1. Lecture 7 Introduction to Hypothesis Testing

2. Lecture Goals After completing this lecture, you should be able to: Formulate null and alternative hypotheses for applications involving a single population mean or proportion Formulate a decision rule for testing a hypothesis Know how to use the test statistic, critical value, and p-value approaches to test the null hypothesis Know what Type I and Type II errors are Compute the probability of a Type II error

3. What is a Hypothesis? A hypothesis is a claim (assumption) about a population parameter: –population mean Example: The mean DMF score for all 12 years old children in this city is claim to be 3 DMFT per child.

4. The Null Hypothesis, H 0 States the assumption (numerical) to be tested Example: The average number of DMF score for 12 years old children in Riyadh is at least three ( ) Is always about a population parameter, not about a sample statistic

5. The Null Hypothesis, H 0 Begin with the assumption that the null hypothesis is true –Similar to the notion of innocent until proven guilty Refers to the status quo Always contains “=”, “≤” or “  ” sign May or may not be rejected (continued)

6. The Alternative Hypothesis, H A Is the opposite of the null hypothesis –e.g.: The average number of DMF score for 12 years old children in Riyadh is at most les than three DMFT per child (H A :  < 3) Challenges the status quo Never contains the “=”, “≤” or “  ” sign May or may not be accepted Is generally the hypothesis that is believed (or needs to be supported) by the researcher

7. Population Claim: the population mean age is 50. (Null Hypothesis: REJECT Suppose the sample mean age is 20: x = 20 Sample Null Hypothesis 20 likely if  = 50?  Is Hypothesis Testing Process If not likely, Now select a random sample H 0 :  = 50 ) x

8. Sampling Distribution of x  = 50 If H 0 is true If it is unlikely that we would get a sample mean of this value...... then we reject the null hypothesis that  = 50 Reason for Rejecting H 0 20... if in fact this were the population mean… x

9. Level of Significance,  Defines unlikely values of sample statistic if null hypothesis is true –Defines rejection region of the sampling distribution Is designated by , (level of significance) –Typical values are.01,.05, or.10 Is selected by the researcher at the beginning Provides the critical value(s) of the test

10. Level of Significance and the Rejection Region H 0 : μ ≥ 3 H A : μ < 3 0 H 0 : μ ≤ 3 H A : μ > 3 H 0 : μ = 3 H A : μ ≠ 3   /2 Represents critical value Lower tail test Level of significance =  0 0  /2  Upper tail test Two tailed test Rejection region is shaded

11. Errors in Making Decisions Type I Error –Reject a true null hypothesis –Considered a serious type of error The probability of Type I Error is  Called level of significance of the test Set by researcher in advance

12. Errors in Making Decisions Type II Error –Fail to reject a false null hypothesis The probability of Type II Error is β Note: Power of the test = 1- β (continued)

13. Outcomes and Probabilities State of Nature Decision Do Not Reject H 0 No error (1 - )  Type II Error ( β ) Reject H 0 Type I Error ( )  Possible Hypothesis Test Outcomes H 0 False H 0 True Key: Outcome (Probability) No Error ( 1 - β )

14. Type I & II Error Relationship  Type I and Type II errors can not happen at the same time  Type I error can only occur if H 0 is true  Type II error can only occur if H 0 is false If Type I error probability (  ), then Type II error probability ( β )

15. Factors Affecting Type II Error All else equal, – β when the difference between hypothesized parameter and its true value – β when  – β when σ – β when n

16. Critical Value Approach to Testing Convert sample statistic (e.g.: ) to test statistic ( Z or t statistic ) Determine the critical value(s) for a specified level of significance  from a table or computer If the test statistic falls in the rejection region, reject H 0 ; otherwise do not reject H 0

17. Reject H 0 Do not reject H 0  The cutoff value, or, is called a critical value Lower Tail Tests  -z α xαxα xαxα 0 μ H 0 : μ ≥ 3 H A : μ < 3

