Question

Fat contents (in percentage) for 10 randomly selected hot dogs were given in the article "Sensory and Mechanical Assessment of the Quality of Frankfurters". Use the following data to construct a 90% confidence interval for the true mean fat percentage of hot dogs: (Give the answers to two decimal places.)
(  ,  )

26.0

22.1

23.6

17.0

30.6

21.8

26.3

16.0

20.9

19.5

Answer 1

TRADITIONAL METHOD
given that,
sample mean, x =22.38
standard deviation, s =4.4482
sample size, n =10
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 4.4482/ sqrt ( 10) )
= 1.41
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 9 d.f is 1.833
margin of error = 1.833 * 1.41
= 2.58
III.
CI = x ± margin of error
confidence interval = [ 22.38 ± 2.58 ]
= [ 19.8 , 24.96 ]
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DIRECT METHOD
given that,
sample mean, x =22.38
standard deviation, s =4.4482
sample size, n =10
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 9 d.f is 1.833
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 = [ 22.38 ± t a/2 ( 4.4482/ Sqrt ( 10) ]
= [ 22.38-(1.833 * 1.41) , 22.38+(1.833 * 1.41) ]
= [ 19.8 , 24.96 ]
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interpretations:
1) we are 90% sure that the interval [ 19.8 , 24.96 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean
Answer:
90% sure that the interval [ 19.8 , 24.96 ]%

margin of error =2.58%