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The Standard Deviation as a Ruler and the Normal Model Chapter 6 Objectives: • • • • • • • • Standardized values Z-score Transforming data Normal Distribution Standard Normal Distribution 68-95-99.7 rule Normal precentages Normal probability plot The Standard Deviation as a Ruler • The trick in comparing very differentlooking values is to use standard deviations as our rulers. • The standard deviation tells us how the whole collection of values varies, so it’s a natural ruler for comparing an individual to a group. • As the most common measure of variation, the standard deviation plays a crucial role in how we look at data. Standardizing with z-scores • We compare individual data values to their mean, relative to their standard deviation using the following formula: z y y s • We call the resulting values standardized values, denoted as z. They can also be called z-scores. Slide 6 - 4 Standardizing with z-scores (cont.) • Standardized values have no units. • z-scores measure the distance of each data value from the mean in standard deviations. • A negative z-score tells us that the data value is below the mean, while a positive z-score tells us that the data value is above the mean. Benefits of Standardizing • Standardized values have been converted from their original units to the standard statistical unit of standard deviations from the mean (z-score). • Thus, we can compare values that are measured on different scales, with different units, or from different populations. WHY STANDARDIZE A VALUE? • Gives a common scale. • We can compare two different distributions with different means and standard deviations. 2.15 SD Z=-2.15 This Z-Score tells us it is 2.15 Standard Deviations from the mean • Z-Score tells us how many standard deviations the observation falls away from the mean. • Observations greater than the mean are positive when standardized and observations less than the mean are negative. Example: Standardizing • The men’s combined skiing event in the in the winter Olympics consists of two races: a downhill and a slalom. In the 2006 Winter Olympics, the mean slalom time was 94.2714 seconds with a standard deviation of 5.2844 seconds. The mean downhill time was 101.807 seconds with a standard deviation of 1.8356 seconds. Ted Ligety of the U.S., who won the gold medal with a combined time of 189.35 seconds, skied the slalom in 87.93 seconds and the downhill in 101.42 seconds. • On which race did he do better compared with the competition? Solution: z y y s • Slalom time (y): 87.93 sec. Slalom mean y : 94.2714 sec. Slalom standard deviation (s): 5.2844 sec. 87.93 94.2714 zSlalom 1.2 5.2844 • Downhill time (y): 101.42 sec. Downhill mean y : 101.807 sec. Downhill standard deviation (s): 1.8356 sec. 101.42 101.807 0.21 1.8356 The z-scores show that Ligety’s time in the slalom is farther below the mean than his time in the downhill. Therefore, his performance in the slalom was better. zDownhill • Your Turn: WHO SCORED BETTER? • Timmy gets a 680 on the math of the SAT. The SAT score distribution is normal with a mean of 500 and a standard deviation of 100. Little Jimmy scores a 27 on the math of the ACT. The ACT score distribution is normal with a mean of 18 and a standard deviation of 6. • Who does better? (Hint: standardize both scores then compare z-scores) TIMMY DOES BETTER • Timmy: • Little Jimmy: 27 18 680 500 1.5 z 1 .8 z 6 100 • Timmy’s z score is further away from the mean so he does better than Little Jimmy who’s only 1.5 SD’s from the mean • Little Jimmy does better than average and is 1.5 SD’s from the mean but Timmy beats him because he is .3 SD further. Combining z-scores • Because z-scores are standardized values, measure the distance of each data value from the mean in standard deviations and have no units, we can also combine z-scores of different variables. Example: Combining z-scores • In the 2006 Winter Olympics men’s combined event, Ted Ligety of the U.S. won the gold medal with a combined time of 189.35 seconds. Ivica Kostelic of Croatia skied the slalom in 89.44 seconds and the downhill in 100.44 seconds, for a combined time of 189.88 seconds. • Considered in terms of combined z-scores, who should have won the gold medal? Solution • Ted Ligety: zSlalom zDownhill 87.93 94.2714 1.2 5.2844 101.42 