Question

3. For a normal population with known variance σ, answer the following questions: (a) What is the confidence level for the interval: R-1.85o/ym μ + 1.85o/vn? (b) What is the confidence level for the interval: μ R + 1.4o/

Answer 1

3.

a.
Assumed values
standard deviation, σ =10
sample mean, x =32
population size (n)=25
TRADITIONAL METHOD
given that,
standard deviation, σ =10
sample mean, x =32
population size (n)=25
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 10/ sqrt ( 25) )
= 2
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, α = 0.064
from standard normal table, two tailed z α/2 =1.85
since our test is two-tailed
value of z table is 1.85
margin of error = 1.85 * 2
= 3.7
III.
CI = x ± margin of error
confidence interval = [ 32 ± 3.7 ]
= [ 28.3,35.7 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =10
sample mean, x =32
population size (n)=25
level of significance, α = 0.064
from standard normal table, two tailed z α/2 =1.85
since our test is two-tailed
value of z table is 1.85
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 = [ 32 ± Z a/2 ( 10/ Sqrt ( 25) ) ]
= [ 32 - 1.85 * (2) , 32 + 1.85 * (2) ]
= [ 28.3,35.7 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 93.6% sure that the interval [28.3 , 35.7 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 93.6% of these intervals will contains the true population mean
b.
TRADITIONAL METHOD
given that,
standard deviation, σ =10
sample mean, x =32
population size (n)=25
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 10/ sqrt ( 25) )
= 2
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, α = 0.15
from standard normal table, two tailed z α/2 =1.4
since our test is two-tailed
value of z table is 1.4
margin of error = 1.4 * 2
= 2.8
III.
CI = x ± margin of error
confidence interval = [ 32 ± 2.8 ]
= [ 29.2,34.8 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =10
sample mean, x =32
population size (n)=25
level of significance, α = 0.15
from standard normal table, two tailed z α/2 =1.4
since our test is two-tailed
value of z table is 1.4
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 = [ 32 ± Z a/2 ( 10/ Sqrt ( 25) ) ]
= [ 32 - 1.4 * (2) , 32 + 1.4 * (2) ]
= [ 29.2,34.8 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 85% sure that the interval [29.2 , 34.8 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 85% of these intervals will contains the true population mean

Answer:
confidence interval upper bound = 34.8