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Session 6: Basic Statistics Part 1 (and how not to be frightened by the word). Prof Neville Yeomans Director of Research, Austin LifeSciences. So now we’ve got some results? How can we make sense out of them?. What I will cover (over 2 sessions, August 28 and September 25).
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Session 6: Basic Statistics Part 1 (and how not to be frightened by the word) Prof Neville Yeomans Director of Research, Austin LifeSciences
So now we’ve got some results? How can we make sense out of them?
What I will cover(over 2 sessions, August 28 and September 25) • Sampling populations • Describing the data in the samples • How accurately do those data reflect the ‘real’ population the samples were taken from? • We’ve compared two groups. Are they really from different populations, or are they samples from the same population and the measured differences are just due to chance?
What I will cover (contd.) • Tests to answer the question ‘Are the differences likely to be just due to chance?’ • Data consisting of values (e.g. hemoglobin concentration)(‘continuous variables’) • Data consisting of whole numbers – frequencies, proportions (percentages) • Tests for just two groups; tests for multiple groups • Tests that examine relationships between two or more variables (correlation, regression analysis, life-table)
What I will cover (contd.) • How many subjects should I study to find the answer to my question? (Power calculations) • Statistical packages and other resources
We’ve got some numbers. How are we going to describe them to others?Suppose we’ve measured heights of a number of females (‘a sample’) picked off the street in Heidelberg
How could we more concisely describe these data – using just one or two numbers that would give us useful information about the sample as a whole? A measure of ‘central tendency’ 2. A measure of how widely the values are spread
The median (middle value) = 162.5 The range (150-179) a poor measure for describing the whole population because it depends on sample size – range is likely to be wider with larger samples Interquartile range (25th percentile to 75th percentile of values: 159-168) what we should always use with the median – it’s largely independent of sample size
(amalgamated into 3cm ranges: e.g.141-143, 144-146 etc.) the Mean (average)(Ʃx/N) = 163.3 cm the Standard Deviation* ± 7.2 cm • doesn’t vary much with sample size (except very small samples) • approx. 67% of values will lie within ± 1 SD either side of mean** • approx 95% of values will lie within ± 2 SD either side of mean** ** Provided the population is ‘normally distributed’ *In Excel, enter formula ‘=STDEV(range of cells)’
The ‘Normal distribution’ Mean = Median in a ‘perfect’ normal distribution Standard deviations away from mean
We measured the mean height of our sample of 25 women ... (it was 163.3 cm) • But what is the average height of the whole population – of ALL Heidelberg women? • We didn’t have time or resources to track them all down – that’s why we just took what we hoped was a representative sample. • What I’m asking is: how good an estimate of the true population mean is our sample mean? • This is where the Standard Error of the Mean* (or just Standard Error, SE) comes in. *It’s sometimes called the Standard ESTIMATE of the Error of the mean
The Standard Error (contd.) • The mean height of our sample of 25 women was 163.3 cm • We calculated the Standard Deviation (SD) of the sample to be 7.3 cm (that value, on either side of the mean, that should contain about 2/3 of those measured) • Standard Error of the mean = SD/√N , i.e. 7.3/ √25 = 7.3/5 = 1.46 • So now we can express our results for the height of our sample as 163.3 ± 1.5 (Mean ± SEM) ...... But what does this really tell us? • The actual true mean height of the whole population of women has a 67% likelihood of lying within 1.5 cm (i.e. 1 SEM) either side of the mean we found in our sample; and a 95% likelihood of lying within 3 cm (i.e. 2 x SEM) either side of that sample mean. (It’s actually 1.96xSEM for a reasonably large sample - e.g. roughly 30 – and wider for small samples, but let’s keep it simple).
