1. Statistics for clinicians Biostatistics course by Kevin E. Kip, Ph.D., FAHA Professor and Executive Director, Research Center University of South Florida, College of Nursing Professor, College of Public Health Department of Epidemiology and Biostatistics Associate Member, Byrd Alzheimer’s Institute Morsani College of Medicine Tampa, FL, USA 1 SECTION 3.8 Guidelines and primary steps involved in hypothesis testing, including “null” and the “alternative” hypothesis

2. Hypothesis Formulation Scientific Method (not unique to health sciences) ---Formulate a hypothesis ---Test the hypothesis

3. Basic Strategy of Analytical Epidemiology 1.Identify variables you are interested in: Exposure Outcome 2.Formulate a hypothesis 3.Compare the experience of two groups of subjects with respect to the exposure and outcome

4. Basic Strategy of Analytical Epidemiology Hypothesis Testing Two competing hypotheses ---“Null” hypothesis (no association) – typically, but not always assumed ---“Alternative” hypothesis (postulates there is an association) Hypothesis testing is based on probability theory and the Central Limit Theorem

5. Hypothesis Formulation The “Biostatistician’s” way H 0 :“Null” hypothesis (assumed) H 1 :“Alternative” hypothesis The “Epidemiologist’s” way Direct risk estimate (e.g. best estimate of risk of disease associated with the exposure).

6. Hypothesis Formulation Biostatistician: H 0 :There is no association between the exposure and disease of interest H 1 :There is an association between the exposure and disease of interest (beyond what might be expected from random error alone)

7. Hypothesis Formulation Epidemiologist: What is the best estimate of the risk of disease in those who are exposed compared to those who are unexposed (i.e. exposed are at XX times higher risk of disease). This moves away from the simple dichotomy of yes or no for an exposure/disease association – to the estimated magnitude of effect irrespective of whether it differs from the null hypothesis.

8. Hypothesis Formulation “Association” Statistical dependence between two variables: Exposure(risk factor, protective factor, predictor variable, treatment) Outcome(disease, event)

9. Hypothesis Formulation “Association” The degree to which the rate of disease in persons with a specific exposure is either higher or lower than the rate of disease among those without that exposure.

10. Hypothesis Formulation Ways to Express Hypotheses: 1.Suggest possible events… The incidence of tuberculosis will increase in the next decade.

11. Hypothesis Formulation Ways to Express Hypotheses: 2.Suggest relationship between specific exposure and health-related event… A high cholesterol intake is associated with the development (risk) of coronary heart disease.

12. Hypothesis Formulation Ways to Express Hypotheses: 3.Suggest cause-effect relationship…. Cigarette smoking is a cause of lung cancer

13. Hypothesis Formulation Ways to Express Hypotheses: 4.“One-sided” vs. “Two-sided” One-sided example: Helicobacter pylori infection is associated with increased risk of stomach ulcer Two-sided example: Weight-lifting is associated with risk of lower back injury

14. Hypothesis Formulation Guidelines for Developing Hypotheses: 1.State the exposure to be measured as specifically as possible. 2.State the health outcome as specifically as possible. Strive to explain the smallest amount of ignorance

15. Hypothesis Formulation Example Hypotheses: POOR Eating junk food is associated with the development of cancer. GOOD The human papilloma virus (HPV) subtype 16 is associated with the development of cervical cancer.

16. SECTION 3.9 Parameters used in hypothesis testing

17. Level of significance: A fixed value of the probability of rejecting the null hypothesis (in favor of the alternative) when the null hypothesis is actually true (i.e. type I or alpha (α) error rate) Common levels of significance: 0.10, 0.05, 0.01, 0.001 α = 0.05: The probability of incorrectly rejecting the null hypothesis in favor of the alternative is 5% when the null hypothesis is true. P-value: A calculated probability of obtaining a test statistic at least as extreme as the one that was actually observed, assuming that the null hypothesis is true. A low p-value means that it is unlikely that the null hypothesis is actually true, and the alternative hypothesis should be considered (e.g. “reject” the null hypothesis).

