An educational program on healthy eating is given to individuals at risk for hea
ID: 3319825 • Letter: A
Question
An educational program on healthy eating is given to individuals at risk for heart disease. Systolic blood pressures are measured at baseline before the program and again 4 weeks later. Calculate a two-sided 95% confidence interval for the mean difference in the scores. Give an interpretation of the confidence interval. You MUST show your work to receive full credit. Partial credit is available.
Baseline
4 weeks
120
122
145
142
130
135
160
158
152
155
143
140
126
130
Baseline
4 weeks
120
122
145
142
130
135
160
158
152
155
143
140
126
130
Explanation / Answer
TRADITIONAL METHOD
given that,
mean(x)=139.4286
standard deviation , s.d1=14.5586
number(n1)=7
y(mean)=140.2857
standard deviation, s.d2 =12.9192
number(n2)=7
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((211.953/7)+(166.906/7))
= 7.357
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, = 0.05
from standard normal table, two tailed and
value of |t | with min (n1-1, n2-1) i.e 6 d.f is 2.447
margin of error = 2.447 * 7.357
= 18.002
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (139.4286-140.2857) ± 18.002 ]
= [-18.859 , 17.145]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=139.4286
standard deviation , s.d1=14.5586
sample size, n1=7
y(mean)=140.2857
standard deviation, s.d2 =12.9192
sample size,n2 =7
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 = [( 139.4286-140.2857) ± t a/2 * sqrt((211.953/7)+(166.906/7)]
= [ (-0.857) ± t a/2 * 7.357]
= [-18.859 , 17.145]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-18.859 , 17.145] 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
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