The concept of the standard error of the mean (SEM) – e.g. serum sodium values True population mean 142 mmol/L (SD=4.0) Sample mean =142.8 mmol/L (SD of sample = 3.4) 1 x SEM = (i.e. 3.4/√10) = 1.1 mmol/L 2 x SEM (~’95% confidence interval’) = 2.2 mmol/L Random sample of 10 normal individuals That means: ‘There is a 95% chance (19 chances out of 20) that the actual population mean, estimated from our random sample lies between 140.6 and 145.0 mmol/L)’ 130 134 138 142 146 150 154 mmol/L
Why does Standard Error depend on population SD and sample size? SE = SD/√N A: Narrow population spread (i.e. small SD) B: Wide population spread (i.e. large SD) Increasing N decreases SE of mean. i.e. increases accuracy of our estimate of the population mean based on results of our sample
Testing significance of differences 95% confidence intervals Women: Mean = 163.3 SE = 1.46 Men: Mean = 176.5 SE = 1.43 On a quick, rough, check we can see that: the 95% confidence interval for our estimate of the height of women is 160.3-166.4 cm (approximately mean ± 2SE). (b) our estimate of the mean height of the men sampled is quite a lot outside the 95% confidence interval (range)for the women, so it looks improbable that they are from the same population
Testing significance of differences ....How likely is it that the two random samples came from the same population? Student’s t-test Composite frequency distribution, created by pooling data from both samples Women Mean: 170.7 SD: 10.0 cm Men How likely is it that these two samples (the pink and the blue) were taken from the SAME population? [this is called the NULL HYPOTHESIS] Tested for statistical significance of difference: p<0.001 i.e. there is less than 1 chance in 1000 that these two samples came from the same population -3 -2 0 1 2 3 -1 Standard deviation either side of mean
In fact, though, running a Student’st-test on the two samples of height in the Heidelberg men and women slide, gives this error message:
Assumptions to be met before testing significance of differences with PARAMETRIC TESTS* – i.e. tests that use the mathematics of the normal curve distribution • The combined data should approximate a normal curve distribution • in this instance the male data were skewed (not evenly distributed around the mean) and spread a bit too far out into the tails of the frequency-distribution curve • The variances (=SD2)of the groups should not differ significantly from each other *Student’s t-test, Paired t-test, Analysis of Variance
An example of data where groups have different variance (spread), and one group is skewed The equal variance test failed (P<0.05), and the normality test almost failed (P=0.08) – so we should not use a parametric test such as t-test Means Means Lower limit of ‘normal’ range Medians
So what do we do if we can’t use a parametric test to check for significances of differences? • Use a non-parametric test • These tests, instead of using the actual numerical values of the data, put the data from each group into ascending order and assign a rank number for their place in the combined groups. • The maths of the test is then done on these ranks • Examples: Rank sum test, Wilcoxon Rank Test, Mann Whitney rank test, etc. • (the P value for our slide of heights of Heidelberg men and women was calculated using Wilcoxon test)
Example of how a rank-sum (non-parametric) test is constructed manually* *In reality, these days you’ll just feed the raw data into a program to do it for you
Tests to examine significance of differences between 3 or more groups • Parametric tests (tests based on the mathematics of the ‘normal curve’) • Analysis of Variance (1-way, 2-way, factorial, etc.) • Non-parametric tests (rank-sum tests) • Kruskal-Wallis test [strictly this should read ... ‘tests to decide how likely are data from 3 or more samples to come from the same population’]
One variable (dose), compared across 3 groups .... So this gets tested with one-way ANOVA
This tells us that it is very unlikely the three groups belong to the same population. . But which differ from which?