18. 1.0 H 0 0.750.500.250.02.0 3.0 4.05.0 H 1 H 1 Null hypothesis Alternative hypothesis Alternative hypothesis Relative Risk Relative Risk A B C D E P A P D > P E

19. Refer to the figure below. Assume that 3 studies (A, B, and C) are conducted all with the same sample size. Which of the following is most likely true?  Study B will have a lower p-value than Study C  Study C will have the lowest p-value  None of the studies will be statistically significant  None of the above

20. Refer to the figure below. Assume that 3 studies (A, B, and C) are conducted all with the same sample size. Which of the following is most likely true?  Study B will have a lower p-value than Study C  Study C will have the lowest p-value  None of the studies will be statistically significant  None of the above

21. Interpreting Results The p-value is NOT the index of causality It is an arbitrary quantity with no direct relationship to biology

22. SECTION 3.10 Type I and Type II error and factors that impact statistical power

23. Interpreting Results DECISION H 0 True (No assoc.) H 1 True (Yes assoc.) Do not reject H 0 (not stat. sig.) Correct decision Type II (beta error) Reject H 0 (stat. sig.) Type I (alpha error) Correct decision Four possible outcomes of any epidemiologic study:

24. Interpreting Results When evaluating the incidence of disease between the exposed and non-exposed groups, we need guidelines to help determine whether there is a true difference between the two groups, or perhaps just random variation from the study sample.

25. Interpreting Results DECISIONH 0 TrueH 1 True Do not reject H 0 (not stat. sig.) Reject H 0 (stat. sig.) Type I (alpha error) “Conventional” Guidelines: Set the fixed alpha level (Type I error) to 0.05 This means, if the null hypothesis is true, the probability of incorrectly rejecting it is 5% of less. The “p-value” is a measure of the compatibility of the data and the null hypothesis.

26. Interpreting Results D+D+ D-D- E+E+ 1585 E-E- 1090 Example: I E+ = 15 / (15 + 85) = 0.15 I E- = 10 / (10 + 90) = 0.10 RR = I E+ /I E- = 1.5, p = 0.30 Although it appears that the incidence of disease may be higher in the exposed than in the non-exposed (RR=1.5), the p-value of 0.30 exceeds the fixed alpha level of 0.05. This means that the observed data are relatively compatible with the null hypothesis. Thus, we do not reject H 0 in favor of H 1 (alternative hypothesis).

27. Interpreting Results DECISIONH 0 TrueH 1 True Do not reject H 0 (not stat. sig.) Type II (beta error) Reject H 0 (stat. sig.) Conventional Guidelines: Set the fixed beta level (Type II error) to 0.20 This means, if the null hypothesis is false, the probability of not rejecting it is 20% of less. The “power” of a study is (1 – beta). This means having 80% probability to reject H 0 when H 1 is true.

28. Interpreting Results Example: With the above sample size of 400, and if the alternative hypothesis is true, we need to expect a RR of about 2.1 (power = 82%) or higher to be able to reject the null hypothesis in favor of the alternative hypothesis. NIncid. E-E- 2000.10 E+E+ 2000.180.210.24 RR1.82.12.4 Power58%82%95%

29. Interpreting Results Factors that affect the power of a study: 1.The fixed alpha level (the lower the level, the the lower the power). 2.The total and within group sample sizes (the smaller the sample size, the lower the power -- unbalanced groups have lower power than balanced groups). 3.The anticipated effect size (the higher the expected/observed effect size, the higher the power).

30. Interpreting Results Trade-offs between fixed alpha and beta levels: Reducing the fixed alpha level (e.g. to < 0.01) is considered “conservative.” This reduces the likelihood of a type I error (erroneously rejecting the null hypothesis), but at the expense of increasing the probability of a type II error if the alternative hypothesis is true.

31. Interpreting Results Trade-offs between fixed alpha and beta levels: Increasing the fixed alpha level (e.g. to < 0.10) reduces the probability of a type II error (failing to reject H 0 when H 1 is true), but at the expense of increasing the probability of a type I error if the null hypothesis is true.

