Outliers In some data sets, there are values (observed data points) called outliers . Outliers are observed data points that are far from the least squares line. They have large "errors", where the "error" or residual is the vertical distance from the line to the point. 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 and should remain included in the data. The key is to examine carefully what causes a data point to be an outlier. Besides outliers, a sample may contain one or a few points that are called influential points . Influential points are observed data points that are far from the other observed data points in the horizontal direction. These points may have a big effect on the slope of the regression line. To begin to identify an influential point, you can remove it from the data set and see if the slope of the regression line is changed significantly. Computers and many calculators can be used to identify outliers from the data. Computer output for regression analysis will often identify both outliers and influential points so that you can examine them. Identifying Outliers We could guess at outliers by looking at a graph of the scatterplot and best fit-line. However, we would like some guideline as to how far away a point needs to be in order to be considered an outlier. As a rough rule of thumb, we can flag any point that is located further than two standard deviations above or below the best-fit line as an outlier . The standard deviation used is the standard deviation of the residuals or errors. We can do this visually in the scatter plot by drawing an extra pair of lines that are two standard deviations above and below the best-fit line. Any data points that are outside this extra pair of lines are flagged as potential outliers. Or we can do this numerically by calculating each residual and comparing it to twice the standard deviation. On the TI-83, 83+, or 84+, the graphical approach is easier. The graphical procedure is shown first, followed by the numerical calculations. You would generally need to use only one of these methods. In the third exam/final exam example , you can determine if there is an outlier or not. If there is an outlier, 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. Graphical Identification of Outliers With the TI-83, 83+, 84+ graphing calculators, it is easy to identify the outliers graphically and visually. 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 were equal to 2 s or more, then we would consider the data point to be "too far" from the line of best fit. We need to find and graph the lines that are two standard deviations below and above the regression line. Any points that are outside these two lines are outliers. We will call these lines Y2 and Y3: As we did with the equation of the regression line and the correlation coefficient, we will use technology to calculate this standard deviation for us. Using the LinRegTTest with this data, scroll down through the output screens to find s = 16.412 . Line Y2 = –173.5 + 4.83 x –2(16.4) and line Y3 = –173.5 + 4.83 x + 2(16.4) where ŷ = –173.5 + 4.83 x is the line of best fit. Y2 and Y3 have the same slope as the line of best fit. Graph the scatterplot with the best fit line in equation Y1, then enter the two extra lines as Y2 and Y3 in the "Y="equation editor and press ZOOM 9. You will find that the only data point that is not between lines Y2 and Y3 is the point x = 65, y = 175. On the calculator screen it is just barely outside these lines. The outlier is the student who had a grade of 65 on the third exam and 175 on the final exam; this point is further than two standard deviations away from the best-fit line. Sometimes a point is so close to the lines used to flag outliers on the graph that it is difficult to tell if the point is between or outside the lines. On a computer, enlarging the graph may help; on a small calculator screen, zooming in may make the graph clearer. Note that when the graph does not give a clear enough picture, you can use the numerical comparisons to identify outliers. Try It Identify the potential outlier in the scatter plot. The standard deviation of the residuals or errors is approximately 8.6. Numerical Identification of Outliers In , the first two columns are the third-exam and final-exam data. The third column shows the predicted ŷ values calculated from the line of best fit: ŷ = –173.5 + 4.83 x . The residuals, or errors, have been calculated in the fourth column of the table: observed y value−predicted y value = y − ŷ . s is the standard deviation