One Way Analysis of VarianceThursday, June 28, 2012, 3:39:25 PM Normality Test: Passed (P = 0.786) Equal Variance Test: Passed (P = 0.694) Group Name N Missing Mean Std Dev SEM Control 12 0 126.500 7.243 2.091 Normopress 0.5 mg 12 0 125.083 5.551 1.602 Normopress 2.0 mg 12 0 114.000 5.625 1.624 Source of Variation DF SS MS F P Between Groups 2 1124.389 562.194 14.679 <0.001 Residual 33 1263.917 38.301 Total 35 2388.306 The differences in the mean values among the treatment groups are greater than would be expected by chance; there is a statistically significant difference (P = <0.001). All Pairwise Multiple Comparison Procedures (Holm-Sidak method): Overall significance level = 0.05 Comparisons for factor: Comparison Diff of Meanst Unadjusted PCriticalSignificant? • Level Control vs. Normopress 2.0 mg 12.500 4.947<0.001 0.017Yes Normopress 0.5 mg vs. Normopress 2.0 mg 11.083 4.387<0.001 0.025Yes Control vs. Normopress 0.5 mg 1.417 0.5610.579 0.050No
Before and after data – paired tests • Create a paired analysis of length of hair after going to hairdresser. • Hypothesis: cutting hair makes it shorter
Two independent groupsDifference between means tested for significance with Student’s t test P=0.58 9.7±1.9* 8.2±1.7 *mean ± SE
Our actual data Difference between means tested for significance with paired Student’s t test The variation within each individual is much less than between individuals P=0.025 The paired t-test examines the mean and standard error of the changes In each individual, and tests how likely are the changes due to chance Before After
So far we have been dealing with ‘CONTINUOUS VARIABLES’ – numbers such as heights, laboratory values, velocities, temperatures etc. that could have any value (e.g. many decimal points) if we could measure accurately enough. Now we’ll look at ...‘DISCONTINUOUS VARIABLES’ – whole numbers, most often as proportions or percentages.
Rates and proportions • In 1969, a home for retired pirates has 93 inmates, 42 of whom have only one leg. • In 2004, a subsequent survey finds there are now 62 inmates, 6 of whom have only one leg. Has there been a ‘real’ change (i.e. a change unlikely to be due to chance) in the proportion of one-legged pirates in the home between the two surveys?
Expected 29 19 Chi-square= 20.282 with 1 degrees of freedom. (P = <0.001) i.e. The likelihood that the difference in proportions of 1-legged inmates between 1969 and 2004 is due to chance ... is less than 1:1000
One trap with chi-square tests and small numbers .... Penicillin treatment for pneumonia Fisher’s exact test: P = 10! x 13! x 14! x 9! = 0.029 9! x 5! x 1! x 8! x 23!
Correlation • Fairly straightforward concept of how likely are two variables to be related to each other • Examples: • Do children’s heights vary with their age, and if so is the relation direct (i.e. get bigger as get older) or converse (get smaller as get older)? • Does respiratory rate increase as pulse rate increases during exertion? • The correlation coefficient, R, tells us how closely the two variables ‘travel together’ • P value is calculated to tell us how likely the relationship is to be ‘only’ by chance
Examples of regression (correlation) data R = 0.965 P<0.0001 R = - 0.37 P = 0.30
Some other common statistical analyses • Life-table analyses • Observing and comparing events developing over time; allows us to compensate for dropouts at varying times during the study • Multiple linear and multivariate regression analyses • Looking for relationships between multiple variables
Life table analyses Advanced lung cancer. Trial compared motesanib + 2 conventional chemo drugs ... with placebo plus the two other drugs Scagliotti et al. J ClinOncol 2012; 30: 2829
Multiple regression analysis • Examines the possible effect of more than one variable on the thing we are measuring (the ‘dependent variable’) Perret JL et al. The Interplay between the Effects of Lifetime Asthma, Smoking, and Atopy on Fixed Airflow Obstruction in Middle Age. Am J RespirCrit Med 2013; 187: 42-8. ...from Institute of Breathing and Sleep (Austin), University of Melbourne, Monash University, Alfred Hospital, And others
Sample Size Calculations How many patients, subjects, mice etc. do we need to study to reliably* find the answer to our research question? *We can never be certain to do this, but should aim to be considerably more likely than not to find out the truth about the question
