1. Topic 8 Hypothesis Testing Mathematics & Statistics Statistics

2. Topic Goals After completing this topic, you should be able to: Formulate null and alternative hypotheses for applications involving a population mean from a normal distribution a population proportion (large samples) Formulate a decision rule for testing a hypothesis Know how to use the critical value and p-value approaches to test the null hypothesis (for both mean and proportion problems) Know what Type I and Type II errors are Assess the power of a test

3. What is a Hypothesis? A hypothesis is a claim (assumption) about a population parameter: population mean population proportion Example: The mean monthly cell phone bill in Dublin is more at least €42, i.e. μ ≥ 42 Example: The proportion of Irish voters in favour of a Dublin metro is 70%, i.e. p =.70

4. Hypothesis testing Procedure Comparable to trial procedures Start with two hypotheses Null hypothesis (defendent innocent) Alternative hypothesis (denfendent guilty) Initially assume that the null hypothesis is true innocent until proven guilty Gather data (evidence) and compute sample statistics Decide to reject null hypothesis or not How to make such a decision?

5. Hypothesis Tests for the Mean  Known (Z-test)  Unknown (T-test) Hypothesis Tests for 

6. The Null Hypothesis, H 0 States the assumption (numerical) to be tested Example: The mean monthly cell phone bill in Dublin is μ ≥ €42 ( ) Is always about a population parameter, not about a sample statistic

7. The Null Hypothesis Refers to the status quo Always contains “=”, “≤” or “  ” sign May or may not be rejected Is generally the hypothesis that the researcher is trying to challenge (continued)

8. The Alternative Hypothesis, H 1 Is the opposite of the null hypothesis e.g., The average phone bill in Dublin is less than €42 ( H 1 : μ < 42 ) Challenges the status quo Never contains the “=”, “≤” or “  ” sign May or may not be supported Is generally the hypothesis that the researcher is trying to support

9. Population Claim: the population mean phone bill is ≥ €42. (Null Hypothesis: REJECT Suppose the sample mean bill is €20: X = 20 Sample Null Hypothesis 20 likely if μ ≥ 42?  Is Hypothesis Testing Process If not likely, Now select a random sample H 0 : μ ≥ 42 ) X

10. Sampling Distribution of X μ = 42 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 μ ≥ 42. Reason for Rejecting H 0 20... if in fact this were the population mean… X

11. Main idea If X “too small”: reject H 0 :μ ≥ 42

12. Example Even if μ=42 is the true mean it is still possible that we observe a sample mean of 20 by sheer chance Suppose the amount of a phone bill in Dublin (call it X) is normal with (known) st. dev. σ=30. If n=9, we know that the sample mean is normal with So,

13. Example So, if we find a sample mean of 20 and if the population mean is 42......we have observed an event that occurs with probability.0139. Do we deem this “unlikely”? Below what probability will we call an event “unlikely”? This threshold probability is chosen by the researcher and is called the.... (continued)

14. Level of Significance,  The threshold probability below which we will judge observations to be “unlikely” Defines a rejection region of the sampling distribution Values for the sample mean that are deemed “unlikely”. Is denoted by , (level of significance) Typical values are.01,.05, or.10 Is selected by the researcher before the test takes place Provides the critical value(s) of the test

15. The Rejection Region H 0 : μ ≥ μ 0 H 1 : μ < μ 0 μ0μ0  c represents critical value Lower-tail test Level of significance =  Rejection region is shaded Reject if X too small c

16. The Rejection region So, given a level of significance α we find a value c for the sample mean sunch that smaller values occur with a (conditional) probability equal to α......given that the null hypothesis H 0 is true. In mathematical notation: (continued)

17. The Decision Rule If we observe a value X < c then we know that: either the null hypothesis is not true or we have observed an event that occurs with a probability less than α But we find events that occur with a probability less than α unlikely......and therefore we choose to reject H 0 instead. If X > c......we do not reject H 0.

18. How to find c? Recall that Let z α be such that Then we find that So, reject H 0 if

19. Example H 0 : μ ≥ 42 vs. H 1 : μ < 42 Choose level α = 0.05 Suppose X = 20, σ = 30, n = 9 Use Table 1 (or 8) to find that P(Z<-z α )=α ↔ F(-z.05 ) =.05 ↔ -z.05 = -1.645 Note that or So, we reject H 0 and say thatthe mean monthly phone bill in Dublin is significantly lower than 42 at the 5% level

20. Summary: Test of Hypothesis for the Mean (σ Known) Convert sample result ( ) to a z value The decision rule is: σ Knownσ Unknown Hypothesis Tests for  Consider the test (Assume the population is normal)