of all the y − ŷ = ε values where n = the total number of data points. If each residual is calculated and squared, and the results are added, we get the SSE. The standard deviation of the residuals is calculated from the SSE as: s = S S E n − 2 NOTE We divide by ( n – 2) because the regression model involves two estimates. Rather than calculate the value of s ourselves, we can find s using the computer or calculator. For this example, the calculator function LinRegTTest found s = 16.4 as the standard deviation of the residuals 35 –17 16 –6 –19 9 3 –1 –10 –9 –1 . x y ŷ y – ŷ 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 We are looking for all data points for which the residual is greater than 2 s = 2(16.4) = 32.8 or less than –32.8. Compare these values to the residuals in column four of the table. The only such data point is the student who had a grade of 65 on the third exam and 175 on the final exam; the residual for this student is 35. How does the outlier affect the best fit line? Numerically and graphically, we have identified the point (65, 175) as an outlier. We should re-examine the data for this point to see if there are any problems with the data. If there is an error, we should fix the error if possible, or delete the data. If the data is correct, we would leave it in the data set. For this problem, we will suppose that we examined the data and found that this outlier data was an error. Therefore we will continue on and delete the outlier, so that we can explore how it affects the results, as a learning experience. Compute a new best-fit line and correlation coefficient using the ten remaining points: On the TI-83, TI-83+, TI-84+ calculators, delete the outlier from L1 and L2. Using the LinRegTTest, the new line of best fit and the correlation coefficient are: ŷ = –355.19 + 7.39 x and r = 0.9121 The new line with r = 0.9121 is a stronger correlation than the original ( r = 0.6631) because r = 0.9121 is closer to one. This means that the new line is a better fit to the ten remaining data values. The line can better predict the final exam score given the third exam score. Numerical Identification of Outliers: Calculating s and Finding Outliers Manually If you do not have the function LinRegTTest, then you can calculate the outlier in the first example by doing the following. First, square each | y – ŷ | The squares are 35 2 17 2 16 2 6 2 19 2 9 2 3 2 1 2 10 2 9 2 1 2 Then, add (sum) all the | y – ŷ | squared terms using the formula Σ i = 1 11 ( | y i − y ^ i | ) 2 = Σ i = 1 11 ε i 2 (Recall that y i – ŷ i = ε i .) = 35 2 + 17 2 + 16 2 + 6 2 + 19 2 + 9 2 + 3 2 + 1 2 + 10 2 + 9 2 + 1 2 = 2440 = SSE . The result, SSE is the Sum of Squared Errors. Next, calculate s , the standard deviation of all the y – ŷ = ε values where n = the total number of data points. The calculation is s = SSE n – 2 . For the third exam/final exam problem, s = 2440 11 – 2 = 16.47 . Next, multiply s by 2: (2)(16.47) = 32.94 32.94 is 2 standard deviations away from the mean of the y – ŷ values. 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 is at least 2 s , then we would consider the data point to be "too far" from the line of best fit. We call that point a potential outlier . For the example, if any of the | y – ŷ | values are at least 32.94, the corresponding ( x , y ) data point is a potential outlier. For the third exam/final exam problem, all the | y – ŷ |'s are less than 31.29 except for the first one which is 35. 35 > 31.29 That is, | y – ŷ | ≥ (2)(s) The point which corresponds to | y – ŷ | = 35 is (65, 175). Therefore, the data point (65,175) is a potential outlier. For this example, we will delete it. (Remember, we do not always delete an outlier.) NOTE When outliers are deleted, the researcher should either record that data was deleted, and why, or the researcher should provide results both with and without the deleted data. If data is erroneous and the correct values are known (e.g., student one actually scored a 70 instead of a 65), then this correction can be made to the data. The next step is to compute a new best-fit line using the ten remaining points. The new line of best fit and the correlation coefficient are: ŷ = –355.19 + 7.39 x and r = 0.9121 Using this new line of best fit (based on the remaining ten data points in the third exam/final exam example ), what would a student who receives a 73 on the third exam expect to receive on the final exam? Is this the same as the prediction made using the original line? Using the new line of best fit, ŷ = –355.19 + 7.39(73) = 184.28. A student who scored 73 points on the third exam would expect to earn 184 points on the final exam. The original line predicted ŷ = –173.51 + 4.83(73) = 179.08 so the prediction using the new line with the outlier eliminated differs from the original prediction. Try It The data points for the graph from the third exam/final exam example are as follows: (1, 5), (2, 7), (2, 6), (3, 9), (4, 12), (4, 13), (5, 18), (6, 19), (7, 12), and (7, 21). Remove the outlier and recalculate the line of best fit. Find the value of ŷ when x = 10. 