Sample size calculations (1) First we need to grapple with two types of ‘error’ in interpreting differences between means and/or medians of groups: • Type 1 (or α) error: ... that we think the difference is ‘real’ (data are from 2 or more different populations) when it is not • This is what we’ve dealt with so far, and the P-values assess how likely the differences are due to chance • Type 2 (or β) error: ... that our experiment, and the stats test we’ll apply to the results, will FAIL to show a significant difference when there REALLY IS ONE
Sample size calculations (2) • If we end up with a Type 2 () error, it will be because our sample size(s) was too small to persuade us that the actual difference between means was unlikely due to chance (i.e. P<0.05) • The smaller the real difference between population means, the larger the sample size needs to be to detect it as being statistically significant
Sample size calculations (3) How do we go about it? Most of the good statistical packages have a function for calculating sample sizes • Decide what statistical test will be appropriate to apply to primary endpoint when study completes • Estimate the likely size of difference between groups, if the hypothesis is correct • Decide how confident you want to be that the difference(s) you observe is unlikely due to chance • Decide how much you want to risk missing a true difference (i.e. what power you want the study to have) Note: We really should have done a sample size calculation before we started our experiments, but for this course we needed to deal with the basics of stats tests first
Sample size calculations A worked example (i) • We want to see whether drug X will reduce the incidence of peptic ulcer in patients taking aspirin for 6 months 1. Decide what statistical test: chi square, to compare differences in frequencies in 2 groups
Sample size calculations A worked example (i) • We want to see whether drug X will reduce the incidence of peptic ulcer in patients taking aspirin for 6 months • We expect a 10% incidence of ulcers in the controls • We hypothesize that a 50% reduction (i.e. 5%) in those treated with X would be clinically worthwhile 2. We’ve now decided the size of the difference between groups we are interested to look for
Sample size calculations (4) A worked example (i) • We want to see whether drug X will reduce the incidence of peptic ulcer in patients taking aspirin for 6 months • We expect a 10% incidence of ulcers in the controls • We hypothesize that a 50% reduction (i.e. 5%) in those treated with X would be clinically worthwhile • We decide to be happy with a likelihood of only 1:20 that difference observed is due to chance 3. That is to say, we want to set P0.05 as the level of α (alpha) risk (the risk of concluding the difference is real when it’s actually due to chance)
Sample size calculations (4) A worked example (i) • We want to see whether drug X will reduce the incidence of peptic ulcer in patients taking aspirin for 6 months • We expect a 10% incidence of ulcers in the controls • We hypothesize that a 50% reduction (i.e. 5%) in those treated with X would be clinically worthwhile • We decide to be happy with a likelihood of only 1:20 that difference observed is due to chance • We would like to have at least an 80% chance of finding that 50% reduction (20% of missing it, i.e. of β risk) 4. That is, set Power of the study ≥80% (1-β) to detect such a difference (if it exists)
Sample size calculations A worked example (i) Summary of sample size calculation setting: • Estimated ulcer incidence in controls = 10% • Estimated incidence in group receiving drug X = 5% • For P(α) 0.05, and Power (1-β) ≥80% • Data tested by chi square ...................................................................... Calculated required sample size = 449 in each group
Sample size calculations A worked example (ii) • We hypothesize that removing the spleen in rats will result in an increase in haemoglobin (Hb) from the normal mean of 14.0 g/L to 15.0 g/L • We already know that the SD (Standard deviation) of Hb values in normal rats is 1.2 g/L (if we don’t know we’ll have to guess!) • Testing will be with Student’s t test • We’ll set α (likelihood observed difference due to chance) at 0.05 • We want a power (1-β) of at least 80% to minimize risk of missing such a difference if its real* *More correctly, we should say if the samples really are from different populations
Sample size calculations A worked example (ii) Summary of sample size calculation setting: • Control mean = 14.0 g/L; Operated mean = 15.0 g/L • Estimated SD in both groups = 1.2 g/L • For P(α) 0.05, and Power (1-β) ≥80% • Data tested by Student t ...................................................................... Calculated required sample size = 24 in each group