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### Hypothesis - Biostatistics

1. 1. Biostatistics Lecture 8
2. 2. Lecture 7 Review– Using confidence intervals and p-values to interpret the results of statistical analyses • • • Null hypothesis P-value Interpretation of confidence intervals & p- values
3. 3. Null hypothesis • A null hypothesis is one that proposes there is no difference in outcomes • We commonly design research to disprove a null hypothesis
4. 4. P-value:- comparing two groups What is the probability (P-value) of finding the observed difference How likely is it we would see a difference this big IFIF The null hypothesis is true? There was NO real difference between the populations?
5. 5. Interpretation of p-values 1! Weak evidence against the null hypothesis0.1! Increasing evidence against the null hypothesis with decreasing P-value 0.01! 0.001! Strong evidence against the null hypothesis 0.0001! P-value!
6. 6. Objective To assess the effect of combined hormone replacement therapy on health related quality of life. Design Randomised placebo controlled double blind trial. (HRT) Table 3 EuroQoL Visual Analogue Scores (EQ-VAS) by treatment group. Figures are means (SE) one year (95% Combined HRT (n=1043*) Placebo (n=1087*) Adjusted difference at CI) P-value EQ-VAS 77.9 (0.5) 78.5 (0.4) -0.59 (-1.66 to 0.47) 0.28
7. 7. Five trials of drugs to reduce serum cholesterol A reduction of 0.5 mmol/L or more corresponds to a clinically important effect of the drug Trial Drug Cost No. of patients per group Observed difference in mean cholesterol (mmol/L) s.e. of difference (mmol/L) 95% CI for population difference in mean cholesterol P-value 1 A Cheap 30 -1.00 1.00 -2.96 to 0.96 0.32 2 A Cheap 3000 -1.00 0.10 -1.20 to -0.80 <0.001 3 B Cheap 40 -0.50 0.83 -2.13 to 1.13 0.55 4 B Cheap 4000 -0.05 0.083 -0.21 to 0.11 0.55 5 C Expensive 5000 -0.125 0.05 -0.22 to -0.03 0.012
8. 8. Lecture 8 – Proportions and intervals Binary variables (RECAP) confidence • • Single proportion – Standard error, confidence interval • Incidence & prevalence • Difference in two proportions – Standard error, confidence interval
9. 9. Categorical variables - Binary Binary variable – two categories only (also termed – dichotomous variable) Examples:-  Outcome – Diseased or Healthy; Alive or Dead…  Exposure - Male or Female; Smoker or non- smoker; Treatment or control group….
10. 10. Inference Proportion of population diseased – π?? Proportion of sample diseased, p=d/n Number of subjects who do experience outcome (diseased) = d Number of subjects who do not experience outcome (healthy) = h Total number in sample = n = h + d
11. 11. Inference - example Proportion of population with vivax malaria - π Proportion of sample with vivax p = d/n = 15/100 = 0.15 (15%) malaria, Number of sample with vivax malaria = d = 15 Number of sample without vivax malaria = h = 85 Total number in sample = n = 15 + 85 = 100
12. 12. Single proportion - Inference • Obtain a sample estimate, p, of the population proportion, π • REMEMBER different samples would give different estimates of π (e.g. sample 1 p1, sample 2 p2,…) • Derive: – Standard error – Confidence interval
13. 13. Standard error & confidence interval of a single proportion • Standard error (SE) for single proportion:- (from the Binomial distribution) π (1−π ) p(1− p) s.e.( p) = ~ n n • 95% CI for single proportion:- (approximate method based on the normal distribution) – – Lower limit = p - 1.96×s.e.(p) Upper limit = p + 1.96×s.e.(p)
14. 14. Standard error & confidence interval a single proportion – malaria exampleof • • Estimated proportion of vivax Standard error of p malaria (p) = 15/100 = 0.15 p(1− p) 0.15(1−0.15) s e ( p). . = = 0.036= n 100 • 95% Confidence interval for population proportion (π) – – Lower limit = p - 1.96×s.e.(p) = 0.15 – 1.96×0.036 = 0.079 Upper limit = p + 1.96×s.e.(p) = 0.15 + 1.96×0.036 = 0.221 Interpretation.. “We are 95% confident, the population proportion of people vivax malaria is between 0.079 and 0.221 (or between 7.9% and 22.1%)” with
15. 15. Definition of a confidence REMEMBER….. interval If we were to draw several independent, random samples (of equal size) from the sample population and calculate 95% confidence intervals for each of them, 0. 4 0.3 5 0. 3 Populatio n0.2 5 then on average 19 out of every 20 (95%) such confidence intervals would contain the true population proportion (π), and one of every 20 0. 2 0.1 5 0. 1 (5%) would not. 0.0 5 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Sample Sampleproportionand95%CI proportion = 0.16 (16%)
16. 16. WARNING…. Confidence Interval of a single proportion The normal approximation method breaks down 1) 2) if: Sample Sample size (n) is small proportion (p) is close to 0 or 1 Require: np ≥ 10 or n(1-p) ≥ 10 Stata lets you calculate an ‘exact’ CI
17. 17. Confidence Interval for a single proportion in Stata • cii 100 15 • • • • -- Binomial Exact -- [95% Conf. Interval]Variable | Obs Mean Std. Err. -------------+-------------------------------------------------------------- - | 100 .15 .0357071 .0864544 .2353075 • cii 100 1 • • • • -- Binomial Exact -- [95% Conf. Interval]Variable | Obs Mean Std. Err. -------------+-------------------------------------------------------------- - | 100 .01 .0099499 .0002531 .0544594
18. 18. Interpretation of proportions: Incidence versus Prevalence
19. 19. Prevalence Proportion of people in a defined population that have a given disease at a specified point in time • Prevalence = no. of people with the disease at particular point in time no. of people in the population at a particular point in time Examples:- • Prevalence living in the Prevalence Thailand. Prevalence of chronic pain among people aged 25+ years and Grampian region, UK. of typhoid among villagers living in Tak province,• • of diagnosed asthma in individuals aged 15 to 50 years, registered with a particular general practice in Carlton.
