Post on 02-Jan-2016
description
1
Dr. Jerrell T. Stracener
EMIS 7370 STAT 5340
Probability and Statistics for Scientists and Engineers
Department of Engineering Management, Information and Systems
SMU BOBBY B. LYLESCHOOL OF ENGINEERING
EMIS - SYSTEMS ENGINEERING PROGRAM
Tests of HypothesisTests of Means and Variances
2
A company produces and markets coffee in canswhich 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 salesmay decrease. To monitor the process, 25 cansof coffee are randomly selected during each day’sproduction. The process will be adjusted if there is evidence to indicate that the true mean amountof coffee is not 1 pound.
Example
3
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.
Example
4
The decision rule is:
Action 1 - adjust process if
or if
797.21
51,
2
0
nt
s
X
n
sX
t
797.21,
2
n
tt
Example - solution
5
The decision rule is:
Action 2 - do not adjust process if
797.21
5797.2
s
X
Example - solution
6
Suppose that for a given day
and
s = 0.012
Then t =
006.1X
012.0
15
X
5.2012.0
1006.15
Example - solution
7
so that - 2.797 < 2.5 < 2.797
and Action 2: no adjustment, is taken. Weconclude that the true mean weight of coffee per can is 1 pound. We have thus tested the statisticalhypothesis that = 1 pound versus the alternativehypothesis that does not equal 1 pound at the 1% level of significance.
Example - solution
8
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 HypothesisH0: = 0 , a given or specified value
against the appropriate Alternative Hypothesis
1. HA: < 0 ,or
2. HA: > 0 ,or
3. HA: 0 ,
Test of Means
9
at the 100 % level of significance. Calculate the value of the test statistic
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.
n
sX
t 0
Test of Means
10
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)
Test on Two Means
11
1. H1: µ1 - µ2 < do, where do 0,
or2. H1: µ1 - µ2 > do, where do 0,
or3. H1: µ1 - µ2 do, where do 0,
at the 100% level of significance, calculate the value of the test statistic.
Test on Two Means
12
Calculate the value of the test statistic
2
22
1
21
021
ns
ns
dXXt'
Test on Two Means
13
Reject Ho if
1. t' < -tν
or 2. t' > tν
or 3. t' < -tν
or t' > tν depending on the alternative
hypothesis, where
1
ns
1
ns
ns
ns
2
2
2
22
1
2
1
21
2
2
22
1
21
nn
Test on Two Means
14
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.
Example - Test on Two Means
15
Test
H0: 1 = 2 or 1 - 2 = 0.
Vs.
H1: 1 2 or 1 - 2 0.
With a 10% level of significance, i.e.,
Then
07.2
ns
ns
dXXt'
2
22
1
21
021
Example
16
where
and
17
1
ns
1
ns
ns
ns
2
2
2
22
1
2
1
21
2
2
22
1
21
nn
725.1tt 17,05.02,2α
Example
The calculate Critical Region is:t’ < -1.725 and t’ > 1.725,
17
Since t’ = 2.07, we can reject H0 and conclude that
the two materials do not exhibit the same abrasive
wear.
Example
18
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 HypothesisH0: 2 = o
2 , a specified value
against the appropriate Alternative Hypothesis
1. HA: 2 < o2
,or
2. HA: 2 > o2
,or
3. HA: 2 o2
,
Test of Variances
19
at the 100 % level of significance. Calculate the value of the test statistic
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.
20
22 1
s
n
Test of Variances
20
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:
against the appropriate alternative hypothesis
22
21 σσ
21
1. H1:
or2. H1:
or3. H1:
at the 100% level of significance, calculate the value of the test statistic.
σσ 22
21
22
21 σσ
22
21 σσ
22
21
S
SF
Test on Two Variances
22
Reject Ho if
or
or
depending on the alternative hypothesis,
and where
and
),(FF 211 vv
)(FF 21α ,vv
),(FFor ),(FF 212/212/1 vvvv
1
1
22
11
nv
nv
Test on Two Variances
23
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.
Example - Test on Variances
24
H0: 12
= 22
H1: 12
22
With a 10% level of significance, i.e.,
Critical region: From the graph we see that
F0.05(11,9) = 3.11
Example - Test of Variances
0 0.34
0.05 0.05
3.11x
v1 = 11 and v2 = 9
f (x)
25
Therefore, the null hypothesis is rejected when F < 0.34 or
F > 3.11.
Decision:
Do not reject H0. Conclude that there is insufficient evidence
that the variances differ.
0.34(9,11)F
1(11,9)F
0.050.95
0.6425
16
S
SF
22
21
Example - Test of Variances