18. Reject H 0 Do not reject H 0 The cutoff value, or, is called a critical value Upper Tail Tests  zαzα xαxα zαzα xαxα 0 μ H 0 : μ ≤ 3 H A : μ > 3

19. Do not reject H 0 Reject H 0 There are two cutoff values (critical values): or Two Tailed Tests  /2 -z α/2 x α/2 ± z α/2 x α/2 0 μ0μ0 H 0 : μ = 3 H A : μ  3 z α/2 x α/2 Lower Upper x α/2 LowerUpper  /2

20. Critical Value Approach to Testing Convert sample statistic ( ) to a test statistic ( Z or t statistic ) x  Known Large Samples  Unknown Hypothesis Tests for  Small Samples

21.  Known Large Samples  Unknown Hypothesis Tests for μ Small Samples The test statistic is: Calculating the Test Statistic

22.  Known Large Samples  Unknown Hypothesis Tests for  Small Samples The test statistic is: Calculating the Test Statistic But is sometimes approximated using a z: (continued)

23.  Known Large Samples  Unknown Hypothesis Tests for  Small Samples The test statistic is: Calculating the Test Statistic (The population must be approximately normal) (continued)

24. Review: Steps in Hypothesis Testing 1. Specify the population value of interest 2.Formulate the appropriate null and alternative hypotheses 3. Specify the desired level of significance 4. Determine the rejection region 5. Obtain sample evidence and compute the test statistic 6. Reach a decision and interpret the result

25. Hypothesis Testing Example Test the claim that The average number of DMF score for 12 years old children in Riyadh is at least three DMFT per child (Assume σ = 0.8) 1. Specify the population value of interest The mean number of DMF score for 12 years old children in Riyadh is at least three DMFT per child 2. Formulate the appropriate null and alternative hypotheses H 0 : μ  3 H A : μ < 3 (This is a lower tail test) 3. Specify the desired level of significance Suppose that  =.05 is chosen for this test

26. Reject H 0 Do not reject H 0  4. Determine the rejection region  =.05 -z α = -1.645 0 This is a one-tailed test with  =.05. Since σ is known, the cutoff value is a z value: Reject H 0 if z < z  = -1.645 ; otherwise do not reject H 0 Hypothesis Testing Example (continued)

27.  5. Obtain sample evidence and compute the test statistic Suppose a sample is taken with the following results: n = 100, x = 2.84 (  = 0.8 is assumed known) –Then the test statistic is: Hypothesis Testing Example

28. Reject H 0 Do not reject H 0  =.05 -1.6450  6. Reach a decision and interpret the result -2.0 Since z = -2.0 < -1.645, we reject the null hypothesis that the DMF score for 12 years old children in Riyadh is at least three DMFT per child Hypothesis Testing Example (continued) z

29. Reject H 0  =.05 2.8684 Do not reject H 0 3 An alternate way of constructing rejection region: 2.84 Since x = 2.84 < 2.8684, we reject the null hypothesis Hypothesis Testing Example (continued) x Now expressed in x, not z units

30. p-Value Approach to Testing Convert Sample Statistic (e.g. ) to Test Statistic ( Z or t statistic ) Obtain the p-value from a table or computer Compare the p-value with  –If p-value < , reject H 0 –If p-value  , do not reject H 0 x

31. p-Value Approach to Testing  p-value: Probability of obtaining a test statistic more extreme ( ≤ or  ) than the observed sample value given H 0 is true –Also called observed level of significance –Smallest value of  for which H 0 can be rejected (continued)

32. p-value =.0228  =.05 p-value example Example: How likely is it to see a sample mean of 2.84 (or something further below the mean) if the true mean is  = 3.0? 2.86843 2.84 x

33. Compare the p-value with  –If p-value < , reject H 0 –If p-value  , do not reject H 0 Here: p-value =.0228  =.05 Since.0228 <.05, we reject the null hypothesis (continued) p-value example p-value =.0228  =.05 2.86843 2.84

