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Objective: To determine whether or not a curved relationship can be salvaged and re-expressed into a linear relationship. If so, complete the re-expression.

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We cannot use a linear model unless the relationship between the two variables is linear. Often re-expression can save the day, straightening bent relationships so that we can fit and use a simple linear model. Two simple ways to re-express data are with logarithms and reciprocals.

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Original Scatterplot The relationship between fuel efficiency (in miles per gallon) and weight (in pounds) for late model cars looks fairly linear at first: Residual Plot A look at the residuals plot shows a problem:

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Re-Expressed Scatterplot We can re-express fuel efficiency as gallons per hundred miles (a reciprocal- multiplying each by 100) and eliminate the bend in the original scatterplot: Re-Expressed Residual Plot A look at the residuals plot for the new model seems more reasonable:

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Ladder of Powers There is a family of simple re-expressions that move data toward our goals in a consistent way. This collection of re- expressions is called the Ladder of Powers. The Ladder of Powers orders the effects that the re- expressions have on data. o The farther you move from “1” in the ladder of powers, the greater the effect on the original data.

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Ratios of two quantities (e.g., mph) often benefit from a reciprocal. The reciprocal of the data –1 An uncommon re-expression, but sometimes useful. Reciprocal square root –1/2 Measurements that cannot be negative often benefit from a log re-expression. We’ll use logarithms here “0” Counts often benefit from a square root re- expression. Square root of data values ½ Data with positive and negative values and no bounds are less likely to benefit from re- expression. Raw data1 Try with unimodal distributions that are skewed to the left. Square of data values 2CommentNamePower

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Using the ladder of powers, what transformations might you use as a starting point? 1. You want to model the relationship between the number of birds counted at a nesting site and the temperature (in degrees Celsius). The scatterplot of counts vs. temperature shows an upwardly curving pattern, with more birds spotted at higher temperatures. 2. You want to model the relationship between prices for various items in Paris and in Hong Kong. The scatterplot of Hong Kong prices vs. Parisian prices shows a generally straight pattern with a small amount of scatter. 3. You want to model the population growth of the US over the past 200 years. The scatterplot shows a strongly upwardly curved pattern.

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When none of the data values is zero or negative, logarithms can be a helpful ally in the search for a useful model. Try taking the logs of both the x- and y-variable. Then re-express the data using some combination of x or log(x) vs. y or log(y).

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When none of the data values is zero or negative, logarithms can be a helpful ally in the search for a useful model. Try taking the logs of both the x- and y-variable. Then re-express the data using some combination of x or log(x) vs. y or log(y).

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If there’s a curve in the scatterplot, why not just fit a curve to the data?

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The mathematics and calculations for “curves of best fit” are considerably more difficult than “lines of best fit.” Besides, straight lines are easy to understand. o We know how to think about the slope and the y- intercept.

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Let’s try an example to note the techniques to re-express that tuition rates verses the years at Arizona State University:

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Beware of multiple modes. o Re-expression cannot pull separate modes together. Watch out for scatterplots that turn around. o Re-expression can straighten many bent relationships, but not those that go up then down, or down then up.

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Watch out for negative data values. o It’s impossible to re-express negative values by any power that is not a whole number on the Ladder of Powers or to re-express values that are zero for negative powers. Watch for data far from 1. o Data values that are all very far from 1 may not be much affected by re-expression unless the range is very large. If all the data values are large (e.g., years), consider subtracting a constant to bring them back near 1.

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Day 1: Day 1: 9.5 Problem Set Online # 1, 2, 5, 6, 7 Day 2: Day 2: 9.5 Problem Set Online # 14, 15, 16

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