32. Interpreting Results Expos- ure NIncid. Risk ratio P- value Power * RR ** None10000.101.0--- Low5000.151.50.00677%1.52 Medium2500.151.50.0260%1.64 High1000.151.50.1227%2.08 *Power with given sample size and risk ratio (  = 0.05) **Risk ratio needed for 80% power with given sample size

33. SECTION 3.11 Calculate and interpret sample hypotheses – one sample continuous outcome

34. General Steps for Hypothesis Testing: 1)Set up the hypothesis and determine the level of statistical significance (including 1 versus 2-sided hypothesis). 2)Select the appropriate test statistic 3)Set up the decision rule 4)Compute the test statistic 5)Conclusion (interpretation)

35. 1. Hypothesis Testing – One Sample Continuous Outcome  Compare a “historical control” mean (µ 0 ) from a population to a “sample” mean. Example: Average annual health care expenses per person in the year 2005 (n=100, X = $3,190, s = $890) are lower than “historical” control costs in the year 2002 ($3,302) 1)Set up the hypothesis and determine the level of statistical significance (including 1 versus 2-sided hypothesis). H 0 : µ = $3,302 H 1 : µ < $3,302 (one-sided hypothesis, lower-tailed test) α = 0.05

36. 1. Hypothesis Testing – One Sample Continuous Outcome Example: Average annual health care expenses per person in the year 2005 (n=100, X = $3,190, s = $890) are lower than “historical” control costs in the year 2002 ($3,302) 2)Select the appropriate test statistic. If (n < 30), then use t If (n > 30), then use z

37. 1. Hypothesis Testing – One Sample Continuous Outcome Example: Average annual health care expenses per person in the year 2005 (n=100, X = $3,190, s = $890) are lower than “historical” control costs in the year 2002 ($3,302) 3)Set up the decision rule (look up z value – Table 1c). Reject H 0 if z < -1.645 4)Compute the test statistic: 3190 - 3302 z = ---------------= -1.26 890 / 100 5)Conclusion: -1.26 > -1.645 (critical value): Do not reject H 0 Note: we cannot confirm the null hypothesis because perhaps the sample size was too small for a conclusive result (i.e. low power)

38. 1. One Sample Continuous Outcome (Practice)  Compare historical “control” mean (µ 0 ) from a population to a “sample” mean. Example: Total cholesterol levels in the Framingham Heart Study in the year 2002 (n=3,310, X = 200.3, s = 36.8) are different than the national average in 2002 (203.0) 1)Set up the hypothesis and determine the level of statistical significance (including 1 versus 2-sided hypothesis). H 0 : µ = ______ H 1 : µ ______ (???-sided hypothesis, ???-tailed test) α = 0.05

39. 1. One Sample Continuous Outcome (Practice)  Compare historical “control” mean (µ 0 ) from a population to a “sample” mean. Example: Total cholesterol levels in the Framingham Heart Study in the year 2002 (n=3,310, X = 200.3, s = 36.8) are different than the national average in 2002 (203.0) 1)Set up the hypothesis and determine the level of statistical significance (including 1 versus 2-sided hypothesis). H 0 : µ = 203 H 1 : µ = 203 (2-sided hypothesis, 2-tailed test) α = 0.05

40. 1. One Sample Continuous Outcome (Practice) Example: Total cholesterol levels in the Framingham Heart Study in the year 2002 (n=3,310, X = 200.3, s = 36.8) are different than the national average in 2002 (203.0) 2)Select the appropriate test statistic. If (n 30), then use z 3)Set up the decision rule (look up ??? value – Table 1c). Reject H 0 if ______________________________ 4)Compute the test statistic: 5)Conclusion:

41. 1. One Sample Continuous Outcome (Practice) Example: Total cholesterol levels in the Framingham Heart Study in the year 2002 (n=3,310, X = 200.3, s = 36.8) are different than the national average in 2002 (203.0) 2)Select the appropriate test statistic. If (n 30), then use z 3)Set up the decision rule (look up z value – Table 1c). Reject H 0 if z 1.96 4)Compute the test statistic: 200.3 - 203 z = ---------------= -4.22 36.8 / 3310 5)Conclusion: Reject H 0 because -4.22 < -1.96 What about statistical versus clinical significance?

