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Student's T-test, Paired T-Test, ANOVA & Proportionate Test

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Student's T-test, Paired T-Test, ANOVA & Proportionate Test

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Student's T-test, Paired T-Test, ANOVA & Proportionate Test

  1. 1. FK6163 T Test, ANOVA & Proportionate TestAssoc. Prof . Dr Azmi Mohd Tamil Dept of Community Health Universiti Kebangsaan Malaysia ©drtamil@gmail.com 2012
  2. 2. T-TestIndependent T-Test Student’s T-Test Paired T-Test ANOVA ©drtamil@gmail.com 2012
  3. 3. Student’s T-test William Sealy Gosset @“Student”, 1908. The Probable Error of Mean. Biometrika. ©drtamil@gmail.com 2012
  4. 4. Student’s T-Test4 To compare the means of two independent groups. For example; comparing the mean Hb between cases and controls. 2 variables are involved here, one quantitative (i.e. Hb) and the other a dichotomous qualitative variable (i.e. case/control).4 t= ©drtamil@gmail.com 2012
  5. 5. Examples: Student’s t- test4 Comparing the level of blood cholestrol (mg/dL) between the hypertensive and normotensive.4 Comparing the HAMD score of two groups of psychiatric patients treated with two different types of drugs (i.e. Fluoxetine & Sertraline ©drtamil@gmail.com 2012
  6. 6. Example Group Statistics DRUG N Mean Std. Deviation DHAMAWK6 F 35 4.2571 3.12808 S 32 3.8125 4.39529 Independent Samples Test t-test for Equality of Means Sig. Mean t df (2-tailed) DifferenceDHAMAWK6 Equal variances .48 65 .633 .4446 assumed ©drtamil@gmail.com 2012
  7. 7. Assumptions of T test4 Observations are normally distributed in each population. (Explore)4 The population variances are equal. (Levene’s Test)4 The 2 groups are independent of each other. (Design of study) ©drtamil@gmail.com 2012
  8. 8. Manual Calculation4 Sample size > 30 4 Small sample size, equal variance X1 − X 2 t= X1 − X 2 1 1t= s0 + n1 n2 2 2 s s + 1 2 n1 n2 (n1 − 1) s12 + (n2 − 1) s2 2 s0 = 2 (n1 − 1) + (n2 − 1) ©drtamil@gmail.com 2012
  9. 9. Example – compare cholesterol level4 Hypertensive : 4 Normal : Mean : 214.92 Mean : 182.19 s.d. : 39.22 s.d. : 37.26 n : 64 n : 36• Comparing the cholesterol level between hypertensive and normal patients.• The difference is (214.92 – 182.19) = 32.73 mg%.• H0 : There is no difference of cholesterol level between hypertensive and normal patients.• n > 30, (64+36=100), therefore use the first formula. ©drtamil@gmail.com 2012
  10. 10. Calculation X1 − X 2 t= 2 2 s s 1 + 2 n1 n24t = (214.92- 182.19)________ ((39.222/64)+(37.262/36))0.54 t = 4.1374 df = n1+n2-2 = 64+36-2 = 984 Refer to t table; with t = 4.137, p < 0.001 ©drtamil@gmail.com 2012
  11. 11. If df>100, can refer Table A1.We don’t have 4.137 so weuse 3.99 instead. If t = 3.99,then p=0.00003x2=0.00006Therefore if t=4.137,p<0.00006.
  12. 12. Or can refer to Table A3. We don’t have df=98, so we use df=60 instead. t = 4.137 > 3.46 (p=0.001)Therefore if t=4.137, p<0.001.
