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Statistical methods and data analysis
- 1. STATISTICS - II STAT(401)
- 2. UMAR DARAZ Entomologist BZU Multan Statistic Anynumerical value calculated from sample is known as statistic. Sample mean ( ) Sample variance (s2) Sample standard deviation (s)
- 3. UMAR DARAZ Entomologist BZU Multan Parameter Anynumerical value calculated from population is known as parameter. Population mean (µ) Population variance ( ) Population standard deviation ( )
- 4. UMAR DARAZ Entomologist BZU Multan Testingof hypothesis 1) Hypothesis (Null and alternative hypothesis) 2) Level of significance (α= 1%, 2%, 5%, 10%) 3) Test-statistic (Formula) 4) Calculation (Value) 5) Critical region (Frequency curve) 6) Conclusion (Result) Formulas: z-test , t-test , χ -test , F-test
- 5. UMAR DARAZ Entomologist BZU Multan Statisticalhypothesis A statistical hypothesis is a statement about one or more parameters of population, this statement may or may not be true. Null hypothesis Alternative hypothesis
- 6. UMAR DARAZ Entomologist BZU Multan Nullhypothesis A null hypothesis is a hypothesis which is to be tested for possible rejection under the assumption that is true. It is denoted by Ho For example, suppose we think that the average age of ICS student is 16, then null hypothesis is written as Ho : µ = 16
- 7. UMAR DARAZ Entomologist BZU Multan Alternativehypothesis An alternative hypothesis is a hypothesis which is accepted when null hypothesis is rejected. It is denoted by H1 . It is written as H1 : µ ≠ 16
- 8. UMAR DARAZ Entomologist BZU Multan Threehypothesis Null hypothesis Ho : µ = 16 Ho : µ = 16 Ho : µ = 16 Alternative hypothesis H1 : µ ≠ 16 H1 : µ > 16 H1 : µ < 16
- 9. UMAR DARAZ Entomologist BZU Multan Simplehypothesis A hypothesis in which all parameters of population are specified is known as simple hypothesis. Only sign will be used = , ≠ Ho : µ = 16 H1 : µ ≠ 16
- 10. UMAR DARAZ Entomologist BZU Multan Compositehypothesis A hypothesis in which all parameters of population are not specified is known as composite hypothesis. Only sign will be used > or < Ho : µ = 16 Ho : µ = 16 H1 : µ > 16 H1 : µ <16
- 11. UMAR DARAZ Entomologist BZU Multan Type-1-error If we reject the null hypothesis when it is actually true is known as type-1-error. E.g. A deserving player is not selected by team. Type-2-error If we accept the null hypothesis when it is falls is known as type-2-error. E.g. A non-deserving player is selected by team. Level of significance Probability of making type-1-error is known as level of significance. It is denoted by α
- 12. UMAR DARAZ Entomologist BZU Multan Test-statistic A test-statistic is a function or formula of sample observation that provides basis for testing null hypothesis. Most commonly used test-statistic are z-test, t- test, F-test and χ -test z-test, t-test and F-test are used for quantitative data χ -test is used for qualitative data
- 13. UMAR DARAZ Entomologist BZU Multan Onesided test/One tailed test If the rejection region is located in only one side of test-statistic is known as one sided test. It is also called composite hypothesis For > sign Ho : µ = 16 H1 : µ > 16 For < sign Ho : µ = 16 H1 : µ < 16 - Table value 0 + Table value 0
- 14. UMAR DARAZ Entomologist BZU Multan Twosided test / Two tailed test If the rejection region is located in both side of sampling distribution of test-statistic is known as two sided test. It is also called simple hypothesis Ho : µ = 16 H1 : µ ≠ 16 + Table value
- 15. UMAR DARAZ Entomologist BZU Multan Rejectionregion Rejection region is the part of sampling distribution which leads to the rejection of null hypothesis. Acceptance region Acceptance region is the part of sampling distribution which leads to the acceptance of null hypothesis.