21. Upper Tail Test H 0 : μ ≤ μ 0 H 1 : μ > μ 0 0  c represents critical value Upper-tail test Level of significance =  Rejection region is shaded Reject if X too large c

22. Upper Tail Test This is the mirror image of the Lower Tail test. So, reject H 0 if Z is large. How large? (continued)

23. Example: Upper-Tail Z Test for Mean (  Known) A phone industry manager thinks that customer monthly cell phone bill have increased, and now average over €52 per month. The company wishes to test this claim. (Assume  = 10 is known) H 0 : μ ≤ 52 the average is not over €52 per month H 1 : μ > 52 the average is greater than €52 per month (i.e., sufficient evidence exists to support the manager’s claim) Form hypothesis test:

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

25. Obtain sample and compute the test statistic Suppose a sample is taken with the following results: n = 64, x = 53.1 (  =10 was assumed known) Using the sample results, Example: Sample Results (continued)

26. 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 at the 10% level that the mean bill is over €52 z = 0.88 Reach a decision and interpret the result: (continued)

27. Test of Hypothesis for the Mean (σ Known) Convert sample result ( ) to a z value The decision rule is: σ Knownσ Unknown Hypothesis Tests for  Consider the test (Assume the population is normal)

28. 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

29. p-Value Approach to Testing Convert sample result (e.g., ) to test statistic (e.g., z statistic ) Obtain the p-value For an upper tail test: Decision rule: compare the p-value to  If p-value < , reject H 0 If p-value  , do not reject H 0 (continued)

30. Example: Upper-Tail Z Test for Mean (  Known) A phone industry manager thinks that customer monthly cell phone bill have increased, and now average over €52 per month. The company wishes to test this claim. (Assume  = 10 is known) H 0 : μ ≤ 52 the average is not over €52 per month H 1 : μ > 52 the average is greater than €52 per month (i.e., sufficient evidence exists to support the manager’s claim) Form hypothesis test:

31. Reject H 0  =.10 Do not reject H 0 1.28 0 Reject H 0 Z =.88 Calculate the p-value and compare to  (assuming that μ = 52.0) (continued) p-value =.1894 Example: p-Value Solution Do not reject H 0 since p-value =.1894 >  =.10

32. Do not reject H 0 Reject H 0 There are two critical values, defining the two regions of rejection Two-Tail Tests  /2 0 H 0 : μ = μ 0 H 1 : μ  μ 0  /2 Lower critical value Upper critical value μ0μ0 z x -z  /2 +z  /2 In some settings, the alternative hypothesis does not specify a unique direction

33. Two Tail Test This is a combination of the Upper and Lower Tail tests. So, reject H 0 either if Z is small or large. How small or large? (continued)

34. Hypothesis Testing Example Test the claim that the true mean # of TV sets in Irish homes is equal to 3. (Assume σ = 0.8) State the appropriate null and alternative hypotheses H 0 : μ = 3, H 1 : μ ≠ 3 (This is a two tailed test) Specify the desired level of significance Suppose that  =.05 is chosen for this test Choose a sample size Suppose a sample of size n = 100 is selected

35. Hypothesis Testing Example Determine the appropriate technique σ is known so this is a z test Set up the critical values For  =.05 the critical z values are ±1.96 Collect the data and compute the test statistic Suppose the sample results are n = 100, x = 2.84 (σ = 0.8 is assumed known) So the test statistic is: (continued)

36. Reject H 0 Do not reject H 0 Is the test statistic in the rejection region?  =.05/2 -z = -1.96 0 Reject H 0 if z 1.96; otherwise do not reject H 0 Hypothesis Testing Example (continued)  =.05/2 Reject H 0 +z = +1.96 Here, z = -2.0 < -1.96, so the test statistic is in the rejection region

37. Reach a decision and interpret the result -2.0 Since z = -2.0 < -1.96, we reject the null hypothesis and conclude that there is sufficient evidence at the 5% level that the mean number of TVs in Irish homes is not equal to 3 Hypothesis Testing Example (continued) Reject H 0 Do not reject H 0  =.05/2 -z = -1.96 0  =.05/2 Reject H 0 +z = +1.96

38. .0228  /2 =.025 Example: p-Value Example: How likely is it to see a sample mean of 2.84 (or something further from the mean, in either direction) if the true mean is  = 3.0? -1.960 -2.0 Z1.96 2.0 x = 2.84 is translated to a z score of z = -2.0 p-value =.0228 +.0228 =.0456.0228  /2 =.025

