1. Copyright © Cengage Learning. All rights reserved. 8 Introduction to Statistical Inferences

2. Copyright © Cengage Learning. All rights reserved. 8.4 Hypothesis Test of Mean  (  Known): A Probability-Value Approach

3. 3 The assumption for hypothesis tests about mean  using a known  The sampling distribution of has a normal distribution.

4. 4 Hypothesis Test of Mean  (  Known): A Probability-Value Approach The hypothesis test is a well-organized, step-by-step procedure used to make a decision. Two different formats are commonly used for hypothesis testing. The probability-value approach, or simply p-value approach, is the hypothesis test process that has gained popularity in recent years, largely as a result of the convenience and the “number-crunching” ability of the computer. This approach is organized as a five-step procedure.

5. 5 Hypothesis Test of Mean  (  Known): A Probability-Value Approach The Probability-value Hypothesis Test: A Five-step Procedure Step 1 The Set-Up: a. Describe the population parameter of interest. The population parameter of interest is the mean , the mean shearing strength of (or mean force required to shear) the rivets being considered for purchase.

6. 6 Hypothesis Test of Mean  (  Known): A Probability-Value Approach b. State the null hypothesis (H o ) and the alternative hypothesis (H a ). The null hypothesis and the alternative hypothesis are formulated by inspecting the problem or statement to be investigated and first formulating two opposing statements about the mean . For our example, these two opposing statements are (A) “The mean shearing strength is less than 925” (  < 925, the aircraft manufacturer’s concern), and

7. 7 Hypothesis Test of Mean  (  Known): A Probability-Value Approach (B) “The mean shearing strength is at least 925” (  = 925, the rivet supplier’s claim and the aircraft manufacturer’s spec). The three possible combinations of signs and hypotheses are shown in Table 8.5 The Three Possible Statements of Null and Alternative Hypotheses Table 8.5

8. 8 Example 13 – Writing Null and Alternative Hypotheses (One-Tailed Situation) Suppose the Environmental Protection Agency is suing the city of Rochester for noncompliance with carbon monoxide standards. Specifically, the EPA would want to show that the mean level of carbon monoxide in downtown Rochester’s air is dangerously high, higher than 4.9 parts per million. State the null and alternative hypotheses.

9. 9 Example 13 – Solution To state the two hypotheses, we first need to identify the population parameter in question: the “mean level of carbon monoxide in Rochester.” The parameter  is being compared with the value 4.9 parts per million, the specific value of interest. The EPA is questioning the value of  and wishes to show it is higher than 4.9 (i.e.,  > 4.9).

10. 10 Example 13 – Solution The three possible relationships— (1)   4.9, (2)  = 4.9, and (3)  > 4.9 —must be arranged to form two opposing statements: one states the EPA’s position, “The mean level is higher than 4.9 (  > 4.9),” and the other states the negation, “The mean level is not higher than 4.9 (   4.9).” cont’d

11. 11 Example 13 – Solution One of these two statements will become the null hypothesis, H o, and the other will become the alternative hypothesis, H a. We know that there are two rules for forming the hypotheses: (1) the null hypothesis states that the parameter in question has a specified value (“H o must contain the equal sign”), and (2) the EPA’s contention becomes the alternative hypothesis (“higher than”). Both rules indicate: H o :  = 4.9 (  ) and H a :  > 4.9 cont’d

12. 12 Example 15 – Writing Null and Alternative Hypotheses (Two-tailed Situation) Job satisfaction is very important to worker productivity. A standard job satisfaction questionnaire was administered by union officers to a sample of assembly-line workers in a large plant in hopes of showing that the assembly workers’ mean score on this questionnaire would be different from the established mean of 68. State the null and alternative hypotheses.

13. 13 Example 15 – Solution Either the mean job-satisfaction score is different from 68 (   68) or the mean is equal to 68 (  = 68). Therefore, H o :  = 68 and H a :   68 Notes 1. The alternative hypothesis is referred to as being “two-tailed” when H a is “not equal.” 2. When “less than” is combined with “greater than,” they become “not equal to.”

14. 14 Hypothesis Test of Mean  (  Known): A Probability-Value Approach Step 2 The Hypothesis Test Criteria: a. Check the assumptions. Assume that the standard deviation of the shearing strength of rivets is known from past experience to be  = 18. Variables like shearing strength typically have a mounded distribution; therefore, a sample of size 50 should be large enough for the CLT to apply and ensure that the SDSM will be normally distributed.