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. In the following table, x is the year and y is the CPI. Data x y x y 1915 10.1 1979 72.6 1926 17.7 1980 82.4 1935 13.7 1986 109.6 1940 14.7 1991 130.7 1947 24.1 1999 166.6 1952 26.5 2010 219.2 1964 31.0 2020 258 1969 36.7 2023 299.2 1975 49.3 Draw a scatterplot of the data. Calculate the least squares line. Write the equation in the form ŷ = a + bx . Draw the line on the scatterplot. Find the correlation coefficient. Is it significant? What is the average CPI for the year 1990? See . ŷ = –3204 + 1.662 x 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. ŷ = –3204 + 1.662(1990) = 103.4 CPI Using the calculator LinRegTTest, we find that s = 25.4 ; graphing the lines Y2 = –3204 + 1.662X – 2(25.4) and Y3 = –3204 + 1.662X + 2(25.4) shows that no data values are outside those lines, identifying no outliers. (Note that the year 1999 was very close to the upper line, but still inside it.) NOTE In the example, notice the pattern of the points compared to the line. Although the correlation coefficient is significant, the pattern in the scatterplot indicates that a curve would be a more appropriate model to use than a line. In this example, a statistician should prefer to use other methods to fit a curve to this data, rather than model the data with the line we found. In addition to doing the calculations, it is always important to look at the scatterplot when deciding whether a linear model is appropriate. If you are interested in seeing more years of data, visit the Bureau of Labor Statistics CPI website ftp://ftp.bls.gov/pub/special.requests/cpi/cpiai.txt; our data is taken from the column entitled "Annual Avg." (third column from the right). For example you could add more current years of data. Try adding the more recent years: 2010: CPI = 219.2; 2020: CPI = 258.0; 2023: CPI = 299.2. See how it affects the model. (Check: y ⏜ = - 5030 + 2 . 598 x ; r = 0 . 9067 . Is r significant? Is the fit better with the addition of the new points?) Try It The following table shows economic development measured in per capita income PCINC. Year PCINC Year PCINC 1870 340 1920 1050 1880 499 1930 1170 1890 592 1940 1364 1900 757 1950 1836 1910 927 1960 2132 What are the independent and dependent variables? Draw a scatter plot. Use regression to find the line of best fit and the correlation coefficient. Interpret the significance of the correlation coefficient. Is there a linear relationship between the variables? Find the coefficient of determination and interpret it. What is the slope of the regression equation? What does it mean? Use the line of best fit to estimate PCINC for 1900, for 2000. Determine if there are any outliers. 95% Critical Values of the Sample Correlation Coefficient Table Degrees of Freedom: n – 2 Critical Values: (+ and –) 1 0.997 2 0.950 3 0.878 4 0.811 5 0.754 6 0.707 7 0.666 8 0.632 9 0.602 10 0.576 11 0.555 12 0.532 13 0.514 14 0.497 15 0.482 16 0.468 17 0.456 18 0.444 19 0.433 20 0.423 21 0.413 22 0.404 23 0.396 24 0.388 25 0.381 26 0.374 27 0.367 28 0.361 29 0.355 30 0.349 40 0.304 50 0.273 60 0.250 70 0.232 80 0.217 90 0.205 100 0.195 References Data from the House Ways and Means Committee, the Health and Human Services Department. Data from Microsoft Bookshelf. Data from the United States Department of Labor, the Bureau of Labor Statistics. Data from the Physician’s Handbook, 1990. Data from the United States Department of Labor, the Bureau of Labor Statistics. Chapter Review To determine if a point is an outlier, do one of the following: Input the following equations into the TI 83, 83+,84, 84+: y 1 = a + b x y 2 = a + b x + 2 s y 3 = a + b x − 2 s where s is the standard deviation of the residuals If any point is above y 2 or below y 3 then the point is considered to be an outlier. Use the residuals and compare their absolute values to 2 s where s is the standard deviation of the residuals. If the absolute value of any residual is greater than or equal to 2 s , then the corresponding