  13. 13. Conclusion• Therefore p < 0.05, null hypothesis rejected.• There is a significant difference of cholesterol level between hypertensive and normal patients.• Hypertensive patients have a significantly higher cholesterol level compared to normotensive patients. ©drtamil@gmail.com 2012
  14. 14. Exercise (try it)• Comparing the mini test 1 (2012) results between UKM and ACMS students.• The difference is 11.255• H0 : There is no difference of marks between UKM and ACMS students.• n > 30, therefore use the first formula. ©drtamil@gmail.com 2012
  15. 15. Exercise (answer)4 Nullhypothesis rejected4 There is a difference of marks between UKM and ACMS students. UKM marks higher than AUCMS ©drtamil@gmail.com 2012
  16. 16. T-Test In SPSS4 For this exercise, we will be using the data from the CD, under Chapter 7, sga-bab7.sav4 This data came from a case-control study on factors affecting SGA in Kelantan.4 Open the data & select - >Analyse >Compare Means >Ind-Samp T Test… ©drtamil@gmail.com 2012
  17. 17. T-Test in SPSS4 We want to see whether there is any association between the mothers’ weight and SGA. So select the risk factor (weight2) into ‘Test Variable’ & the outcome (SGA) into ‘Grouping Variable’.4 Now click on the ‘Define Groups’ button. Enter • 0 (Control) for Group 1 and • 1 (Case) for Group 2.4 Click the ‘Continue’ button & then click the ‘OK’ button. ©drtamil@gmail.com 2012
  18. 18. T-Test Results Group Statistics Std. Error SGA N Mean Std. Deviation MeanWeight at first ANC Normal 108 58.666 11.2302 1.0806 SGA 109 51.037 9.3574 .89634 Compare the mean+sd of both groups. • Normal 58.7+11.2 kg • SGA 51.0+ 9.4 kg4 Apparently there is a difference of weight between the two groups. ©drtamil@gmail.com 2012
  19. 19. Results & Homogeneity of Variances Independent Samples Test Levenes Test for Equality of Variances t-test for Equality of Means 95% Confidence Interval of the Mean Std. Error Difference F Sig. t df Sig. (2-tailed) Difference Difference Lower UpperWeight at first ANC Equal variances 1.862 .174 5.439 215 .000 7.629 1.4028 4.8641 10.3940 assumed Equal variances 5.434 207.543 .000 7.629 1.4039 4.8612 10.3969 not assumed 4 Look at the p value of Levene’s Test. If p is not significant then equal variances is assumed (use top row). 4 If it is significant then equal variances is not assumed (use bottom row). 4 So the t value here is 5.439 and p < 0.0005. The difference is significant. Therefore there is an association between the mothers weight and SGA. ©drtamil@gmail.com 2012
  20. 20. How to present the result?Group N Mean test pNormal 108 58.7+11.2 kg T test <0.0005 t = 5.439 SGA 109 51.0+ 9.4 ©drtamil@gmail.com 2012
  21. 21. Paired t-test“Repeated measurement on the same individual” ©drtamil@gmail.com 2012
  22. 22. Paired T-Test4 “Repeated measurement on the same individual”4 t= ©drtamil@gmail.com 2012
  23. 23. Formula d −0t= sd n (∑ d ) 2 ∑d i 2 − nsd = n −1df = n p − 1 ©drtamil@gmail.com 2012
  24. 24. Examples of paired t-test4 Comparing the HAMD score between week 0 and week 6 of treatment with Sertraline for a group of psychiatric patients.4 Comparing the haemoglobin level amongst anaemic pregnant women after 6 weeks of treatment with haematinics. ©drtamil@gmail.com 2012
  25. 25. Example Paired Samples Statistics Mean N Std. Deviation Pair DHAMAWK0 13.9688 32 6.48315 1 DHAMAWK6 3.8125 32 4.39529 Paired Samples Test Paired Differences Std. Sig. Mean Deviation t df (2-tailed)Pair DHAMAWK0 - 10.1563 6.75903 8.500 31 .0001 DHAMAWK6 ©drtamil@gmail.com 2012
  26. 26. Manual Calculation4 The measurement of the systolic and diastolic blood pressures was done two consecutive times with an interval of 10 minutes. You want to determine whether there was any difference between those two measurements.4 H0:There is no difference of the systolic blood pressure during the first (time 0) and second measurement (time 10 minutes). ©drtamil@gmail.com 2012
  27. 27. Calculation4 Calculate the difference between first & second measurement and square it. Total up the difference and the square. ©drtamil@gmail.com 2012
  28. 28. Calculation4∑ d = 112 ∑ d2 = 1842 n = 364 Mean d = 112/36 = 3.114 sd = ((1842-1122/36)/35)0.5 d −0 t= sd sd = 6.53 n4 t = 3.11/(6.53/6) t = 2.858 (∑ d ) 24 df = np – 1 = 36 – 1 = 35. ∑ d i2 − n sd = n −14 Refer to t table; df = n p − 1 ©drtamil@gmail.com 2012
  29. 29. Refer to Table A3. We don’t have df=35, so we use df=30 instead. t = 2.858, larger than 2.75(p=0.01) but smaller than 3.03(p=0.005). 3.03>t>2.75 Therefore if t=2.858, 0.005<p<0.01.
