Survey

Survey

Transcript

Chapter 6: The Standard Deviation as a Ruler and the Normal Model Women’s Heptathlon The women’s heptathlon in the Olympics consists of seven track and field events: the 200m and 800-m runs, 100-m high hurdles, shot put, javelin, high jump, and long jump. Somehow, the performances in all seven events have to be combined into one score. How can the performances in such different events be compared? They don’t even have the same units; the races are recorded in seconds and the throwing and jumping events in meters. More Heptathlon In the 2000 Olympics, the best 800-m time, run by Getrud Bacher of Italy, was 8 seconds faster than the mean. The winning long jump by the Russian Yelena Prokhorova was 60 centimeters longer than the mean. Which performance deserves more points? The trick to comparing very different-looking values is to use standard deviations. The standard deviation tells us how the collection of values varies, so it’s a natural ruler for comparing an individual value to the group. More Heptathlon How many standard deviations better than the mean is each woman’s results? Bacher’s winning 800-m time of 129 seconds was 8 seconds faster than the mean of 137 seconds. The standard deviation of qualifying times was 5 seconds, so her jump was (129 – 137)/5 = –8/5 = –1.6, or 1.6 standard deviations better than the mean. Prokhorova’s winning long jump was 60 cm longer than the average 6-m jump. The standard deviation was (60/30) = 2 standard deviations better than the mean. Prokhorova’s performance was better because it was a greater improvement over the mean long jump than Bacher’s improvement over the mean 800-m run. Standardizing with z-scores observation y y z s standard mean deviation We call the resulting values standardized values, and denote them with the letter z. Usually, we just call them z-scores. Z-Scores • No units – measures the distance of each data value from the mean in standard deviations • A z-score of 2 tells us that a data value is 2 standard deviations above the mean. • A z-score of -1.6 tells us that a data value is 1.6 standard deviations below the mean. More about Z-Scores When standardizing with z-scores, we do two things: 1. Shift the data by subtracting the mean 2. Rescale the values by dividing by the standard deviation Some Questions… • How does shifting or rescaling the data work? • What happens to the grade distribution if everyone gets a five-point bonus? • If we switch from feet to meters, what happens to the distribution of heights of students in our class? Shifting Data Since the 1960’s the Center for Disease Control’s National Center for Health Statistics has been collecting health and nutritional information on people of all ages and backgrounds. A recent survey, the National Health and Nutrition Examination Survey (NHANES) 2001-2002, measured a wide variety of variables, including body measurements, cardiovascular fitness, blood chemistry, and demographic information on more than 11,000 individuals. Included in this group were 80 men between 19 and 24 years old of average height (between 5’8” and 5’10” tall). # of Men NHANES Weight (kg) # of Men NHANES Rescaled Kg Above Recommended Weight What Do We Notice? • When adding/subtracting a constant to each value, all measures of position (center, percentiles, min, max) will increase/decrease by the same constant • Measures of spread (range, IQR, and standard deviation) do not change # of Men What about Pounds? Weight (pounds) What Do We Notice? • When we multiply/divide all the values by any constant, all measures of position (such as mean, median, and percentiles) are multiplied/divided by that same constant. • Measures of spread (range, IQR, and standard deviation) are also multiplied/divided by that same constant. Just Checking 1. Your statistics teacher has announced that the lower of your two tests will be dropped. You got a 90 on test 1 and an 80 on test 2. You’re all set to drop the 80 until she announces that she grades “on a curve.” She standardized the scores in order to decide which is the lower one. If the mean on the first test is 88 with a standard deviation of 4 and the mean on the second is a 75 with a standard deviation of 5, a) Which one will be dropped? b) Does this seem “fair?” Just Checking 2. In 1995 the Education Testing Service (ETS) adjusted the scores of SAT tests. Before ETS recentered the SAT Verbal test, the mean of all test scores was 450. a) How would adding 50 points to each score affect the mean? b) The standard deviation was 100 points. What would the standard deviation be after adding 50 points? c) Suppose we drew boxplots of test takers’ scores a year before and a year after the re-centering. How would the boxplots of the two years differ? Just Checking 3. A company manufactures wheels for roller