- 16. UMAR DARAZ Entomologist BZU Multan Writea general procedure for hypothesis testing single population mean(µ) 1) Hypothesis Ho : µ = µo H1 : µ ≠ µo Ho : µ = µo H1 : µ > µo Ho : µ = µo H1 : µ < µo
- 17. UMAR DARAZ Entomologist BZU Multan CONT… 2)Level of significance α= 1%, 2%, 5%, 10% 3) Test-statistic Conditions… If “ ” is given and the value of “n” is small or large then we apply z-test z µ
- 18. UMAR DARAZ Entomologist BZU Multan CONT… If “ ” is not given and n ≥ 30 then we apply z- test z µ If “ ” is not given and n ≤ 30 then we apply t- test t µ
- 19. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation S 5) Critical region Frequency curve
- 20. UMAR DARAZ Entomologist BZU Multan CONT… 6)Conclusion If calculated value lie in rejection region, then we reject the null hypothesis and accept the alternative hypothesis. If calculated value lie in acceptance region, then we accept the null hypothesis and reject the alternative hypothesis.
- 21. UMAR DARAZ Entomologist BZU Multan Example: 1 A random sample of size n=36 is taken from a population with variance =25 , sample mean =42.6 test a null hypothesis is Ho : µ = 45 , H1 : µ < 45 with α= 5% and table value is 1.645 1) Hypothesis Ho : µ = 45 H1 : µ < 45 2) Level of significance α= 5% 3) Test-statistic z µ One sided test
- 22. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation =42.6 µ = 45 =25 , =5 n=36 z 45 z z z
- 23. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion Calculated value lie in rejection region so, we reject the null hypothesis and accept the alternative hypothesis. -2.88 -1.645
- 24. UMAR DARAZ Entomologist BZU Multan Example: 2 A random sample of nine values are taken from a population 66, 68, 62, 61, 59, 67, 64, 66, 63, test a hypothesis that population mean is 60, =3, α= 10% and table value is 1.645 1) Hypothesis Ho : µ = 60 H1 : µ ≠ 60 2) Level of significance α= 10% 3) Test-statistic z µ Two sided test
- 25. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation = 64 µ = 60 =3 n=9 z z z
- 26. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion Calculated value lie in rejection region so, we reject the null hypothesis and accept the alternative hypothesis. -4 -1.645 +1.645 +4
- 27. UMAR DARAZ Entomologist BZU Multan Example: 3 A random sample of size n=10 is taken from a population with standard deviation =1.2 , sample mean =27 test a null hypothesis is Ho : µ = 26.3 , H1 : µ > 26.3 with α= 5% and table value is 1.645 1) Hypothesis Ho : µ = 26.3 H1 : µ > 26.3 2) Level of significance α= 5% 3) Test-statistic z µ One sided test
- 28. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation =27 µ = 26.3 =1.2 n=10 z z z z
- 29. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion Calculated value lie in rejection region so, we reject the null hypothesis and accept the alternative hypothesis. +1.645 +1.84
- 30. UMAR DARAZ Entomologist BZU Multan Example: 4 A random sample of size n=49 is taken from a population with standard deviation =4.2 , sample mean =125 test a null hypothesis is Ho : µ = 123.5 , H1 : µ > 123.5 with α= 1% and table value is 2.33 1) Hypothesis Ho : µ = 123.5 H1 : µ > 123.5 2) Level of significance α= 1% 3) Test-statistic z µ One sided test
- 31. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation =125 µ = 123.5 =4.2 n=49 z z z z
- 32. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion Calculated value lie in rejection region so, we reject the null hypothesis and accept the alternative hypothesis. +2.33 +2.5