  30. 30. Conclusionwith t = 2.858, 0.005<p<0.01Therefore p < 0.01.Therefore p < 0.05, null hypothesisrejected.Conclusion: There is a significantdifference of the systolic blood pressurebetween the first and secondmeasurement. The mean average of firstreading is significantly higher comparedto the second reading. ©drtamil@gmail.com 2012
  31. 31. Paired T-Test In SPSS4 For this exercise, we will be using the data from the CD, under Chapter 7, sgapair.sav4 This data came from a controlled trial on haematinic effect on Hb.4 Open the data & select - >Analyse >Compare Means >Paired-Samples T Test… ©drtamil@gmail.com 2012
  32. 32. Paired T-Test In SPSS4 We want to see whether there is any association between the prescription on haematinic to anaemic pregnant mothers and Hb.4 We are comparing the Hb before & after treatment. So pair the two measurements (Hb2 & Hb3) together.4 Click the ‘OK’ button. ©drtamil@gmail.com 2012
  33. 33. Paired T-Test Results Paired Samples Statistics Std. Error Mean N Std. Deviation Mean Pair HB2 10.247 70 .3566 .0426 1 HB3 10.594 70 .9706 .11604 Thisshows the mean & standard deviation of the two groups. ©drtamil@gmail.com 2012
  34. 34. Paired T-Test Results Paired Samples Test Paired Differences 95% Confidence Interval of the Std. Error Difference Mean Std. Deviation Mean Lower Upper t df Sig. (2-tailed)Pair 1 HB2 - HB3 -.347 .9623 .1150 -.577 -.118 -3.018 69 .004 4 This shows the mean difference of Hb before & after treatment is only 0.347 g%. 4 Yet the t=3.018 & p=0.004 show the difference is statistically significant. ©drtamil@gmail.com 2012
  35. 35. How to present the result? Mean D Group N Test p (Diff.) Beforetreatment Paired T- (HB2) vs 70 0.35 + 0.96 test 0.004 After t = 3.018treatment (HB3) ©drtamil@gmail.com 2012
  36. 36. ANOVA ©drtamil@gmail.com 2012
  37. 37. ANOVA – Analysis of Variance4 Extension of independent-samples t test4 Comparesthe means of groups of independent observations • Don’t be fooled by the name. ANOVA does not compare variances.4 Can compare more than two groups ©drtamil@gmail.com 2012
  38. 38. One-Way ANOVA F-Test4 Tests the equality of 2 or more population means4 Variables • One nominal scaled independent variable – 2 or more treatment levels or classifications (i.e. Race; Malay, Chinese, Indian & Others) • One interval or ratio scaled dependent variable (i.e. weight, height, age)4 Used to analyse completely randomized experimental designs ©drtamil@gmail.com 2012
  39. 39. Examples4 Comparing the blood cholesterol levels between the bus drivers, bus conductors and taxi drivers.4 Comparing the mean systolic pressure between Malays, Chinese, Indian & Others. ©drtamil@gmail.com 2012
  40. 40. One-Way ANOVA F-Test Assumptions4 Randomness & independence of errors • Independent random samples are drawn4 Normality • Populations are normally distributed4 Homogeneity of variance • Populations have equal variances ©drtamil@gmail.com 2012
  41. 41. Example DescriptivesBirth weight N Mean Std. Deviation Minimum MaximumHousewife 151 2.7801 .52623 1.90 4.72Office work 23 2.7643 .60319 1.60 3.96Field work 44 2.8430 .55001 1.90 3.79Total 218 2.7911 .53754 1.60 4.72 ANOVA Birth weight Sum of Squares df Mean Square F Sig. Between Groups .153 2 .077 .263 .769 Within Groups 62.550 215 .291 Total 62.703 217 ©drtamil@gmail.com 2012
  42. 42. Manual Calculation ANOVA ©drtamil@gmail.com 2012
  43. 43. 24.93Example:Time To CompleteAnalysis45 samples wereanalysed using 3 differentblood analyser (Mach1,Mach2 & Mach3). 22.6115 samples were placedinto each analyser.Time in seconds wasmeasured for each 20.59sample analysis.
  44. 44. 24.93Example:Time To CompleteAnalysisThe overall mean of theentire sample was 22.71seconds. 22.71 22.61This is called the “grand”mean, and is oftendenoted by X .If H0 were true then we’dexpect the group means 20.59to be close to the grandmean.