101.807 0.21 1.8356 • Combined z-score: -1.41 • Ivica Kostelic: zSlalom 89.44 94.2714 0.91 5.2844 zDownhill 100.44 101.807 0.74 1.8356 • Combined z-score: -1.65 • Using standardized scores, Kostelic would have won the gold. Your Turn: • The distribution of SAT scores has a mean of 500 and a standard deviation of 100. The distribution of ACT scores has a mean of 18 and a standard deviation of 6. Jill scored a 680 on the math part of the SAT and a 30 on the ACT math test. Jack scored a 740 on the math SAT and a 27 on the math ACT. • Who had the better combined SAT/ACT math score? Solution: • Jill zSAT z ACT 680 500 1.8 100 30 18 2.0 6 • Combined math score: 3.8 • Jack zSAT 740 500 2.4 100 z ACT 27 18 1.5 6 • Combined math score: 3.9 • Jack did better with a combined math score of 3.9, to Jill’s combined math score of 3.8. Linear Transformation of Data • Linear transformation • Changes the original variable x into the new variable xnew given by xnew = a + bx • Adding the constant a shifts all values of x upward or downward by the same amount. • Multiplying by the positive constant b changes the size of the values or rescales the data. Shifting Data • Shifting data: • Adding (or subtracting) a constant amount to each value just adds (or subtracts) the same constant to (from) the mean. This is true for the median and other measures of position too. • In general, adding a constant to every data value adds the same constant to measures of center and percentiles, but leaves measures of spread unchanged. Example: Adding a Constant • Given the data: 2, 4, 6, 8, 10 • Center: mean = 6, median = 6 • Spread: s = 3.2, IQR = 6 • Add a constant 5 to each value, new data 7, 9, 11, 13, 15 • New center: mean = 11, median = 11 • New spread: s = 3.2, IQR = 6 • Effects of adding a constant to each data value • Center increases by the constant 5 • Spread does not change • Shape of the distribution does not change Shifting Data (cont.) • The following histograms show a shift from men’s actual weights to kilograms above recommended weight: Rescaling Data • Rescaling data: • When we divide or multiply all the data values by any constant value, all measures of position (such as the mean, median and percentiles) and measures of spread (such as the range, IQR, and standard deviation) are divided and multiplied by that same constant value. Example: Multiplying by a Constant • Given the data: 2, 4, 6, 8, 10 • Center: mean = 6, median = 6 • Spread: s = 3.2, IQR = 6 • Multiple a constant 3 to each value, new data: 6, 12, 18, 24, 30 • New center: mean = 18, median = 18 • New spread: s = 9.6, IQR = 18 • Effects of multiplying each value by a constant • Center increases by a factor of the constant (times 3) • Spread increases by a factor of the constant (times 3) • Shape of the distribution does not change Rescaling Data (cont.) • The men’s weight data set measured weights in kilograms. If we want to think about these weights in pounds, we would rescale the data: Summary of Effect of a Linear Transformation • Multiplying each observation by a positive number b multiples both measures of center (mean and median) and measures of spread (IQR and standard deviation) by b. • Adding the same number a (either positive or negative) to each observation adds a to measures of center and to quartiles, but does not change measures of spread. • Linear transformations do not change the shape of a distribution. Back to z-scores • Standardizing data into z-scores shifts the data by subtracting the mean and rescales the values by dividing by their standard deviation. • Standardizing into z-scores does not change the shape of the distribution. • Standardizing into z-scores changes the center by making the mean 0. • Standardizing into z-scores changes the spread by making the standard deviation 1. Standardizing Data into z-scores Standardizing Data into z-scores When Is a z-score BIG? • A z-score gives us an indication of how unusual a value is because it tells us how far it is from the mean. • A data value that sits right at the mean, has a z-score equal to 0. • A z-score of 1 means the data value is 1 standard deviation above the mean. • A z-score of –1 means the data value is 1 standard deviation below the mean. When Is a z-score BIG? • How far from 0 does a z-score have to be to be interesting or unusual? • There is