- 33. UMAR DARAZ Entomologist BZU Multan Example: 5 The marks obtain by students at a large number of colleges are known to be normally distribution of mean 25 and a random sample of 36 student has an average marks of 27 with standard deviation of 5 at 5% level of significance, what conclusion should be taken, table value is 1.96 1) Hypothesis Ho : µ = 25 H1 : µ ≠ 25 2) Level of significance α= 5% 3) Test-statistic z µ Two sided test
- 34. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation = 27 µ = 25 s = 5 n= 36 z z z z
- 35. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion Calculated value lie in rejection region so, we reject the null hypothesis and accept the alternative hypothesis. -2.40 -1.96 +1.96 +2.40
- 36. UMAR DARAZ Entomologist BZU Multan Example: 6 A random sample of 12 graduate pass students from a certain typing distribute was taken with a average speed 72.6 words/minute and is 17.64 words/minute use α= 1% test the claim that institute graduate average less than 75….table value is 2.718 1) Hypothesis Ho : µ = 75 H1 : µ < 75 2) Level of significance α= 1% 3) Test-statistic t µ One sided test
- 37. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation =72.6 µ = 75 =17.64 , =4.2 n=12 t 75 t t t
- 38. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion Calculated value lie in acceptance region so, we accept the null hypothesis and reject the alternative hypothesis. -2.718 - 1.98
- 39. UMAR DARAZ Entomologist BZU Multan Example: 7 A random sample of 100 workers with children in day care shows a mean day care cost of Rs.2600 and standard deviation of Rs.500 verify the department claim that the mean exceed Rs.2500 at 5% level of significance with in this information and table value is 1.645 1) Hypothesis Ho : µ = 2500 H1 : µ > 2500 2) Level of significance α= 5% 3) Test-statistic z µ One sided test
- 40. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation =2600 µ = 2500 =500 n=100 z z z z
- 41. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion Calculated value lie in rejection region so, we reject the null hypothesis and accept the alternative hypothesis. +1.645 +2
- 42. UMAR DARAZ Entomologist BZU Multan Example: 8 A random sample of 100 observation from a population produced sample value =182 and =299 test the hypothesis that Ho : µ = 180, H1 : µ > 180 and α= 5% ,table value is 1.645 1) Hypothesis Ho : µ = 180 H1 : µ > 180 2) Level of significance α= 5% 3) Test-statistic z µ One sided test
- 43. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation =182 µ = 180 =299 , =17.29 n=100 z z z z
- 44. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion Calculated value lie in acceptance region so, we accept the null hypothesis and reject the alternative hypothesis. +1.16 +1.645
- 45. UMAR DARAZ Entomologist BZU Multan Writea general procedure for hypothesis testing of difference between population mean(µ1 µ2) 1) Hypothesis Ho : µ1 µ2 = 0 H1 : µ1 µ2 ≠ 0 Ho : µ1 µ2 = 0 H1 : µ1 µ2 > 0 Ho : µ1 µ2 = 0 H1 : µ1 µ2 < 0
- 46. UMAR DARAZ Entomologist BZU Multan CONT… 2)Level of significance α= 1%, 2%, 5%, 10% 3) Test-statistic Conditions… If “ 1 , 2” is given and the value of “ 1 , 2” is small or large then we apply z-test ( 1 2 (µ1 µ2) 1 2
- 47. UMAR DARAZ Entomologist BZU Multan CONT… If “ 1 , 2” is not given and 1 , 2 ≥ 30 then we apply z-test ( 1 2 (µ1 µ2) 1 2 If “ 1 , 2” is not given and 1 , 2 ≤ 30 then we apply t-test t ( 1 2 (µ1 µ2) SP 1 2 SP=Pooled standard deviation
- 48. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation SP SP 5) Critical region Frequency curve
- 49. UMAR DARAZ Entomologist BZU Multan CONT… 6)Conclusion If calculated value lie in rejection region, then we reject the null hypothesis and accept the alternative hypothesis. If calculated value lie in acceptance region, then we accept the null hypothesis and reject the alternative hypothesis.
