Question

Tests are run on 16 samples of certain metal alloy. The average
strength was 27.3 ksi with a standard deviation of 1.20 ksi.
1) What is the range in strength with 95% certainty?
2) What is the range in strength with 99% certainty?
3) Can you be 100% sure of having a certain range?

Answer 1

1.
TRADITIONAL METHOD
given that,
sample mean, x =27.3
standard deviation, s =1.2
sample size, n =16
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 1.2/ sqrt ( 16) )
= 0.3
II.
margin of error = t α/2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 15 d.f is 2.131
margin of error = 2.131 * 0.3
= 0.639
III.
CI = x ± margin of error
confidence interval = [ 27.3 ± 0.639 ]
= [ 26.661 , 27.939 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =27.3
standard deviation, s =1.2
sample size, n =16
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 15 d.f is 2.131
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 = [ 27.3 ± t a/2 ( 1.2/ Sqrt ( 16) ]
= [ 27.3-(2.131 * 0.3) , 27.3+(2.131 * 0.3) ]
= [ 26.661 , 27.939 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 26.661 , 27.939 ] 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
2.
TRADITIONAL METHOD
given that,
sample mean, x =27.3
standard deviation, s =1.2
sample size, n =16
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 1.2/ sqrt ( 16) )
= 0.3
II.
margin of error = t α/2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 15 d.f is 2.947
margin of error = 2.947 * 0.3
= 0.884
III.
CI = x ± margin of error
confidence interval = [ 27.3 ± 0.884 ]
= [ 26.416 , 28.184 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =27.3
standard deviation, s =1.2
sample size, n =16
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 15 d.f is 2.947
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 = [ 27.3 ± t a/2 ( 1.2/ Sqrt ( 16) ]
= [ 27.3-(2.947 * 0.3) , 27.3+(2.947 * 0.3) ]
= [ 26.416 , 28.184 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 26.416 , 28.184 ] 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

3.
There is no 100% sure of having a certain range but only range is possible for 90%,95%,99% etc.