1. 1.1 - Populations, Samples and Processes 1.2 - Pictorial and Tabular Methods in Descriptive Statistics 1.3 - Measures of Location 1.4 - Measures of Variability 1 Chapter 1 Overview and Descriptive Statistics

2. Example: Suppose the random variable is X = Age (years) in a certain population of individuals, and we select the following random sample of n = 20 ages. In published journal articles, the original data are almost never shown, but displayed in tabular form as above. This summary is called “grouped data.” 4 values 8 values5 values 2 values 1 value From these values, we can construct a table which consists of the frequencies of each age-interval in the dataset, i.e., a frequency table. {18, 19, 19, 19, 20, 21, 21, 23, 24, 24, 26, 27, 31, 35, 35, 37, 38, 42, 46, 59}{18, 19, 19, 19, 20, 21, 21, 23, 24, 24, 26, 27, 31, 35, 35, 37, 38, 42, 46, 59} 4 8 2 5 1 Frequency Histogram Suggests population may be skewed to the right (i.e., positively skewed). Class IntervalFrequency [10, 20)4 [20, 30)8 [30, 40)5 [40, 50)2 [50, 60)1 Totaln = 20 “Endpoint convention” Here, the left endpoint is included, but not the right. Note!... Stay away from “10-20,” “20-30,” “30-40,” etc. 2

3. Class IntervalFrequency [10, 20)4 [20, 30)8 [30, 40)5 [40, 50)2 [50, 60)1 Totaln = 20 Relative Frequency 4/20 = 0.20 8/20 = 0.40 5/20 = 0.25 2/20 = 0.10 1/20 = 0.05 20/20 = 1.00 Example: Suppose the random variable is X = Age (years) in a certain population of individuals, and we select the following random sample of n = 20 ages. {18, 19, 19, 19, 20, 21, 21, 23, 24, 24, 26, 27, 31, 35, 35, 37, 38, 42, 46, 59} Often though, it is preferable to work with proportions, i.e., relative frequencies… Divide frequencies by n = 20. ↓ Relative frequencies are always between 0 and 1, and sum to 1. Relative Frequency Histogram.20.40.10.25.05 3 0.4 0.3 0.2 0.1 0.0

4. Class IntervalFrequency [10, 20)4 [20, 30)8 [30, 40)5 [40, 50)2 [50, 60)1 Totaln = 20 Relative Frequency 4/20 = 0.20 8/20 = 0.40 5/20 = 0.25 2/20 = 0.10 1/20 = 0.05 20/20 = 1.00 Example: Suppose the random variable is X = Age (years) in a certain population of individuals, and we select the following random sample of n = 20 ages. {18, 19, 19, 19, 20, 21, 21, 23, 24, 24, 26, 27, 31, 35, 35, 37, 38, 42, 46, 59} Often though, it is preferable to work with proportions, i.e., relative frequencies… Divide frequencies by n = 20. ↓ Relative frequencies are always between 0 and 1, and sum to 1. Relative Frequency Histogram.20.40.10.25.05 4 0.4 0.3 0.2 0.1 0.0 “0.20 of the sample is under 20 yrs old” “0.60 of the sample is under 30 yrs old” “0.85 of the sample is under 40 yrs old” “0.95 of the sample is under 50 yrs old” “1.00 of the sample is under 60 yrs old” “0.00 of the sample is under 10 yrs old” Cumulative (0.00) 0.20 0.60 0.85 0.95 1.00

5. Example: Exactly what proportion of the sample is under 34 years old? Approximately Class IntervalFrequency [10, 20)4 [20, 30)8 [30, 40)5 [40, 50)2 [50, 60)1 Totaln = 20 Relative Frequency 4/20 = 0.20 8/20 = 0.40 5/20 = 0.25 2/20 = 0.10 1/20 = 0.05 20/20 = 1.00 Example: Suppose the random variable is X = Age (years) in a certain population of individuals, and we select the following random sample of n = 20 ages. {18, 19, 19, 19, 20, 21, 21, 23, 24, 24, 26, 27, 31, 35, 35, 37, 38, 42, 46, 59} Often though, it is preferable to work with proportions, i.e., relative frequencies… Divide frequencies by n = 20. ↓ Relative frequencies are always between 0 and 1, and sum to 1. Relative Frequency Histogram.20.40.10.25.05 5 0.4 0.3 0.2 0.1 0.0 Cumulative (0.00) 0.20 0.60 0.85 0.95 1.00 Cumulative relative frequencies always increase from 0 to 1. Solution: [30, 34) contains 4/10 of 0.25 = 0.1, [0, 30) contains 0.6, 0.7 sum = 0.7