- 50. UMAR DARAZ Entomologist BZU Multan Example: 9 A random sample of size 36 from a population with variance 24 and 1=15 , a second sample size of 28 from an other population with variance 80 and 2=13 test Ho : µ1 µ2 = 0, H1 : µ1 µ2 ≠ 0 at 5% of level of significance and table value is 1.96 1) Hypothesis Ho : µ1 µ2 = 0 H1 : µ1 µ2 ≠ 0 2) Level of significance α= 5% 3) Test-statistic ( 1 2 (µ1 µ2) 1 2 Two sided test
- 51. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation 1 = 36 2 = 28 = 24 = 80 1 =15 2 =13 𝑧 = (15 − 13) − (0) 24 36 + 80 28 𝑧 = 2 − 0 0.67 + 2.86 𝑧 = 2 3.53 𝑧 = . = ±1.06
- 52. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion Calculated value lie in acceptance region so, we accept the null hypothesis and reject the alternative hypothesis. -1.96 -1.06 +1.06 +1.96
- 53. UMAR DARAZ Entomologist BZU Multan Example: 10 A random sample of size 40 from a population produce the sample values 1=70.4 , = 31.40 and other random sample size of 50 from a second population produce the sample values 2=65.3, = 44.82 test Ho : µ1 µ2 = 2, H1 : µ1 µ2 > 2 at 5% of level of significance and table value is 1.645 1) Hypothesis Ho : µ1 µ2 = 2 H1 : µ1 µ2 > 2 2) Level of significance α= 5% 3) Test-statistic ( 1 2 (µ1 µ2) 1 2 One sided test
- 54. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation 1 = 40 2 = 50 = 31.40 = 44.82 1 =70.4 2 =65.3 𝑧 = (70.4 − 65.3) − (2) 31.40 40 + 44.82 50 𝑧 = 5.1 − 2 0.79 + 0.90 𝑧 = 3.1 1.69 𝑧 = . . = 2.38
- 55. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion Calculated value lie in rejection region so, we reject the null hypothesis and accept the alternative hypothesis. +1.645 +2.38
- 56. UMAR DARAZ Entomologist BZU Multan Example: 11 On an examination, in a statistics course the average marks of 12 boys was 78 with a standard deviation of 6 while average marks of 15 girls was 74 with a standard deviation of 8 test the hypothesis that there is no difference between perform of boys and girls, level of significance is 1% and table value is 2.06 1) Hypothesis Ho : µ1 µ2 = 0 H1 : µ1 µ2 ≠ 0 2) Level of significance α= 1% 3) Test-statistic t ( 1 2 (µ1 µ2) SP 1 2 Two sided test
- 57. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation 1 = 12 2 = 15 = 6 = 36 = 8 = 64 1 =78 2 =74 SP t ( ̅1 ̅2) (µ1 µ2) SP 1 2 SP t ( ) ( ) 7.18 SP t 7.18 . . SP t 7.18 . SP =7.18 t = ±1.44
- 58. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion Calculated value lie in acceptance region so, we accept the null hypothesis and reject the alternative hypothesis. -2.06 -1.44 +1.44 +2.06
- 59. UMAR DARAZ Entomologist BZU Multan Example: 12 The test was given to a group of 100 scouts and to a group of 144 guides, the mean score for scout was 27.53 and mean score for guide was 26.81 assuming a common population standard deviation of 3.48 using 5% level of significance without scout performance was better than that of guides, table value is 1.645 1) Hypothesis Ho : µ1 µ2 = 0 H1 : µ1 µ2 > 0 2) Level of significance α= 5% 3) Test-statistic ( 1 2 (µ1 µ2) 1 2 One sided test
- 60. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation 1 = 100 2 = 144 = 3.48 = 12.11 1 =27.53 2 =26.81 (27.53 26.81) ( ) 12.11 12.11 . . . . . . . = 1.6
- 61. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion Calculated value lie in acceptance region so, we accept the null hypothesis and reject the alternative hypothesis. +1.6 +1.645
