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Data Analysis Using R: 2. Descriptive Statistics

Data Analysis Using R: 2. Descriptive Statistics. Tuan V. Nguyen Garvan Institute of Medical Research, Sydney, Australia. Overview. Measurements Population vs sample Summary of data: mean, variance, standard deviation, standard error Graphical analyses Transformation.

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Data Analysis Using R: 2. Descriptive Statistics

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  1. Data Analysis Using R:2. Descriptive Statistics Tuan V. Nguyen Garvan Institute of Medical Research, Sydney, Australia

  2. Overview • Measurements • Population vs sample • Summary of data: mean, variance, standard deviation, standard error • Graphical analyses • Transformation

  3. Scales of Measurement • In general, most observable behaviors can be measured on a ratio-scale • In general, many unobservable psychological qualities (e.g., extraversion), are measured on interval scales • We will mostly concern ourselves with the simple categorical (nominal) versus continuous distinction (ordinal, interval, ratio) variables categorical continuous ordinal interval ratio

  4. Ordinal Measurement • Ordinal: Designates an ordering; quasi-ranking • Does not assume that the intervals between numbers are equal. • finishing place in a race (first place, second place) 1st place 2nd place 3rd place 4th place 1 hour 2 hours 3 hours 4 hours 5 hours 6 hours 7 hours 8 hours

  5. Interval and Ratio Measurement • Interval: designates an equal-interval ordering • The distance between, for example, a 1 and a 2 is the same as the distance between a 4 and a 5 • Example: Common IQ tests are assumed to use an interval metric • Ratio: designates an equal-interval ordering with a true zero point (i.e., the zero implies an absence of the thing being measured) • Example: number of intimate relationships a person has had • 0 quite literally means none • a person who has had 4 relationships has had twice as many as someone who has had 2

  6. Statististics: Enquiry to the unknown

  7. Estimate the population mean Population height mean = 160 cm Standard deviation = 5.0 cm ht <- rnorm(10, mean=160, sd=5) mean(ht) ht <- rnorm(10, mean=160, sd=5) mean(ht) ht <- rnorm(100, mean=160, sd=5) mean(ht) ht <- rnorm(1000, mean=160, sd=5) mean(ht) ht <- rnorm(10000, mean=160, sd=5) mean(ht) hist(ht) The larger the sample, the more accurate the estimate is!

  8. Estimate the population proportion Population proportion of males = 0.50 Take n samples, record the number of k males rbinom(n, k, prob) males <- rbinom(10, 10, 0.5) males mean(males) males <- rbinom(20, 100, 0.5) males mean(males) males <- rbinom(1000, 100, 0.5) males mean(males) The larger the sample, the more accurate the estimate is!

  9. Summary of Continuous Data • Measures of central tendency: • Mean, median, mode • Measures of dispersion or variability: • Variance, standard deviation, standard error • Interquartile range R commands length(x), mean(x), median(x), var(x), sd(x) summary(x)

  10. R example height <- rnorm(1000, mean=55, sd=8.2) mean(height) [1] 55.30948 median(height) [1] 55.018 var(height) [1] 68.02786 sd(height) [1] 8.2479 summary(height) Min. 1st Qu. Median Mean 3rd Qu. Max. 28.34 49.97 55.02 55.31 60.78 85.05

  11. Graphical Summary: Box plot boxplot(height) 95% percentile 75% percentile Median, 50% perc. 25% percentile 5% percentile

  12. Strip chart

  13. Histogram

  14. Implications of the mean and SD • “In the Vietnamese population aged 30+ years, the average of weight was 55.0 kg, with the SD being 8.2 kg.” • What does this mean? • 68% individuals will have height between 55 +/- 8.2*1 = 46.8 to 63.2 kg • 95% individuals will have height between 55 +/- 8.2*1.96 = 38.9 to 71.1 kg

  15. Implications of the mean and SD • The distribution of weight of the entire population can be shown to be: 1.96SD 1SD

  16. Summary of Categorical Data • Categorical data: • Gender: male, female • Race: Asian, Caucasian, African • Semi-quantitative data: • Severity of disease: mild, moderate, severe • Stages of cancer: I, II, III, IV • Preference: dislike very much, dislike, equivocal, like, like very much

