2. A machine produces metal rods used in an automobile suspension system. A rand
ID: 3338766 • Letter: 2
Question
2. A machine produces metal rods used in an automobile suspension system. A random sample of 12 rods is selected and the diameter is measured. The resulting data in millimeters are shown here: 8.23, 8.29, 8.19, 8.14, 8.31, 8.19, 8.29, 8.32, 8.42, 8.24, 8.30, and 8.40. a) Check the assumption of normality for rod diameter. b) Is there strong evidence to indicate that mean rod diameter exceeds 8.20 mm using =0.05. c) Find the P-value for this test. d) Find a 95% lower CI on mean rod diameter and draw conclusions.
Explanation / Answer
2.
Given that,
population mean(u)=8.2
sample mean, x =8.2767
standard deviation, s =0.0836
number (n)=12
null, Ho: =8.2
alternate, H1: >8.2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.796
since our test is right-tailed
reject Ho, if to > 1.796
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =8.2767-8.2/(0.0836/sqrt(12))
to =3.178
| to | =3.178
critical value
the value of |t | with n-1 = 11 d.f is 1.796
we got |to| =3.178 & | t | =1.796
make decision
hence value of | to | > | t | and here we reject Ho
p-value :right tail - Ha : ( p > 3.1782 ) = 0.00439
hence value of p0.05 > 0.00439,here we reject Ho
ANSWERS
---------------
a.
null, Ho: =8.2
alternate, H1: >8.2
test statistic: 3.178
critical value: 1.796
decision: reject Ho
b.
strong evidence to indicate that mean rod diameter exceeds 8.20 mm
c.
p-value: 0.00439
d.
TRADITIONAL METHOD
given that,
sample mean, x =8.2767
standard deviation, s =0.0836
sample size, n =12
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.0836/ sqrt ( 12) )
= 0.024
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table,right tailed value of |t /2| with n-1 = 11 d.f is 1.796
margin of error = 1.796 * 0.024
= 0.043
III.
CI = x ± margin of error
confidence interval = [ 8.2767 ± 0.043 ]
= [ 8.233 , 8.32 ]
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DIRECT METHOD
given that,
sample mean, x =8.2767
standard deviation, s =0.0836
sample size, n =12
level of significance, = 0.05
from standard normal table,right tailed value of |t /2| with n-1 = 11 d.f is 1.796
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 = [ 8.2767 ± Z a/2 ( 0.0836/ Sqrt ( 12) ]
= [ 8.2767-(1.796 * 0.024) , 8.2767+(1.796 * 0.024) ]
= [ 8.233 , 8.32 ]
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interpretations:
1) we are 95% sure that the interval [ 8.233 , 8.32 ] 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
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