42. SECTION 3.12 Calculate and interpret sample hypotheses – one sample dichotomous outcome

43. 2. Hypothesis Testing – One Sample Dichotomous Outcome  Compare a “historical control” proportion (p) from a population to a “sample” proportion. Note: The example below assumes a “large” sample defined as: np 0 > 5 and n(1-p 0 ) > 5 If not, then “exact” methods must be used. Example: The prevalence of smoking in the 2002 Framingham Heart Study (n=3,536, p = 482/3,536 = 0.1363) is lower than a national (“historical”) report in the year 2002 (p=0.211) 1)Set up the hypothesis and determine the level of statistical significance (including 1 versus 2-sided hypothesis). H 0 : p = 0.211 H 1 : p < 0.211 (one-sided hypothesis, lower-tailed test) α = 0.05

44. 2. Hypothesis Testing – One Sample Dichotomous Outcome Example: The prevalence of smoking in the 2002 Framingham Heart Study (n=3,536, p = 482/3,536 = 0.1363) is lower than a national (“historical”) report in the year 2002 (p=0.211) 2)Select the appropriate test statistic. p – p 0 z = ---------------- p 0 (1-p 0 ) / n

45. 2. Hypothesis Testing – One Sample Dichotomous Outcome Example: The prevalence of smoking in the 2002 Framingham Heart Study (n=3,536, p = 482/3,536 = 0.1363) is lower than a national (“historical”) report in the year 2002 (p=0.211) 3)Set up the decision rule (look up z value – Table 1c). Reject H 0 if z < -1.645 4)Compute the test statistic: p – p 0 z = ---------------- p 0 (1-p 0 ) / n 0.136 – 0.211 z = ---------------------------- = -10.93 0.211(1-0.211) / 3,536 5)Conclusion: -10.93 < -1.645 (critical value): Reject H 0

46. 2. One Sample Dichotomous Outcome (Practice)  Compare historical “control” proportion (p) from a population to a “sample” mean. Example: The prevalence of diabetes in a Tampa-based study of adults in 2002 (n=1,240, p = 108/1,240 = 0.0871) is different than a national (“historical”) report in the year 2002 (p=0.1082) 1)Set up the hypothesis and determine the level of statistical significance (including 1 versus 2-sided hypothesis). H 0 : p = ______ H 1 : p ______ (???-sided hypothesis, ???-tailed test) α = 0.05

47. 2. One Sample Dichotomous Outcome (Practice)  Compare historical “control” proportion (p) from a population to a “sample” mean. Example: The prevalence of diabetes in a Tampa-based study of adults in 2002 (n=1,240, p = 108/1,240 = 0.0871) is different than a national (“historical”) report in the year 2002 (p=0.1082) 1)Set up the hypothesis and determine the level of statistical significance (including 1 versus 2-sided hypothesis). H 0 : p = 0.1082 H 1 : p = 0.1082 (2-sided hypothesis, 2-tailed test) α = 0.05

48. 2. One Sample Dichotomous Outcome (Practice) Example: The prevalence of diabetes in a Tampa-based study of adults in 2002 (n=1,240, p = 108/1,240 = 0.0871) is different than a national (“historical”) report in the year 2002 (p=0.1082) 2)Select the appropriate test statistic (assume z – large sample) 3)Set up the decision rule (look up z value – Table 1c). Reject H 0 if _______________________ 4)Compute the test statistic: p – p 0 z = ---------------- p 0 (1-p 0 ) / n 5)Conclusion:

49. 2. One Sample Dichotomous Outcome (Practice) Example: The prevalence of diabetes in a Tampa-based study of adults in 2002 (n=1,240, p = 108/1,240 = 0.0871) is different than a national (“historical”) report in the year 2002 (p=0.1082) 2)Select the appropriate test statistic (assume z – large sample) 3)Set up the decision rule (look up z value – Table 1c). Reject H 0 if z 1.96 4)Compute the test statistic: p – p 0 z = ---------------- p 0 (1-p 0 ) / n 0.0871 – 0.1082 z = ---------------------------- = -2.39 0.1082(1-0.1082)/1,240 5)Conclusion: -2.39 < -1.96 (critical value): Reject H 0

50. SECTION 3.13 Calculate and interpret sample hypotheses – one sample categorical/ordinal outcome

51. 3. One Sample Categorical/Ordinal Outcome  Goal is to compare the proportion of subjects in 3 or more categories to proportions from a known distribution  Forms the basis of comparing “observed” (O) versus “expected” (E) frequency counts  Expected frequencies (counts) are determined by multiplying the observed sample size (n) by the proportions specified in the null hypothesis (p 10, p 20,…. p k0 )  The test is a comparison of distribution of responses to a χ 2 distribution and “goodness of fit” test.

