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HAPPY TUESDAY 1. 2. 3. 4. 5. 6. Yes Okay Maybe Perhaps No ? 17% 17% 17% 17% 17% 17% Slide 1- 1 1 2 3 4 5 6 UPCOMING IN CLASS Homework #3 due Monday at 5pm Part 2 of the Data Project due February 7th Quiz #2 in class February 9th (open book) CHAPTER 6 The Standard Deviation as a Ruler and the Normal Model WHAT ABOUT SPREAD? THE STANDARD DEVIATION (CONT.) The variance, notated by s2, is found by summing the squared deviations and (almost) averaging them: y y 2 s 2 n 1 The variance will play a role later in our study, but it is problematic as a measure of spread—it is measured in squared units! WHAT ABOUT SPREAD? THE STANDARD DEVIATION (CONT.) The standard deviation, s, is just the square root of the variance and is measured in the same units as the original data. y y 2 s n 1 THE STANDARD DEVIATION AS A RULER The trick in comparing very different-looking 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. NORMAL MODEL The following shows what the 68-95-99.7 Rule tells us: STANDARDIZING WITH Z-SCORES We compare individual data values to their mean, relative to their standard deviation using the following formula: y y z s We call the resulting values standardized values, denoted as z. They can also be called z-scores. Slide 1- 8 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. Thus, we can compare values that are measured on different scales, with different units, or from different populations. STANDARDIZING DATA By calculating z-scores for each observation, we change the distribution of the data by Shifting the data Rescaling the data SHIFTING DATA Shifting data: Adding (or subtracting) a constant to every data value adds (or subtracts) the same constant to measures of position. Adding (or subtracting) a constant to each value will increase (or decrease) measures of position: center, percentiles, max or min by the same constant. Its shape and spread - range, IQR, standard deviation - remain unchanged. SHIFTING DATA The following histograms show a shift from men’s actual weights to kilograms above recommended weight: RESCALING DATA Rescaling data: When we multiply (or divide) all the data values by any constant, all measures of position (such as the mean, median, and percentiles) and measures of spread (such as the range, the IQR, and the standard deviation) are multiplied (or divided) by that same constant. 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: Slide 1- 15 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. Slide 1- 16 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. The larger a z-score is (negative or positive), the more unusual it is. We use the theory of the Normal Model to see. NORMAL MODEL The following shows what the 68-95-99.7 Rule tells us: Slide 1- 18 NORMAL MODEL 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. Slide 1- 19 NORMAL MODEL 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 do not come from data—they are numbers (called parameters) that specify the model. Slide 1- 20 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 y y z s 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. Slide 1- 22 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 Thursday). 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. Slide 1- 24 TWO STANDARDIZED TESTS, A AND B USE VERY DIFFERENT SCALES OF SCORES. THE FORMULA A=50*B+200 APPROXIMATES THE RELATIONSHIP BETWEEN SCORES ON THE TWO TWO TEST. USE THE SUMMARY STATISTICS WHO TOOK TEST B TO DETERMINE THE SUMMARY STATISTICS FOR EQUIVALENT SCORES ON TEST A. Lowest = 18 Mean = 26 St. Dev=5 Q3=28 Median=30 IQR = 6 Slide 1- 25 WHAT IS THE LOWEST SCORE FOR TEST A? 1. 2. 3. 4. 5. 6. 7. 18 50 200 250 1100 1500 2000 14% 14% 14% 14% 14% 14% 14% Slide 1- 26 1 2 3 4 5 6 7 WHAT IS THE MEAN FOR TEST A? 1. 2. 3. 4. 5. 6. 7. 26 50 200 250 1100 1500 2000 14% 14% 14% 14% 14% 14% 14% Slide 1- 27 1 2 3 4 5 6 7 WHAT IS THE STANDARD DEVIATION? 1. 2. 3. 4. 5. 6. 200 250 450 500 1100 1500 17% 17% 17% 17% 17% 17% Slide 1- 28 1 2 3 4 5 6 WHAT IS THE Q3 FOR TEST A? 1. 2. 3. 4. 5. 1000 1400 1500 1600 2000 20% 20% 20% 20% 20% Slide 1- 29 1 2 3 4 5 WHAT IS THE MEDIAN FOR TEST A? 1. 2. 3. 4. 5. 1000 1400 1500 1700 2000 20% 20% 20% 20% 20% Slide 1- 30 1 2 3 4 5 WHAT IS THE IQR FOR TEST A? 1. 2. 3. 4. 5. 200 250 300 500 1000 20% 20% 20% 20% 20% Slide 1- 31 1 2 3 4 5 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 (z-scores). 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. Slide 1- 32 TWO STUDENTS TAKE LANGUAGE EXAMS Anna score 87 on both Megan scores 76 on first, and 91 on the second Overall student scores on the first exam Mean=83 St. Dev. 5 Second exam Mean = 70 St. Dev. 14 TO QUALIFY FOR LANGUAGE HONORS, A STUDENT MUST AVERAGE AT LEAST 85 ACROSS 25% 25% 25% 25% ALL COURSE. DO ANNA AND MEGAN QUALIFY? 1. 2. 3. 4. Only Anna qualifies Both qualify Neither qualify Only Megan Slide 1- 34 1 2 3 4 WHO PERFORMED BETTER OVERALL? 1. 2. Anna Megan 50% 50% Slide 1- 35 1 2 COW PROBLEM The mean weight of a group of yearling steers is 1153 pounds. Suppose that weights of all such animals can be described with a Normal model with a standard deviation of 86 pounds. Cattle buyers hope that yearling steers will weight AT LEAST 1000 pounds. To see how much over or under that goal the cattle are, we could subtract 1000 pounds from all the weights. What would the new mean and standard deviation be? COW PROBLEM (CONT.) Suppose such cattle sell at auction for $0.35 a pound. Find the mean and standard deviation of their sale price. COW PROBLEM (CONT) Assume the Normal Model applies to the sale price for steers. Use the 68-95-99.7 rule to draw the model for the price of steers. NEXT TIME… The benefits of symmetry in the Normal Model