I. (10 pts.) Test of mean with population variance unknown: ASTM Standard E23 de
ID: 3331702 • Letter: I
Question
I. (10 pts.) Test of mean with population variance unknown: ASTM Standard E23 defines standard test methods for notched bar impact testing of metallic materials. The Charpy V notch (CVN) technique measures impact energy and is often used to determine whether or not a material experiences a ductile-to-brittle transition with decreasing temperature. Ten measurements of impact energy (J) on specimens of A238 steel cut at 60°C are as follows: 64.1, 64.7, 64.5, 64.6, 64.5, 64.3, 64.6, 64.8, 64.2, and 64.3. Assume that impact energy is normally distributed. Suppose it is believed that the impact energy would be 65 J Ho: = 65 (a) Given the above hypothesis, calculate the average impact energy from the sample and the sample standard deviation and test the hypothesis at a-0.05. (b) Construct a 90% confidence interval for the true population mean.Explanation / Answer
1.
a.
Given that,
population mean(u)=65
sample mean, x =64.46
standard deviation, s =0.2271
number (n)=10
null, Ho: =65
alternate, H1: <65
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.833
since our test is left-tailed
reject Ho, if to < -1.833
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =64.46-65/(0.2271/sqrt(10))
to =-7.519
| to | =7.519
critical value
the value of |t | with n-1 = 9 d.f is 1.833
we got |to| =7.519 & | t | =1.833
make decision
hence value of | to | > | t | and here we reject Ho
p-value :left tail - Ha : ( p < -7.5193 ) = 0.00002
hence value of p0.05 > 0.00002,here we reject Ho
ANSWERS
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null, Ho: =65
alternate, H1: <65
test statistic: -7.519
critical value: -1.833
decision: reject Ho
p-value: 0.00002
we have enough evidence to support the claim
b.
TRADITIONAL METHOD
given that,
sample mean, x =64.6
standard deviation, s =0.2271
sample size, n =10
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.2271/ sqrt ( 10) )
= 0.072
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 9 d.f is 1.833
margin of error = 1.833 * 0.072
= 0.132
III.
CI = x ± margin of error
confidence interval = [ 64.6 ± 0.132 ]
= [ 64.468 , 64.732 ]
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DIRECT METHOD
given that,
sample mean, x =64.6
standard deviation, s =0.2271
sample size, n =10
level of significance, = 0.1
from standard normal table, two tailed value of |t /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 = [ 64.6 ± Z a/2 ( 0.2271/ Sqrt ( 10) ]
= [ 64.6-(1.833 * 0.072) , 64.6+(1.833 * 0.072) ]
= [ 64.468 , 64.732 ]
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interpretations:
1) we are 90% sure that the interval [ 64.468 , 64.732 ] 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