I am having difficulty with part b only. A data set ncudes 108 body temperatures

Question

I am having difficulty with part b only.

A data set ncudes 108 body temperatures healthy adut humans ftrnhon or and Ss] br3c ?e anti e intenal estimate afitte mEm boity tepertue af all heathy humans De the coidence nten contain 986F? What does the sample suggest sbout the use uf 3 as the mean body tamgermure What is the confidence mnal estimate me pour non meate Round to three decimal glaces as needed Do the confidence intenal limts coman968 O Yes What does ths suggest about the use of986 as the meantoty temperatue? Question ls complete Tap on the red ndicabns see noerem arewers

Explanation / Answer

b.

TRADITIONAL METHOD
given that,
sample mean, x =98.2
standard deviation, s =0.64
sample size, n =108
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.64/ sqrt ( 108) )
= 0.062
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 107 d.f is 2.623
margin of error = 2.623 * 0.062
= 0.162
III.
CI = x ± margin of error
confidence interval = [ 98.2 ± 0.162 ]
= [ 98.038 , 98.362 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =98.2
standard deviation, s =0.64
sample size, n =108
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 107 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 = [ 98.2 ± t a/2 ( 0.64/ Sqrt ( 108) ]
= [ 98.2-(2.623 * 0.062) , 98.2+(2.623 * 0.062) ]
= [ 98.038 , 98.362 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 98.038 , 98.362 ] 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

confidence interval limits do not contain 98.6 degree F mean body temparature

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