6. Class IntervalFrequency [10, 20)4 [20, 30)8 [30, 40)5 [40, 50)2 [50, 60)1 Totaln = 20 Relative Frequency 4/20 = 0.20 8/20 = 0.40 5/20 = 0.25 2/20 = 0.10 1/20 = 0.05 20/20 = 1.00 Example: Suppose the random variable is X = Age (years) in a certain population of individuals, and we select the following random sample of n = 20 ages. {18, 19, 19, 19, 20, 21, 21, 23, 24, 24, 26, 27, 31, 35, 35, 37, 38, 42, 46, 59} Often though, it is preferable to work with proportions, i.e., relative frequencies… Divide frequencies by n = 20. ↓ Relative frequencies are always between 0 and 1, and sum to 1. Relative Frequency Histogram.20.40.10.25.05 6 0.4 0.3 0.2 0.1 0.0 Cumulative (0.00) 0.20 0.60 0.85 0.95 1.00 Cumulative relative frequencies always increase from 0 to 1. Solution: [30, 34) contains 4/10 of 0.25 = 0.1, [0, 30) contains 0.6, 0.7 sum = 0.7 Example: Approximately what proportion of the sample is under 34 years old?Exactly But alas, there is a major problem….

7. Relative Frequency Histogram.20.40.10.25.05 Suppose that, for the purpose of the study, we are not primarily concerned with those 30 or older, and wish to “lump” them into a single class interval. {18, 19, 19, 19, 20, 21, 21, 23, 24, 24, 26, 27, What effect will this have on the histogram? Class IntervalFrequency [10, 20)4 [20, 30)8 [30, 40)5 [40, 50)2 [50, 60)1 Totaln = 20 Relative Frequency 4/20 = 0.20 8/20 = 0.40 5/20 = 0.25 2/20 = 0.10 1/20 = 0.05 20/20 = 1.00 4 values 8 values 31, 35, 35, 37, 38, 42, 46, 59} Class Interval [10, 20) [20, 30) [30, 60) Total Relative Frequency 4/20 = 0.20 8/20 = 0.40 20/20 = 1.00.40 The skew no longer appears. The histogram is distorted because of the presence of an outlier (59) in the data, creating the need for unequal class widths. 8 values 7 0.4 0.3 0.2 0.1 0.0

8. What are they? Informally, an outlier is a sample data value that is either “much” smaller or larger than the other values. How do they arise? o experimental error o measurement error o recording error o not an error; genuine What can we do about them? o double-check them if possible o delete them? o include them… somehow o perform analysis both ways (A Pain in the Tuches) 8

9. IDEA: Instead of having height of each class rectangle = relative frequency, make... area of each class rectangle = relative frequency. Class Interval Relative Frequency [10, 20) 0.20 [20, 30) 0.40 [30, 60) 0.40 Total20/20 = 1.00 Density (= height) 0.20/10 = 0.020 0.40/10 = 0.040 0.40/30 = 0.013 height“Density” = relative frequency × width/ width = 10 width = 30 Density Histogram 0.02 0.04 0.0133… 0.20 0.40 Total Area = 1! 9 The outlier is included, and the overall skewed appearance is restored. Exercise: What if the outlier was 99 instead of 59?

10. 1.1 - Populations, Samples and Processes 1.2 - Pictorial and Tabular Methods in Descriptive Statistics 1.3 - Measures of Location 1.4 - Measures of Variability 10 Chapter 1 Overview and Descriptive Statistics

11. “Measures of ” Example: Sample exam scores = {70, 80, 80, 80, 80, 90, 90, 100, 100, 100} sample mode most frequent value = 80 sample median “middle” value = (80 + 90) / 2 = 85 sample mean average value = 11 Data values x i Frequencies f i 701 804 902 1003 Totaln = 10 i = 1 i = 2 i = 3 i = 4 (70)(1) + (80)(4) + (90)(2) + (100)(3) x =  x i f i = 87 (Quartiles are found similarly: Q 1 =, Q 2 = 85, Q 3 = )80100 Center 1/10

12. sample mode most frequent value = 80 sample median “middle” value = (80 + 90) / 2 = 85 sample mean average value = “Measures of Center” Example: Sample exam scores = {70, 80, 80, 80, 80, 90, 90, 100, 100, 100} 12 Data values x i Frequencies f i 701 804 902 1003 Totaln = 10 (70)(1) + (80)(4) + (90)(2) + (100)(3)1/10 = 87 x =  x i f i

13. sample mean “Measures of Center” Example: Sample exam scores = {70, 80, 80, 80, 80, 90, 90, 100, 100, 100} 13 Data values x i Frequencies f i 701 804 902 1003 Totaln = 10 Relative Frequencies f (x i ) = f i /n 1/10 = 0.1 4/10 = 0.4 2/10 = 0.2 3/10 = 0.3 10/10 = 1.0 (70)(1) + (80)(4) + (90)(2) + (100)(3)1/10 x =  x i f (x i ) “Notation, notation, notation.” 1 10 4 10 2 10 3 10 (70)(1) + (80)(4) + (90)(2) + (100)(3) =1/10 87 x =  x i f i “weighted” sample mean