no universal standard, but the larger a z-score is (negative or positive), the more unusual it is. • Remember that a negative z-score tells us that the data value is below the mean, while a positive z-score tells us that the data value is above the mean. When Is a z-score Big? (cont.) • There is no universal standard for zscores, but there is a model that shows up over and over in Statistics. • This model is called the Normal model (You may have heard of “bell-shaped curves.”). • Normal models are appropriate for distributions whose shapes are unimodal and roughly symmetric. • These distributions provide a measure of how extreme a z-score is. Smooth Curve (model) vs Histogram • Sometimes the overall pattern is so regular that it can be described by a Smooth Curve. • Can help describe the location of individual observations within the distribution. Smooth Curve (model) vs Histogram • The distribution of a histogram depends on the choice of classes, while with a smooth curve it does not. • Smooth curve is a mathematical model of the distribution. • How? • The smooth curve describes what proportion of the observations fall in each range of values, not the frequency of observations like a histogram. • Area under the curve represents the proportion of observations in an interval. • The total area under the curve is 1. Smooth Curve or Mathematical Model • Always on or above the horizontal axis. • Total Area under curve = 1 Area underneath curve=1 Normal Distributions (normal Curves) • One Particular class of distributions or model. 1. Symmetric 2. Single Peaked 3. Bell Shaped • All have the same overall shape. DESCRIBING A NORMAL DISTRIBUTION The exact curve for a particular normal distribution is described by its Mean (μ) and Standard Deviation (σ). μ located at the center of the symmetrical curve σ controls the spread Notation: N(μ,σ) More Normal Distribution • The Mean (μ) is located at the center of the single peak and controls location of the curve on the horizontal axis. • The standard deviation (σ) is located at the inflection points of the curve and controls the spread of the curve. Inflection Points • The point on the curve where the curve changes from falling more steeply to falling less steeply (change in curvature – concave down to concave up). Inflection point Inflection point • Located one standard deviation (σ) from the mean (μ). Are not Normal Curves • Why a) b) c) d) Normal curve gets closer and closer to the horizontal axis, but never touches it. Normal curve is symmetrical. Normal curve has a single peak. Normal curve tails do not curve away from the horizontal axis. When Is a z-score Big? (cont.) • There is a Normal model for every possible combination of mean and standard deviation. • We write N(μ,σ) to represent a Normal model with a mean of μ and a standard deviation of σ. • We use Greek letters because this mean and standard deviation are not numerical summaries of the data. They are part of the model. They don’t come from the data. They are numbers that we choose to help specify the model. • Such numbers are called parameters of the model. When Is a z-score Big? (cont.) • Summaries of data, like the sample mean and standard deviation, are written with Latin letters. Such summaries of data are called statistics. • When we standardize Normal data, we still call the standardized value a z-score, and we write z y When Is a z-score Big? (cont.) • Once we have standardized, we need only one model: • The N(0,1) model is called the standard Normal model (or the standard Normal distribution). • Be careful—don’t use a Normal model for just any data set, since standardizing does not change the shape of the distribution. Standardizing Normal Distributions • All normal distributions are the same general shape and share many common properties. • Normal distribution notation: N(μ,σ). • We can make all normal distributions the same by measuring them in units of standard deviation (σ) about the mean (μ). • This is called standardizing and gives us the Standard Normal Curve. Standardizing & Z - SCORES • We can standardize a variable that has a normal distribution to a new variable that has the standard normal distribution using the formula: Substitute your variable as y z BAM! Pops out your z-score y Then divide by your Standard Deviation Subtract the mean from