ENVE-GEOE 224 - 3 - Location Statistics - Handouts

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Oct 30, 2023

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2021-01-09 1 ENVE/GEOE 224: Probability & Statistics Location Statistics Prof. Philip J. Schmidt Department of Civil & Environmental Engineering, University of Waterloo Winter 2021 Learning Objectives Learn how to calculate the mean, median, and mode of a set of data Be able to describe how these central tendency measures compare in unimodal distributions Be able to identify unimodal, bimodal, and polymodal distributions Learn how to calculate percentiles from raw data Derive an equation applying linear interpolation 2 of 15 ENVE/GEOE 224 (W2021) – P. Schmidt 1 2
2021-01-09 2 The Mean Sample mean , arithmetic mean Population mean , expected value For a sample of size 𝑛 from a For all members of a population larger population: of size 𝑁 : 𝑥̅ = 𝑥 ௜ୀଵ μ = 𝑥 ௜ୀଵ e.g., the mean height of 10 e.g., the mean height of all the students from a class students in a class The most common meaning of the ambiguous term “average” 3 of 15 ENVE/GEOE 224 (W2021) – P. Schmidt Example 1.1 – Mean Consider an evaluation of the sensitivity of several analytical techniques to determine various water quality parameters (ion concentrations) such as NO 3 - (nitrate). Using one analytical technique, you measure NO 3 - concentration 12 times. Your recorded data (in mg/L) are: 𝑥̅ = 𝑥 ௜ୀଵ = ଵଶ 0.63 + 0.65 + ⋯ + 0.72 + 0.74 = 0.695 mg/L 4 of 15 ENVE/GEOE 224 (W2021) – P. Schmidt 0.63 0.65 0.71 0.69 0.73 0.68 0.66 0.68 0.68 0.77 0.72 0.74 This is a sample! 3 4
2021-01-09 3 The Median The median ( 𝑚 ) is the middle of the ordered set of data 𝑥 , 𝑥 , … , 𝑥 For odd 𝑛 : 𝑚 = 𝑥 ௡ାଵ For even 𝑛 : 𝑚 = 𝑥 ௡ ଶ + 𝑥 ௡ ଶ ାଵ The median is also the 50 th percentile. More on this soon… 5 of 15 ENVE/GEOE 224 (W2021) – P. Schmidt Example 1.1 – Median Ordered data ( 𝑛 = 12 ): 0.63, 0.65, 0.66, 0.68, 0.68, 0.68, 0.69, 0.71, 0.72, 0.73, 0.74, 0.77 𝑚 = 𝑥 ௡ ଶ + 𝑥 ௡ ଶ ାଵ = ା௫ = ଴.଺଼ା଴.଺ଽ = 0.685 mg/L 6 of 15 ENVE/GEOE 224 (W2021) – P. Schmidt 5 6
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2021-01-09 4 Example 1.2 A small engineering firm employs four junior engineers who each earn $32,000 per year and an owner who earns $192,000 per year. Comment on the claim that the average salary of the company is $64,000 per year, and hence, the company pays well. Ordered data ( 𝑛 = 5 ): $32,000, $32,000, $32,000, $32,000, $192,000 𝑚 = 𝑥 ௡ାଵ = 𝑥 = $32,000 The $64,000 mean is misleading; most employees are not well paid! 7 of 15 ENVE/GEOE 224 (W2021) – P. Schmidt The Mode The mode is the most frequently occurring value Often no single value occurs more than once, in which case the data can be grouped and the mode can be reported as the midpoint of the class interval with the highest frequency Distributions can be unimodal, bimodal, or polymodal 8 of 15 ENVE/GEOE 224 (W2021) – P. Schmidt 7 8
2021-01-09 5 Example 1.1 – Mode Although concentration data are conceptually continuous (they can take any non-negative value with any number of decimal places), these data are practically discrete because they have only two decimal places. Mode = 0.68 mg/L (the only value that occurs more than once) With grouped data as follows, the mode is 0.675 mg/L 9 of 15 ENVE/GEOE 224 (W2021) – P. Schmidt Class Intervals Frequency 0.60 < NO 3 - ≤ 0.65 2 0.65 < NO 3 - ≤ 0.70 5 0.70 < NO 3 - ≤ 0.75 4 0.75 < NO 3 - ≤ 0.80 1 Comparing the Mean, Median, and Mode Three scenarios for central tendency of a unimodal distribution: 10 of 15 ENVE/GEOE 224 (W2021) – P. Schmidt mean = median = mode mode < median < mean mean < median < mode 9 10
2021-01-09 6 Percentiles A percentile ( 𝑄 ) is the value below which fraction 𝑝 of the data lie The median is the 50 th percentile The 25 th , 50 th , and 75 th percentiles are called quartiles 1) Sort the n data from least to greatest, assigning each a unique index number I 2) 𝑘 is the integer part of 𝑛 + 1 𝑝 , 𝑑 is the remainder 𝑛 + 1 𝑝 − 𝑘 e.g., if 𝑛 + 1 𝑝 = 3.25 , then 𝑘 = 3 and 𝑑 = 0.25 3) 𝑄 lies between 𝑋 and 𝑋 ௞ାଵ : 𝑄 = 𝑋 + 𝑑 𝑋 ௞ାଵ − 𝑋 11 of 15 ENVE/GEOE 224 (W2021) – P. Schmidt Percentiles Why do we use 𝑛 + 1 𝑝? If we simply use 𝑛𝑝 , the data are not centred around 0.5 Using 𝑛 + 1 𝑝 is one solution to this problem (but not the only one) 12 of 15 ENVE/GEOE 224 (W2021) – P. Schmidt X 1 X 3 X 2 1/3 3/3 2/3 0 X 1 1/4 X 3 3/4 X 2 2/4 0 1 11 12
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2021-01-09 7 Derivation Using Linear Interpolation The equation 𝑄 = 𝑋 + 𝑑 𝑋 ௞ାଵ − 𝑋 applies linear interpolation 13 of 15 ENVE/GEOE 224 (W2021) – P. Schmidt (k+1)/(n+1) k/(n+1) p X k X k+1 Q p 𝑄 − 𝑋 𝑋 ௞ାଵ − 𝑋 = 𝑝 − 𝑘 𝑛 + 1 𝑘 + 1 𝑛 + 1 − 𝑘 𝑛 + 1 𝑄 − 𝑋 𝑋 ௞ାଵ − 𝑋 = 𝑛 + 1 𝑝 − 𝑘 = 𝑑 𝑄 = 𝑋 + 𝑑 𝑋 ௞ାଵ − 𝑋 Data X 1 X 2 X 3 X 4 X n-2 X n-1 X n Fraction 1/(n+1) 2/(n+1) 3/(n+1) 4/(n+1) (n-2)/(n+1) (n-1)/(n+1) n/(n+1) Suppose 𝑛 + 1 𝑝 = 3.25 Example 1.1 – 25 th Percentile Ordered data ( 𝑛 = 12 ): 0.63, 0.65, 0.66, 0.68, 0.68, 0.68, 0.69, 0.71, 0.72, 0.73, 0.74, 0.77 𝑛 + 1 𝑝 = 13 × 0.25 = 3.25 ∴ 𝑘 = 3 , 𝑑 = 0.25 , 𝑋 = 0.66 mg/L, and 𝑋 ௞ାଵ = 0.68 mg/L 𝑄 = 𝑋 + 𝑑 𝑋 ௞ାଵ − 𝑋 𝑄 ଴.ଶହ = 0.66 + 0.25 0.68 − 0.66 = 0.665 mg/L 14 of 15 ENVE/GEOE 224 (W2021) – P. Schmidt 13 14
2021-01-09 8 Online Lecture In this week’s online lecture, we will apply these statistics to some real environmental data using features in Excel! Learn how to use built-in functions to calculate basic statistics What was the most common daily maximum temperature at the University of Waterloo Weather Station in 2012? What are the 5 th and 95 th percentiles of daily maximum temperature? 15 of 15 ENVE/GEOE 224 (W2021) – P. Schmidt 15

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