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Statistics- by Dr.Asma Rahim and Dr.Bindu Vasudevan

Statistics- by Dr.Asma Rahim and Dr.Bindu Vasudevan

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- 1. Interpretation of statistical values & fundamentals of epidemiology Dr.Asma Rahim Dr.Bindhu vasudevan Dept. of Community Medicine
- 2. What you are expected to Know?• Mean• What is SD ?• What is SE?• What is Confidence limits as noted in many journals?
- 3. • What is P value ?How to interpret it?• Which are the different statistical tests to be applied on different situations?• Study designs in Medical research.• Measurements of risk in clinical research• What is sensitivity ,Specificity, Predictive value of a test?
- 4. Dilemma of a PG Student!!!•DNB exams more stress on Original work.•Methodology of your work is important.•Look ahead for statistical queries.•Examiners familiar with research designs•OSCE stations have questions onStatistics.
- 5. Types of variables• Qualitative – Dichotomous – Nominal – Ordinal• Quantitative – Discrete – Continuous
- 6. 1. Which is a qualitative variable• a) BMI• b) S. bilirubin• c) Name of residing place• d) Blood urea
- 7. 2. Which is a quantitative variable• Causes of deaths• Religious distribution• Age group distribution• Age distribution
- 8. 4. Which is an ordinal variable• A)Blood pressure• B)Name of residing place• C)Grading of carcinoma• D) temperature
- 9. 5. Which is not a nominal scale variable• A)Causes of death• B) religion• C)diagnosis• D)visual analogue scale
- 10. Quantitative data Qualitative dataHb in gm% Anemic/non anemicHeight in cm Tall/shortB.P in mm of Hg Hypo/normo/ hypertensives
- 11. In a group of 100 under five childrenattending IMCH O.P the mean weight is 15kg. The standard deviation is 2.1.In what range 95% of children’s weight will lie in the sample? 2. In what range the mean weight of all children who are attending IMCH OP will lie?
- 12. Range in which 95% children’s weight in the sample will lie: 95% reference range = mean +/- 2SD = 13-17KgRange in which 95% children’s weight attending IMCH O.P will lie: 95% Confidence interval = mean +/- 2SE( Standard error)-
- 13. Standard Error 17 19 16 17kg 1815
- 14. Central limit Theorem• Central limit theorem states that• The random sampling distribution of sample means will be normal distribution• Means of random sample means will be equal to population mean• The standard deviation of sample means from population mean is the standard error
- 15. • The PEFR of 100, 11 year old girls follow a normal distribution with a mean of 300 1/min, standard deviation 20 l/min and standrd error of 2 l/min• What will be the range in which 95% of the girl’s PEFR will lie in the sample?• What will be the range in which mean of the population will lie from which the sample was taken?
- 16. Range in which 95% of girls PEFR in the sample will lie: mean +/- 2SD = 260 - 340Range in which mean PEFR Value will lie: mean +/- 2SE( Standard error)- 95% Confidence interval = 298-304
- 17. Normal distribution curve•
- 18. Sample size• Calculate the sample size to find out the prevalence of a disease after implementing a control programme with 10% allowable error. Prevalence of the disease before implementing the programme was 80 %
- 19. Sample size• Qualitative data N = 4pq/L2• P = positive factor /prevalence/proportion • Q = 100 – p • L = allowable error or precision or variability• Quantitative data N = 4SD2/L2
- 20. • N= 4 x 80 x 20/8 x 8 = 100
- 21. • Determine the sample size to find out the Vitamin A requirement in the under five children of Calicut district . From the existing literature the mean daily requirement of the same was documented as 930 I.U with a SD of 90 I.U. Consider the precision as 9.
- 22. • N = 4SD2/L2• 4 x 90 x 90 /9 x9 = 400
- 23. • Determine the sample size to prove that drug A is better than drug B in reducing the S.Cholesterol. The findings from a previous study is given Drug Mean SD A 215 20 B 240 30
- 24. • Quantitative data N = (Zα + Zβ )2 x S2 x 2 /d2Zα = Z value for α level = 1.96 at α 0.05Zβ = Z value for β level =1.28 for β at 10%S = average SDd = difference between the two means
- 25. • Qualitative data N = (Zα + Zβ )2 p x q /d2Zα = Z value for α level = 1.96 at α 0.5Zβ = Z value for β level =1.28 for β at 10%P = average prevalenced = difference between the prevalence
- 26. Reject Accept Null hypothesis Null hypothesisNull hypothesis Type 1 error Correcttrue (alpha error) decisionNull hypothesis Correct Type 2 errorfalse decision (Beta error)
- 27. • Alpha = 1.96.• Beta = 0.1 to 0.2 or 10 to 20%.• Power of the study = 1- beta error• Strength at which we conclude there is no difference between the two groups.