your variable Standardize a Normal Curve to the Standard Normal Curve y y The Standard Normal Distribution • • • • • Shape – normal curve Mean (μ) = 0 Standard Deviation (σ) = 1 Horizontal axis scale – Z score No vertical axis Z-SCORE z y Standard Normal Distribution N(μ,σ) When Is a z-score Big? (cont.) • When we use the Normal model, we are assuming the distribution is Normal. • We cannot check this assumption in practice, so we check the following condition: • Nearly Normal Condition: The shape of the data’s distribution is unimodal and symmetric. • This condition can be checked with a histogram or a Normal probability plot (to be explained later). The 68-95-99.7 Rule (Empirical Rule) • Normal models give us an idea of how extreme a value is by telling us how likely it is to find one that far from the mean. • We can find these numbers precisely, but until then we will use a simple rule that tells us a lot about the Normal model… The 68-95-99.7 Rule (cont.) • It turns out that in a Normal model: • about 68% of the values fall within one standard deviation of the mean; (µ – σ to µ + σ) • about 95% of the values fall within two standard deviations of the mean; (µ – 2σ to µ + 2σ ) and, • about 99.7% (almost all!) of the values fall within three standard deviations of the mean. (µ – 3σ to µ + 3σ) The 68-95-99.7 Rule (cont.) • The following shows what the 68-95-99.7 Rule tells us: More 68-95-99.7% Rule Using the 68-95-99.7 Rule • SOUTH AMERICAN RAINFALL • The distribution of rainfall in South American countries is approximately normal with a (mean) µ = 64.5 cm and (standard deviation) σ = 2.5 cm. • The next slide will demonstrate the empirical rule of this application. N(64.5,2.5) • 68% of the countries receive rain fall between 64.5(μ) – 2.5(σ) cm (62) and 64.5(μ)+2.5(σ) cm (67). • 68% = 62 to 67 • 95% of the countries receive rain fall between 64.5(μ) – 5(2σ) cm (59.5) and 64.5 (μ) + 5(2σ) cm (69.5). • 95% = 59.5 to 69.5 • 99.7% of the countries receive rain fall between 64.5(μ) – 7.5(3σ) cm (57) and 64.5(μ) + 7.5(3σ) cm (72). • 99.7% = 57 to 72 The middle 68% of the countries (µ ± σ) have rainfall between 62 – 67 cm The middle 95% of the countries (µ ± 2σ) have rainfall between 59.5 – 69.5 cm Almost all of the data (99.7%) is within 57 – 72 cm (µ ± 3σ) Example: IQ Test • The scores of a referenced population on the IQ Test are normally distributed with μ=100 and σ=15. 1) Approximately what percent of scores fall in the range from 70 to 130? 2) A score in what range would represent the top 16% of the scores? Example: IQ Test μ=100 σ=15 1) 70 to 130 is μ±2σ, therefore it would 95% of the scores. 2) The top 16% of the scores is one σ above the μ, therefore the score would be 115. Your Turn: • Runner’s World reports that the times of the finishes in the New York City 10-km run are normally distributed with a mean of 61 minutes and a standard deviation of 9 minutes. 1) Find the percent of runners who take more than 70 minutes to finish. 16% 2) Find the percent of runners who finish in less than 43 minutes. 2.5% The First Three Rules for Working with Normal Models • Make a picture. • Make a picture. • Make a picture. • And, when we have data, make a histogram to check the Nearly Normal Condition to make sure we can use the Normal model to model the distribution. Finding Normal Percentiles by Hand • When a data value doesn’t fall exactly 1, 2, or 3 standard deviations from the mean, we can look it up in a table of Normal percentiles. • Table Z in Appendix D provides us with normal percentiles, but many calculators and statistics computer packages provide these as well. Finding Normal Percentiles by Hand (cont.) • Table Z is the standard Normal table. We have to convert our data to z-scores before using the table. • The figure shows us how to find the area to the left when we have a z-score of 1.80: Standard Normal Distribution Table • Gives area under the curve to the left of a positive z-score. • Z-scores are in the 1st column and the 1st row • 1st column – whole number and first decimal place • 1st row – second decimal place Standard Normal Distribution Table • Also gives areas to the left of negative z-scores. • The curve is symmetrical, therefore the area to the left of a negative z-score is the same as the area to the right of the same positive z-score. Table Z • The table entry for each value z is the area under the curve to the LEFT of z. USING THE Z TABLE • 1. 