point is an outlier. Note: The calculator function LinRegTTest (STATS TESTS LinRegTTest) calculates s . Use the following information to answer the next four exercises. The scatter plot shows the relationship between hours spent studying and exam scores. The line shown is the calculated line of best fit. The correlation coefficient is 0.69. Do there appear to be any outliers? Yes, there appears to be an outlier at (6, 58). A point is removed, and the line of best fit is recalculated. The new correlation coefficient is 0.98. Does the point appear to have been an outlier? Why? What effect did the potential outlier have on the line of best fit? The potential outlier flattened the slope of the line of best fit because it was below the data set. It made the line of best fit less accurate as a predictor for the data. Are you more or less confident in the predictive ability of the new line of best fit? The Sum of Squared Errors for a data set of 18 numbers is 49. What is the standard deviation? s = 1.75 The Standard Deviation for the Sum of Squared Errors for a data set is 9.8. What is the cutoff for the vertical distance that a point can be from the line of best fit to be considered an outlier? Homework The height (sidewalk to roof) of notable tall buildings in America is compared to the number of stories of the building (beginning at street level). Height (in feet) Stories 1,050 57 428 28 362 26 529 40 790 60 401 22 380 38 1,454 110 1,127 100 700 46 Using “stories” as the independent variable and “height” as the dependent variable, make a scatter plot of the data. Does it appear from inspection that there is a relationship between the variables? Calculate the least squares line. Put the equation in the form of: ŷ = a + bx Find the correlation coefficient. Is it significant? Find the estimated heights for 32 stories and for 94 stories. Based on the data in , is there a linear relationship between the number of stories in tall buildings and the height of the buildings? Are there any outliers in the data? If so, which point(s)? What is the estimated height of a building with six stories? Does the least squares line give an accurate estimate of height? Explain why or why not. Based on the least squares line, adding an extra story is predicted to add about how many feet to a building? What is the slope of the least squares (best-fit) line? Interpret the slope. Ornithologists, scientists who study birds, tag sparrow hawks in 13 different colonies to study their population. They gather data for the percent of new sparrow hawks in each colony and the percent of those that have returned from migration. Percent return: 74; 66; 81; 52; 73; 62; 52; 45; 62; 46; 60; 46; 38 Percent new: 5; 6; 8; 11; 12; 15; 16; 17; 18; 18; 19; 20; 20 Enter the data into your calculator and make a scatter plot. Use your calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot from part a. Explain in words what the slope and y-intercept of the regression line tell us. How well does the regression line fit the data? Explain your response. Which point has the largest residual? Explain what the residual means in context. Is this point an outlier? An influential point? Explain. An ecologist wants to predict how many birds will join another colony of sparrow hawks to which 70% of the adults from the previous year have returned. What is the prediction? a. and b. Check student’s solution. c. The slope of the regression line is -0.3031 with a y-intercept of 31.93. In context, the y-intercept indicates that when there are no returning sparrow hawks, there will be almost 32% new sparrow hawks, which doesn’t make sense since if there are no returning birds, then the new percentage would have to be 100% (this is an example of why we do not extrapolate). The slope tells us that for each percentage increase in returning birds, the percentage of new birds in the colony decreases by 30.3%. d. If we examine r2, we see that only 57.52% of the variation in the percent of new birds is explained by the model and the correlation coefficient, r = –.7584 only indicates a somewhat strong correlation between returning and new percentages. e. The ordered pair (66, 6) generates the largest residual of 6.0. This means that when the observed return percentage is 66%, our observed new percentage, 6%, is almost 6% less than the predicted new value of 11.98%. If we remove this data pair, we see only an adjusted slope of -.2789 and an adjusted intercept of 30.9816. In other words, even though this data generates the largest residual, it is not an outlier, nor