Question

Construct the confidence interval for the population mean μ. c =0.90​, x=9.8​, σ =0.4​, and n =46

A 90​% confidence interval for μ is ?

Answer 1

TRADITIONAL METHOD
given that,
standard deviation, σ =0.4
sample mean, x =9.8
population size (n)=46
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 0.4/ sqrt ( 46) )
= 0.059
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
value of z table is 1.645
margin of error = 1.645 * 0.059
= 0.097
III.
CI = x ± margin of error
confidence interval = [ 9.8 ± 0.097 ]
= [ 9.703,9.897 ]
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DIRECT METHOD
given that,
standard deviation, σ =0.4
sample mean, x =9.8
population size (n)=46
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
value of z table is 1.645
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 9.8 ± Z a/2 ( 0.4/ Sqrt ( 46) ) ]
= [ 9.8 - 1.645 * (0.059) , 9.8 + 1.645 * (0.059) ]
= [ 9.703,9.897 ]
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interpretations:
1. we are 90% sure that the interval [9.703 , 9.897 ] 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