2. 3. You found your z-score to be 1.40 and you want to find the area to the left of 1.40. Find 1.4 in the left-hand column of the Table Find the remaining digit 0 as .00 in the top row The entry opposite 1.4 and under .00 is 0.9192. This is the area we seek: 0.9192 Other Types of Tables Using Left-Tail Style Table 1. For areas to the left of a specified z value, use the table entry directly. 2. For areas to the right of a specified z value, look up the table entry for z and subtract the area from 1. (can also use the symmetry of the normal curve and look up the table entry for –z). 3. For areas between two z values, z1 and z2 (where z2 > z1), subtract the table area for z1 from the table area for z2. More using Table Z (left tailed table) Use table directly Example: Find Area Greater Than a Given Z-Score • Find the area from the standard normal distribution that is greater than -2.15 THE ANSWER IS 0.9842 • Find the corresponding Table Z value using the z-score -2.15. • The table entry is 0.0158 • However, this is the area to the left of -2.15 • We know the total area of the curve = 1, so simply subtract the table entry value from 1 • 1 – 0.0158 = 0.9842 • The next slide illustrates these areas Practice using Table A to find areas under the Standard Normal Curve 1. z<1.58 2. z<-.93 3. z>-1.23 4. z>2.48 5. .5<z<1.89 6. -1.43<z<1.43 1. .9429 (directly from table) 2. .1762 (directly from table) 3. .8907 (1-.1093 z<-1.23 or use symmetry z<1.23) 4. .0066 (1-.9934 z<2.48 or use symmetry z<-2.48) 5. .2791 (z<1.89=.9706 – z<.5=.6915) 6. .8472 (z<1.43=.9236 – z<1.43=>0764) CAUTION! • The average statistics student will look up a z-value in Table Z and use the entry corresponding to that z-value, not paying attention to if the problem asks for the area to the right or to the left of that z-value • BUT, YOU as an AP stats student should always be more meticulous and make sure your answer is reasonable in the context of the problem Using the TI-83/84 to Find the Area Under the Standard Normal Curve • Under the DISTR menu, the 2nd entry is “normalcdf”. • Calculates the area under the Standard Normal Curve between two z-scores (-1.43<z<.96). • Syntax normalcdf(lower bound, upper bound). Upper and lower bounds are z-scores. • If finding the area > or < a single z-score use a large positive value for the upper bound (ie. 100) and a large negative value for the lower bound (ie. -100) respectively. Practice use the TI-83/84 to find areas under the standard normal curve 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. z>-2.35 and z<1.52 .85<z<1.56 -3.5<z<3.5 0<z<1 z<1.63 z>.85 z>2.86 z<-3.12 z>1.5 z<-.92 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. .9264 .1383 .9995 .3413 .9484 .1977 .0021 .0009 .0668 .1789 Using TI-83/84 to Find Areas Under the Standard Normal Curve Without Z-Scores • The TI-83/84 can find areas under the standard normal curve without first changing the observation x to a z-score • normalcdf(lower bound, upper bound, mean, standard deviation) If finding area < or > use very large observation value for the lower and upper bound receptively. • Example: N(136,18) 100<x<150 • Answer: .7589 • Example: N(2.5,.42) x>3.21 • Answer: .0455 Procedure for Finding Normal Percentiles 1. State the problem in terms of the observed variable y. • Example : y > 24.8 2. Standardize y to restate the problem in terms of a z-score. • Example: z > (24.8 - μ)/σ, therefore z > ? 3. Draw a picture to show the area under the standard normal curve to be calculated. 4. Find the required area using Table Z or the TI-83/84 calculator. Example 1: • The heights of men are approximately normally distributed with a mean of 70 and a standard deviation of 3. What proportion of men are more than 6 foot tall? Answer: 1. State the problem in terms of y. (6’=72”) y 72 2. Standardize and state in terms of z. y 72 70 z z .67 3 3. Draw a picture of the area under the curve to be calculated. 4. Calculate the area under the curve. Example 2: • Suppose family incomes in a town are normally distributed with a mean of $1,200 and a standard deviation of $600 per month. What are the percentage of families that have income between $1,400 and $2,250 per month? Answer: 1. State the problem in terms of y. 1400 y 2250 2. Standardize and state in terms of z. 1400 1200 2250 1200 z 600 600 .33 z 1.75 3. Draw a picture. 