Systems Engineering Program
Department of Engineering Management, Information and Systems
EMIS 7370/5370 STAT 5340 :
PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS
Tests of Hypothesis
Tests of Means and Variances
Dr. Jerrell T. Stracener, SAE Fellow
Leadership in Engineering
1
Example
A company produces and markets coffee in cans
which are advertised as containing one pound
of coffee. What this means is that the true mean
weight of coffee per can is 1 pound. If the true
mean weight of coffee per can exceeds 1 pound,
the company’s profit will suffer. On the other
hand, if the true mean weight is very much less
than 1 pound, consumers will complain and sales
may decrease. To monitor the process, 25 cans
of coffee are randomly selected during each day’s
production. The process will be adjusted if there
is evidence to indicate that the true mean amount
of coffee is not 1 pound.
2
Example
A decision rule is desired so that the probability
of adjusting the process when the true mean
weight of coffee per can is equal to 1 pound is
1%. Assume that weight of coffee per can has a
normal distribution with unknown mean and
standard deviation.
3
Example - solution
The decision rule is:
Action 1 - adjust process if
 X 1
X  0
  t  2.797
t
 5

s
s

 2 ,n 1
n
or if
t  t
2
, n 1
 2.797
4
Example - solution
The decision rule is:
Action 2 - do not adjust process if
 X 1
  2.797
 2.797  5

s


5
Example - solution
Suppose that for a given day
X  1.006
and
s = 0.012
 X 1 

Then t = 5

0
.
012


 1.006  1 
 5
  2.5
 0.012 
6
Example - solution
so that
- 2.797 < 2.5 < 2.797
and Action 2: no adjustment, is taken. We
conclude that the true mean weight of coffee per
can is 1 pound. We have thus tested the statistical
hypothesis that  = 1 pound versus the alternative
hypothesis that  does not equal 1 pound at the
1% level of significance.
7
Test of Means
Let X1, …, Xn, be a random sample of size n, from
a normal distribution with mean  and standard
deviation , both unknown.
To test the Null Hypothesis
H0:  = 0 , a given or specified value
against the appropriate Alternative Hypothesis
or
or
1. HA:  < 0 ,
2. HA:  > 0 ,
3. HA:   0 ,
8
Test of Means
at the 100  % level of significance. Calculate the
value of the test statistic
t
X  0
s
n
Reject H0 if
1. t < -t, n-1 ,
2. t > t, n-1 ,
3. t < -t/2, n-1 , or if t > t/2, n-1 ,
depending on the Alternative Hypothesis.
9
Test on Two Means
Let X11, X12, …, X1n1 be a random sample of size n1 from N(1,
1) and X21, X22, …, X2n2 be a random sample of size n2
from N(2, 2), where 1, 1, 2 and 2 are all unknown.
To test against the appropriate alternative
hypothesis
H0:
µ1 - µ2 = do, where do  0 (usually do=0)
10
Test on Two Means
1.
H1: µ1 - µ2 < do, where do  0,
2.
H1: µ1 - µ2 > do, where do  0,
3.
H1: µ1 - µ2  do, where do  0,
or
or
at the   100% level of significance, calculate the
value of the test statistic.
11
Test on Two Means
Calculate the value of the test statistic

X
t' 
1
 X2  d0
s12 s 22

n1 n 2
12
Test on Two Means
Reject Ho if
1.
t' < -t,ν
or
2.
t' > t, ν
or
3.
t' < -t/2, ν
or t' > t/2, ν depending on the alternative
hypothesis, where
2
s
s 
  
n1 n 2 


2
2
 s12   s 22 
   
 n1    n 2 
n1  1 n2  1
2
1
2
2
13
Example - Test on Two Means
An experiment was performed to compare the
abrasive wear of two different laminated materials.
Twelve pieces of material 1 were tested, by
exposing each piece to a machine measuring wear.
Ten pieces of material 2 were similarly tested. In
each case, the depth of wear was observed. The
samples of material 1 gave an average (coded)
wear of 85 units with a standard deviation of 4,
while samples of material 2 gave an average of 81
and a standard deviation of 5. Test the hypothesis
that the two types of material exhibit the same
mean abrasive wear at the 0.10 level of
significance. Assume the populations to be
approximately normal.
14
Example
Test
H0: 1 = 2
or
1 - 2
H1: 1  2
or
1 - 2  0.
=
0.
Vs.
With a 10% level of significance, i.e.,   0.10
Then