  45. 45. 24.93Example:Time To CompleteAnalysisThe ANOVA test isbased on the combineddistances from X . 22.71If the combined 22.61distances are large, thatindicates we shouldreject H0. 20.59
  46. 46. The Anova StatisticTo combine the differences from the grand mean we • Square the differences • Multiply by the numbers of observations in the groups • Sum over the groups ( )2 ( ) 2 ( SSB = 15 X Mach1 − X + 15 X Mach 2 − X + 15 X Mach3 − X ) 2where the X * are the group means. “SSB” = Sum of Squares Between groups
  47. 47. The Anova StatisticTo combine the differences from the grand mean we • Square the differences • Multiply by the numbers of observations in the groups • Sum over the groups ( )2 ( ) 2 ( SSB = 15 X Mach1 − X + 15 X Mach 2 − X + 15 X Mach3 − X ) 2where the X * are the group means. “SSB” = Sum of Squares Between groupsNote: This looks a bit like a variance.
  48. 48. Sum of Squares Between ( ) 2 ( )2 ( SSB = 15 X Mach1 − X + 15 X Mach 2 − X + 15 X Mach3 − X )24 Grand Mean = 22.714 Mean Mach1 = 24.93; (24.93-22.71)2=4.92844 Mean Mach2 = 22.61; (22.61-22.71)2=0.014 Mean Mach3 = 20.59; (20.59-22.71)2=4.49444 SSB = (15*4.9284)+(15*0.01)+(15*4.4944)4 SSB = 141.492 ©drtamil@gmail.com 2012
  49. 49. How big is big?4 For the Time to Complete, SSB = 141.4924 Is that big enough to reject H0?4 As with the t test, we compare the statistic to the variability of the individual observations.4 InANOVA the variability is estimated by the Mean Square Error, or MSE
  50. 50. MSE Mean Square Error The Mean Square Error is a measure of the variability after the group effects have been taken into account. ∑∑ (x − X j) 1 2MSE = ij N −K j i where xij is the ith observation in the jth group.
  51. 51. MSE Mean Square Error 24.93 The Mean Square Error is a measure of the variability after the group effects have been taken into 22.61 account. ∑∑ (x − X j) 1 2MSE = ij N −K j i 20.59 where xij is the ith observation in the jth group.
  52. 52. MSE Mean Square Error 24.93 The Mean Square Error is a measure of the variability after the group effects have been taken into 22.61 account. ∑∑ (x − X j) 1 2MSE = ij N −K j i 20.59
  53. 53. ∑∑ (xij − X j ) 1 2 MSE = N −K j iMach1 (x-mean)^2 Mach2 (x-mean)^2 Mach3 (x-mean)^223.73 1.4400 21.5 1.2321 19.74 0.722523.74 1.4161 21.6 1.0201 19.75 0.705623.75 1.3924 21.7 0.8281 19.76 0.688924.00 0.8649 21.7 0.8281 19.9 0.476124.10 0.6889 21.8 0.6561 20 0.348124.20 0.5329 21.9 0.5041 20.1 0.240125.00 0.0049 22.75 0.0196 20.3 0.084125.10 0.0289 22.75 0.0196 20.4 0.036125.20 0.0729 22.75 0.0196 20.5 0.008125.30 0.1369 23.3 0.4761 20.5 0.008125.40 0.2209 23.4 0.6241 20.6 0.000125.50 0.3249 23.4 0.6241 20.7 0.012126.30 1.8769 23.5 0.7921 22.1 2.280126.31 1.9044 23.5 0.7921 22.2 2.592126.32 1.9321 23.6 0.9801 22.3 2.9241SUM 12.8380 9.4160 11.1262 ©drtamil@gmail.com 2012
  54. 54. ∑∑ (xij − X j ) 1 2 MSE = N −K j i4 Note that the variation of the means (141.492) seems quite large (more likely to be significant???) compared to the variance of observations within groups (12.8380+9.4160+11.1262=33.3802).4 MSE = 33.3802/(45-3) = 0.7948 ©drtamil@gmail.com 2012
  55. 55. Notes on MSE4 Ifthere are only two groups, the MSE is equal to the pooled estimate of variance used in the equal-variance t test.4 ANOVA assumes that all the group variances are equal.4 Other options should be considered if group variances differ by a factor of 2 or more.4 (12.8380 ~ 9.4160 ~ 11.1262)
  56. 56. ANOVA F Test4 The ANOVA F test is based on the F statistic SSB (K − 1) F= MSE where K is the number of groups.4 Under H0 the F statistic has an “F” distribution, with K-1 and N-K degrees of freedom (N is the total number of observations)
  57. 57. Time to Analyse: F test p-valueTo get a p-value wecompare our F statisticto an F(2, 42)distribution.