4. Calculate the area. Your Turn: • The Chapin Social Insight (CSI) Test evaluates how accurately the subject appraises other people. In the reference population used to develop the test, scores are approximately normally distributed with mean 25 and standard deviation 5. The range of possible scores is 0 to 41. 1. What percent of subjects score above a 32 on the CSI Test? 2. What percent of subjects score at or below a 13 on the CSI Test? 3. What percent of subjects score between 16 and 34 on the CSI Test? Solution: 1) What percent of subjects score above a 32 on the CSI Test? 1. y>32 32 25 1.4 2. z 5 3. Picture 4. 8.1% Solution: 2) What percent of subjects score at or below a 13 on the CSI Test? 1) y≤13 13 25 2.4 2) z 5 3) Picture 4) .82% Solution: 3) What percent of subjects score between 16 and 34 on the CSI Test? 1) 16<y<34 2) 16 25 z 34 25 , 5 5 3) Picture 4) 92.8% 1.8 z 1.8 From Percentiles to Scores: z in Reverse • Sometimes we start with areas and need to find the corresponding z-score or even the original data value. • Example: What z-score represents the first quartile in a Normal model? z in Reverse • Given a normal distribution proportion (area under the standard normal curve), find the corresponding observation value. • Table Z – find the area in the table nearest the given proportion and read off the corresponding z-score. • TI-83/84 Calculator – Use the DISTR menu, 3rd entry invNorm. Syntax for invNorm(area,[μ,σ]) is the area to the left of the z-score (or Observation y) wanted (left-tail area). From Percentiles to Scores: z in Reverse (cont.) • Look in Table Z for an area of 0.2500. • The exact area is not there, but 0.2514 is pretty close. • This figure is associated with z = –0.67, so the first quartile is 0.67 standard deviations below the mean. Inverse Normal Practice Proportion (area under curve, left tail) Using Table Z 1. .3409 2. .7835 3. .9268 4. .0552 Using TI-83/84 1. .3409 2. .7835 3. .9268 4. .0552 Z-Score Using Table Z 1. Z = -.41 2. Z = .78 3. Z = 1.45 4. Z = -1.60 Using the TI-83/84 1. Z = -.4100 2. Z = .7841 3. Z = 1.4524 4. Z = -1.5964 Procedure for Inverse Normal Proportions 1. Draw a picture showing the given proportion (area under the curve). 2. Find the z-score corresponding to the given area under the curve. 3. Unstandardize the z-score. 4. Solve for the observational value y and answer the question. Example 1: SAT VERBAL SCORES • SAT Verbal scores are approximately normal with a mean of 505 and a standard deviation of 110 • How high must a student score in order to place in the top 10% of all students taking the verbal section of the SAT. Analyze the Problem and Picture It. • The problem wants to know the SAT score y with the area 0.10 to its right under the normal curve with a mean of 505 and a standard deviation of 110. Well, isn't that the same as finding the SAT score y with the area 0.9 to its left? Let's draw the distribution to get a better look at it. 1. Draw a picture showing the given proportion (area under the curve). y=505 y=? 2. Find Your Z-Score 1. Using Table Z - Find the entry closest to 0.90. It is 0.8997. This is the entry corresponding to z = 1.28. So z = 1.28 is the standardized value with area 0.90 to its left. 2. Using TI-83/84 – DISTR/invNorm(.9). It is 1.2816. 3. Unstandardize • Now, you will need to unstandardize to transform the solution from the z, back to the original y scale. We know that the standardized value of the unknown y is z = 1.28. So y itself satisfies: y 505 1.28 110 4. Solve for y and Summarize • Solve the equation for y: y 505 (1.28)(110) 645.8 • The equation finds the y that lies 1.28 standard deviations above the mean on this particular normal curve. That is the "unstandardized" meaning of z = 1.28. • Answer: A student must score at least 646 to place in the highest 10% Example 2: • A four-year college will accept any student ranked in the top 60 percent on a national examination. If the test score is normally distributed with a mean of 500 and a standard deviation of 100, what is the cutoff score for acceptance? Answer: 1. Draw picture of given proportion. 