Question

In a randomized comparative experiment on the effect of dietary calcium on blood pressure, researchers divided 60 healthy white males at random into two groups. One group received calcium; the other, a placebo. At the beginning of the study, the researchers measured many variables on the subjects. The paper reporting the study gives x⎯⎯⎯=x¯=117.8 and s = 9.2 for the seated systolic blood pressure of the 30 members of the placebo group.

Give a 96% confidence interval for the mean blood pressure in the population from which the subjects were recruited is.

The confidence interval (±±0.1) is from  to

Answer 1

TRADITIONAL METHOD
given that,
sample mean, x =117.8
standard deviation, s =9.2
sample size, n =30
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 9.2/ sqrt ( 30) )
= 1.68
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.04
from standard normal table, two tailed value of |t α/2| with n-1 = 29 d.f is 2.15
margin of error = 2.15 * 1.68
= 3.61
III.
CI = x ± margin of error
confidence interval = [ 117.8 ± 3.61 ]
= [ 114.19 , 121.41 ]
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DIRECT METHOD
given that,
sample mean, x =117.8
standard deviation, s =9.2
sample size, n =30
level of significance, α = 0.04
from standard normal table, two tailed value of |t α/2| with n-1 = 29 d.f is 2.15
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 117.8 ± t a/2 ( 9.2/ Sqrt ( 30) ]
= [ 117.8-(2.15 * 1.68) , 117.8+(2.15 * 1.68) ]
= [ 114.19 , 121.41 ]
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interpretations:
1) we are 96% sure that the interval [ 114.19 , 121.41 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 96% of these intervals will contains the true population mean