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

2. Copyright © Cengage Learning. All rights reserved. 8.5 Hypothesis Test of Mean  (  known): A Classical Approach (Optional)

3. 3 The assumption for hypothesis tests about mean  using a known  The sampling distribution of has a normal distribution. The information we need to ensure that this assumption is satisfied is contained in the sampling distribution of sample means and in the central limit theorem.

4. 4 Hypothesis Test of Mean  (  known): A Classical Approach (Optional) 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 classical approach is the hypothesis test process that has enjoyed popularity for many years. This approach is organized as a five-step procedure. The Classical Hypothesis Test: A Five-Step Procedure Step 1 The Set-Up: a. Describe the population parameter of interest. b. State the null hypothesis (H o ) and the alternative hypothesis (H  ).

5. 5 Hypothesis Test of Mean  (  known): A Classical Approach (Optional) Step 2 The Hypothesis Test Criteria: a. Check the assumptions. b. Identify the probability distribution and the test statistic to be used. c. Determine the level of significance, . Step 3 The Sample Evidence: a. Collect the sample information. b. Calculate the value of the test statistic.

6. 6 Hypothesis Test of Mean  (  known): A Classical Approach (Optional) Step 4 The Probability Distribution: a. Determine the critical region and critical value(s). b. Determine whether or not the calculated test statistic is in the critical region. Step 5 The Results: a. State the decision about H o. b. State the conclusion about H .

7. 7 Hypothesis Test of Mean  (  known): A Classical Approach (Optional) 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. b. State the null hypothesis (H o ) and the alternative hypothesis (H  ). 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 .

8. 8 Hypothesis Test of Mean  (  known): A Classical Approach (Optional) For our example, these two opposing statements are: (A)“The mean shearing strength is less than 925” (   925, the aircraft manufacturer’s concern), and (B) “The mean shearing strength is at least 925” (  = 925, the rivet supplier’s claim and the aircraft manufacturer’s spec).

9. 9 Hypothesis Test of Mean  (  known): A Classical Approach (Optional) The Three Possible Statements of Null and Alternative Hypotheses Table 8.9

10. 10 Example 18 – Writing Null and Alternative Hypotheses (One-Tailed Situation) A consumer advocate group would like to disprove a car manufacturer’s claim that a specific model will average 24 miles per gallon of gasoline. Specifically, the group would like to show that the mean miles per gallon is considerably less than 24. State the null and alternative hypotheses. Solution: To state the two hypotheses, we first need to identify the population parameter in question: The “mean mileage attained by this car model.” The parameter  is being compared with the value 24 miles per gallon, the specific value of interest.

11. 11 Example 18 – Solution The advocates are questioning the value of  and wish to show it to be less than 24 (i.e.,   24).There are three possible relationships: (1)   24, (2)  = 24, and (3)  > 24. These three cases must be arranged to form two opposing statements: one states what the advocates are trying to show, “The mean level is less than 24 (   24),” whereas the “negation” is “The mean level is not less than 24 (   24).” cont’d

12. 12 Example 18 – Solution One of these two statements will become the null hypothesis H o, and the other will become the alternative hypothesis H . Note: Recall 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 consumer advocate group’s contention becomes the alternative hypothesis (“less than”). Both rules indicate: H o :  = 24 (  ) and H  :   24 cont’d

13. 13 Example 20 – 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. Solution: Either the mean job-satisfaction score is different from 68 (  ≠ 68) or the mean score is equal to 68 (  = 68). Therefore, H o :  = 68 and H  :  ≠ 68

14. 14 Hypothesis Test of Mean  (  known): A Classical Approach (Optional) Common Phrases and Their Negations Table 8.10

15. 15 Hypothesis Test of Mean  (  known): A Classical Approach (Optional) 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 satisfy the assumption; the SDSM is normally distributed. 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 or approximately normal distribution.

16. 16 Hypothesis Test of Mean  (  known): A Classical Approach (Optional) Test Statistic for Mean 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. The (“star”) is to remind us that this is the calculated value of the test statistic. The test statistic to be used is z (8.4)

17. 17 Hypothesis Test of Mean  (  known): A Classical Approach (Optional) Step 3 The Sample Evidence: a. Collect the sample information. We are ready for the data. 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. 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 the known quantity.)

