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P Values - part 2 Samples & Populations. Robin Beaumont 2011 With much help from Professor Chris Wilds material University of Auckland. probability. Aspects of the P value. P Value. sampling. statistic. Rule. A P value is a conditional probability considering a range of outcomes.
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P Values - part 2Samples & Populations Robin Beaumont 2011 With much help from Professor Chris Wilds material University of Auckland
probability Aspects of the P value P Value sampling statistic Rule
A P value is a conditional probability considering a range of outcomes Resume Sample value P value = P(observed summary value + those more extreme |population value = x) Hypothesised population value
Populations and samples Ever constant at least for your study! = Parameter estimate = statistic
Standard deviation - individual level Area! Wait and see But does not take into account sample size = t distribution = measure of variability 'Standard Normal distribution' Area: Defined by sample size aspect ~ df 95% 68% Total Area = 1 SD value = 2 1 0 Between + and - three standard deviations from the mean = 99.7% of area Therefore only 0.3% of area(scores) are more than 3 standard deviations ('units') away. -
Sampling level -‘accuracy’ of estimate Talking about means here We can predict the accuracy of your estimate (mean) by just using the SEM formula. From a single sample = 5/√5 = 2.236 SEM = 5/√25 = 1 From: http://onlinestatbook.com/stat_sim/sampling_dist/index.html
Example - Bradford Hill, (Bradford Hill, 1950 p.92) • mean systolic blood pressure for 566 males around Glasgow = 128.8 mm. Standard deviation =13.05 • Determine the ‘precision’ of this mean. • “We may conclude that our observed mean may differ from the true mean by as much as ± 2.194 (.5485 x 4) but not more than that in around 95% of samples. page 93. [edited] All possible values of POPULATION mean
Sampling summary • The SEM formula allows us to: • predict the accuracy of your estimate ( i.e. the mean value of our sample) • From a single sample • Assumes Random sample
Variation what have we ignored! Onto Probability now
Probabilities are rel. frequencies All outcomes at any one time = 1
Probability Density Function 11 The total area = 1 total 48 scores 10 9 8 7 6 Probability 5 4 Density 3 B A 2 1 0 33 37 43 47 53 57 63 67 73 77 83 87 Scores p(score<45) = area A p(score > 50) = area B Multiple outcomes at any one time P(score<45 and score >50) = Just add up the individual outcomes
What happens in the past affects the present = Conditional Probability Disease X P(disease x |male) Disease X AND Male Male P(male) P(disease AND male) = P(male) x P(disease x | male) No Disease X Disease X female P(disease AND male) /P(male) = P(disease x | male) No Disease X Multiple each branch of the tree to get end value
Screening Example 0.1% of the population carry a particular faulty gene. A test exists for detecting whether an individual is a carrier of the gene. In people who actually carry the gene, the test provides a positive result with probability 0.9. In people who don’t carry the gene, the test provides a positive result with probability 0.01. Let G = person carries gene P = test is positive for gene N = test is negative for gene If someone gets a positive result when tested, find the probability that they actually are a carrier of the gene. We want to find P(P) = P(G and P) + P(G' and P) = 0.0009 + 0.00999 = 0.01089 P( P | G) Errors P(P | G) ≠ P (G | p) ORDER MATTERS
Survival analysis • Each years survival depends on previous ones or does it?