  17. Mean and variance of a proportion • For an individual i consumer, the probability he/she prefers A is pi. Assuming that all consumers are independent, then pi = p. • Variance of pi is var(pi) = p(1-p) • For a sample of n consumers, the estimated probability of preference for A is: and the variance of p_bar is:

  18. Normal approximation of a binomial distribution • For an individual i consumer, the probability he/she prefers A is pi. Assuming that all consumers are independent, then pi = p. • Variance of pi is var(pi) = p(1-p) • For a sample of n consumers, the estimated probability of preference for A is: and the variance of p_bar is: and standard deviation:

  19. Normal approximation of a binomial distribution - example • 10 consumbers, 8 preferred product A. • Proportion of preference for A: p = 0.8 • Variance: var(p) = 0.8(0.2)/10 = 0.016 • Standard deviation of p: s = 0.126 • 95% CI of p: 0.8 + 1.96(0.126) = 0.55 to 1.00

  20. Descriptive AnalysesContinuous data

  21. Paired t-test • Continuous data • Normally distributed • Two samples are NOT independent

  22. Paired t-test – an example • The problem: Viewing certain meats under red light might enhance judges preferences for meat. 12 judges were asked to score the redness of meat under red light and white light Results: Judge Red White 1 20 22 2 18 19 3 19 17 4 22 18 5 17 21 6 20 23 7 19 19 8 16 20 9 21 22 10 17 20 11 23 27 12 18 24

  23. Paired t-test – analysis Mean difference: 1.83, SD: 0.81 Standard error (SE): SD/sqrt(n) = 0.81/sqrt(10) = 0.81 T-test = (1.83 – 0)/0.81 = 2.23 P-value = 0.0459 Conclusion: there was a significant effect of light colour.

  24. Paired t-test – R analysis red < -c(20,18,19,22,17,20,19,16,21,17,23,18) white < -c(22,19,17,18,21,23,19,20,22,20,27,24) t.test(red, white, paired=TRUE) data: red and white t = -2.2496, df = 11, p-value = 0.04592 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -3.6270234 -0.0396433 sample estimates: mean of the differences -1.833333

  25. Two-sample t-test Mean difference: D = x – y Variance of D: Sample Group 1 Group2 1 x1y1 2 x2y2 3 x3y3 4 x4y4 5 x5y5 … … n xnyn Sample size n1n2 Mean xy SD sxsy T-statistic: 95% Confidence interval:

  26. Two-group comparison: an example 20 consumers rated their preference for two rice desserts (A and B) ID A B 1 3 3 2 7 1 3 1 2 4 9 4 5 3 5 6 4 2 7 1 2 8 2 5 9 6 3 10 7 2 ID A B 11 5 3 12 8 4 13 5 2 14 9 3 15 4 5 16 6 4 17 4 3 18 3 1 19 9 3 20 5 2

  27. Unpaired t-test using R a<-c(3,7,1,9,3,4,1,2,6,7,5,8,5,9,4,6,4,3,9,5) b<-c(3,1,2,4,5,2,2,5,3,2,3,4,2,3,5,4,3,1,3,2) t.test(red,white) Welch Two Sample t-test data: a and b t = 3.3215, df = 27.478, p-value = 0.002539 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 0.8037895 3.3962105 sample estimates: mean of x mean of y 5.05 2.95

  28. Transformation of data: multiplicative effects • The following data represent lysozyme levels in the gastric juice of 29 patients with peptic ulcer and of 30 normal controls. It was interested to know whether lysozyme levels were different between two groups. Group 1: 0.2 0.3 0.4 1.1 2.0 2.1 3.3 3.8 4.5 4.8 4.9 5.0 5.3 7.5 9.8 10.4 10.9 11.3 12.4 16.2 17.6 18.9 20.7 24.0 25.4 40.0 42.2 50.0 60.0 Group 2: 0.2 0.3 0.4 0.7 1.2 1.5 1.5 1.9 2.0 2.4 2.5 2.8 3.6 4.8 4.8 5.4 5.7 5.8 7.5 8.7 8.8 9.1 10.3 15.6 16.1 16.5 16.7 20.0 20.7 33.0