15. 15 Hypothesis Test of Mean  (  Known): A Probability-Value Approach b. Identify the probability distribution and the test statistic to be used. The standard normal probability distribution is used because is expected to have a normal distribution. The resulting calculated value is identified as z (“z star”) because it is expected to have a standard normal distribution when the null hypothesis is true and the assumptions have been satisfied.

16. 16 Hypothesis Test of Mean  (  Known): A Probability-Value Approach The (“star”) is to remind us that this is the calculated value of the test statistic. The test statistic to be used is with  = 18. c. Determine the level of significance, . The type I error occurs when a true null hypothesis is rejected. This would occur when the manufacturer tested rivets that did meet the specs and rejected them.

17. 17 Hypothesis Test of Mean  (  Known): A Probability-Value Approach Undoubtedly this would lead to the rivets not being purchased even though they do meet the specs. In order for the manager to set a level of significance, related information is needed—namely, how soon is the new supply of rivets needed? If they are needed tomorrow and this is the only vendor with an available supply, waiting a week to find acceptable rivets could be very expensive; therefore, rejecting good rivets could be considered a serious error.

18. 18 Hypothesis Test of Mean  (  Known): A Probability-Value Approach On the other hand, if the rivets are not needed until next month, then this error may not be very serious. Only the manager will know all the ramifications, and therefore the manager’s input is important here. After much consideration, the manager assigns the level of significance:  =0.05.

19. 19 Hypothesis Test of Mean  (  Known): A Probability-Value Approach Step 3 The Sample Evidence: a. Collect the sample information. The sample must be a random sample drawn from the population whose mean  is being questioned. A random sample of 50 rivets is selected, each rivet is tested, and the sample mean shearing strength is calculated: = 921.18 and n = 50.

20. 20 Hypothesis Test of Mean  (  Known): A Probability-Value Approach b. Calculate the value of the test statistic. The sample evidence ( and n found in Step 3a) is next converted into the calculated value of the test statistic, z, using formula (8.4). (  is 925 from H o, and  = 18 is a known quantity.) We have

21. 21 Hypothesis Test of Mean  (  Known): A Probability-Value Approach Step 4 The Probability Distribution: a. Calculate the p-value for the test statistic. Probability value, or p-value The probability that the test statistic could be the value it is or a more extreme value (in the direction of the alternative hypothesis) when the null hypothesis is true. (Note: The symbol P will be used to represent the p-value, especially in algebraic situations.)

22. 22 Hypothesis Test of Mean  (  Known): A Probability-Value Approach To find the p-value, you may use any one of the three methods outlined here. The method you use is not the important thing, because each method is just the tool of choice to help you find the p-value. Method 1: Use Table 3 in Appendix B to determine the tabled area related to the left of z = –1.50: p-value = P (z < z ) = P (z < –1.50) = 0.0668

23. 23 Hypothesis Test of Mean  (  Known): A Probability-Value Approach Method 2: Use Table 5 in Appendix B and the symmetry property: Table 5 is set up to allow you to read the p-value directly from the table. Since P (z 1.50), simply locate z = 1.50 on Table 5 and read the p-value: P (z < –1.50) = 0.0668

24. 24 Hypothesis Test of Mean  (  Known): A Probability-Value Approach Method 3: Use the cumulative probability function on a computer or calculator to find the p-value: P (z < –1.50) = 0.0668 b. Determine whether or not the p-value is smaller than . The p-value (0.0668) is not smaller than  (0.05).

25. 25 Hypothesis Test of Mean  (  Known): A Probability-Value Approach Step 5 The Results: a. State the decision about H o. Is the p-value small enough to indicate that the sample evidence is highly unlikely in the event that the null hypothesis is true? In order to make the decision, we need to know the decision rule.

26. 26 Hypothesis Test of Mean  (  Known): A Probability-Value Approach Decision rule a. If the p-value is less than or equal to the level of significance , then the decision must be reject H o. b. If the p-value is greater than the level of significance a, then the decision must be fail to reject H o. Decision about H o : Fail to reject H o.

