1. Chapter 7 Hypothesis Testing

2. Define a hypothesis and hypothesis testing. Describe the five step hypothesis testing procedure. Distinguish between a one-tailed and a two-tailed test of hypothesis. Conduct a test of hypothesis about a population mean. Conduct a test of hypothesis about a population proportion. Define Type I and Type II errors. GOALS:

3. Hypothesis testing Definitions & steps hypothesis testing One-tailed test Two-tailed test p-value Type I & Type II error

4. What is a Hypothesis? A statement about the value of a population parameter Data are then used to check the reasonableness of the statement Examples  The mean monthly income for systems analysts is Rs15,000.  Twenty percent of all customers at Pizza Hut return for another meal within a month.

5. What is Hypothesis Testing? A procedure, based on sample evidence and probability theory, used to determine whether the hypothesis is a reasonable statement and should not be rejected, or is unreasonable and should be rejected.

6. Definitions Null Hypothesis (H 0 ): A statement about the value of a population parameter Alternative Hypothesis (H 1 ): A statement that is accepted if the sample data provide evidence that the null hypothesis is false. Level of Significance: The probability of rejecting the null hypothesis when it is actually true.

7. Definitions Test statistic: A value, determined from sample information, used to determine whether or not to reject the null hypothesis. Critical value: The dividing point between the region where the null hypothesis is rejected and the region where it is not rejected.

8. Five step procedure for testing a hypothesis Step 1: State null & alternate hypothesis Step 2: Select a level of significance Step 3: Identify the test statistic Step 4: Formulate a decision rule Step 5: Take a sample to arrive at a decision Do not reject null Reject null

9. State null & alternate hypothesis Example 1: A recent article indicated the mean age of US commercial aircraft is 15 years The null hypothesis represents the current or reported condition. It is written as: H 0 :  = 15 The alternative hypothesis is that the statement is not true, that is: H 1 :  15

10. State null & alternate hypothesis Example 2: A recent article indicated the mean age of US commercial aircraft is no more than15 years The null hypothesis represents the current or reported condition. It is written as: H 0 :   15 The alternative hypothesis is that the statement is not true, that is: H 1 :  > 15

11. Select a level of significance The level of significance (sometimes called the level of risk or critical value) is designated as  (Greek letter alpha)

12. Select the test statistic for mean (for large sample) For the mean (  ) when  is known or the sample size is large (n  30), use Z distribution as test statistic Z value is based on the sampling distribution of X, which is normally distributed when the sample is reasonably large with a mean (  x ) equal to  and a standard deviation  x, which is equal to  /  n

13. Select the test statistic for mean (for small sample) When the sample is less than 30 and the population standard deviation is not known we use the t distribution

14. One-Tailed Tests A test is one-tailed when the alternate hypothesis, H 1, states a direction  H1: The mean yearly commissions earned by full- time Insurance agent is more than Rs35,000. (µ>Rs35,000)  H1: The mean speed of trucks traveling on A1 route is less than 60 km per hour. (µ<60)  H1: Less than 20 percent of the customers pay cash for their gasoline purchase. (µ<.20) Note; when region of rejection is only in one tail/end

15. One-Tailed Tests Sampling Distribution for the test statistic (Z) for a one- Tailed Test (0.05 Level of Significance) [H 0 :  h & H 1 :  >  h ] 1.65 0 0.95 0.05 Rejection region Do not reject H 0 Critical value

16. One-Tailed Tests Sampling Distribution for the test statistic (Z) for a one- Tailed Test (0.05 Level of Significance) [H 0 :  h & H 1 :  <  h ] 1.65 0 0.95 0.05 Rejection region Do not reject H 0 Critical value

17. One-tailed test (Rejection of null hypothesis) H 0 :  h & H 1 :  >  h when + critical value < + test statistic H 0 :  h & H 1 :  - test statistic 0 Critical value Test statistic 0 Critical value Test statistic