- 62. UMAR DARAZ Entomologist BZU Multan Example: 13 The two sample A and B taken from population of standard deviation 0.8 test the difference of mean are low, A= 10.5, 11.6, 12.7, 12.9, 13.5, 13.6, 14.8 and B= 11.3, 12.4, 12.4, 13.9, 14.2, 14.7, 14.9, 15.6 and level of significance is 5% and table value is 1.96 1) Hypothesis Ho : µ1 µ2 = 0 H1 : µ1 µ2 < 0 2) Level of significance α= 5% 3) Test-statistic ( 1 2 (µ1 µ2) 1 2 One sided test
- 63. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation 1 = 7 2 = 8 = 0.8 = 0.64 1 = 1 10.5+11.6+12.7+12.9+13.5+13.6+14.8 = . =12.8 2 = 1 11.3+12.4+12.4+13.9+14.2+14.7+14.9+15.6 = . =13.7 𝑧 = (12.8 13.7) ( ) 0.64 0.64 𝑧 = . . . 𝑧 = . . 𝑧 = . . = − 2.19
- 64. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion Calculated value lie in rejection region so, we reject the null hypothesis and accept the alternative hypothesis. -2.19 -1.96
- 65. UMAR DARAZ Entomologist BZU Multan Writea general procedure for hypothesis testing about the difference between mean when the observations are paired Sometimes in testing hypothesis about two means the two samples are dependent such samples are known as pair sample. Suppose, we take a sample of 10 patients then we get two measurements from each patient blood pressure before medicine and blood pressure after medicine and hence we obtain paired data. Always be used t-test
- 66. UMAR DARAZ Entomologist BZU Multan Hypothesis testing technique in paired observation is as follows… 1) Hypothesis Ho : µd = 0 H1 : µd ≠ 0 Ho : µd = 0 H1 : µd > 0 Ho : µd = 0 H1 : µd < 0
- 67. UMAR DARAZ Entomologist BZU Multan CONT… 2)Level of significance α= 1%, 2%, 5%, 10% 3) Test-statistic t µ s 4) Calculation = s d= difference =After Before
- 68. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion If calculated value lie in rejection region, then we reject the null hypothesis and accept the alternative hypothesis. If calculated value lie in acceptance region, then we accept the null hypothesis and reject the alternative hypothesis.
- 69. UMAR DARAZ Entomologist BZU Multan Example: 14 The follows are the weight of persons when they started weight reducing diet and two week later, Before =212,193,241,218,205,216,215,298,200,233,209,198 After = 195,185,225,199,194,193,205,176,188,224,201,195 and test a hypothesis mean weight loss of person two weeks on this diet is at most 10 pound and level of significance is 1% and table value is 2.718 1) Hypothesis Ho : µd = 10 H1 : µd < 10 2) Level of significance at most – (<) α= 1% at least – (≥) 3) Test-statistic t µd sd One sided test
- 70. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation = 12 µ 𝒅 = 10 Before After d=After Before 𝟐 212 195 -17 289 193 185 -8 64 241 225 -16 256 218 199 -19 361 205 194 -11 121 216 193 -23 529 215 205 -10 100 298 176 -122 14884 200 188 -12 144 233 224 -9 81 209 201 -8 64 198 195 -3 9 𝟐
- 71. UMAR DARAZ Entomologist BZU Multan CONT… == sd t µ s sd t sd t sd t sd t sd
- 72. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion Calculated value lie in rejection region so, we reject the null hypothesis and accept the alternative hypothesis. -3.39 -2.718
- 73. UMAR DARAZ Entomologist BZU Multan Example: 15 10 young recruits were put through physical training program by the army their weight were recorded before and after the training with the following results… Before = 125,195,160,171,140,201,170,176,195,139 After=136,201,158,184,145,195,175,190,190,145 and test the hypothesis at the level of significance 5% and table value is 2.262 1) Hypothesis Ho : µd = 0 H1 : µd 0 2) Level of significance α= 5% 3) Test-statistic t µd sd Two sided test
