1. Department of Engineering Management, Information and Systems
EMIS 7370 STAT 5340
Department of Engineering Management, Information and Systems
Probability and Statistics for Scientists and Engineers
Tests of Hypothesis
Tests of Means and Variances
Dr. Jerrell T. Stracener

2. 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.

3. 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.

4. Example - solution The decision rule is: Action 1 - adjust process if
or if

5. Example - solution The decision rule is:
Action 2 - do not adjust process if

6. Example - solution
Suppose that for a given day
and
s = 0.012
Then t =

7. 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.

8. 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
1. HA:  < 0 , or
2. HA:  > 0 ,
3. HA:   0 ,

9. Test of Means 1. t < -t, n-1 , 2. t > t, n-1 ,
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.
Test of Means

10. Test on Two Means
Let X11, X12, …, X1n1 be a random sample of size n1 from N(m1, s1) and X21, X22, …, X2n2 be a random sample of size n2 from N(m2, s2), where m1, s1, m2 and s2 are all unknown.
To test against the appropriate alternative hypothesis
H0: µ1 - µ2 = do, where do  0 (usually do=0)

11. Test on Two Means 1. H1: µ1 - µ2 < do, where do  0, or
at the a  100% level of significance, calculate the value of the test statistic.

12. Test on Two Means
Calculate the value of the test statistic

13. Test on Two Means Reject Ho if 1. t' < -ta,ν or 2. t' > ta, ν
or t' > ta/2, ν depending on the alternative hypothesis, where

14. 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.

15. Example Test H0: m1 = m2 or m1 - m2 = 0. Vs.
With a 10% level of significance, i.e., a = 0.10
Then

16. Example
where
and
The calculate Critical Region is: t’ < and t’ > 1.725,

17. Example
Since t’ = 2.07, we can reject H0 and conclude that the two materials do not exhibit the same abrasive wear.

18. 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
1. HA: 2 < o2 , or
2. HA: 2 > o2 ,
3. HA: 2  o2 ,

19. Test of Variances 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. Test on Two Variances To test H0:
Let X11, X12, …, X1n1 be a random sample of size n1 from N(m1, s1) and X21, X22, …, X2n2 be a random sample of size n2 from N(m2, s2), where m1, s1, m2 and s2 are all unknown.
To test
H0:
against the appropriate alternative hypothesis

21. Test on Two Variances 1. H1: or 2. H1: 3. H1:
at the a  100% level of significance, calculate the value of the test statistic.

22. Test on Two Variances Reject Ho if or
depending on the alternative hypothesis,
and where
and

23. 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.

24. Example - Test of Variances
H0: s12 = s22
H1: s12  s22
With a 10% level of significance, i.e., a = 0.10
Critical region: From the graph we see that F0.05(11,9) = 3.11
0.34
0.05
3.11 x
v1 = 11 and v2 = 9
f (x)

25. Example - Test of Variances
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.