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1. 1. Interpretation of statistical values & fundamentals of epidemiology Dr.Asma Rahim Dr.Bindhu vasudevan Dept. of Community Medicine
2. 2. What you are expected to Know? • Mean • What is SD ? • What is SE & its applications • What is Confidence limits as noted in many journals?
3. 3. • What is P value ?How to interpret it? • Which are the different statistical tests to be applied on different situations?
4. 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 on Statistics.
5. 5. • Descriptive Statistics • Inferential statistics
6. 6. Types of variables • Qualitative – Dichotomous – Nominal – Ordinal • Quantitative – Discrete – Continuous
7. 7. 1. Which is a qualitative variable • a) BMI • b) S. bilirubin • c) Name of residing place • d) Blood urea
8. 8. 2. Which is a quantitative variable • Causes of deaths • Religious distribution • Age group distribution • Age distribution
9. 9. 4. Which is an ordinal variable • A)Blood pressure • B)Name of residing place • C)Grading of carcinoma • D) temperature
10. 10. 5. Which is not a nominal scale variable • A)Causes of death • B) religion • C)diagnosis • D)visual analogue scale
11. 11. Quantitative data Qualitative data Hb in gm% Anemic/non anemic Height in cm Tall/short B.P in mm of Hg Hypo/normo/ hypertensives
12. 12. Measures of central Tendency • Qualitative data – Proportion • Quantitative data – Mean,Median,Mode
13. 13. In a group of 100 under five children attending 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?
14. 14. Range in which 95% children’s weight in the sample will lie: 95% reference range = mean +/- 2SD = 11-19Kg What is the mean Birth weight of all the children attending IMCH O.P 95% Confidence interval = mean +/- 2SE( Standard error)-
15. 15. Central limit theorem • Difficult to study the whole population • Researcher wants to extrapolate the study findings to population
16. 16. In a group of 100 under five children attending IMCH O.P the mean weight is 15kg. The standard deviation is 2. what will be the mean weight of all children who are attending IMCH OP will lie?
17. 17. 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
18. 18. 17kg 19 16 15 18 17 Standard Error
19. 19. 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
20. 20. 17kg 19 16 15 18 17 Standard Error
21. 21. Applications of SE • To find out the range in which the population mean will lie ( 95% confidence interval- sample mean +/- 2SE) • To know whether the sample is representative of the population if population mean is known • To find the observed difference of two samples is statistically significant
22. 22. • In a group of 100 children the mean weight is 15kg. The standard error is 0.02. In what range the population mean will lie.
23. 23. 95% Confidence interval • Range in which the mean population value will lie • Mean +/ - 2 SE
24. 24. • 95% confidence limits – sample mean +/- 2 SE • 95% CI =15+/- 2x0.02=14.96-15.04kg
25. 25. • 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 PEFR of the population will lie from which the sample was taken?
26. 26. Range in which 95% of girls PEFR in the sample will lie: mean +/- 2SD = 260 - 340 Range in which mean PEFR Value will lie: mean +/- 2SE( Standard error)- 95% Confidence interval = 296-304
27. 27. Normal distribution curve •
28. 28. Applications of SE • To find out the range in which the population mean will lie ( 95% confidence interval- sample mean +/- 2SE) • To know whether the sample is representative of the population if population mean is known • To find the observed difference of two samples is statistically significant
29. 29. • In a village the percentage of male population is 52%. In a sample of 100 people the male percentage was 40 with a standard error of 5.Is this sample representing the population
30. 30. Answer • SE = 5 95% CI= sample proportion +/- 2 SE = 40 +/- 2 x 5 =30- 50 52% is higher than this range
31. 31. Applications of SE • To find out the range in which the population mean will lie ( 95% confidence interval- sample mean +/- 2SE) • To know whether the sample is representative of the population if population mean is known • To find the observed difference of two samples is statistically significant
32. 32. Height of 100 boys & 100 girls gave the following values. Do these two groups differ significantly Mean height SE Girls 150cm 2 boys 160cm 3
33. 33. Answer Girls – 95% CI = 150 +/- 2 x 2 =146 -154 • Boys – 95% CI = 160 +/- 2x 3 =154-166 • Overlapping is present among the 95% CI • Both groups can have the same population mean
34. 34. 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 %
35. 35. Sample size • Qualitative data N = 4pq/L2 • P = positive factor /prevalence/proportion • Q = 100 – p • L = allowable error or precision or variability • 4 = 1.962(Alpha error) 2 • Quantitative data N = 4SD2/L2
36. 36. 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 %
37. 37. • N= 4 x 80 x 20/8 x 8 = 100
38. 38. • 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.