39. Compare the p-value with  If p-value < , reject H 0 If p-value  , do not reject H 0 Here: p-value =.0456  =.05 Since.0456 <.05, we reject the null hypothesis (continued) Example: p-Value.0228  /2 =.025 -1.960 -2.0 Z1.96 2.0.0228  /2 =.025

40. Summary Z-test (σ known) H 0 : μ ≥ μ 0 H 1 : μ < μ 0 H 0 : μ ≤ μ 0 H 1 : μ > μ 0 H 0 : μ = μ 0 H 1 : μ ≠ μ 0 Reject H 0 if

41. Hypothesis Tests for the Mean  Known (Z-test)  Unknown (T-test) Hypothesis Tests for 

42. General idea Exactly the same as before......except that we don’t know σ......and therefore cannot use Z. However, we do know that

43. T-test for the mean of a normal T-test (σ unknown) H 0 : μ ≥ μ 0 H 1 : μ < μ 0 H 0 : μ ≤ μ 0 H 1 : μ > μ 0 H 0 : μ = μ 0 H 1 : μ ≠ μ 0 Reject H 0 if

44. Example: Two-Tail Test (  Unknown) The average cost of a hotel room in New York is said to be $168 per night. A random sample of 25 hotels resulted in x = $172.50 and s = $15.40. Test at the  = 0.05 level. (Assume the population distribution is normal) H 0 : μ  = 168 H 1 : μ  168

45.  = 0.05 n = 25  is unknown, so use a t statistic Critical Value: t 24,.025 = ± 2.0639 Example Solution: Two-Tail Test Do not reject H 0 : not sufficient evidence at the 5% level that true mean cost is different from $168 Reject H 0  /2=.025 -t n-1,α/2 Do not reject H 0 0  /2=.025 -2.0639 2.0639 1.46 H 0 : μ  = 168 H 1 : μ  168 t n-1,α/2

46. Tests of the Population Proportion Involves categorical variables Two possible outcomes “Success” (a certain characteristic is present) “Failure” (the characteristic is not present) Fraction or proportion of the population in the “success” category is denoted by p Assume sample size is large

47. Proportions Sample proportion in the success category is denoted by When np(1 – p) > 9, can be approximated by a normal distribution with mean and standard deviation (continued)

48. The sampling distribution of is approximately normal, so the test statistic is a z value: Hypothesis Tests for Proportions nP(1 – P) > 9 Hypothesis Tests for P Not discussed in this course nP(1 – P) < 9

49. Summary Z-test (proportions) H 0 : p ≥ p 0 H 1 : p < p 0 H 0 : p ≤ p 0 H 1 : p > p 0 H 0 : p = p 0 H 1 : p ≠ p 0 Reject H 0 if

50. Example: Z Test for Proportion A marketing company claims that it receives 8% responses from its mailing. To test this claim, a random sample of 500 were surveyed with 25 responses. Test at the  =.05 significance level. Check: Our approximation for p is = 25/500 =.05 np(1 - p) = (500)(.05)(.95) = 23.75 > 9

51. Z Test for Proportion: Solution  =.05 n = 500, =.05 Reject H 0 at  =.05 H 0 : p =.08 H 1 : p .08 Critical Values: ± 1.96 Test Statistic: Decision: Conclusion: z 0 Reject.025 1.96 -2.47 There is evidence at the 5% level to reject the company’s claim of 8% response rate. -1.96

52. Do not reject H 0 Reject H 0  /2 =.025 1.96 0 Z = -2.47 Calculate the p-value and compare to  (For a two sided test the p-value is always two sided) (continued) p-value =.0136: p-Value Solution Reject H 0 since p-value =.0136 <  =.05 Z = 2.47 -1.96  /2 =.025.0068

53. Errors in Making Decisions Type I Error Reject a true null hypothesis Considered a serious type of error Since we reject H 0 if the statistic is in the rejection region, we know that: The probability of Type I Error is  That’s how we chose α in the first place!

54. Errors in Making Decisions Type II Error Fail to reject a false null hypothesis The probability of Type II Error is β (continued)

55. Outcomes and Probabilities Actual Situation 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 )

56. 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 )

57. Power of the Test The power of a test is the probability of rejecting a null hypothesis that is false i.e., Power = P(Reject H 0 | H 1 is true) Power of the test increases as the sample size increases

58. Recall the possible hypothesis test outcomes: Actual Situation Decision Do Not Reject H 0 No error (1 - )  Type II Error ( β ) Reject H 0 Type I Error ( )  H 0 False H 0 True Key: Outcome (Probability) No Error ( 1 - β ) β denotes the probability of Type II Error 1 – β is defined as the power of the test Power = 1 – β = the probability that a false null hypothesis is rejected Power of the Test