  29. Unpaired t-test by R g1 <- c( 0.2, 0.3, 0.4, 1.1, 2.0, 2.1, 3.3, 3.8, 4.5, 4.8, 4.9, 5.0, 5.3, 7.5, 9.8, 10.4, 10.9, 11.3, 12.4, 16.2, 17.6, 18.9, 20.7, 24.0, 25.4, 40.0, 42.2, 50.0, 60) g2 <- c(0.2, 0.3, 0.4, 0.7, 1.2, 1.5, 1.5, 1.9, 2.0, 2.4, 2.5, 2.8, 3.6, 4.8, 4.8, 5.4, 5.7, 5.8, 7.5, 8.7, 8.8, 9.1, 10.3, 15.6, 16.1, 16.5, 16.7, 20.0, 20.7, 33.0) t.test(g1, g2) data: g1 and g2 t = 2.0357, df = 40.804, p-value = 0.04831 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 0.05163216 13.20239083 sample estimates: mean of x mean of y 14.310345 7.683333

  30. Exploration of data par(mfrow=c(1,2)) hist(g1) hist(g2) Group 1: mean(g1) = 14.3 sd(g1) = 15.7 Group 2: mean(g2) = 7.7 sd(g2) = 7.8

  31. Re-analysis of lysozyme data log.g1 <- log(g1) log.g2 <- log(g2) t.test(log.g1, log.g2) data: log.g1 and log.g2 t = 1.406, df = 55.714, p-value = 0.1653 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -0.2182472 1.2453165 sample estimates: mean of x mean of y 1.921094 1.407559 exp(1.921-1.407) = 1.67 Group 1’s mean is 67% higher than group 2’s

  32. Descriptive analysisCategorical data

  33. Comparison of two proportions - theory Group 1 2 ____________________________________________ Sample size n1n2 Number of events e1e2 Proportion of events p1p2 Difference: D = p1 – p2 SE difference: SE = [p1(1–p1)/n1 + p2(1–p2)/n2]1/2 Z = D / SE 95% CI: D+ 1.96(SE) With (n1 + n2) > 20, and if Z > 2, it is possible to reject the null hypothesis.

  34. Comparison of two proportions - example Thirty-day mortality rate (%) of 100 rats who had been exposed to heroine or cocain. Analysis Difference: D = 0.90 – 0.36 = 0.54 SE (D) = [0.9(0.1)/100 + 0.36(0.64)/100]1/2 = 0.057 Z = 0.54 / 0.057 = 9.54 95% CI: 0.54 + 1.96(0.057) 0.43 to 0.65 Conclusion: reject the null hypothesis. Group Heroine Cocaine __________________________________________ Sample size 100 100 Number of deaths 90 36 Mortality rate 0.90 0.36

  35. Comparison of two proportions - R events <- c(90, 36) total <- c(100, 100) prop.test(events, total) 2-sample test for equality of proportions with continuity correction data: deaths out of total X-squared = 60.2531, df = 1, p-value = 8.341e-15 alternative hypothesis: two.sided 95 percent confidence interval: 0.4190584 0.6609416 sample estimates: prop 1 prop 2 0.90 0.36

  36. Comparison of >2 proportions – Chi square analysis table(sex, ethnicity) ethnicity sex African Asian Caucasian Others Female 4 43 22 0 Male 4 17 8 2 females <- c(4, 43, 22, 0) total <- c(8, 60, 30, 2) prop.test(females, total)

  37. Comparison of >2 proportions – Chi square analysis 4-sample test for equality of proportions without continuity correction data: females out of total X-squared = 6.2646, df = 3, p-value = 0.09942 alternative hypothesis: two.sided sample estimates: prop 1 prop 2 prop 3 prop 4 0.5000000 0.7166667 0.7333333 0.0000000 Warning message: Chi-squared approximation may be incorrect in: prop.test(females, total)

  38. Summary • Examine the distribution of data • Mean and variance: systematic difference? • Normally distributed ? • Transformation? • Present confidence intervals (and p-values)

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