Variance Variance: Standard deviation: The average squared distance of observations from the mean. Its corresponding measure of central tendency is the mean. Standard deviation: The square root of the variance. Variance is in squared units, so can be hard to interpret. Standard deviation is in the original units of the measurements. These measure how much the data deviate from the mean. They use all the data in their calculations. Draw distributions with large and small variances on the board.

Variance Two versions depending on the context of data Population: all subjects of interest Every single person in the United States Entire UCM student body Calculation: 𝜎 2 = 𝑖=1 𝑁 𝑥 𝑖 −𝜇 2 𝑁 = 𝑆𝑆 𝑁 Population standard deviation: 𝜎= 𝜎 2 Sample: subset of population 100 random people in the US 100 random students from UCM 𝑠 2 = 𝑖=1 𝑛 𝑥 𝑖 − 𝑥 2 𝑛−1 = 𝑆𝑆 𝑛−1 Sample standard deviation: 𝑠= 𝑠 2 Point out what a deviation is. Make note that SS stands for sum of squares. Make a slight allusion to why s^2 is calculated with n-1 instead of n.

Population Standard Deviation (σ) Measure of how spread out the values are in a data set consisting of the entire population of interest (i.e., how much do the population values deviate from the mean?) Step 1: Calculate the deviation (distance) of each value (𝒙 𝒊 ) from the population mean (µ) 1 2 3 4 5 µ = 3 Step 2: Square each deviation, then add them up (−2) 2 + (−1) 2 + (0) 2 +(1) 2 +(2) 2 =10 Step 3: Divide the sum of squared deviations by the number of observations (N) 10/5=2 (population variance = σ 2 ) Step 4: Take the square root of the “average squared deviation” (population variance) 2 =1.41=𝝈 (3-3) = 0 (2-3) = -1 (4-3) = 1 (1-3) = -2 (5-3) = 2 You square these values because otherwise they will add up to 0 BY DEFINITION. Show why that is.

𝜎= (1−3) 2 + (2−3) 2 + (3−3) 2 + (4−3) 2 + (5−3) 2 5 =1.41 Formulas Population Standard Deviation 𝜎= 𝑖=1 𝑁 ( 𝑥 𝑖 −𝜇) 2 𝑁 𝜎= (1−3) 2 + (2−3) 2 + (3−3) 2 + (4−3) 2 + (5−3) 2 5 =1.41

Sample Standard Deviation (s) Measure of how spread out the values are in a data set consisting of a sample of the population of interest (i.e., how much do the sample values deviate from the mean?) Step 1: Calculate the deviation (distance) of each value ( 𝒙 𝒊 ) from the sample mean ( 𝒙 ) 1 2 3 4 5 𝑥 =3 Step 2: Square each deviation, then add them up (−2) 2 + (−1) 2 + (0) 2 +(1) 2 +(2) 2 =10 Step 3: Divide the sum of squared deviations by number of observations minus 1 (n-1) 10/(5−1)=10/4=2.5 (sample variance = 𝑠 2 ) Step 4: Take the square root of the “corrected average squared deviation” (sample variance) 2.5 =1.58=𝒔 (3-3) = 0 (2-3) = -1 (4-3) = 1 (1-3) = -2 (5-3) = 2

𝑠= (1−3) 2 + (2−3) 2 + (3−3) 2 + (4−3) 2 + (5−3) 2 5−1 =1.58 Formulas Sample Standard Deviation 𝑠= 𝑖=1 𝑛 ( 𝑥 𝑖 − 𝑥 ) 2 𝑛−1 𝑠= (1−3) 2 + (2−3) 2 + (3−3) 2 + (4−3) 2 + (5−3) 2 5−1 =1.58 Write the computational formula/proof on the board.

Simplified Formula Using the rules for summations, a simpler formula for the standard deviation can be found. 𝑠= 𝑖=1 𝑛 ( 𝑥 𝑖 − 𝑥 ) 2 𝑛−1 = 𝑖=1 𝑛 ( 𝑥 𝑖 2 −2 𝑥 𝑖 𝑥 + 𝑥 2 ) 𝑛−1 = 𝑖=1 𝑛 𝑥 𝑖 2 − 𝑖=1 𝑛 2 𝑥 𝑥 𝑖 + 𝑖=1 𝑛 𝑥 2 𝑛−1 = 𝑖=1 𝑛 𝑥 𝑖 2 −2 𝑥 𝑖=1 𝑛 𝑥 𝑖 +𝑛 𝑥 2 𝑛−1 = 𝑖=1 𝑛 𝑥 𝑖 2 −2𝑛 𝑥 2 +𝑛 𝑥 2 𝑛−1 = 𝑖=1 𝑛 𝑥 𝑖 2 −𝑛 𝑥 2 𝑛−1 𝑠= ( 1 2 + 2 2 + 3 2 + 4 2 + 5 2 )−5× 3 2 5−1 =1.58

Another Example Suppose we have a sample of 6 scores on a quiz: 5, 10, 3, 7, 2, 3 What is the mode for these data? 3 What is the median for these data? 4 What is the mean for these data? 5 What is the sample variance for these data? 46 5 =9.2 What is the sample standard deviation for these data? 9.2 ≈3.033

Population versus Sample SD The sample SD seeks to estimate the population SD, in the same way that the sample mean seeks to estimate the population mean The distribution of the sample mean is such that it may incorrectly estimate the population mean, but it’s no more likely to overestimate than it is to underestimate It is unbiased The sample SD, on the other hand, will consistently underestimate the population SD, so we divide it by a smaller number (n-1 instead of n) to correct for this underestimation error. Without correction, the sample SD and sample variance would be biased estimators