blades. The diameter of the wheels has a mean of 3 inches and a standard deviation of 0.1 inches. Because so many of its customers use the metric system, the company decided to report their production statistics in millimeters (1 inch = 25.4mm). They report that the standard deviation is now 2.54 mm. A corporate executive is worried about this increase in variation. Should they be concerned? Explain. Back to Z-Scores • All we do when standardizing z-scores is shift the data by the mean and rescale by the standard deviation. It does not change the shape of the distribution of a variable. • It changes the center by making the mean 0 • It changes the spread by making the standard deviation 1 Think, Show, Tell Many colleges and universities require applicants to submit scores on standardized tests such as the SAT Writing, Math, and Critical Reading tests. The college your little sister wants to apply to says that while there is no minimum score required, the middle 50% of their students have combined SAT scores between 1530 and 1850. You’d feel confident if you knew her score was in the top 25%, but unfortunately, she took the ACT test. How high does her ACT need to be to make it into the top quarter of equivalent SAT scores? For college-bound seniors, the average combined SAT scores is about 1500 and the standard deviation is about 250 points. For the same group, the ACT average is 20.8 with a standard deviation of 4.8 Think I want to know what ACT score corresponds to the upper quartile SAT score. I know the mean and standard deviation for both the SAT and ACT scores based on all test takers, but I have no individual data values. Quantitative Variable Condition – Scores for both tests are quantitative, but have no meaningful units other than points. Show The middle 50% of SAT scores at this college fall between 1530 and 1850 points. To be in the top quarter, my sister would have to have a score of at least 1850. That’s a z-score of y y 1850 1500 z 1.40 s 250 So an SAT score of 1850 is 1.40 standard deviations above the mean of all test takers. For the ACT, 1.40 standard deviations about the mean is y y sz 20.8 4.81.40 27.52 Note: the formula z y y can be rearranged to solve for a different variable. s Tell To be in the top quarter of applicants in terms of combined SAT score, she’d need to have an ACT score of at least 27.52. When is a Z-Score BIG? • How far from 0 does a z-score have to be to be “interesting?” • We need to reference a model, but carefully – No model will ever be “perfect” for the data, but is usually “good enough” • While there is no universal standard for z-scores, there is a model that shows up frequently in statistics. Normal Models • “bell-shaped curves” • Appropriate for distributions whose shapes are unimodal and roughly symmetric • For the normal model, the mean is s and the standard deviation is m • We write Nm, s to represent a normal model • Why the Greek? These numbers are parameters and are part of the model. They are NOT from numerical summaries (“actual data”) Z-Scores y y Remember this formula?: z s What if we standardize z-scores for the normal model? Replace y with m and s with s z ym s Standard Normal Model • Usually it’s easier to standardize data first (using its mean and standard deviation). Then we need only the model N(0, 1) • The normal model with mean 0 and standard deviation 1 is called the standard Normal model (or the standard Normal distribution) Normality Assumption • In order to use the Normal model, we must assume the distribution is “Normal” • The Nearly Normal Condition satisfies this assumption. If the shape of the data’s distribution is unimodal and symmetric, the condition is met. • ALWAYS check this condition first!! 68-95-99.7 Rule In a Normal model, about 68% of all values fall within 1 standard deviation of the mean, 95% of values fall within 2 standard deviations of the mean, and 99.7% of values fall within 3 standard deviations of the mean. Sometimes called the “Empirical Rule” in books 68-95-99.7 Rule Another view Just Checking 1. As a group, the Dutch are among the tallest in the world. The average Dutch man is 184 cm tall – just over 6 feet. If a Normal model is appropriate and the standard deviation for men is about 8 cm, what percent of all Dutch men will be over 2 meters (6’6”) tall? Just Checking 2. Suppose it takes you 20 minutes, on average, to drive to school, with a standard deviation of 2 minutes. Suppose a Normal model is appropriate for the distributions of driving times. a) How often will you arrive at school in less than 22 minutes? b) How often will it take you more than 24 minutes? c) Do you think the distribution of your driving times is unimodal and symmetric? d) What does this say about the accuracy of your predictions? Explain. Think, Show, Tell The SAT Reasoning Test has three parts: Writing, Math, and Critical Reading. Each part has a distribution that is roughly unimodal and symmetric and is designed to have an overall mean of about 500 and a standard deviation of 100 for all test takers. In any one year, the mean and standard deviation may differ from these targets by a small amount, but they are also a good overall approximation. Suppose you earned a 600 on one part of your SAT. Where do you stand among all students who took that test? Think I want to see how my SAT score compares with all other students. To do that, I’ll need to model the distribution. a Nearly Normal Condition is satisfied because we’re told the data is roughly symmetric and unimodal (we’d check a histogram if we had actual data points) Show We will model SAT score with a N(500, 100) model. 200 300 400 500 600 700 800 The score of 600 is 1 standard deviation above the mean. That corresponds to the 68-95-99.7% Rule. Tell The score of 600 is higher than about 84% of all scores on this portion of the SAT. Finding Normal Percentiles by Hand An SAT score of 600 is easy to assess because it’s exactly one standard deviation away from the mean. What if we wanted to see how a person stands against the population if his score was 680? Normal Percentiles by Hand • Step 1: Calculate the z-score by hand. z ym s 680 500 1.8 100 • Step 2: Look up the z-score using a standard Normal table (we have one on pg A-118 in the back of the book) The student scored higher than 96.41% of all SAT test takers (or that 3.59% scored higher) Finding Normal Percentiles on the Calculator To access the Normal model on your calculator: 2nd DISTR to access various distributions and models We’re going to primarily use 2: normalcdf( and 3: invNorm( *We rarely (if ever) use 1: normalpfd( in this course! normalcdf( normalcdf( finds the area between two z-score cut points, by specifying a lower bound and an upper bound normalcdf(zLeft, zRight) normalcdf(-0.5, 1.0) = 0.5328 z = -0.5 z = 1.0 So, 53.28% of the data lies between ½ standard deviation below the mean and 1 standard deviation above the mean normalcdf( What about our SAT question? Where we had a score of 680… (remember, z = 1.8) -3 -2 -1 0 1 normalcdf(-999999, 1.8) = 0.9640 normancdf(1.8, 9999999) = .0359 2 3 What do we notice here? Think, Show, Tell What proportion of SAT scores fall between 450 and 580? Assume a Normal model is appropriate a mean of 500 and standard deviation of 100. THINK: We want to know the proportion of SAT scores between 450 and 580. The nearly Normal condition is satisfied because the data follows a Normal model. Show z ym s 580 500 0.8 100 z ym s 450 500 0.5 100 Area (z < 0.8) = 0.7881 normalcdf(-0.5, 0.8) = .4796 Area (z < -0.5) = 0.3085 Area (-0.5 < z < 0.8) = 0.7881-0.3085 = 0.4796 Tell The Normal model estimates that about 47.96% of SAT scores fall between 450 and 580. From Percentiles to Z-Scores • What if we know the areas/percentiles and want to find the corresponding z-scores? • A college says it admits only people with SAT Verbal test scores among the top 10%. How high a score does it take to be eligible? In Reverse • By hand: find the percent in the table and give the corresponding z-score; give answer in context • By calculator: 2nd DISTR and choose 3: invNorm( Plug in the percent you want to find Back to that 10% A college says it admits only people with SAT Verbal test scores among the top 10%. How high a score does it take to be eligible? What do we want to find on the table? Does .1000 make sense? (z = -1.28; y zs m 1.28100 500 372 ) *ALWAYS read the area to the left on the critical point) The Correct way So, if we’re finding the top 10%, then these students score better that the lower 90%. .9000 is the % we’re looking for; z = 1.28 (the fact that this is symmetric with z = -1.28 is NOT coincidental!) The necessary SAT score is 628 Normal Probability Plots If the distribution of a data set is roughly Normal, a Normal probability plot will be roughly a diagonal straight line. Deviations from a straight line indicate the distribution is not Normal In the calculator – plot using the last plot option under “Type” Always use the Y axis How Does it Work? The plot is really a scatter plot of actual data values compared to their z-scores under Normal conditions. If they’re the same, they’ll lie on the line y = x. Therefore, the closer the points are to forming a diagonal line (the y = x line), the closer the data is to being Normal. What Can Go Wrong? • Don’t use a Normal model when the distribution is not unimodal and symmetric. – ALWAYS check the data with a histogram/normal probability plot first • Don’t use the mean/standard deviation when there are outliers • Don’t round off too soon (or in the middle of a calculation) • Don’t worry about minor differences in results