Question

Give an example of a confidence interval for the mean of a population when you do not know the population standard deviation by describing the population, the variable, and how you would go about finding the interval. You may want to mention a sample size, a confidence level, and a possible interval.

Answer 1

TRADITIONAL METHOD
Assumed data, because not given in the data,
standard deviation, σ =7
sample mean, x =40
population size (n)=40
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 7/ sqrt ( 40) )
= 1.107
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 1.107
= 2.169
III.
CI = x ± margin of error
confidence interval = [ 40 ± 2.169 ]
= [ 37.831,42.169 ]
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DIRECT METHOD
given that,
standard deviation, σ =7
sample mean, x =40
population size (n)=40
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
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 = [ 40 ± Z a/2 ( 7/ Sqrt ( 40) ) ]
= [ 40 - 1.96 * (1.107) , 40 + 1.96 * (1.107) ]
= [ 37.831,42.169 ]
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interpretations:
1. we are 95% sure that the interval [37.831 , 42.169 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean