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Jan. 19 Statistic for the day: Number of Wisconsin’s 33 Senators who voted in favor of a 1988 bill that allows the blind to hunt: 27 Source: Legislative Hotline, Madison, Wisc. Assignment: Read Chapter 8 and do exercises 1, 2, 3, 8, 10 These slides were created by Tom Hettmansperger and in some cases modified by David Hunter Age at Death of English Rulers Revisited 60, 50, 47, 53, 48, 33, 71, 43, 65, 34, 56, 59, 49, 81, 67, 68, 49, 16, 86, 67 Turning this data into information. Shape: Histogram Age at death of a sample of 20 rulers of England 5 Frequency 4 3 2 1 0 10 20 30 40 50 age 60 70 80 90 Histogram but different shape (The number of intervals has been changed from 8 to 10.) Sample of Rulers of England n = 20 The number of intervals has been changed to 10. 6 Frequency 5 4 3 2 1 0 11 19 27 35 43 51 age 59 67 75 83 91 Alternatives to median and IQR: Mean or average of the data. Standard deviation of the data. The mean is easy. Just add up the numbers and divide by the sample size. The standard deviation is a pain. Generally you will use a calculator or computer. Rough way to approximate the standard deviation: Look at the histogram and estimate the range of the middle 95% of the data. The standard deviation is about ¼ of this range Age at death of a sample of 20 rulers of England 5 Frequency 4 3 2 1 0 10 20 30 40 50 60 70 80 90 age Take 95% range = 85 – 15 = 70 Estimate of standard deviation = .25x70 = 17.5 Using the calculation formula std dev = 16.7 Percent 10 5 0 15 20 25 HandSpan Range of middle 95% is roughly 24 – 16 = 8 Standard deviation is roughly .25x8 = 2 Standard deviation from formula 1.927 In Summary: The standard deviation is roughly .25 times the range of the middle 95% of the data. Look at the histogram or stem and leaf. The mean is the arithmetic average. If you want to visualize a histogram and you only know the mean and the standard deviation: 1. Put the center at the mean or the median. 2. Go out 2 standard deviations on either side of the center. 3. Draw a histogram humped up at the mean and dropping off on either side within part 2. Smoothing the histogram: The Normal Curve (Chapt 8) The histogram is rough and noisy. Replace it with a bell shaped curve. Center the bell at the mean. The middle 95% of the bell should be 4 standard deviations. Makes more accurate predictions provided the bell shape is appropriate for the underlying population. Histogram of HandSpan, with Normal Curve Frequency 20 10 0 15 20 HandSpan Mean = 20.86 Standard deviation = 1.927 25 Histogram of HandSpan, with Normal Curve Frequency 20 10 0 15 20 HandSpan Mean = 20.86 Standard deviation = 1.927 25 Histogram of Height, with Normal Curve Frequency 30 20 10 0 60 70 Height Mean = 68 inches or 5 feet 8 inches Standard deviation = 4 inches 80 Research Question 1: How high should I build my doorways so that 99% of the people will not have to duck? Secondary Question 2: If I built my doors 75 inches (6 feet 3 inches) high, what percent of the people would have to duck? Histogram of Height, with Normal Curve Frequency 30 20 10 0 60 70 80 Height Question 2 Question 1 Find the value at Question 1 so that 99% of the distribution is below it. The value at Question 2 is 75; find the amount of distribution above it. Z-Scores: Measurement in Standard Deviations Given the mean (68), the standard deviation (4), and a value (height say 75) compute 75 mean 75 68 Z 1.75 SD 4 This says that 75 is 1.75 standard deviations above the mean. Compute your Z-score. 1. How many standard deviations are you above or below the mean. Use: Mean = 68 inches Standard deviation = 4 inches 2. Now use the table from the book to determine what percentile you are. Answer to Question 2: What percent of people would have to duck if I built my doors 75 inches high? Recall: 75 has a Z-score of 1.75 From the standard normal table in the book: .96 or 96% of the distribution is below 1.75. Hence, .04 or 4% is above 1.75. So 4% of the distribution is above 75 inches. Histogram of Height, with Normal Curve Frequency 30 20 4% in here 10 0 60 70 75 80 Height Question 2 The value at Question 2 is 75; find the amount of distribution above it. Convert 75 to Z = 1.75 and use Table 8.1 in book. Question 1: What is the value so that 99% of the distribution is below it? Called the 99th percentile. 1. Look up the Z-score that corresponds to the 99th percentile. From the table: Z = 2.33. 2. Now convert it over to inches: h99 68 2.33 4 h99 68 2.33x 4 77.3 Since 77 inches is 6 feet 5 inches, 99% of the distribution is shorter than 77 inches and they will not have to duck. Histogram of Height, with Normal Curve Frequency 30 20 10 99% in here 0 60 70 80 Height Question 1 77.3 inches is the 99th percentile Find the value at Question 1 so that 99% of the distribution is below it. Look up Z-score for 99th percentile and convert it back to inches. Stat 100 students Sp01 Height 80 70 60 Female Male Sex Heights in Inches ( red circle is my doorway 77 inches) 85 78 75 (6-3) 65 Lakers n=14 PSU BB n=12 Steelers n=30 Stat 100 Males n=78