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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

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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

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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

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The decision rule is:

Action 2 - do not adjust process if

797.21

5797.2

s

X

Example - solution

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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

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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

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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

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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

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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

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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

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Calculate the value of the test statistic

2

22

1

21

021

ns

ns

dXXt'

Test on Two Means

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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

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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

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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

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where

and

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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,

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Since t’ = 2.07, we can reject H0 and conclude that

the two materials do not exhibit the same abrasive

wear.

Example

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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

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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

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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

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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

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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

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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

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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)

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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