRADITIONAL METHOD
given that,
mean(x)=1.3
standard deviation , s.d1=2.32
number(n1)=11
y(mean)=-0.96
standard deviation, s.d2 =1.58
number(n2)=10
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((5.382/11)+(2.496/10))
= 0.86
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 9 d.f is 2.262
margin of error = 2.262 * 0.86
= 1.944
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (1.3--0.96) ± 1.944 ]
= [0.316 , 4.204]
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DIRECT METHOD
given that,
mean(x)=1.3
standard deviation , s.d1=2.32
sample size, n1=11
y(mean)=-0.96
standard deviation, s.d2 =1.58
sample size,n2 =10
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 = [( 1.3--0.96) ± t a/2 * sqrt((5.382/11)+(2.496/10)]
= [ (2.26) ± t a/2 * 0.86]
= [0.316 , 4.204]
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interpretations:
1. we are 95% sure that the interval [0.316 , 4.204] 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
b.
Given that,
mean(x)=1.3
standard deviation , s.d1=2.32
number(n1)=11
y(mean)=-0.96
standard deviation, s.d2 =1.58
number(n2)=10
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.262
since our test is two-tailed
reject Ho, if to < -2.262 OR if to > 2.262
c.
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =1.3--0.96/sqrt((5.3824/11)+(2.4964/10))
to =2.6291
| to | =2.6291
critical value
the value of |t α| with min (n1-1, n2-1) i.e 9 d.f is 2.262
we got |to| = 2.62906 & | t α | = 2.262
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.6291 ) = 0.027
hence value of p0.05 > 0.027,here we reject Ho
ANSWERS
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d.
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 2.6291
critical value: -2.262 , 2.262
decision: reject Ho
p-value: 0.027
we have enough evidence to support the claim that the difference in means between the two groups