14. sample mean 14 Data values x i Frequencies f i 701 804 902 1003 Totaln = 10 … but how do we measure the “spread” of a set of values? First attempt: sample range = x n – x 1 = 100 – 70 = 30. Simple, but… Spread “Measures of ” Example: Sample exam scores = {70, 80, 80, 80, 80, 90, 90, 100, 100, 100} Ignores all of the data except the extreme points, thus far too sensitive to outliers to be of any practical value. Example: Company employee salaries, including CEO Can modify with… sample interquartile range (IQR) = Q 3 – Q 1 = 100 – 80 = 20. We would still prefer a measure that uses all of the data.

15. Deviations from mean x i – x 70 – 87 = –17 80 – 87 = –7 90 – 87 = +3 100 – 87 = +13 sample mean 15 Data values x i Frequencies f i 701 804 902 1003 Totaln = 10 … but how do we measure the “spread” of a set of values? Better attempt: Calculate the average of the “deviations from the mean.” 1/10 [ (–17)(1) + (–7)(4) + (3)(2) + (13)(3) ] = 0. ???????? This is not a coincidence – the deviations always sum to 0* – so it is not a good measure of variability. Spread “Measures of ” Example: Sample exam scores = {70, 80, 80, 80, 80, 90, 90, 100, 100, 100}  (x i – x) f i = * Physically, the sample mean is a “balance point” for the data.

16. Deviations from mean x i – x 70 – 87 = –17 80 – 87 = –7 90 – 87 = +3 100 – 87 = +13 sample mean 16 Data values x i Frequencies f i 701 804 902 1003 Totaln = 10  (x i – x) 2 f i [ (–17) 2 (1) + (–7) 2 (4) + (3) 2 (2) + (13) 2 (3) ] Calculate the “Measures of Spread” Example: Sample exam scores = {70, 80, 80, 80, 80, 90, 90, 100, 100, 100} s 2 = sample variance sample standard deviation s = 1/9 = 112.22 average of the “squared deviations from the mean.” s = 10.59 a modified “typical” sample value “typical” distance from mean

17. Grouped Data - revisited 17 Class Interval Absolute Frequency [10, 20)4 [20, 30)8 [30, 60)8 Use the interval midpoints for

18. Grouped Data - revisited 18 Class Interval Absolute Frequency [10, 20)4 [20, 30)8 [30, 60)8 15 25 45 Use the interval midpoints for Compare this “grouped mean” with the actual mean.

19. Class Interval Absolute Frequency [10, 20)4 [20, 30)8 [30, 60)8 Grouped Data - revisited 19 Use the interval midpoints for median Q 2 = ? Compare this “grouped mean” with the actual mean. Class Interval Absolute Frequency Relative Frequency Density [10, 20)40.200.020 [20, 30)80.400.040 [30, 60)80.400.01333 0.020.040.0133… 0.20 0.40 Step 1. Identify the interval & rectangle. Step 2. Split the rectangle so that 0.5 area lies above and below. 0.3 0.1

20. 00 0.1 0.3 Grouped Data - revisited Use the interval midpoints for median Q 2 = ? Compare this “grouped mean” with the actual mean. Step 1. Identify the interval & rectangle. Step 2. Split the rectangle so that 0.5 area lies above and below. Step 3. Observe that this rectangle can be split into 4 strips of 0.1 each. 0.1 22.52527.5 Step 4. Thus, split the interval into 4 equal parts, each of width (30 – 20 )/4. …OR…

21. 00 0.3 0.1 Grouped Data - revisited Use the interval midpoints for median Q 2 = ? Compare this “grouped mean” with the actual mean. Step 1. Identify the interval & rectangle. Step 2. Split the rectangle so that 0.5 area lies above and below. Step 3. Set up a proportion and solve for Q: Label as shown, and use the formula. …OR… Other percentiles are done similarly. Solve using cumul dist, w/o histogram. Solve for areas, given Q. See posted Lecture Notes! Other percentiles are done similarly. Solve using cumul dist, w/o histogram. Solve for areas, given Q. See posted Lecture Notes! …OR…

22. Comments is an unbiased estimator of the population mean , s 2 is an unbiased estimator of the population variance  2. (Their “expected values” are  and  2, respectively.) Beware of roundoff error!!! There is an alternate, more computationally stable formula for sample variance s 2. The numerator of s 2 is called a sum of squares (SS); the denominator “n – 1” is the number of degrees of freedom (df) of the n deviations x i –, because they must satisfy a constraint (sum = 0), hence 1 degree of freedom is “lost.” A natural setting for these formulas and concepts is geometric, specifically, the Pythagorean Theorem: a 2 + b 2 = c 2. See lecture notes appendix… 22 a c b