Probability summary • All outcomes at any one time add up to 1 • Probability histogram = area under curve =1 • -> specific areas = set of outcomes • -> specific areas = ‘equal to or more extreme’ • Conditional probability – present dependent on past – ORDER MATTERS
sampling Putting it all together P Value probability statistic Rule
Statistics • Summary measure – SEM, Average etc • T statistic – different types, simplest: So when t = 0 means 0/anything = estimated and hypothesised population mean are equal So when t = 1 observed different same as SEM So when t = 10 observed different much greater than SEM
T statistic example Serum amylase values from a random sample of 15 apparently healthy subjects. The mean = 96 SD= 35 units/100 ml. How likely would such a sample be obtained from a population of serum amylase determinations with a mean of 120. (taken from Daniel 1991 p.202 adapted) GIVEN the population value = the null hypothesis This looks like a rare occurrence? But for what
9.037 n =15 t density: s = x 96 Original units: 120 Shaded area=0.0188 0 2.656 0 -2.656 t Given that the sample was obtained from a population with a mean of 120 a sample with a T(n=15) statistic of -2.656 or 2.656 or one more extreme will occur 1.8% of the time = just under two samples per hundred on average. . . . . Given that the sample was obtained from a population with a mean of 120 a sample of 15 producing a mean of 96 (120-x where x=24) or 144 (120+x where x=24) or one more extreme will occur 1.8% of the time, that is just under two samples per hundred on average. What does the shaded area mean! Serum amylase values from a random sample of 15 apparently healthy subjects. mean =96 SD= 35 units/100 ml. How likely would such a sample be obtained from a population of serum amylase determinations with a mean of 120. (taken from Daniel 1991 p.202 adapted) But it this not a P value P value= 2 · P(t(n−1) < t| Ho is true) = 2 · [area to the left of t under a t distribution with df= n − 1]
P value and probability for t statistic p value = 2 x P(t(n-1) values more extreme than t(n-1) | Ho is true) = 2 · [area to the left of t under a t distribution with n − 1 shape] A p value is a special type of probability with: Multiple outcomes + conditional upon the specified parameter value
sampling Putting it all together P Value probability statistic Rule Do we need it!
9.037 n =15 t density: s = x 96 Original units: 120 Shaded area=0.0188 0 2.656 0 -2.656 t Say one in twenty 1/20 = Or 1/100 Or 1/1000 or . . . . Rules Set a level of acceptability = critical value (CV)! If our result has a P value of less than our level of acceptability. Reject the parameter value. Say 1 in 20 (i.e.CV=0.5) Given that the sample was obtained from a population with a mean (parameter value) of 120 a sample with a T(n=15) statistic of -2.656 or 2.656 or one more extreme with occur 1.8% of the time, This is less than one in twenty therefore we dismiss the possibility that our sample came from a population mean of 120 . . . . What do we replace it with?
Fisher – only know and only consider the model we have i.e. The parameter we have used in our model – when we reject it we accept that any value but that one can replace it. Neyman and Pearson + Gossling Must have an alternative specified value for the parameter
Power – sample size • Affect size • – indication of clinical importance: If there is an alternative - what is it – another distribution! Serum amylase values from a random sample of 15 apparently healthy subjects. mean =96 SD= 35 units/100 ml. How likely would such a sample be obtained from a population of serum amylase determinations with a mean of 120. (taken from Daniel 1991 p.202 adapted)
α = the reject region = 96 = 120 Correct decisions incorrect decisions
Insufficient power – never get a significant result even when effect size large Too much power get significant result with trivial effect size
Life after P values • Confidence intervals • Effect size • Description / analysis • Bayesian statistics - qualitative approach by the back door! • Planning to do statistics for your dissertation? • see: My medical statistics courses: • Course 1: • www.robin-beaumont.co.uk/virtualclassroom/stats/course1.html • YouTube videos to accompany course 1: • http://www.youtube.com/playlist?list=PL9F0EBD42C0AB37D0 • Course 2: • www.robin-beaumont.co.uk/virtualclassroom/stats/course2.html • YouTube videos to accompany course 2: • http://www.youtube.com/playlist?list=PL05FC4785D24C6E68
Students bloomers • The p value did not indicate much statistic significance • Given that the population comes from one population • The p value is 0.003 thus rejecting the null hypothesis and there is a statistical significance • Correlation = 0.25 (p<0.001) indicating that assuming that the data come from a bivariate normal distribution with a correlation of zero you would obtain a correlation of <0.000. There is 95% chance that the relationship among the variables is not due to chance