- 28. Statistical test chosen depends on----• Whether comparison is between independent or related groups.• Whether proportions or means are being compared.• Whether more than 2 groups are compared.
- 29. Deciding statistical tests?• In a clinical trial of a micronutrient on growth, the weight was measured before and after giving the micronutrient.. Which test will you use for comparison?• paired t test• F test• T test• Chi square test
- 30. Parametric and Nonparametric testsParametric: When the data is normally distributed.Nonparametric : When data is not normally distributed,usually with small sample size.
- 31. Common statistical testsDesign Nature of variable Statistical test Statistic derivedTwo independent Qualitative (nominal) Chi square Chi squareGroups Quantitative (continous) Student t test tTwo relatedgroups Qualitative (nominal) Chi square Chi square Quantitative (continous) Paired t test tMore than 2 Qualitative (nominal Chi square Chi squareIndependent Quantitative (continous) Anova Fgroups
- 32. Difference in proportion Chi-square test, Z test,Difference in mean(Before Paired t testand after comparison-samegroup)Difference in mean (two Unpaired t test, If sampleindependent groups) > 30-Z testMore than 2 means(> 2 Anovagroups)Association Spearman correlationPrediction regression
- 33. Non parametric testsChi-square testFishers test,Mc Nemar testWilcoxon Signed rank test Paired t testWilcoxon test , Mann- independent t testWhitney U , KolmogrovKruskal-wallis test Anova
- 34. The most appropriate test for comparing Hb values in the adultwomen in two different population of size 150 and 200 is • A) t test • B) Anova • C) Z test • D) Chi square test
- 35. Answer• C – Two groups – >30 – Continuous variable – Comparing mean
- 36. The most appropriate test to compare birth weight in 3 different regions is• A) t test• B) Anova• C) Z test• D) Chi square test
- 37. Answer• B – Continuous variable – Compare means – > 2 groups
- 38. The most appropriate test to compare BMI in two different adult population of size 24 and 30 is• A) Two sampled t test• B) Paired t test• C) Z test• D) Chi square test
- 39. Answer• A – Two different groups – Continuous variable – Size <30
- 40. The association between smoking status and MI is tested by• A) t test• B) Anova• C) F test• D) Chi square test
- 41. Standard drug used 40% of patients responded and a new drug when used 60% of patients responded. Which of the following tests of parametric significance is most useful in this study?• A) Fishers t Test• B) Independent sample t test• C) Paired t test• D) Chi square test.
- 42. • A consumer group would like to evaluate the success of three different commercial weight loss programmes. Subjects are assigned to one of three programmes (Group A , Group B ,GROUP C) . Each group follows different diet regimen. At first time and at the end of 6 weeks subjects are weighed an their BP measurements recorded.
- 43. Test to detect mean difference inbody weight between Group A & Group B• T-TEST• Difference between means of two samples
- 44. Is there a significant difference in body weight in Group A at Time 1 and Time 2?• Paired T Test• Same people sampled on two Occasions.
- 45. Is the difference in body weight of subjects in Group A,GROUP b ,group C significantly different at Time 2• Analysis of variance
- 46. Is there any relation between blood pressure and body weight of these subjects?• Correlation
- 47. Correlation coefficient• Shows the relation between two quantitative variable• Shows the rate of change of one variable as the other variable change• The value lies between –1 to + 1• Correlation coefficient of zero means that there is no relationship
- 48. No.of deaths in 8 villages due to water borne diseases before &after installation of water supply system• Villages: 1 2 3 4 5 6 7 8• Before :13 6 12 13 4 13 9 10• After :15 4 10 9 1 11 8 13
- 49. Did the Installation of water supply system significantly reduce deaths• Small sample size• Distribution is not normal• Non parametric test• Wilcoxon signed rank test
- 50. For treatment of Hepatitis A 7 patients treated with herbal medicines& 7 patients treated withAllopathic symptomatic management.S.Br values after 10 days of treatment is given below• Herbal :9 6 10 3 6 3 2• Allopathy: 6 3 5 6 2 4 8
- 51. Is herbal treatment is better than allopathic treatment?• Small sample size• Distribution is not normal• Non parametric test• Mann- Whitney test
- 52. After applying a statistical test an investigator get the p value as 0.01. It means that• A)The probability of finding a significant difference is 1%• B) The probability of finding a significant difference when there is no difference is 1%• C) The difference is not significant 1% times and significant 99% times• D) The power of the test used is 99%
- 53. Answer• B• Null hypothesis states there is no difference,If there is any difference it is due to chance• P value = If the null hypothesis is true the probability of the sample variation to occur by chance• P value 0.05= probability of the sample variation by chance is only 5% if null hypothesis was true• 95% the sample variation is not due to chance,& there is a difference. So we will reject NH
