Linear Regression and Correlation: Outliers 1.6 2008/06/23 16:55:41 GMT-5 2008/10/27 18:20:14.664 GMT-5 Susan Dean deansusan@deanza.edu Barbara Illowsky illowskybarbara@deanza.edu Susan Dean deansusan@deanza.edu Barbara Illowsky illowskybarbara@deanza.edu Connexions cnx@cnx.org elementary statistics This module provides an overview of Linear Regression and Correlation: Outliers as a part of Collaborative Statistics collection (col10522) by Barbara Illowsky and Susan Dean. In some data sets, there are values (points) called outliers. Outliers are points that are far from the least squares line. They have large "errors." Outliers need to be examined closely. Sometimes, for some reason or another, they should not be included in the analysis of the data. It is possible that an outlier is a result of erroneous data. Other times, an outlier may hold valuable information about the population under study. The key is to carefully examine what causes a data point to be an outlier. In the third exam/final exam example, you can determine if there is an outlier or not. If there is one, as an exercise, delete it and fit the remaining data to a new line. For this example, the new line ought to fit the remaining data better. This means the SSE should be smaller and the correlation coefficient ought to be closer to 1 or -1. Computers and many calculators can determine outliers from the data. However, as an exercise, we will go through the steps that are needed to calculate an outlier. In the table below, the first two columns are the third exam and the final exam data. The third column shows the y-hat values calculated from the line of best fit. x y y^ 65    175   140   67 133 150 71 185 169 71 163 169 66 126 145 75 198 189 67 153 150 70 163 164 71 159 169 69 151 160 69 159 160 A Residual is the Actual y value−predicted y value=y−y^ Calculate the absolute value of each residual. Calculate each |y-y^|: x y y^ |y-y^| 65 175 140 |175-140|=35 67 133 150 |133-150|=17 71 185 169 |185-169|=16 71 163 169 |163-169|=6 66 126 145 |126-145|=19 75 198 189 |198-189|=9 67 153 150 |153-150|=3 70 163 164 |163-164|=1 71 159 169 |159-169|=10 69 151 160 |151-160|=9 69 159 160 |159-160|=1 Square each |y-y^|: 352 172 162 62 192 92 32 12 102 92 12 Then, add (sum) all the |y-y^| squared terms: Σ i = 1 11 (|y-y^|)2 = Σ i = 1 11 ε2 (Recall that |yi-y^i|=εi.)=352 +172 +162 +62 +192 +92 +32 +12 +102 +92 +12 =2440= SSENext, calculate s, the standard deviation of all the |y-y^|=ε values where n = the total number of data points. (Calculate the standard deviation of 35 17 16 6 19 9 3 1 10 9 1 .)s= SSE n-2 For the third exam/final exam problem, s= 2440 11-2 =16.47 Next, multiply s by 1.9 and get (1.9)⋅(16.47)=31.29 (the value 31.29 is almost 2 standard deviations away from the mean of the |y-y^| values.)The number 1.9s is equal to 1.9 standard deviations. It is a measure that is almost 2 standard deviations. If we were to measure the vertical distance from any data point to the corresponding point on the line of best fit and that distance was equal to 1.9s or greater, then we would consider the data point to be "too far" from the line of best fit. We would call that point a potential outlier.For the example, if any of the |y-y^| values are at least 31.29, the corresponding (x,y) point (data point) is a potential outlier.Mathematically, we say that if |y-y^|≥(1.9)⋅(s), then the corresponding point is an outlier.For the third exam/final exam problem, all the |y-y^|'s are less than 31.29 except for the first one which is 35.35>31.29 That is, |y-y^|≥(1.9)⋅(s)The point which corresponds to |y-y^|=35 is (65,175). Therefore, the point (65,175) is an outlier. For this example, we will delete it. (Remember, we do not always delete an outlier.) The next step is to compute a new best-fit line using the 10 remaining points. The new line of best fit and the correlation coefficient are: y^ =-355.19+7.39x and r=0.9121 If you compare r=0.9121 to its critical value 0.632, 0.9121>0.632. Therefore, r is significant. In fact, r=0.9121 is a better r than the original (0.6631) because r=0.9121 is closer to 1. This means that the 10 points fit the line better. The line can better predict the final exam score given the third exam score. Using the new line of best fit (calculated with 10 points), what would a student who receives a 73 on the third exam expect to receive on the final exam? 184.28 (From The Consumer Price Indexes Web site) The Consumer Price Index (CPI) measures the average change over time in the prices paid by urban consumers for consumer goods and services. The CPI affects nearly all Americans because of the many ways it is used. One of its biggest uses is as a measure of inflation. By providing information about price changes in the Nation's economy to government, business, and labor, the CPI helps them to make economic decisions. The President, Congress, and the Federal Reserve Board use the CPI's trends to formulate monetary and fiscal policies. Data: x y 1915   10.1   1926 17.7 1935 13.7 1940 14.7 1947 24.1 1952 26.5 1964 31.0 1969 36.7 1975 49.3 1979 72.6 1980 82.4 1986 109.6 1991 130.7 1999 166.6 Make a scatterplot of the data. Calculate the least squares line. Write the equation in the form y^=a+bx. Draw the line on the scatterplot. Find the correlation coefficient. Is it significant? What is the average CPI for the year 1990? Scatter plot and line of best fit. y^=-3204+1.662x is the equation of the line of best fit. r=0.8694 The number of data points is n=14. Use the 95% Critical Values of the Sample Correlation Coefficient table at the end of Chapter 12. n-2=12. The corresponding critical value is 0.532. Since 0.8694>0.532, r is significant. y^=-3204+1.662(1990)=103.4 CPI

Outlier An observation that does not fit the rest of the data.