52. 3. One Sample Categorical/Ordinal Outcome Example: Compare proportions of levels of exercise among a sample of college students at USF to a national sample (historical) No regular exercise Sporadic exercise Regular exerciseTotal n6,0002,5001,50010,000 p0.600.250.151.0 National Sample No regular exercise Sporadic exercise Regular exerciseTotal n25512590470 p0.54260.26600.19151.0 USF 1)Set up the hypothesis and determine the level of statistical significance H 0 : p 1 = 0.60, p 2 = 0.25, p 3 = 0.15 H 0 : Distribution of responses is 0.60, 0.25, 0.15 H 1 : H 0 is False (distribution of responses are similar) α = 0.05

53. 3. One Sample Categorical/Ordinal Outcome No regular exercise Sporadic exercise Regular exerciseTotal n6,0002,5001,50010,000 p0.600.250.151.0 National Sample No regular exercise Sporadic exercise Regular exerciseTotal n25512590470 p0.54260.26600.19151.0 USF 2)Select the appropriate test statistic 3)Set up the decision rule d.f. = k(categories) – 1, so 3 categories – 1 = 2; α = 0.05 Refer to Table 3 in Appendix: critical value = 5.99 Reject H 0 if χ 2 > 5.99

54. 3. One Sample Categorical/Ordinal Outcome No regular exercise Sporadic exercise Regular exerciseTotal n6,0002,5001,50010,000 p0.600.250.151.0 National Sample No regular exercise Sporadic exercise Regular exerciseTotal n25512590470 p0.54260.26600.19151.0 e470x0.6 (282) 470x0.25 (117.5) 470x0.15 (70.5) USF 4)Compute the test statistic (255 - 282) 2 (125 – 117.5) 2 (90 – 70.5) 2 χ 2 = -------------- + ----------------- + --------------- 282 117.5 70.5 χ 2 = 2.59 + 0.48 + 5.39 = 8.46 5) Conclusion: 8.46 > 5.99; Reject H 0 --- proportions are different Note expected frequencies above

55. 3. One Sample Categorical/Ordinal Outcome (Practice) Example: Compare proportions of categories of smoking among a sample of graduate students at USF to a national sample (historical) Never Smoker Former Smoker Current SmokerTotal n6,8002,0001,20010,000 p0.680.200.121.0 National Sample Never Smoker Former Smoker Current SmokerTotal n3749848520 p0.71920.18850.09231.0 USF 1)Set up the hypothesis and determine the level of statistical significance H 0 : H 1 : α = 0.05

56. 3. One Sample Categorical/Ordinal Outcome (Practice) Example: Compare proportions of categories of smoking among a sample of graduate students at USF to a national sample (historical) Never Smoker Former Smoker Current SmokerTotal n6,8002,0001,20010,000 p0.680.200.121.0 National Sample Never Smoker Former Smoker Current SmokerTotal n3749848520 p0.71920.18850.09231.0 USF 1)Set up the hypothesis and determine the level of statistical significance H 0 : p 1 = 0.68, p 2 = 0.20, p 3 = 0.12 H 0 : Distribution of responses is 0.68, 0.20, 0.12 H 1 : H 0 is False (distribution of responses are similar) α = 0.05

57. 3. One Sample Categorical/Ordinal Outcome (Practice) Never Smoker Former Smoker Current SmokerTotal n6,8002,0001,20010,000 p0.680.200.121.0 National Sample Never Smoker Former Smoker Current SmokerTotal n3749848520 p0.71920.18850.09231.0 USF 2)Select the appropriate test statistic 3)Set up the decision rule d.f. = _____________; α = 0.05 Refer to Table 3 in Appendix: critical value = _____________ Reject H 0 if χ 2 > _________________

58. 3. One Sample Categorical/Ordinal Outcome (Practice) Never Smoker Former Smoker Current SmokerTotal n6,8002,0001,20010,000 p0.680.200.121.0 National Sample Never Smoker Former Smoker Current SmokerTotal n3749848520 p0.71920.18850.09231.0 USF 2)Select the appropriate test statistic 3)Set up the decision rule d.f. = k(categories) – 1, so 3 categories – 1 = 2; α = 0.05 Refer to Table 3 in Appendix: critical value = 5.99 Reject H 0 if χ 2 > 5.99

