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Do Olympians Get Too Much Exercise? - The New York Times

This article in the NY Times is to some degree an example of the problem of survivor bias, or generalizing from the few successful individuals to all individuals Do Olympians Get Too Much Exercise?

My comment:
One obvious problem with this type of study, is that athletes that might have developed issues might no longer engage in, and might have previously given up, said activities. If you only look at the winners, and not the losers, one gets an unfair picture of the effect or traits on all participants. Yes, you know that the winners did not have problems, but what about those that gave the activity up, or died?

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Decision Tree in R, with Graphs: Predicting State Politics from Big Five Traits

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Links to sample data, as well as to source references, are at the end of this entry.

Example Code

# Decision Tree - Big Five and Politics library("rpart") # grow tree input.dat <- read.table("BigFiveScoresByState.csv", header = TRUE, sep = ",") fit <- rpart(Liberal ~ Openness + Conscientiousness + Neuroticism + Extraversion + Agreeableness, data = input.dat, method="poisson") # display the results printcp(fit) # visualize cross-validation results plotcp(fit) # detailed summary of splits summary(fit) # plot tree plot(fit, uniform = TRUE, main = "Classific…

Mean Median, and Mode with R, using Country-level IQ Estimates

Reusing the code posted for Correlations within with Hofstede's Cultural Values, Diversity, GINI, and IQ, the same data can be used for mean, median, and mode. Additionally, the summary function will return values in addition to mean and median, Min, Max, and quartile values:

Example Code
oecdData <- read.table("OECD - Quality of Life.csv", header = TRUE, sep = ",") v1 <- oecdData$IQ # Mean with na.rm = TRUE removed NULL avalues mean(v1, na.rm = TRUE) # Median with na.rm = TRUE removed NULL values median(v1, na.rm = TRUE) # Returns the same data as mean and median, but also includes distribution values: # Min, Quartiles, and Max summary(v1) # Mode does not exist in R, so we need to create a function getmode <- function(v) { uniqv <- unique(v) uniqv[which.max(tabulate(match(v, uniqv)))] } #returns the mode getmode(v1)
Example Results
> oecdData <- read.table("OECD - Quality of L…

Chi-Square in R on by State Politics (Red/Blue) and Income (Higher/Lower)

This is a significant result, but instead of a logistic regression looking at the income average per state and the likelihood of being a Democratic state, it uses Chi-Square. Interpreting this is pretty straightforward, in that liberal states typically have cities and people that earn more money. When using adjusted incomes, by cost of living, this difference disappears.

Example Code
# R - Chi Square rm(list = ls()) stateData <- read.table("CostByStateAndSalary.csv", header = TRUE, sep = ",") # Create vectors affluence.median <- median(stateData$Y2014, na.rm = TRUE) affluence.v <- ifelse(stateData$Y2014 > affluence.median, 1, 0) liberal.v <- stateData$Liberal # Solve pol.Data = table(liberal.v, affluence.v) result <- chisq.test(pol.Data) print(result) print(pol.Data)
Example Results
Pearson's Chi-squared test with Yates' continuity correction data: pol.Data X-squared = 12.672, df …