34. Example: Upper Tail z Test for Mean (  Known) A demographic manager thinks that the average age of Saudi population have decreased, and now average over 52 years. A researcher wishes to test this claim. (Assume  = 10 is known) H 0 : μ ≤ 52 the average is not over 52 years H A : μ > 52 the average is greater than 52 years (i.e., sufficient evidence exists to support the manager’s claim) Form hypothesis test:

35. Reject H 0 Do not reject H 0 Suppose that  =.10 is chosen for this test Find the rejection region:  =.10 z α =1.28 0 Reject H 0 Reject H 0 if z > 1.28 Example: Find Rejection Region (continued)

36. Review: Finding Critical Value - One Tail Z.07.09 1.1.3790.3810.3830 1.2.3980.4015 1.3.4147.4162.4177 z 01.28. 08 Standard Normal Distribution Table (Portion) What is z given  = 0.10?  =.10 Critical Value = 1.28.90.3997.10.40.50

37. Obtain sample evidence and compute the test statistic Suppose a sample is taken with the following results: n = 64, x = 53.1 (  =10 was assumed known) –Then the test statistic is: Example: Test Statistic (continued)

38. Reject H 0 Do not reject H 0 Example: Decision  =.10 1.28 0 Reject H 0 Do not reject H 0 since z = 0.88 ≤ 1.28 i.e.: there is not sufficient evidence that the mean age of Saudi population is over 52 years z =.88 Reach a decision and interpret the result: (continued)

39. Reject H 0  =.10 Do not reject H 0 1.28 0 Reject H 0 z =.88 Calculate the p-value and compare to  (continued) p-value =.1894 p -Value Solution Do not reject H 0 since p-value =.1894 >  =.10

40. Example: Two-Tail Test (  Unknown) The average cost of a privet dentistry clinic visit in Riyadh is said to be 168 SR per visit. A random sample of 25 Dentistry Clinics resulted in SR x = 172.50 and s = 15.40 Test at the  = 0.05 level. (Assume the population distribution is normal) H 0 : μ  = 168 H A : μ  168

41.   = 0.05  n = 25   is unknown, so use a t statistic  Critical Value: t 24 = ± 2.0639 Example Solution: Two-Tail Test Do not reject H 0 : not sufficient evidence that true mean cost is different than 168 SR Reject H 0  /2=.025 -t α/2 Do not reject H 0 0 t α/2  /2=.025 -2.0639 2.0639 1.46 H 0 : μ  = 168 H A : μ  168

42. Reject H 0 : μ  52 Do not reject H 0 : μ  52 Type II Error  Type II error is the probability of failing to reject a false H 0 5250 Suppose we fail to reject H 0 : μ  52 when in fact the true mean is μ = 50 

43. Reject H 0 :   52 Do not reject H 0 :   52 Type II Error  Suppose we do not reject H 0 :   52 when in fact the true mean is  = 50 5250 This is the true distribution of x if  = 50 This is the range of x where H 0 is not rejected (continued)

44. Reject H 0 : μ  52 Do not reject H 0 : μ  52 Type II Error  Suppose we do not reject H 0 : μ  52 when in fact the true mean is μ = 50  5250 β Here, β = P( x  cutoff ) if μ = 50 (continued)

45. Reject H 0 : μ  52 Do not reject H 0 : μ  52  Suppose n = 64, σ = 6, and  =.05  5250 So β = P( x  50.766 ) if μ = 50 Calculating β (for H 0 : μ  52) 50.766

46. Reject H 0 : μ  52 Do not reject H 0 : μ  52  Suppose n = 64, σ = 6, and  =.05  5250 Calculating β (continued) Probability of type II error: β =.1539

47. Summary Addressed hypothesis testing methodology Performed z Test for the mean (σ known) Discussed p–value approach to hypothesis testing Performed one-tail and two-tail tests... Performed t test for the mean (σ unknown) Discussed type II error and computed its probability