- 74. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation = 10 µ 𝒅 = 0 Before After d=After Before 𝟐 125 136 11 121 195 201 6 36 160 158 -2 4 171 184 13 169 140 145 5 25 201 195 -6 36 170 175 5 25 176 190 14 196 195 190 -5 25 139 145 6 36 𝟐
- 75. UMAR DARAZ Entomologist BZU Multan CONT… == sd t µ s sd t sd t sd t sd t sd
- 76. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion Calculated value lie in acceptance region so, we accept the null hypothesis and reject the alternative hypothesis. -2.262 -2.18 +2.18 +2.262
- 77. UMAR DARAZ Entomologist BZU Multan Writea general procedure for testing of hypothesis about variance of population 1) Hypothesis Ho : = H1 : ≠ Ho : = H1 : > Ho : = H1 : <
- 78. UMAR DARAZ Entomologist BZU Multan CONT… 2)Level of significance α= 1%, 2%, 5%, 10% 3) Test-statistic 4) Calculation
- 79. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion If calculated value lie in rejection region, then we reject the null hypothesis and accept the alternative hypothesis. If calculated value lie in acceptance region, then we accept the null hypothesis and reject the alternative hypothesis.
- 80. UMAR DARAZ Entomologist BZU Multan Example: 16 Given two random sample of size 1=12 and 2=10 from two population with 1=2.3 and 2=1.5 test the hypothesis at 5% level of significance, Ho : = H1 : > and table value is 3.10 1) Hypothesis Ho : = H1 : > 2) Level of significance α= 5% 3) Test-statistic One sided test
- 81. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation 1 = 2.3 = 5.29 2 = 1.5 = 2.25 5.29 2.25
- 82. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion Calculated value lie in acceptance region so, we accept the null hypothesis and reject the alternative hypothesis. +2.35 +3.10
- 83. UMAR DARAZ Entomologist BZU Multan Example: 17 Two random sample from two population are =20, 16,26,27,23, 22, 18,24,15,19 and =27,33,42,35,32,34,38,28,41,43,30,37 obtain estimate variance of population and test the two population have same variance at 5% level of significance and table value is 3.92 1) Hypothesis Ho : = H1 : ≠ 2) Level of significance α= 5% 3) Test-statistic Two sided test
- 84. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation = 10 = 12 𝟐 𝟐 20 400 27 729 16 256 33 1089 26 676 42 1764 27 729 35 1225 23 529 32 1024 22 484 34 1156 18 324 38 1444 24 576 28 784 15 225 41 1681 19 361 43 1849 30 900 37 1369 𝟐 𝟐
- 85.
- 86. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion Calculated value lie in acceptance region so, we accept the null hypothesis and reject the alternative hypothesis. -3.92 -0.58 +0.58 +3.92
- 87. UMAR DARAZ Entomologist BZU Multan Example: 18 In an experiments on reaction times in seconds of two individuals A&B measured under ideal conditions. The following results are obtained… A=0.41, 0.38, 0.37, 0.42, 0.35, 0.38 and B=0.32, 0.36, 0.38, 0.33,0.38 test the hypothesis Ho : = , H1 : ≠ at 5% level of significance and table value is 9.36 1) Hypothesis Ho : = H1 : ≠ 2) Level of significance α= 5% 3) Test-statistic Two sided test
- 88. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation = 6 = 5 0.41 0.1681 0.32 0.1024 0.38 0.1444 0.36 0.1296 0.37 0.1369 0.38 0.1444 0.42 0.1764 0.33 0.1089 0.35 0.1225 0.38 0.1444 0.38 0.1444 𝑩 𝑩
- 89.
