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

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 Test Assoc. Prof . Dr Azmi Mohd Tamil Dept of Community Health Universiti Kebangsaan Malaysia ©drtamil@gmail.com 2012
2. 2. T-Test Independent 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-Test 4 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- test 4 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) Difference DHAMAWK6 Equal variances .48 65 .633 .4446 assumed ©drtamil@gmail.com 2012
7. 7. Assumptions of T test 4 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 Calculation 4 Sample size > 30 4 Small sample size, equal variance X1 − X 2 t= X1 − X 2 1 1 t= 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 level 4 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 n2 4t = (214.92- 182.19)________ ((39.222/64)+(37.262/36))0.5 4 t = 4.137 4 df = n1+n2-2 = 64+36-2 = 98 4 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 we use 3.99 instead. If t = 3.99, then p=0.00003x2=0.00006 Therefore 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 rejected 4 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 SPSS 4 For this exercise, we will be using the data from the CD, under Chapter 7, sga-bab7.sav 4 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 SPSS 4 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 Mean Weight at first ANC Normal 108 58.666 11.2302 1.0806 SGA 109 51.037 9.3574 .8963 4 Compare the mean+sd of both groups. • Normal 58.7+11.2 kg • SGA 51.0+ 9.4 kg 4 Apparently there is a difference of weight between the two groups. ©drtamil@gmail.com 2012
19. 19. Results & Homogeneity of Variances Independent Samples Test Levene's 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 Upper Weight 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 p Normal 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-Test 4 “Repeated measurement on the same individual” 4 t= ©drtamil@gmail.com 2012
23. 23. Formula d −0 t= sd n (∑ d ) 2 ∑d i 2 − n sd = n −1 df = n p − 1 ©drtamil@gmail.com 2012
24. 24. Examples of paired t-test 4 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 .000 1 DHAMAWK6 ©drtamil@gmail.com 2012
26. 26. Manual Calculation 4 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. Calculation 4 Calculate the difference between first & second measurement and square it. Total up the difference and the square. ©drtamil@gmail.com 2012
28. 28. Calculation 4∑ d = 112 ∑ d2 = 1842 n = 36 4 Mean d = 112/36 = 3.11 4 sd = ((1842-1122/36)/35)0.5 d −0 t= sd sd = 6.53 n 4 t = 3.11/(6.53/6) t = 2.858 (∑ d ) 2 4 df = np – 1 = 36 – 1 = 35. ∑ d i2 − n sd = n −1 4 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. Conclusion with t = 2.858, 0.005<p<0.01 Therefore p < 0.01. Therefore p < 0.05, null hypothesis rejected. Conclusion: There is a significant difference of the systolic blood pressure between the first and second measurement. The mean average of first reading is significantly higher compared to the second reading. ©drtamil@gmail.com 2012
31. 31. Paired T-Test In SPSS 4 For this exercise, we will be using the data from the CD, under Chapter 7, sgapair.sav 4 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 SPSS 4 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 .1160 4 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.) Before treatment Paired T- (HB2) vs 70 0.35 + 0.96 test 0.004 After t = 3.018 treatment (HB3) ©drtamil@gmail.com 2012
37. 37. ANOVA – Analysis of Variance 4 Extension of independent-samples t test 4 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-Test 4 Tests the equality of 2 or more population means 4 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. Examples 4 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 Assumptions 4 Randomness & independence of errors • Independent random samples are drawn 4 Normality • Populations are normally distributed 4 Homogeneity of variance • Populations have equal variances ©drtamil@gmail.com 2012
41. 41. Example Descriptives Birth weight N Mean Std. Deviation Minimum Maximum Housewife 151 2.7801 .52623 1.90 4.72 Office work 23 2.7643 .60319 1.60 3.96 Field work 44 2.8430 .55001 1.90 3.79 Total 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. 24.93 Example: Time To Complete Analysis 45 samples were analysed using 3 different blood analyser (Mach1, Mach2 & Mach3). 22.61 15 samples were placed into each analyser. Time in seconds was measured for each 20.59 sample analysis.
44. 44. 24.93 Example: Time To Complete Analysis The overall mean of the entire sample was 22.71 seconds. 22.71 22.61 This is called the “grand” mean, and is often denoted by X . If H0 were true then we’d expect the group means 20.59 to be close to the grand mean.