39. 39. • N = 4SD2/L2 • 4 x 90 x 90 /9 x9 = 400
40. 40. • 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
41. 41. • Quantitative data N = (Zα + Zβ )2 x S2 x 2 /d2 Zα = Z value for α level = 1.96 at α 0.05 Zβ = Z value for β level =1.28 for β at 10% & 0.82 at 20% S = average SD d = difference between the two means
42. 42. • 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
43. 43. • N = 10.5 x 25 x 25 x2/ 25 x 25 = 21
44. 44. • Qualitative data N = (Zα + Zβ )2 p x q /d2 Zα = Z value for α level = 1.96 at α 0.5 Zβ = Z value for β level =1.28 for β at 10% P = average prevalence /proportion/positive factor d = difference between the prevalence/proportion/positive factor
45. 45. • In a study conducted on a sample of 400 adults, it was found that mean daily requirement of Vit. A was 900 I.U. From the existing literature the same was documented as 930 I.U with a SE of 4.5 I.U. Does the study finding differ from the existing literature finding significantly?
46. 46. Steps for testing a hypothesis • State Null Hypothesis • State alternate hypothesis • Fix the alpha error • Identify the correct statistical test
47. 47. • In a study conducted on a sample of 400 adults, it was found that mean daily requirement of Vit. A was 900 I.U. From the existing literature the same was documented as 930 I.U with a SE of 4.5 I.U. Does the study finding differ from the existing literature finding significantly?
48. 48. Reject Null hypothesis Accept Null hypothesis Null hypothesis true Type 1 error (alpha error) Correct decision Null hypothesis false Correct decision Type 2 error (Beta error)
49. 49. • Alpha = 5% (0.05) • 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.
50. 50. Difference in proportion Chi-square test, Z test, Difference in mean (Before and after comparison-same group) Paired t test Difference in mean (two independent groups) Unpaired t test, If sample > 30-Z test More than 2 means(> 2 groups) Anova Association b/w 2 quantitative variables Spearman correlation Prediction regression
51. 51. Parametric and Nonparametric tests Parametric: When the data is normally distributed. Nonparametric : When data is not normally distributed,usually with small sample size.
52. 52. Non parametric tests Qualitative data Chi-square test Fishers test, Mc Nemar test Paired t test Wilcoxon Signed rank test independent t test Wilcoxon test , Mann- Whitney U , Kolmogrov independent t test Kruskal-wallis test
53. 53. 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
54. 54. Difference in proportion Chi-square test, Z test, Difference in mean (Before and after comparison-same group) Paired t test Difference in mean (two independent groups) Unpaired t test, If sample > 30-Z test More than 2 means(> 2 groups) Anova Association b/w 2 quantitative variables Spearman correlation Prediction regression
55. 55. The most appropriate test for comparing Hb values in the adult women in two different population of size 150 and 200 is • A) t test • B) Anova • C) Z test • D) Chi square test
56. 56. Difference in proportion Chi-square test, Z test, Difference in mean (Before and after comparison-same group) Paired t test Difference in mean (two independent groups) Unpaired t test, If sample > 30-Z test More than 2 means(> 2 groups) Anova Association b/w 2 quantitative variables Spearman correlation Prediction regression
57. 57. Answer • C – Two groups – >30 – Continuous variable – Comparing mean
58. 58. 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
59. 59. Difference in proportion Chi-square test, Z test, Difference in mean (Before and after comparison-same group) Paired t test Difference in mean (two independent groups) Unpaired t test, If sample > 30-Z test More than 2 means(> 2 groups) Anova Association b/w 2 quantitative variables Spearman correlation Prediction regression
60. 60. Answer • B – Continuous variable – Compare means – > 2 groups
61. 61. 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
62. 62. Difference in proportion Chi-square test, Z test, Difference in mean (Before and after comparison-same group) Paired t test Difference in mean (two independent groups) Unpaired t test, If sample > 30-Z test More than 2 means(> 2 groups) Anova Association b/w 2 quantitative variables Spearman correlation Prediction regression
63. 63. Answer • A – Two different groups – Continuous variable – Size <30
64. 64. The association between smoking status and MI is tested by • A) t test • B) Anova • C) F test • D) Chi square test
65. 65. Difference in proportion Chi-square test, Z test, Difference in mean (Before and after comparison-same group) Paired t test Difference in mean (two independent groups) Unpaired t test, If sample > 30-Z test More than 2 means(> 2 groups) Anova Association b/w 2 quantitative variables Spearman correlation Prediction regression
66. 66. 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.
67. 67. Difference in proportion Chi-square test, Z test, Difference in mean (Before and after comparison-same group) Paired t test Difference in mean (two independent groups) Unpaired t test, If sample > 30-Z test More than 2 means(> 2 groups) Anova Association b/w 2 quantitative variables Spearman correlation Prediction regression
68. 68. • 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.
69. 69. Test to detect mean difference in body weight between Group A & Group B • T-TEST • Difference between means of two samples
70. 70. 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.
71. 71. Is the difference in body weight of subjects in Group A,GROUP b ,group C significantly different at Time 2 • Analysis of variance
72. 72. Is there any relation between blood pressure and body weight of these subjects?