2. Find the z-score. From TI-83/84, invNorm(.4) is z = -.25. y 500 3. Unstandardize: 0.25 100 4. Solve for y and answer the question. y = 475, therefore the minimum score the college will accept is 475. Your Turn: • Intelligence Quotients are normally distributed with a mean of 100 and a standard deviation of 16. Find the 90th percentile for IQ’s. Answer: 1. Draw picture of given proportion. 2. Find the z-score. From TI-83/84, invNorm(.9) is z = 1.28. y 100 3. Unstandardize: 1.28 16 4. Solve for y and answer the question. y = 120.48, what this means; the 90th percentile for IQ’s is 120.48. In other words, 90% of people have IQ’s below 120.48 and 10% have IQ’s above 120.48. Are You Normal? How Can You Tell? • When you actually have your own data, you must check to see whether a Normal model is reasonable. • Looking at a histogram of the data is a good way to check that the underlying distribution is roughly unimodal and symmetric. Are You Normal? How Can You Tell? (cont.) • A more specialized graphical display that can help you decide whether a Normal model is appropriate is the Normal probability plot. • If the distribution of the data is roughly Normal, the Normal probability plot approximates a diagonal straight line. Deviations from a straight line indicate that the distribution is not Normal. Are You Normal? How Can You Tell? (cont.) • Nearly Normal data have a histogram and a Normal probability plot that look somewhat like this example: Are You Normal? How Can You Tell? (cont.) • A skewed distribution might have a histogram and Normal probability plot like this: Summary Assessing Normality (Is The Distribution Approximately Normal) 1. Construct a Histogram or Stemplot. See if the shape of the graph is approximately normal. 2. Construct a Normal Probability Plot (TI-83/84). A normal Distribution will be a straight line. Conversely, non-normal data will show a nonlinear trend. 3. Determine the proportion of observations within one, two, and three standard deviations of the mean and compare with the 68-95-99.7 Rule for normal distributions. Assess the Normality of the Following Data • 9.7, 93.1, 33.0, 21.2, 81.4, 51.1, 43.5, 10.6, 12.8, 7.8, 18.1, 12.7 • Histogram – skewed right • Normal Probability Plot – clearly not linear • 68-95-99.7 Rule – mean = 32.92 & standard deviation = 29 1. μ ± σ = 3.92-61.92 = 10 obs./12 total obs. = 83% 2. μ ± 2σ = -25.08-90.92 = 11/12 = 92% 3. μ ± 3σ = -54-119.92 = 12/12 =100% Distribution doesn’t follow 68-95-99.7 Rule • Distribution is not Normal. What Can Go Wrong? • Don’t use a Normal model when the distribution is not unimodal and symmetric. What Can Go Wrong? (cont.) • Don’t use the mean and standard deviation when outliers are present—the mean and standard deviation can both be distorted by outliers. • Don’t round off too soon. • Don’t round your results in the middle of a calculation. • Don’t worry about minor differences in results. What have we learned? • The story data can tell may be easier to understand after shifting or rescaling the data. • Shifting data by adding or subtracting the same amount from each value affects measures of center and position but not measures of spread. • Rescaling data by multiplying or dividing every value by a constant changes all the summary statistics— center, position, and spread. What have we learned? (cont.) • We’ve learned the power of standardizing data. • Standardizing uses the SD as a ruler to measure distance from the mean (zscores). • With z-scores, we can compare values from different distributions or values based on different units. • z-scores can identify unusual or surprising values among data. What have we learned? (cont.) • We’ve learned that the 68-95-99.7 Rule can be a useful rule of thumb for understanding distributions: • For data that are unimodal and symmetric, about 68% fall within 1 SD of the mean, 95% fall within 2 SDs of the mean, and 99.7% fall within 3 SDs of the mean. What have we learned? (cont.) • We see the importance of Thinking about whether a method will work: • Normality Assumption: We sometimes work with Normal tables (Table Z). These tables are based on the Normal model. • Data can’t be exactly Normal, so we check the Nearly Normal Condition by making a histogram (is it unimodal, symmetric and free of outliers?) or a normal probability plot (is it straight enough?). Assignment • Exercises pg. 129 – 133: #1 – 19 odd, 23, 25, 29, 37, 39, 43, 45, 47 • Read Ch-7, pg. 146 - 163