59. 3. One Sample Categorical/Ordinal Outcome (Practice) Never Smoker Former Smoker Current SmokerTotal n6,8002,0001,20010,000 p0.680.200.121.0 National Sample Never Smoker Former Smoker Current SmokerTotal n3749848520 p0.71920.18850.09231.0 e USF 4)Compute the test statistic χ 2 = ----------------- + ----------------- + --------------- χ 2 = 5) Conclusion: Calculate expected frequencies above

60. 3. One Sample Categorical/Ordinal Outcome (Practice) Never Smoker Former Smoker Current SmokerTotal n6,8002,0001,20010,000 p0.680.200.121.0 National Sample Never Smoker Former Smoker Current SmokerTotal n3749848520 p0.71920.18850.09231.0 e520x0.68 (353.6) 520x0.20 (104) 520x0.12 (62.4) USF 4)Compute the test statistic (374 – 353.6) 2 (98 – 104) 2 (48 – 62.4) 2 χ 2 = ----------------- + ----------------- + --------------- 353.6 104 62.4 χ 2 = 1.177 + 0.3462 + 3.323 = 4.85 5) Conclusion: 4.85 < 5.99; Do not reject H 0 Note expected frequencies above

61. SECTION 3.14 Calculate and interpret sample hypotheses – matched design with continuous outcome

62. 4. One Sample Matched – Continuous Outcome  Comparison of continuous scores of matched (paired) samples, such as clinical symptom scores before and after treatment.  Focus is on difference scores for each subject  Test of hypothesis is based on the mean difference (µ d )  The null hypothesis represents no difference: µ d = 0  Appropriate formula to use is based on the size of the sample (i.e. use z or t formula)

63. 4. One Sample Matched - Continuous Outcome Example: Compare mean systolic blood pressure in 15 randomly selected persons over a 4-year interval: N:15 Mean difference (X d ):-5.3 mmHg SD of difference scores (s d )12.8 1)Set up the hypothesis and determine the level of statistical significance H 0 : µ d = 0 H 1 : µ d = 0 α = 0.05 2)Select the appropriate test statistic (n < 30)

64. 4. One Sample Matched - Continuous Outcome N:15 Mean difference (X d ):-5.3 mmHg SD of difference scores (s d )12.8 3)Set up the decision rule (look up t value – Table 2). Reject H 0 if t 2.145 (d.f. = n-1) 4)Compute the test statistic -5.3 - 0 t = -------------= -1.60 12.8 / 15 5)Conclusion: Do not reject H 0 -2.145 < -1.60 < 2.145

65. 4. One Sample Matched - Continuous Outcome (Practice) Example: Compare mean systolic blood pressure in 12 trial subjects before and after treatment with an anti-hypertensive drug IDBaseline2 WeeksDifference 1158134-24 2162148-14 31441462 4170120-50 5142118-24 6188146-42 7176166-10 8162131-31 9151148-3 10148118-30 11175134-41 12176128-48 Mean162.67136.42-26.25 SD14.6214.7317.33 1)Set up the hypothesis and determine the level of statistical significance H 0 : H 1 : α = 0.05 2)Select the appropriatetest statistic z or t

66. 4. One Sample Matched - Continuous Outcome (Practice) Example: Compare mean systolic blood pressure in 12 trial subjects before and after treatment with an anti-hypertensive drug IDBaseline2 WeeksDifference 1158134-24 2162148-14 31441462 4170120-50 5142118-24 6188146-42 7176166-10 8162131-31 9151148-3 10148118-30 11175134-41 12176128-48 Mean162.67136.42-26.25 SD14.6214.7317.33 1)Set up the hypothesis and determine the level of statistical significance H 0 : µ d = 0 H 1 : µ d = 0 α = 0.05 2)Select the appropriatetest statistic (n < 30)

67. 4. One Sample Matched - Continuous Outcome (Practice) N:12 Mean difference (X d ):-26.25 mmHg SD of difference scores (s d )17.33 3)Set up the decision rule (look up t or z value). Reject H 0 if _______________________________ 4)Compute the test statistic 5)Conclusion:

68. 4. One Sample Matched - Continuous Outcome (Practice) N:12 Mean difference (X d ):-26.25 mmHg SD of difference scores (s d )17.33 3)Set up the decision rule (look up t value – Table 2). Reject H 0 if t 2.201 (d.f. = n-1) 4)Compute the test statistic -26.25 - 0 t = -------------= -5.25 17.33 / 12 5)Conclusion: Reject H 0 -5.25 < -2.201