X
t' 
1
 X2  d0
s12 s 22

n1 n 2
 2.07
15
Example
where
2
s
s 
  
n1 n 2 


2 2
2 2
 s1   s 2 
   
 n1    n 2 
n1  1 n2  1
2
1
2
2
 17
and
t α 2, 2  t 0.05,17  1.725
The calculate Critical Region is:
t’ < -1.725 and t’ > 1.725,
16
Example
Since t’ = 2.07, we can reject H0 and conclude
that the two materials do not exhibit the same
abrasive wear.
17
Test of Variances
Let X1, …, Xn, be a random sample of size n, from
a normal distribution with mean  and standard
deviation , both unknown.
To test the Null Hypothesis
H0: 2 = o2 , a specified value
against the appropriate Alternative Hypothesis
or
or
1. HA: 2 < o2 ,
2. HA: 2 > o2 ,
3. HA: 2  o2 ,
18
Test of Variances
at the 100  % level of significance. Calculate the
value of the test statistic
 2  n  1
s2
 02
Reject H0 if
1. 2 < 21-, n-1 ,
2. 2 > 2, n-1 ,
3. 2 < 21-/2, n-1 , or if 2 > 2/2, n-1 ,
depending on the Alternative Hypothesis.
19
Test on Two Variances
Let X11, X12, …, X1n1 be a random sample of size n1 from
N(1, 1) and X21, X22, …, X2n2 be a random sample of
size n2 from N(2, 2), where 1, 1, 2 and 2 are all
unknown.
To test
H0:
σ12  σ 22
against the appropriate alternative hypothesis
20
Test on Two Variances
1.
H1:
σ12  σ 22
2.
H1:
σ12  σ 22
3.
H1:
σ σ
or
or
2
1
2
2
at the   100% level of significance, calculate the
value of the test statistic.
2
1
2
2
S
F
S
21
Test on Two Variances
Reject Ho if
F  F1 (v1 , v2 )
or
F  Fα (v1,v2 )
or
F  F1 / 2 (v1 , v2 ) or F  F / 2 (v1 , v2 )
depending on the alternative hypothesis,
and where
v1  n1  1
and
v2  n2  1
22
Example - Test on Variances
An experiment was performed to compare the
abrasive wear of two different laminated materials.
Twelve pieces of material 1 were tested, by
exposing each piece to a machine measuring wear.
Ten pieces of material 2 were similarly tested. In
each case, the depth of wear was observed. The
samples of material 1 gave an average (coded)
wear of 85 units with a standard deviation of 4,
while samples of material 2 gave an average of 81
and a standard deviation of 5. Test the hypothesis
that the two types of material exhibit the same
variation in abrasive wear at the 0.10 level of
significance.
23
Example - Test of Variances
H0:
H1:
1 2 = 2 2
1 2  2 2
With a 10% level of significance, i.e.,   0.10
f (x)
v1 = 11 and v2 = 9
0.05
0
0.34
0.05
x
3.11
Critical region: From the graph we see that
F0.05(11,9) = 3.11
24
Example - Test of Variances
F0.95 (11,9) 
1
 0.34
F0.05 (9,11)
Therefore, the null hypothesis is rejected when F < 0.34 or
F > 3.11.
S12 16
F 2 
 0.64
S2 25
Decision:
Do not reject H0. Conclude that there is insufficient evidence
that the variances differ.
25