18. Two-Tailed Tests A test is two-tailed when the alternate hypothesis, H 1, does not state a direction  H1: The mean yearly commissions earned by full-time Insurance agent is not equal Rs35,000. (µ  Rs35,000)  H1: The mean speed of trucks traveling on A1 route is not equal 60 km per hour. (µ  60)

19. Two-Tailed Tests Sampling Distribution for the test statistic (Z) for a one-Tailed Test (0.05 Level of Significance) 1.96 0 0.95 0.025 Rejection region Do not reject H 0 Critical value Rejection region Critical value 0.025 -1.96

20. Two-tailed test (Rejection of null hypothesis) H 0 :  =  h & H 1 :  h when + critical value - test statistic 0 + Critical value - Critical value + Test statistic - Test statistic

21. Example 1 The Glen valley Steel Company manufactures steel bars. If the production process is working properly, it turns out steel bars with mean length of at least 2.8 feet with a standard deviation of 0.20 feet. Longer steel bars can be used or altered but shorter steel bars must be scrapped. A sample of 50 bars is selected from the production line. The sample indicates a mean of length 2.73 feet. The company wants to determine whether the production equipment needs to be adjusted. a) State the null and alternative hypotheses. b) If the company wants to test the hypothesis at the 0.05 level of significance, what decision would be made using the critical value approach to hypothesis testing?

22. Example 1 Step 1 - Null hypothesis & alternative hypothesis H 0 :   2.8 H 1 :  30 use Z test and  = 0.2

23. Example 1 Step 2 – Level of significance level of significance is 0.05 Z value related to 0.05 = -1.645 Step 3 – Identify the test statistic

24. Example 1 Step 4 & 5 – Formulate a decision rule & take a decision -1.645 0 Rejection region Do not reject H 0 -2.47 Critical value Test statistic Decision Rule Test statistic < critical value Reject null hypothesis

25. Example 2 The Glen valley Steel Company manufactures steel bars. If the production process is working properly, it turns out steel bars with mean length of at least 2.8 feet. Longer steel bars can be used or altered but shorter steel bars must be scrapped. A sample of 20 bars is selected from the production line. The sample indicates a mean of length 2.73 feet and a standard deviation of 0.20 feet. The company wants to determine whether the production equipment needs to be adjusted. a) State the null and alternative hypotheses. b) If the company wants to test the hypothesis at the 0.05 level of significance, what decision would be made using the critical value approach to hypothesis testing?

26. Example 2 Step 1 - Null hypothesis & alternative hypothesis H0:   2.8 H1:  < 2.8 One-tailed test because alternative hypothesis shows a direction n = 20 <30 use t-distribution and  = 0.2

27. Example 2 Step 2 – Level of significance level of significance is 0.05 & degree of freedom = 19 t value related to 0.05 = -1.729 Step 3 – Identify the test statistic

28. Example 2 Step 4 & 5 – Formulate a decision rule & take a decision -1.729 0 Rejection region Do not reject H 0 -1.57 Critical value Test statistic Decision Rule Test statistic > critical value Do not reject null hypothesis

29. Exercise 1 The director of manufacturing at a clothing factory needs to determine whether a new machine is producing a particular type of cloth according to the manufacturers specifications, which indicate that the cloth should have a mean breaking strength of 70 pounds and a standard deviation of 3.5 pounds. A sample of 49 pieces of cloth reveals a sample mean breaking strength of 69.1 pounds a) State the null and alternative hypotheses. b) Is there evidence that the machine is not meeting the manufacturers specifications for average breaking strength? (Use a 0.05 level of significance)

30. Exercise 2 The director of manufacturing at a clothing factory needs to determine whether a new machine is producing a particular type of cloth according to the manufacturers specifications, which indicate that the cloth should have a mean breaking strength of 60 pounds and a standard deviation of 2.5 pounds. A sample of 19 pieces of cloth reveals a sample mean breaking strength of 59.1 pounds a) State the null and alternative hypotheses. b) Is there evidence that the machine is not meeting the manufacturers specifications for average breaking strength? (Use a 0.05 level of significance)