  58. 58. Time to Analyse: F test p-valueTo get a p-value wecompare our F statisticto an F(2, 42)distribution.In our example 141.492 2F= = 89.015 33.3802 42We cannot draw the linesince the F value is solarge, therefore the pvalue is so small!!!!!!
  59. 59. Refer to F Dist. Table (α=0.01). We don’t have df=2;42, so we use df=2;40 instead. F = 89.015, larger than 5.18 (p=0.01) Therefore if F=89.015, p<0.01.Why use df=2;42?We have K=3groups so K-1 = 2We have N=45samples thereforeN-K = 42. ©drtamil@gmail.com 2012
  60. 60. Time to Analyse: F test p-valueTo get a p-value wecompare our F statisticto an F(2, 42)distribution.In our example 141.492 2F= = 89.015 33.3802 42The p-value is reallyP(F (2,42) > 89.015) = 0.00000000000008
  61. 61. ANOVA Table Results are often displayed using an ANOVA Table Sum of Mean Squares df Square F Sig.Between 141.492 2 40.746 89.015 p<0.01GroupsWithin Groups 33.380 42 .795Total 174.872 44
  62. 62. ANOVA Table Results are often displayed using an ANOVA Table Sum of Mean Squares df Square F Sig.Between 141.492 2 40.746 89.015 p<0.01GroupsWithin Groups 33.380 42 .795Total 174.872 44 Pop Quiz!: Where are the following quantities presented in this table? Sum of Squares Mean Square F Statistic p value Between (SSB) Error (MSE)
  63. 63. ANOVA Table Results are often displayed using an ANOVA Table Sum of Mean Squares df Square F Sig.Between 141.492 2 40.746 89.015 p<0.01GroupsWithin Groups 33.380 42 .795Total 174.872 44 Sum of Squares Mean Square F Statistic p value Between (SSB) Error (MSE)
  64. 64. ANOVA Table Results are often displayed using an ANOVA Table Sum of Mean Squares df Square F Sig.Between 141.492 2 40.746 89.015 p<0.01GroupsWithin Groups 33.380 42 .795Total 174.872 44 Sum of Squares Mean Square F Statistic p value Between (SSB) Error (MSE)
  65. 65. ANOVA Table Results are often displayed using an ANOVA Table Sum of Mean Squares df Square F Sig.Between 141.492 2 40.746 89.015 p<0.01GroupsWithin Groups 33.380 42 .795Total 174.872 44 Sum of Squares Mean Square F Statistic p value Between (SSB) Error (MSE)
  66. 66. ANOVA Table Results are often displayed using an ANOVA Table Sum of Mean Squares df Square F Sig.Between 141.492 2 40.746 89.015 p<0.01GroupsWithin Groups 33.380 42 .795Total 174.872 44 Sum of Squares Mean Square F Statistic p value Between (SSB) Error (MSE)
  67. 67. ANOVA In SPSS4 For this exercise, we will be using the data from the CD, under Chapter 7, sga-bab7.sav4 This data came from a case-control study on factors affecting SGA in Kelantan.4 Open the data & select - >Analyse >Compare Means >One-Way ANOVA… ©drtamil@gmail.com 2012
  68. 68. ANOVA in SPSS4 We want to see whether there is any association between the babies’ weight and mothers’ type of work. So select the risk factor (typework) into ‘Factor’ & the outcome (birthwgt) into ‘Dependent’.4 Now click on the ‘Post Hoc’ button. Select Bonferonni.4 Click the ‘Continue’ button & then click the ‘OK’ button.4 Then click on the ‘Options’ button. ©drtamil@gmail.com 2012
  69. 69. ANOVA in SPSS4 Select ‘Descriptive’, ‘Homegeneity of variance test’ and ‘Means plot’.4 Click ‘Continue’ and then ‘OK’. ©drtamil@gmail.com 2012
  70. 70. ANOVA Results DescriptivesBirth weight 95% Confidence Interval for Mean N Mean Std. Deviation Std. Error Lower Bound Upper Bound Minimum MaximumHousewife 151 2.7801 .52623 .04282 2.6955 2.8647 1.90 4.72Office work 23 2.7643 .60319 .12577 2.5035 3.0252 1.60 3.96Field work 44 2.8430 .55001 .08292 2.6757 3.0102 1.90 3.79Total 218 2.7911 .53754 .03641 2.7193 2.8629 1.60 4.72 4 Compare the mean+sd of all groups. 4 Apparently there are not much difference of babies’ weight between the groups. ©drtamil@gmail.com 2012