59. Type II Error or The decision rule is: Assume the population is normal and the population variance is known. Consider the test If the null hypothesis is false and the true mean is μ*, then the probability of type II error is

60. Reject H 0 : μ  52 Do not reject H 0 : μ  52 Type II Error Example 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 

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

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

63. 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

64. 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

65. If the true mean is μ* = 50, The probability of Type II Error = β = 0.1539 The power of the test = 1 – β = 1 – 0.1539 = 0.8461 Power of the Test Example Actual Situation Decision Do Not Reject H 0 No error 1 -  = 0.95 Type II Error β = 0.1539 Reject H 0 Type I Error  = 0.05 H 0 False H 0 True Key: Outcome (Probability) No Error 1 - β = 0.8461 (The value of β and the power will be different for each μ*)

66. The Godfather Some of the Dons of important mafia families claim that Don Corleone does not bring in the promised $5mln in “transactions”. Corleone orders his consiglieri to investigate. Tom Hagen asks you You can sample 16 transactions

67. The Godfather Construct hypotheses H 0 : μ ≤ 5 and H 1 : μ > 5 (upper tail test) Choose level of significance α =.005 (you want to be VERY careful) Suppose you observe X = 5.9 mln with st. dev. s X = 1.16 mln. Since variance unknown: use t-test (assume that X is normal) (continued)

68. The Godfather (continued) Reject if: Cut-off point: test statistic: Reject H 0 : there is evidence at the.5% level that the true mean income is larger than 5 mln. Decision:

69. Misdiagnosis Someone claims that the number of misdiagnoses in a particular hospital is significantly higher than elsewhere. You do some research and find that a “normal” rate of misdiagnosis is 1%. You take a sample of 150 cases and find 3 cases of misdiagnosis. Can this be due to chance? Use test for proportions. H 0 : p =.01 and H 1 : p ≠.01 Take α =.1 So z α/2 = z.05 = 1.645

70. Misdiagnosis (continued) There is no evidence at the 10% level that the misdiagnosis rate is significantly different from 1%. So it could be down to “bad luck”

71. Misdiagnosis (continued) 90% confidence interval [.0012,.0388] Note that the 1% rate lies within the interval. That’s why we can’t reject.

72. Errors in Making Decisions Type I Error Reject a true null hypothesis Considered a serious type of error Since we reject H 0 if the statistic is in the rejection region, we know that: The probability of Type I Error is  That’s how we chose α in the first place!

73. Errors in Making Decisions Type II Error Fail to reject a false null hypothesis The probability of Type II Error is denoted by β (continued)

74. Outcomes and Probabilities Actual Situation 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 )

75. 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 )

76. Power of the Test The power of a test is the probability of rejecting a null hypothesis that is false i.e., Power = P(Reject H 0 | H 1 is true) Power of the test increases as the sample size increases

77. Recall the possible hypothesis test outcomes: Actual Situation Decision Do Not Reject H 0 No error (1 - )  Type II Error ( β ) Reject H 0 Type I Error ( )  H 0 False H 0 True Key: Outcome (Probability) No Error ( 1 - β ) β denotes the probability of Type II Error 1 – β is defined as the power of the test Power = 1 – β = the probability that a false null hypothesis is rejected Power of the Test

78. Type II Error The decision rule is: Assume the population is normal and the population variance is known. Consider the (lower-tail) test If the null hypothesis is false and the true mean is μ*, then the probability of type II error is or

79. Reject H 0 : μ  52 Do not reject H 0 : μ  52 Type II Error Example 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 

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

81. Reject H 0 : μ  52 Do not reject H 0 : μ  52 Type II Error Example Suppose we do not reject H 0 : μ  52 when in fact the true mean is μ* = 50  5250 β Here, β = P( X  ) if μ* = 50 (continued) Note: α↑ implies β↓

82. 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

83. 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

84. If the true mean is μ* = 50, The probability of Type II Error = β = 0.1539 The power of the test = 1 – β = 1 – 0.1539 = 0.8461 Power of the Test Example Actual Situation Decision Do Not Reject H 0 No error 1 -  = 0.95 Type II Error β = 0.1539 Reject H 0 Type I Error  = 0.05 H 0 False H 0 True Key: Outcome (Probability) No Error 1 - β = 0.8461 (The value of β and the power will be different for each μ*)

85. Topic Summary Addressed hypothesis testing methodology Performed Z Test for the mean (σ known) Discussed critical value and p-value approaches to hypothesis testing Performed one-tail and two-tail tests Performed t test for the mean (σ unknown) Performed Z test for the proportion Discussed type II error and power of the test