![Weights of Quarters](http://s1.studyres.com/store/data/010016054_1-8a7f6f25ac961b132c1078094b7b4fb6-300x300.png)
Statistics Day 5: Standard Scores in Normal Distribution and More
... mean score for the SAT is 1017 and the standard deviation is 209. (This would require a score at or above the 80th percentile) ...
... mean score for the SAT is 1017 and the standard deviation is 209. (This would require a score at or above the 80th percentile) ...
Descriptive Statistics
... Notice that the sum of the deviations of scores from their mean is zero (as always). If you find the mean of the squared deviations, 10/5 = 2, you have the variance, assuming that these five scores represent the entire population. The population standard deviation is 2 1.414 . Usually we shall con ...
... Notice that the sum of the deviations of scores from their mean is zero (as always). If you find the mean of the squared deviations, 10/5 = 2, you have the variance, assuming that these five scores represent the entire population. The population standard deviation is 2 1.414 . Usually we shall con ...
Bivariate Data Cleaning
... Applying group-specific outlier analysis • Do outlier analyses and trimming/Windsorizing separately for each group • This gets lengthy when working with factorial designs -- remember the purpose of each condition!!! • Some suggest taking an analogous approach with doing regression analyses that invo ...
... Applying group-specific outlier analysis • Do outlier analyses and trimming/Windsorizing separately for each group • This gets lengthy when working with factorial designs -- remember the purpose of each condition!!! • Some suggest taking an analogous approach with doing regression analyses that invo ...