- 90. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion Calculated value lie in acceptance region so, we accept the null hypothesis and reject the alternative hypothesis. -9.36 -0.86 +0.86 +9.36
- 91. UMAR DARAZ Entomologist BZU Multan Example: 19 Two independent random sample of size = 10 and = 7 were observed two have sample variance and using 10% level of significance and test the hypothesis Ho : = ,H1 : ≠ and table value is 4.10 1) Hypothesis Ho : = H1 : ≠ 2) Level of significance α=10 % 3) Test-statistic Two sided test
- 92. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation = 16 = 3 = 10 = 7 16 3
- 93. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion Calculated value lie in rejection region so, we reject the null hypothesis and accept the alternative hypothesis. -5.33 -4.10 +4.10 +5.33
- 94. UMAR DARAZ Entomologist BZU Multan Writea general procedure for testing of hypothesis of checking association between two attributes Always be used χ -test Always be used two sided test 1) Hypothesis Ho : There is no association between two attributes H1 : There is association between two attributes
- 95. UMAR DARAZ Entomologist BZU Multan CONT… 2)Level of significance α= 1%, 2%, 5%, 10% 3) Test-statistic χ It will be used for more than two rows or two column χ | | . It will be used for two rows and two column Observed frequency Expected frequency 4) Calculation
- 96. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion If calculated value lie in rejection region, then we reject the null hypothesis and accept the alternative hypothesis. If calculated value lie in acceptance region, then we accept the null hypothesis and reject the alternative hypothesis.
- 97. UMAR DARAZ Entomologist BZU Multan Example: 20 In an investigation into eye color and left or right handed person, the following results were obtained. Do these results indicate at 5% level of significance and association between eye color and left or right handed person and table value is 3.84 Left handed Right handed Blue 15 85 100 Brown 20 80 100 35 165 200 = n
- 98. UMAR DARAZ Entomologist BZU Multan CONT… 1)Hypothesis Ho : There is no association between eye color and left or right handed person H1 : There is association between eye color and left or right handed person 2) Level of significance α= 5% 3) Test-statistic χ
- 99. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation Expected frequency of 15 = × and so on… χ 15 17.5 2 4 0.23 85 82.5 2 4 0.05 20 17.5 2 4 0.23 80 82.5 2 4 0.05 = 0.56
- 100. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion Calculated value lie in acceptance region so, we accept the null hypothesis and reject the alternative hypothesis. -3.84 -0.56 +0.56 +3.84
- 101. UMAR DARAZ Entomologist BZU Multan Example: 21 Give the following table of hair color and eye color. Test the hypothesis at 5% level of significance that there is no association between hair color and eye color and calculate the co-efficient of contingency. Table value is 9.49 ontingency χ χ Black Brown Grey Black 15 5 20 40 Grey 20 10 20 50 Blue 25 15 20 60 60 30 60 150 = n Hair color Eye color
- 102. UMAR DARAZ Entomologist BZU Multan CONT… 1)Hypothesis Ho : There is no association between eye color and hair color H1 : There is association between eye color and hair color 2) Level of significance α= 5% 3) Test-statistic χ
- 103. UMAR DARAZ Entomologist BZU Multan CONT… 4)Calculation Expected frequency of 15 = × and so on… χ 𝒐 𝒆 𝒐 𝒆 𝒐 𝒆 𝟐 𝒐 𝒆 𝟐 𝒆 15 16 -1 1 0.0625 20 20 0 0 0 25 24 1 1 0.0417 5 8 -3 9 1.125 10 10 0 0 0 15 12 3 9 0.75 20 16 4 16 1 20 20 0 0 0 20 24 -4 16 0.667 = 3.65
- 104. UMAR DARAZ Entomologist BZU Multan CONT… 5)Critical region Frequency curve 6) Conclusion Calculated value lie in acceptance region so, we accept the null hypothesis and reject the alternative hypothesis. -9.49 -3.65 +3.65 +9.49
- 105.