  71. 71. Results & Homogeneity of Variances Test of Homogeneity of Variances Birth weight Levene Statistic df1 df2 Sig. .757 2 215 .4704 Look at the p value of Levene’s Test. If p is not significant then equal variances is assumed. ©drtamil@gmail.com 2012
  72. 72. ANOVA Results ANOVA Birth weight Sum of Squares df Mean Square F Sig. Between Groups .153 2 .077 .263 .769 Within Groups 62.550 215 .291 Total 62.703 2174 Sothe F value here is 0.263 and p =0.769. The difference is not significant. Therefore there is no association between the babies’ weight and mothers’ type of work. ©drtamil@gmail.com 2012
  73. 73. How to present the result?Type of Work Mean+sd Test p Office 2.76 + 0.60 ANOVA Housewife 2.78 + 0.53 0.769 F = 0.263 Farmer 2.84 + 0.55 ©drtamil@gmail.com 2012
  74. 74. Proportionate Test ©drtamil@gmail.com 2012
  75. 75. Proportionate Test4 Qualitativedata utilises rates, i.e. rate of anaemia among males & females4 To compare such rates, statistical tests such as Z-Test and Chi-square can be used. ©drtamil@gmail.com 2012
  76. 76. Formula p1 − p2 • where p1 is the rate forz= event 1 = a1/n1 1 1  p0 q0  +  • p2 is the rate for event 2 = a2/n2  n1 n2  • a1 and a2 are frequencies of event 1 and 2 p1n1 + p2 n2 4 We refer to the normalp0 = distribution table to n1 + n2 decide whether to reject or not the null hypothesis.q0 = 1 − p0 ©drtamil@gmail.com 2012
  77. 77. http://stattrek.com/hypothesis- test/proportion.aspx4 ■The sampling method is simple random sampling.4 ■Each sample point can result in just two possible outcomes. We call one of these outcomes a success and the other, a failure.4 ■The sample includes at least 10 successes and 10 failures.4 ■The population size is at least 10 times as big as the sample size. ©drtamil@gmail.com 2012
  78. 78. Example4 Comparison of worm infestation rate between male and female medical students in Year 2.4 Rate for males ; p1= 29/96 = 0.3024 Rate for females;p2 =24/104 = 0.2314 H0: There is no difference of worm infestation rate between male and female medical students in Year 2 ©drtamil@gmail.com 2012
  79. 79. Cont. p1 p2p0=rate of success q0 p0 q0=rate of failure ©drtamil@gmail.com 2012
  80. 80. Cont.4 p0 = (29/96*96)+(24/104*104) = 0.265 96+1044 q0 = 1 – 0.265 = 0.735 ©drtamil@gmail.com 2012
  81. 81. Cont.4z = 0.302 - 0.231 = 1.1367 ((0.735*0.265) (1/96 + 1/104))0.54 From the normal distribution table (A1), z value is significant at p=0.05 if it is above 1.96. Since the value is less than 1.96, then there is no difference of rate for worm infestatation between the male and female students. ©drtamil@gmail.com 2012
  82. 82. Refer to Table A1.We don’t have 1.1367 so weuse 1.14 instead. If z = 1.14,then p=0.1271x2=0.2542Therefore if z=1.14,p=0.2542. H0 not rejected
  83. 83. Exercise (try it)4 Comparison of failure rate between ACMS and UKM medical students in Year 2 for minitest 1 (MS2 2012).4 Rate for UKM ; p1= 42/196 = 0.2144 Rate for ACMS;p2 = 35/70 = 0.5 ©drtamil@gmail.com 2012
  84. 84. Answer4 P1 = 0.214, p2 = 0.5, p0 = 0.289, q0 = 0.7114 N1 = 196, n2 = 70, Z = 20.470.5 = 4.524 p < 0.00006 ©drtamil@gmail.com 2012

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