Question

A data set includes 106 body temperatures of healthy adult humans having a mean of 98.9 degrees and a standard deviation of 0.62 degrees. Construct a 99​% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6 degrees as the mean body​ temperature?

Answer 1

TRADITIONAL METHOD
given that,
sample mean, x =98.9
standard deviation, s =0.62
sample size, n =106
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.62/ sqrt ( 106) )
= 0.06
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 105 d.f is 2.623
margin of error = 2.623 * 0.06
= 0.158
III.
CI = x ± margin of error
confidence interval = [ 98.9 ± 0.158 ]
= [ 98.742 , 99.058 ]
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DIRECT METHOD
given that,
sample mean, x =98.9
standard deviation, s =0.62
sample size, n =106
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 105 d.f is 2.623
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 = [ 98.9 ± t a/2 ( 0.62/ Sqrt ( 106) ]
= [ 98.9-(2.623 * 0.06) , 98.9+(2.623 * 0.06) ]
= [ 98.742 , 99.058 ]
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interpretations:
1) we are 99% sure that the interval [ 98.742 , 99.058 ] contains the true population mean
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 mean
Answer:
the sample suggest about the use of 98.6 degrees as the mean body​ temperature is not in the confidence interval [ 98.742 , 99.058 ]