45. 45. 24.93 Example: Time To Complete Analysis The ANOVA test is based on the combined distances from X . 22.71 If the combined 22.61 distances are large, that indicates we should reject H0. 20.59
46. 46. The Anova Statistic To 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 ) 2 where the X * are the group means. “SSB” = Sum of Squares Between groups
47. 47. The Anova Statistic To 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 ) 2 where the X * are the group means. “SSB” = Sum of Squares Between groups Note: 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 )2 4 Grand Mean = 22.71 4 Mean Mach1 = 24.93; (24.93-22.71)2=4.9284 4 Mean Mach2 = 22.61; (22.61-22.71)2=0.01 4 Mean Mach3 = 20.59; (20.59-22.71)2=4.4944 4 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 = 141.492 4 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 2 MSE = 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 2 MSE = 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 2 MSE = ij N −K j i 20.59
53. 53. ∑∑ (xij − X j ) 1 2 MSE = N −K j i Mach1 (x-mean)^2 Mach2 (x-mean)^2 Mach3 (x-mean)^2 23.73 1.4400 21.5 1.2321 19.74 0.7225 23.74 1.4161 21.6 1.0201 19.75 0.7056 23.75 1.3924 21.7 0.8281 19.76 0.6889 24.00 0.8649 21.7 0.8281 19.9 0.4761 24.10 0.6889 21.8 0.6561 20 0.3481 24.20 0.5329 21.9 0.5041 20.1 0.2401 25.00 0.0049 22.75 0.0196 20.3 0.0841 25.10 0.0289 22.75 0.0196 20.4 0.0361 25.20 0.0729 22.75 0.0196 20.5 0.0081 25.30 0.1369 23.3 0.4761 20.5 0.0081 25.40 0.2209 23.4 0.6241 20.6 0.0001 25.50 0.3249 23.4 0.6241 20.7 0.0121 26.30 1.8769 23.5 0.7921 22.1 2.2801 26.31 1.9044 23.5 0.7921 22.2 2.5921 26.32 1.9321 23.6 0.9801 22.3 2.9241 SUM 12.8380 9.4160 11.1262 ©drtamil@gmail.com 2012
54. 54. ∑∑ (xij − X j ) 1 2 MSE = N −K j i 4 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 MSE 4 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 Test 4 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-value To get a p-value we compare our F statistic to an F(2, 42) distribution.
58. 58. Time to Analyse: F test p-value To get a p-value we compare our F statistic to an F(2, 42) distribution. In our example 141.492 2 F= = 89.015 33.3802 42 We cannot draw the line since the F value is so large, therefore the p value 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=3 groups so K-1 = 2 We have N=45 samples therefore N-K = 42. ©drtamil@gmail.com 2012
60. 60. Time to Analyse: F test p-value To get a p-value we compare our F statistic to an F(2, 42) distribution. In our example 141.492 2 F= = 89.015 33.3802 42 The p-value is really P(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<0.01 Groups Within Groups 33.380 42 .795 Total 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<0.01 Groups Within Groups 33.380 42 .795 Total 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<0.01 Groups Within Groups 33.380 42 .795 Total 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<0.01 Groups Within Groups 33.380 42 .795 Total 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<0.01 Groups Within Groups 33.380 42 .795 Total 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<0.01 Groups Within Groups 33.380 42 .795 Total 174.872 44 Sum of Squares Mean Square F Statistic p value Between (SSB) Error (MSE)
67. 67. ANOVA In SPSS 4 For this exercise, we will be using the data from the CD, under Chapter 7, sga-bab7.sav 4 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 SPSS 4 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 SPSS 4 Select ‘Descriptive’, ‘Homegeneity of variance test’ and ‘Means plot’. 4 Click ‘Continue’ and then ‘OK’. ©drtamil@gmail.com 2012
70. 70. ANOVA Results Descriptives Birth weight 95% Confidence Interval for Mean N Mean Std. Deviation Std. Error Lower Bound Upper Bound Minimum Maximum Housewife 151 2.7801 .52623 .04282 2.6955 2.8647 1.90 4.72 Office work 23 2.7643 .60319 .12577 2.5035 3.0252 1.60 3.96 Field work 44 2.8430 .55001 .08292 2.6757 3.0102 1.90 3.79 Total 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 .470 4 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 62.703 217 4 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 Test 4 Qualitativedata utilises rates, i.e. rate of anaemia among males & females 4 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 for z= 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 normal p0 = 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.aspx 4 ■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. Example 4 Comparison of worm infestation rate between male and female medical students in Year 2. 4 Rate for males ; p1= 29/96 = 0.302 4 Rate for females;p2 =24/104 = 0.231 4 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 p2 p0=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+104 4 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))0.5 4 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 we use 1.14 instead. If z = 1.14, then p=0.1271x2=0.2542 Therefore 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 = 0.214 4 Rate for ACMS;p2 = 35/70 = 0.5 ©drtamil@gmail.com 2012
84. 84. Answer 4 P1 = 0.214, p2 = 0.5, p0 = 0.289, q0 = 0.711 4 N1 = 196, n2 = 70, Z = 20.470.5 = 4.52 4 p < 0.00006 ©drtamil@gmail.com 2012
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Student's T-test, Paired T-Test, ANOVA & Proportionate Test

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