Download presentation

Presentation is loading. Please wait.

Published byMalachi Wyre Modified over 2 years ago

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

2
2 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 companys 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 days production. The process will be adjusted if there is evidence to indicate that the true mean amount of coffee is not 1 pound. Example

3
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
4 The decision rule is: Action 1 - adjust process if or if Example - solution

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

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

7
7 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. Example - solution

8
8 Let X 1, …, X n, be a random sample of size n, from a normal distribution with mean and standard deviation, both unknown. To test the Null Hypothesis H 0 : = 0, a given or specified value against the appropriate Alternative Hypothesis 1. H A : < 0, or 2. H A : > 0, or 3. H A : 0, Test of Means

9
9 at the 100 % level of significance. Calculate the value of the test statistic Reject H 0 if 1. t < -t, n-1, 2. t > t, n-1, 3. t t/2, n-1, depending on the Alternative Hypothesis. Test of Means

10
10 Let X 11, X 12, …, X 1n 1 be a random sample of size n 1 from N( 1, 1 ) and X 21, X 22, …, X 2n 2 be a random sample of size n 2 from N( 2, 2 ), where 1, 1, 2 and 2 are all unknown. To test against the appropriate alternative hypothesis H 0: µ 1 - µ 2 = d o, where d o 0 (usually d o =0) Test on Two Means

11
11 1. H 1: µ 1 - µ 2 < d o, where d o 0, or 2. H 1: µ 1 - µ 2 > d o, where d o 0, or 3. H 1: µ 1 - µ 2 d o, where d o 0, at the 100% level of significance, calculate the value of the test statistic. Test on Two Means

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

13
13 Reject H o if 1. t' < -t ν or2. t' > t ν or3. t' < -t ν or t' > t ν depending on the alternative hypothesis, where Test on Two Means

14
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
15 Test H 0: 1 = 2 or 1 - 2 = 0. Vs. H 1: 1 2 or 1 - 2 0. With a 10% level of significance, i.e., Then Example

16
16 where and Example The calculate Critical Region is: t 1.725,

17
17 Since t = 2.07, we can reject H 0 and conclude that the two materials do not exhibit the same abrasive wear. Example

18
18 Let X 1, …, X n, be a random sample of size n, from a normal distribution with mean and standard deviation, both unknown. To test the Null Hypothesis H 0 : 2 = o 2, a specified value against the appropriate Alternative Hypothesis 1. H A : 2 < o 2, or 2. H A : 2 > o 2, or 3. H A : 2 o 2, Test of Variances

19
19 at the 100% level of significance. Calculate the value of the test statistic Reject H 0 if 1. 2 < 2 1-, n-1, 2. 2 > 2, n-1, 3. 2 2/2, n-1, depending on the Alternative Hypothesis. Test of Variances

20
20 Test on Two Variances Let X 11, X 12, …, X 1n 1 be a random sample of size n 1 from N( 1, 1 ) and X 21, X 22, …, X 2n 2 be a random sample of size n 2 from N( 2, 2 ), where 1, 1, 2 and 2 are all unknown. To test H 0: against the appropriate alternative hypothesis

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

22
22 Reject H o if or depending on the alternative hypothesis, and where and Test on Two Variances

23
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
24 H 0: 1 2 = 2 2 H 1: 1 2 2 2 With a 10% level of significance, i.e., Critical region: From the graph we see that F 0.05 (11,9) = 3.11 Example - Test of Variances 00.34 0.05 3.11 x v 1 = 11 and v 2 = 9 f (x)

25
25 Therefore, the null hypothesis is rejected when F 3.11. Decision: Do not reject H 0. Conclude that there is insufficient evidence that the variances differ. Example - Test of Variances

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google