- 54. • P = 0.01 - probability of the sample variation by chance is only 1% if null hypothesis was true• 99 % the sample variation is not due to chance,& there is a difference. So we will reject NH• As p value decreases the difference become more significant• For practical purpose p value < 0.05 ; the difference is significant
- 55. In assessing the association between maternal nutritional status and Birthweight of the newborns two investigators A and B studied separately and found significant results with p values 0.02 &0.04 respectively. From this what can you infer about the magnitude of association found by the two investigators
- 56. Type of study Alternative Unit of study nameDescriptive Case series Prevalence Cross sectional study Individual Longitudinal Incidence study CorrelationalAnalytical Ecological Case reference Populationsstudies Case control Follow up Individuals(observational Cohort IndividualsAnalytical studies Randomised Clinical trial Patients(interventional) controlled trial Community Healthy people Field trial intervention Community Community Healthy people trials
- 57. Study questions and appropriate designsType of question Appropriate study designBurden of illness Cross sectional survey Longitudinal surveyCausation, risk and Case control study, Cohort studyprognosisOccupational risk, Ecological studiesenvironmental riskTreatment efficacy RCTDiagnostic test Paired comparative studyevaluationCost effectiveness RCT
- 58. Odd’s ratio• In a study conducted by Gireesh G N etal about the ‘Prevalence of Worm infestation in children”,50 children in anganwadi were examined. Out of this 5 had worm infestation. 2 out of this 5 have a history of pet animals at home while 21 out of the 45 non infested has a history of pet animals at home. Is there any association between pet animals and worm infestations?
- 59. Study design –Case control• Measure of risk –Odd’s ratio
- 60. • Set up a 2x2 table Worm infestation + - a b + 2 21 Pet animals - c d 3 24
- 61. • Odd’s ratio = ad /bc• 2 x 24 = 0.76 21 x3
- 62. Interpretation• OR =1,RISK FACTOR NOT RELATED TO DISEASE• OR <1 ,RISK FACTOR PROTECTIVE• OR >1 RISK FACTOR POSITIVELY ASSOCIATED WITH DISEASE
- 63. Relative risk• In a study to find the effect of Birth weight on subsequent growth of children , 300 children with birth weight 2kg to 2.5 kg were followed till age 1 . A similar number of children with birth weight greater 2.5 kg were followed up too. Anthropometric measurements done in both groups. Results are shown below
- 64. Low birth weight NormalNo.children studied 300 300No.malnourishedAt age one 102 51
- 65. Study design –Cohort study• Measure of risk –Relative risk ,Attributable risk.• Relative risk –Incidence among exposed Incidence among nonexposed = 102/300 = 0.34 = 2 51/ 300 0.17 Inference ?
- 66. • An out break of Pediculosis capitis being investigated in a girls school with 291 pupils.Of 130 Children who live in a nearby housing estate 18 were infested and of 161 who live elsewhere 37 were infested. The Chi square value was found to be 3.93 .• P value = 0.04• Is there a significant difference in the infestation rates between the two groups?
- 67. Results of a screening test Disease Positive Negative Positive TP(a) FP(b)Test Negative FN© TN(d)
- 68. Features of a screening test Sensitivity = a/ a+c Specificity = d/b+d Positive predictive value = a/a+b Negative predictive value = d/c+d False positive rate = bb+d False negative rate = c/a+c
- 69. In a group of patients presenting to a hospital emergency with abdominal pain, 30% of patients have acute appendicitis, 70% of patients with appendicitis have a temperature greater than 37.50c and 40% of patients without appendictis have a temperature greater than 37.50c. Considering these findings which of the following statement is correct ? a) Sensitivity of temperature greater than 37.50c as a marker for appendicitis is 21/49 b) Specificity of temperature grater than 37.50c as a marker for appendicitis is 42/70 c) The positive predictive value of temperature greater than 37.50c as marker for appendicitis is 21/30 d) Specificity of the test will depend upon the prevalence of appendicitis in the population to which it is applied.
- 70. Sensitivity and Specificity + Appendicitis - 21a 28bFever > 37.50c + - 9c 42d 30a+c 70b+d
- 71. • Sensitivity = a/a+c - 21/30=70%• Specificity = d/b+d = 42/70=60%• Positive predictive value = a/a+b = 21/49=43%• Negative predictive value = d/c+d = 42/51
- 72. Exercise 11Disease prevalence in a population of 10,000 was 5%. A urine sugar test with sensitivity of 70% and specificity of 80% was done on the population. The positive predictive value will be : a)15.55% b) 70.08% c) 84.4% d)98.06%
- 73. • Total population = 10,000• Disease prevalence = 5%• No diseased = 500• Applying this to a 2x2 table :
- 74. 2x2 table + -+ TEST 350 a 1900 b 2250 - 150c 7600d 7750 500 9500 10000
- 75. All the Best!!1

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