18. 18 Hypothesis Test of Mean  (  known): A Classical Approach (Optional) We have

19. 19 Hypothesis Test of Mean  (  known): A Classical Approach (Optional) Step 4 The Probability Distribution: a. Determine the critical region and critical value (s). The standard normal variable z is our test statistic for this hypothesis test. Critical region The set of values for the test statistic that will cause us to reject the null hypothesis. The set of values that are not in the critical region is called the noncritical region (sometimes called the acceptance region). Critical value(s) The “first” or “boundary” value(s) of the critical region(s).

20. 20 Hypothesis Test of Mean  (  known): A Classical Approach (Optional) The critical value for our example is –z (0.05) and has the value of –1.65, as found in Table 4A in Appendix B.

21. 21 Hypothesis Test of Mean  (  known): A Classical Approach (Optional) b. Determine whether or not the calculated test statistic is in the critical region. Graphically this determination is shown by locating the value for z on the sketch in Step 4a. The calculated value of z, z = –1.50, is not in the critical region (it is in the unshaded portion of the figure).

22. 22 Hypothesis Test of Mean  (  known): A Classical Approach (Optional) Step 5 The Result: a. State the decision about H o. In order to make the decision, we need to know the decision rule. Decision rule a. If the test statistic falls within the critical region, then the decision must be reject H o. (The critical value is part of the critical region.) b. If the test statistic is not in the critical region, then the decision must be fail to reject H o. The decision is: Fail to reject H o.

23. 23 Hypothesis Test of Mean  (  known): A Classical Approach (Optional) b. State the conclusion about H . There is not sufficient evidence at the 0.05 level of significance to show that the rivets have a mean shearing strength less than 925.“We failed to convict” the null hypothesis. In other words, a sample mean as small as 921.18 is not unlikely to occur (as defined by  ) when the true population mean value is 925.0. Therefore, the resulting action would be to buy the rivets.

24. 24 Hypothesis Test of Mean  (  known): A Classical Approach (Optional) Before we look at another example, let’s summarize briefly some of the details we have seen thus far: 1. The null hypothesis specifies a particular value of a population parameter. 2. The alternative hypothesis can take three forms. Each form dictates a specific location of the critical region(s), as shown in the following table. 3. For many hypothesis tests, the sign in the alternative hypothesis “points” in the direction in which the critical region is located.

25. 25 Hypothesis Test of Mean  (  known): A Classical Approach (Optional) (Think of the not-equal-to sign [≠] as being both less than [  ] and greater than [  ] thus pointing in both directions.)

26. 26 Example 22 – 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. Solution: Step 1 The Set-Up: a. Describe the population parameter of interest. The parameter of interest is the mean  of the population of single digit numbers in Table 1 of Appendix B.

27. 27 Example 22 – Solution b. State the null hypothesis (H o ) and the alternative hypothesis (H  ). H o :  = 4.5 (mean weight is 4.5) H  :  ≠ 4.5 (mean weight is not 4.5) Step 2 The Hypothesis Test Criteria: a. Check the assumptions.  is known. Samples of size 40 should be large enough to satisfy the CLT. cont’d

28. 28 Example 22 – Solution b. Identify the probability distribution and the test statistic to be used. We use the standard normal probability distribution and the test statistic z  = 2.87 c. Determine the level of significance, .  = 0.10 (given in the statement of the problem) cont’d

29. 29 Example 22 – Solution Step 3 The Sample Evidence: a. Collect the sample information. This random sample was drawn from Table 1 in Appendix B. The sample statistics are = 3.975 and n = 40. Random Sample of Single-Digit Numbers [TA08-01] Table 8.11 cont’d

30. 30 Example 22 – Solution b. Calculate the value of the test statistic. Use formula (8.4), information from H o :  = 4.5, and  = 2.87: z : z = –1.156 = –1.16 cont’d

31. 31 Example 22 – Solution Step 4 The Probability Distribution: a. Determine the critical region and critical value(s). A two-tailed critical region will be used, and 0.05 will be the area in each tail. The critical values are  z(0.05) =  1.65 b. Determine whether or not the calculated test statistic is in the critical region. The calculated value of z, z = –1.16, is not in the critical region (shown in red on the figure). cont’d

32. 32 Example 22 – Solution Step 5 The Result: a. State the decision about H o : Fail to reject H o. b. State the conclusion about H . The observed sample mean is not significantly different from 4.5 at the 0.10 significance level. cont’d