73. 73. Difference in proportion Chi-square test, Z test, Difference in mean (Before and after comparison-same group) Paired t test Difference in mean (two independent groups) Unpaired t test, If sample > 30-Z test More than 2 means(> 2 groups) Anova Association b/w 2 quantitative variables Spearman correlation Prediction regression
74. 74. Association b/w 2 quantitative variables •Correlation
75. 75. 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
76. 76. • 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
77. 77. Did the Installation of water supply system significantly reduce deaths Which non parametric test will be used to test the null hypothesis • Small sample size • Distribution is not normal • Non parametric test • Wilcoxon signed rank test
78. 78. Non parametric tests Qualitative data Chi-square test Fishers test, Mc Nemar test Paired t test Wilcoxon Signed rank test independent t test Wilcoxon test , Mann- Whitney U , Kolmogrov independent t test Kruskal-wallis test
79. 79. For treatment of Hepatitis A 7 patients treated with herbal medicines& 7 patients treated with Allopathic 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
80. 80. Is herbal treatment is better than allopathic treatment? • Small sample size • Distribution is not normal • Non parametric test • Mann- Whitney test
81. 81. Non parametric tests Qualitative data Chi-square test Fishers test, Mc Nemar test Paired t test Wilcoxon Signed rank test independent t test Wilcoxon test , Mann- Whitney U , Kolmogrov independent t test Kruskal-wallis test
82. 82. Steps for testing a hypothesis • State Null Hypothesis • State alternate hypothesis • Fix the alpha error • Identify the test statistic
83. 83. • Find out the critical value • Calculate the value for the identified statistical test Difference in means/ SE • If the calculated value is > the table value(critical value)- Reject Null Hypothesis
84. 84. • In a study conducted on a sample of 400 adults, it was found that mean daily requirement of Vit. A was 900 I.U. From the existing literature the same was documented as 930 I.U with a SE of 4.5 I.U. Does the study finding differ from the existing literature finding significantly?
85. 85. Null hypothesis Alpha Error – 5% Test static –Z test SE = 4.5 Z = 930-900/4.5=6.67
86. 86. – For alpha error 5%, critical Z value = 1.96 – 6.67 >1.96 So we will Reject null hypothesis – There is a significant difference – P value
87. 87. • After applying a statistical test an investigator get the p value as 0.01. What does it mean?
88. 88. • 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
89. 89. • 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
90. 90. In assessing the association between maternal nutritional status and Birth weight 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
91. 91. Low birth weight & selected risk factors Risk factor P value Maternal age 0.01 Birth order 0.1 Employment status of mother 0.9 Mean Weight gain during pregnancy 0.002 UTI during pregnancy 0.03 Mean Hb 0.0001
92. 92. • A study was conducted to find out the association between Per Capita National Income and Per Capita Consumer Expenditure from the data given below
93. 93. 230 235 240 245 250 255 260 240 250 260 270 280 290 Series1
94. 94. • . What is the name of this diagram? • What is its use? • From the diagram what is your inference?
95. 95. Type of study Alternative name Unit of study Descriptive Case series Cross sectional Longitudinal Prevalence study Incidence study Individual Analytical studies (observational Ecological Case control Cohort Correlational Case reference Follow up Populations Individuals Individuals Analytical studies (interventional) Randomised controlled trial Field trial Community trials Clinical trial Community intervention Community Patients Healthy people Healthy people
96. 96. Study questions and appropriate designs Type of question Appropriate study design Burden of illness Cross sectional survey Longitudinal survey Causation, risk and prognosis Case control study, Cohort study Occupational risk, environmental risk Ecological studies Treatment efficacy RCT Diagnostic test evaluation Paired comparative study Cost effectiveness RCT
97. 97. 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?
98. 98. Study design –Case control • Measure of risk –Odd’s ratio
99. 99. • Set up a 2x2 table a b 2 21 c d 3 24 Worm infestation + + - -
100. 100. • Odd’s ratio = ad /bc • 2 x 24 = 0.76 21 x3
101. 101. Interpretation • OR =1,RISK FACTOR NOT RELATED TO DISEASE • OR <1 ,RISK FACTOR PROTECTIVE • OR >1 RISK FACTOR POSITIVELY ASSOCIATED WITH DISEASE
102. 102. 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
103. 103. Low birth weight Normal No.children studied 300 300 No.malnourished At age one 102 51
104. 104. 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 ? Rr 0 no association Rr >1 + association
105. 105. • 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?
106. 106. Results of a screening test Disease Positive Negative Positive TP(a) FP(b) Test Negative FN© TN(d)
107. 107. 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
108. 108. 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.
109. 109. Sensitivity and Specificity + - Fever > 37.50c + 21a 28b - 9c 42d 30a+c 70b+d
110. 110. • 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
111. 111. Exercise 11 Disease 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%
112. 112. • Total population = 10,000 • Disease prevalence = 5% • No diseased = 500 • Applying this to a 2x2 table :
113. 113. 2x2 table + - + TEST 350 a 1900 b 2250 - 150c 7600d 7750 500 9500 10000
114. 114. All the Best!!1 http://www.dnbpediatrics.com/
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