20. 20. Incidence risk (Cumulative incidence) Proportion of new cases in a disease free population in a given time period • Incidence risk = no. of new cases of disease in a given time period no. of people disease-free at beginning of time period Examples:- • Incidence risk of death in five years following diagnosis with prostate cancer Incidence risk of breast cancer over 10 years of follow-up in women 40-69 years of age and free from breast cancer in 1990 •
21. 21. Incidence rate (NOT a proportion) Number of new cases in a disease free population per person per unit time • that occur Incidence rate = no. of new cases of disease total person-years of observation Examples:- • Incidence rate of all-cause mortality of men in the Melbourne Collaborative Cohort Study = 9.0 per 1000 men per year ‘9 out of every 1000 men die each year’ (
22. 22. Comparing two proportions
23. 23. Comparing two proportions 2×2 table • • • Proportion Proportion Proportion of all subjects experiencing outcome, p = d/n in exposed group, p1 = d1/n1 in unexposed group, p0 = d0/n0 Be alert (not alarmed): watch for transposing the table and swapping columns or rows With outcome (diseased) Without outcome (disease-free) Total Exposed (group 1) d1 h1 n1 Unexposed (group 0) d0 h0 n0 Total d h n
24. 24. Comparing two proportions Example:- TBM trial (Thwaites GE et al 2004) Adults with tuberculous meningitis randomly allocated into treatment groups: 2 1. 2. Dexamethasone Placebo Outcome measure: Death during nine months following start of treatment. Research question: Can treatment with dexamethasone reduce the risk of death adults with tuberculous meningitis? among
25. 25. Comparing two proportions Example – TBM trial Death during 9 months post start of treatment Treatment group Yes No Total Dexamethasone (group 1) 87 187 274 Placebo (group 0) 112 159 271 Total 199 346 545
26. 26. Difference in two population proportions, π1-π0 Estimate of difference in population proportions = p1 – p0 Example:- TBM trial Dexamethasone p1 = d1/n1 = 87/274 = 0.318 Placebo p0 = d0/n0 = 112/271 = 0.413 p1 – p0 = 0.318 – 0.413 = -0.095 (or -9.5%)
27. 27. Difference in two proportions - Inference • Obtain a sample estimate, p1-p0, of the difference in population proportions, π1Dπ0 • REMEMBER different samples of π1Dπ0 (e.g. sample 1 p11-p10, would give different estimates sample 2 p21-p20,…) • Derive: – Standard error of difference in sample proportions – Confidence interval of difference in population proportions
28. 28. Standard error & confidence interval for difference between two proportions • Standard error (SE) for difference between sample proportions:- [s.e.( p )]2 +[s.e.( p )]2 s.e.( p ) =− p 1 0 1 0 • 95% CI for difference between population Lower limit = (p1-p0) - 1.96×s.e.(p1-p0) Upper limit = (p1-p0) + 1.96×s.e.(p1-p0) proportions:-
29. 29. Standard error & confidence interval for difference between two proportions Example:- TBM trial Estimate of difference in population proportions = p1-p0 = -0.095 s.e.(p1-p0) = 0.041 95% CI for difference in population proportions (π1-π0): -0.095 ± 1.96×0.041 -0.175 up to -0.015 OR -17.5% up to -1.5% Interpretation:- “We are 95% confident, that the difference in population proportions is between -17.5% (dexamethasone reduces the proportion of deaths by a large amount) and -1.5% (dexamethasone marginally reduces the proportion of deaths)”.
30. 30. Comparing proportions using csi 87 112 187 159 Stata | Exposed Unexposed | Total -----------------+------------------------+------------ Cases | Noncases | 87 187 112 159 | | 199 346 -----------------+------------------------+------------ Total | | | | | 274 271 | | | | | 545 Risk .3175182 .4132841 .3651376 Point estimate [95% Conf. Interval] |------------------------+------------------------ Risk difference Risk ratio Prev. frac. ex. Prev. frac. pop | | | | -.0957659 .7682808 .2317192 .1164974 | | | | -.1762352 -.0152966 .6139856 .0386495 .9613505 .3860144 +------------------------------------------------- chi2(1) = 5.39 Pr>chi2 = 0.0202 Remember the warning about how the table is presented -Stata requires presentation with outcome by rows and exposure by columns Results are close to those obtained by hand
31. 31. Difference between two proportions:- Risk difference Example:- TBM trial Outcome measure: Death during nine months treatment. following start of Dexamethasone p1 (incidence risk) = d1/n1 = 87/274 = 0.318 Placebo p0 (incidence risk) = d0/n0 = 112/271 = 0.413 p1 – p0 (risk difference) = 0.318 – 0.413 = -0.095 (or -9.5%)
32. 32. Lecture 8 – Objectives • Define binary variables, prevalence and incidence risk • Calculate and interpret a proportion and 95% confidence interval for the population proportion • Calculate and interpret the difference in sample proportions and 95% confidence interval for difference in population proportions
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