27. 27 Hypothesis Test of Mean  (  Known): A Probability-Value Approach b. State the conclusion about H a. There is not sufficient evidence at the 0.05 level of significance to show that the mean shearing strength of the rivets is less than 925. “We failed to convict” the null hypothesis. In other words, a sample mean as small as 921.18 is likely to occur (as defined by  ) when the true population mean value is 925.0 and is normally distributed. The resulting action by the manager would be to buy the rivets.

28. 28 Hypothesis Test of Mean  (  Known): A Probability-Value Approach Table 8.7 outlines the procedure for all three cases. Table 8.7 Finding p-Values Using the Cumulative Distribution

29. 29 Hypothesis Test of Mean  (  Known): A Probability-Value Approach The fundamental idea of the p-value is to express the degree of belief in the null hypothesis: When the p-value is minuscule (something like 0.0003), the null hypothesis would be rejected by everybody because the sample results are very unlikely for a true H o. When the p-value is fairly small (like 0.012), the evidence against H o is quite strong and H o will be rejected by many.

30. 30 Hypothesis Test of Mean  (  Known): A Probability-Value Approach When the p-value begins to get larger (say, 0.02 to 0.08), there is too much probability that data like the sample involved could have occurred even if H o were true, and the rejection of H o is not an easy decision. When the p-value gets large (like 0.15 or more), the data are not at all unlikely if the H o is true, and no one will reject H o.

31. 31 Hypothesis Test of Mean  (  Known): A Probability-Value Approach The advantages of the p-value approach are as follows: (1) The results of the test procedure are expressed in terms of a continuous probability scale from 0.0 to 1.0, rather than simply on a “reject” or “fail to reject” basis. (2) A p-value can be reported and the user of the information can decide on the strength of the evidence as it applies to his or her own situation. (3) Computers can do all the calculations and report the p-value, thus eliminating the need for tables.

32. 32 Hypothesis Test of Mean  (  Known): A Probability-Value Approach The disadvantage of the p-value approach is the tendency for people to put off determining the level of significance. This should not be allowed to happen, because it is then possible for someone to set the level of significance after the fact, leaving open the possibility that the “preferred” decision will result. However, this is probably important only when the reported p-value falls in the “hard choice” range (say, 0.02 to 0.08), as described previously.

33. 33 Example 17 – Two-tailed Hypothesis Test With Sample Data The mean of single-digit random numbers is 4.5 and the standard deviation is  = 2.87. Draw a random sample of 40 single-digit numbers from Table 1 in Appendix B and test the hypothesis “The mean of the single-digit numbers in Table 1 is 4.5.” Use  = 0.10.

34. 34 Example 17 – Solution Step 1 The Set-Up: a. Describe the population parameter of interest. The population parameter of interest is the mean  of the population of single-digit numbers in Table 1 of Appendix B. b. State the null hypothesis (H o ) and the alternative hypothesis (H a ). H o :  = 4.5 (mean is 4.5) H a :   4.5 (mean is not 4.5)

35. 35 Example 17 – Solution Step 2 The Hypothesis Test Criteria: a. Check the assumptions.  is known. Samples of size 40 should be large enough to satisfy the CLT; see the discussion of this issue on page 370. b. Identify the probability distribution and the test statistic to be used. We use the standard normal probability distribution, and the test statistic is ;  = 2.87. cont’d

36. 36 Example 17 – Solution c. Determine the level of significance, .  = 0.10(given in the statement of the problem) Step 3 The Sample Evidence: a. Collect the sample information. This random sample was drawn from Table 1 in Appendix B [TA08-01]: From the sample = 3.975 and n = 40. cont’d

37. 37 Example 17 – Solution b. Calculate the value of the test statistic. We use formula (8.4), and  is 4.5 from H o, and  = 2.87: = –1.156 = –1.16 cont’d

38. 38 Example 17 – Solution Step 4 The Probability Distribution: a. Calculate the p-value for the test statistic. Because the alternative hypothesis indicates a two-tailed test, we must find the probability associated with both tails. The p-value is found by doubling the area of one tail. z = –1.16. The p-value = P = 2  P (z  –1.16) = 2(0.1230) = 0.2460 cont’d

39. 39 Example 17 – Solution b. Determine whether or not the p-value is smaller than . The p-value (0.2460) is greater than  (0.10). cont’d

40. 40 Example 17 – Solution Step 5 The Results: a. State the decision about H o : Fail to reject H o. b. State the conclusion about H a. The observed sample mean is not significantly different from 4.5 at the 0.10 level of significance. cont’d