is the data pair an influential point. f. If there are 70% returning birds, we would expect to see y = –.2789(70) + 30.9816 = 0.114 or 11.4% new birds in the colony. The following table shows data on average per capita coffee consumption and heart disease rate in a random sample of 10 countries. Yearly coffee consumption in liters 2.5 3.9 2.9 2.4 2.9 0.8 9.1 2.7 0.8 0.7 Death from heart diseases 221 167 131 191 220 297 71 172 211 300 Enter the data into your calculator and make a scatter plot. Use your calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot from part a. Explain in words what the slope and y-intercept of the regression line tell us. How well does the regression line fit the data? Explain your response. Which point has the largest residual? Explain what the residual means in context. Is this point an outlier? An influential point? Explain. Do the data provide convincing evidence that there is a linear relationship between the amount of coffee consumed and the heart disease death rate? Carry out an appropriate test at a significance level of 0.05 to help answer this question. The following table consists of one student athlete’s time (in minutes) to swim 2000 yards and the student’s heart rate (beats per minute) after swimming on a random sample of 10 days: Swim Time Heart Rate 34.12 144 35.72 152 34.72 124 34.05 140 34.13 152 35.73 146 36.17 128 35.57 136 35.37 144 35.57 148 Enter the data into your calculator and make a scatter plot. Use your calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot from part a. Explain in words what the slope and y-intercept of the regression line tell us. How well does the regression line fit the data? Explain your response. Which point has the largest residual? Explain what the residual means in context. Is this point an outlier? An influential point? Explain. Check student’s solution. Check student’s solution. We have a slope of –1.4946 with a y -intercept of 193.88. The slope, in context, indicates that for each additional minute added to the swim time, the heart rate will decrease by 1.5 beats per minute. If the student is not swimming at all, the y -intercept indicates that his heart rate will be 193.88 beats per minute. While the slope has meaning (the longer it takes to swim 2,000 meters, the less effort the heart puts out), the y -intercept does not make sense. If the athlete is not swimming (resting), then his heart rate should be very low. Since only 1.5% of the heart rate variation is explained by this regression equation, we must conclude that this association is not explained with a linear relationship. The point (34.72, 124) generates the largest residual of –11.82. This means that our observed heart rate is almost 12 beats less than our predicted rate of 136 beats per minute. When this point is removed, the slope becomes –2.953 with the y-intercept changing to 247.1616. While the linear association is still very weak, we see that the removed data pair can be considered an influential point in the sense that the y -intercept becomes more meaningful. A researcher is investigating whether population impacts homicide rate. He uses demographic data from Detroit, MI to compare homicide rates and the number of the population that are White men. Population Size Homicide rate per 100,000 people 558,724 8.6 538,584 8.9 519,171 8.52 500,457 8.89 482,418 13.07 465,029 14.57 448,267 21.36 432,109 28.03 416,533 31.49 401,518 37.39 387,046 46.26 373,095 47.24 359,647 52.33 Use your calculator to construct a scatter plot of the data. What should the independent variable be? Why? Use your calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot. Discuss what the following mean in context. The slope of the regression equation The y-intercept of the regression equation The correlation r The coefficient of determination r2. Do the data provide convincing evidence that there is a linear relationship between population size and homicide rate? Carry out an appropriate test at a significance level of 0.05 to help answer this question. School Mid-Career Salary (in thousands) Yearly Tuition Princeton 137 57,410 Harvey Mudd 135 62,817 CalTech 127 60,864 US Naval Academy 122 0 West Point 120 0 MIT 118 57,986 Lehigh University 118 59,930 NYU-Poly 117 58,168 Babson College 117 56,576 Stanford 114 56,169 Using the data to determine the linear-regression line equation with the outliers removed. Is there a linear correlation for the data set with outliers removed? Justify your answer. If we remove the two service academies (the tuition is $0.00), we construct a new regression equation of y = 0.002 x + 0.309 with a correlation coefficient of 0.533 and a coefficient of determination of 0.284. There is not enough evidence to state that there is a strong linear correlation between tuition costs and future salaries, even when the service academies are removed. Bring It Together The average number of people in a family that attended college for various years is given in . Year Number of Family Members Attending College 1969 4.0 1973 3.6 1975 3.2 1979 3.0 1983 3.0 1988 3.0 1991 2.9 Using “year” as the independent variable and “Number of Family Members Attending College” as the dependent variable, draw a scatter plot of the data. Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx Find the correlation coefficient. Is it significant? Pick two years between 1969 and 1991 and find the estimated number of family members attending college. Based on the data in , is there a linear relationship between the year and the average number of family members attending college? Using the least-squares line, estimate the number of family members attending college for 1960 and 1995. Does the least-squares line give an accurate estimate for those years? Explain why or why not. Are there any outliers in the data? What is the estimated average number of family members attending college for 1986? Does the least squares line give an accurate estimate for that year? Explain why or why not. What is the slope of the least squares (best-fit) line? Interpret the slope. The percent of women who are wage and salary workers who are paid hourly rates is given in for a 14-year period. Year Percent of workers paid hourly rates 1 61.2 2 60.7 3 61.3 4 61.3 5 61.8 6 61.7 7 61.8 8 62.0 9 62.7 12 62.8 14 62.9 Using “year” as the independent variable and “percent” as the dependent variable, draw a scatter plot of the data. Does it appear from inspection that there is a relationship between the variables? Why or why not? Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx Find the correlation coefficient. Is it significant? Find the estimated percents for year 13 and year 10. Based on the data, is there a linear relationship between the year and the percent of female wage and salary earners who are paid hourly rates? Are there any outliers in the data? What is the estimated percent for the year 72? Does the least-squares line give an accurate estimate for that year? Explain why or why not. What is the slope of the least-squares (best-fit) line? Interpret the slope. Check student's solution. yes ŷ = −266.8863+0.1656x 0.9448; Yes 62.8233; 62.3265 yes no; (Year 9, 62.7) 72.5937; no slope = 0.1656. As the year increases by one, the percent of workers paid hourly rates tends to increase by 0.1656. Use the following information to answer the next two exercises. The cost of a leading liquid laundry detergent in different sizes is given in . Size (ounces) Cost ($) Cost per ounce 16 3.99 32 4.99 64 5.99 200 10.99 Using “size” as the independent variable and “cost” as the dependent variable, draw a scatter plot. Does it appear from inspection that there is a relationship between the variables? Why or why not? Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx Find the correlation coefficient. Is it significant? If the laundry detergent were sold in a 40-ounce size, find the estimated cost. If the laundry detergent were sold in a 90-ounce size, find the estimated cost. Does it appear that a line is the best way to fit the data? Why or why not? Are there any outliers in the given data? Is the least-squares line valid for predicting what a 300-ounce size of the laundry detergent would you cost? Why or why not? What is the slope of the least-squares (best-fit) line? Interpret the slope. Complete for the cost per ounce of the different sizes. Using “size” as the independent variable and “cost per ounce” as the dependent variable, draw a scatter plot of the data. Does it appear from inspection that there is a relationship between the variables? Why or why not? Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx Find the correlation coefficient. Is it significant? If the laundry detergent were sold in a 40-ounce size, find the estimated cost per ounce. If the laundry detergent were sold in a 90-ounce size, find the estimated cost per ounce. Does it appear that a line is the best way to fit the data? Why or why not? Are there any outliers in the the data? Is the least-squares line valid for predicting what a 300-ounce size of the laundry detergent would cost per ounce? Why or why not? What is the slope of the least-squares (best-fit) line? Interpret the slope. Size (ounces) Cost ($) cents/oz 16 3.99 24.94 32 4.99 15.59 64 5.99 9.36 200 10.99 5.50 Check student’s solution. There is a linear relationship for the sizes 16 through 64, but that linear trend does not continue to the 200-oz size. ŷ = 20.2368 – 0.0819x r = –0.8086 40-oz: 16.96 cents/oz 90-oz: 12.87 cents/oz The relationship is not linear; the least squares line is not appropriate. no outliers No, you would be extrapolating. The 300-oz size is outside the range of x . slope = –0.08194; for each additional ounce in size, the cost per ounce decreases by 0.082 cents. According to a flyer by a Prudential Insurance Company representative, the costs of approximate probate fees and taxes for selected net taxable estates are as follows: Net Taxable Estate ($) Approximate Probate Fees and Taxes ($) 600,000 30,000 750,000 92,500 1,000,000 203,000 1,500,000 438,000 2,000,000 688,000 2,500,000 1,037,000 3,000,000 1,350,000 Decide which variable should be the independent variable and which should be the dependent variable. Draw a scatter plot of the data. Does it appear from inspection that there is a relationship between the variables? Why or why not? Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx . Find the correlation coefficient. Is it significant? Find the estimated total cost for a next taxable estate of $1,000,000. Find the cost for $2,500,000. Does it appear that a line is the best way to fit the data? Why or why not? Are there any outliers in the data? Based on these results, what would be the probate fees and taxes for an estate that does not have any assets? What is the slope of the least-squares (best-fit) line? Interpret the slope. The following are advertised sale prices of hard disk drives on Amazon’s website for various capacity hard drives. Hard Disk Drive Capacity (in terabytes) Sale Price ($) 9 147 20 197 27 297 31 447 35 1177 40 2177 60 2497 Decide which variable should be the independent variable and which should be the dependent variable. Draw a scatter plot of the data. Does it appear from inspection that there is a relationship between the variables? Why or why not? Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx Find the correlation coefficient. Is it significant? Find the estimated sale price for a 32 terabyte hard drive. Find the cost for a 50 terabyte hard drive. Does it appear that a line is the best way to fit the data? Why or why not? Are there any outliers in the data? What is the slope of the least-squares (best-fit) line? Interpret the slope. Size is x , the independent variable, price is y , the dependent variable. Answers may vary. The relationship does not appear to be linear. ŷ = –745.252 + 54.75569 x r = 0.8944, yes it is significant 32-terabyte: $1006.93, 50-terabyte: $1992.53 No, the relationship does not appear to be linear. However, r is significant. no, the 60-terabyte hard drive For each additional terabyte, the price increases by $54.76 shows the average heights for American boys in a specific year. Age (years) Height (cm) birth 50.8 2 83.8 3 91.4 5 106.6 7 119.3 10 137.1 14 157.5 Decide which variable should be the independent variable and which should be the dependent variable. Draw a scatter plot of the data. Does it appear from inspection that there is a relationship between the variables? Why or why not? Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx Find the correlation coefficient. Is it significant? Find the estimated average height for a one-year-old. Find the estimated average height for an eleven-year-old. Does it appear that a line is the best way to fit the data? Why or why not? Are there any outliers in the data? Use the least squares line to estimate the average height for a sixty-two-year-old man. Do you think that your answer is reasonable? Why or why not? What is the slope of the least-squares (best-fit) line? Interpret the slope. State # letters in name Year entered the Union Ranks for entering the Union Area (square miles) Alabama 7 1819 22 52,423 Colorado 8 1876 38 104,100 Hawaii 6 1959 50 10,932 Iowa 4 1846 29 56,276 Maryland 8 1788 7 12,407 Missouri 8 1821 24 69,709 New Jersey 9 1787 3 8,722 Ohio 4 1803 17 44,828 South Carolina 13 1788 8 32,008 Utah 4 1896 45 84,904 Wisconsin 9 1848 30 65,499 We are interested in whether there is a relationship between the ranking of a state and the area of the state. What are the independent and dependent variables? What do you think the scatter plot will look like? Make a scatter plot of the data. Does it appear from inspection that there is a relationship between the variables? Why or why not? Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx Find the correlation coefficient. What does it imply about the significance of the relationship? Find the estimated areas for Alabama and for Colorado. Are they close to the actual areas? Use the two points in part f to plot the least-squares line on your graph from part b. Does it appear that a line is the best way to fit the data? Why or why not? Are there any outliers? Use the least squares line to estimate the area of a new state that enters the Union. Can the least-squares line be used to predict it? Why or why not? Delete “Hawaii” and substitute “Alaska” for it. Alaska is the forty-ninth, state with an area of 656,424 square miles. Calculate the new least-squares line. Find the estimated area for Alabama. Is it closer to the actual area with this new least-squares line or with the previous one that included Hawaii? Why do you think that’s the case? Do you think that, in general, newer states are larger than the original states? Let rank be the independent variable and area be the dependent variable. Answers may vary. There appears to be a linear relationship, with one outlier. ŷ (area) = 24177.06 + 1010.478 x r = 0.50047, r is not significant so there is no relationship between the variables. Alabama: 46407.576 Colorado: 62575.224 Alabama estimate is closer than Colorado estimate. If the outlier is removed, there is a linear relationship. There is one outlier (Hawaii). rank 51: 75711.4; no Alabama 7 1819 22 52,423 Colorado 8 1876 38 104,100 Hawaii 6 1959 50 10,932 Iowa 4 1846 29 56,276 Maryland 8 1788 7 12,407 Missouri 8 1821 24 69,709 New Jersey 9 1787 3 8,722 Ohio 4 1803 17 44,828 South Carolina 13 1788 8 32,008 Utah 4 1896 45 84,904 Wisconsin 9 1848 30 65,499 ŷ = –87065.3 + 7828.532x Alabama: 85,162.404; the prior estimate was closer. Alaska is an outlier. yes, with the exception of Hawaii Outlier an observation that does not fit the rest of the data

Math


Regression (Distance from School) Regression (Distance from School) Class Time: Names: Student Learning Outcomes The student will calculate and construct the line of best fit between two variables. The student will evaluate the relationship between two variables to determine if that relationship is significant. Collect the Data Use eight members of your class for the sample. Collect bivariate data (distance an individual lives from school, the cost of supplies for the current term). Complete the table. Distance from school Cost of supplies this term Which variable should be the dependent variable and which should be the independent variable? Why? Graph “distance” vs. “cost.” Plot the points on the graph. Label both axes with words. Scale both axes. Analyze the Data Enter your data into your calculator or computer. Write the linear equation, rounding to four decimal places. Calculate the following: a = ______ b = ______ correlation = ______ n = ______ equation: ŷ = ______ Is the correlation significant? Why or why not? (Answer in one to three complete sentences.) Supply an answer for the following senarios: For a person who lives eight miles from campus, predict the total cost of supplies this term: For a person who lives eighty miles from campus, predict the total cost of supplies this term: Obtain the graph on your calculator or computer. Sketch the regression line. Discussion Questions Answer each question in complete sentences. Does the line seem to fit the data? Why? What does the correlation imply about the relationship between the distance and the cost? Are there any outliers? If so, which point is an outlier? Should the outlier, if it exists, be removed? Why or why not?

Math


Regression (Textbook Cost) Regression (Textbook Cost) Class Time: Names: Student Learning Outcomes The student will calculate and construct the line of best fit between two variables. The student will evaluate the relationship between two variables to determine if that relationship is significant. Collect the Data Survey ten textbooks. Collect bivariate data (number of pages in a textbook, the cost of the textbook). Complete the table. Number of pages Cost of textbook Which variable should be the dependent variable and which should be the independent variable? Why? Graph “pages” vs. “cost.” Plot the points on the graph in Analyze the Data . Label both axes with words. Scale both axes. Analyze the Data Enter your data into your calculator or computer. Write the linear equation, rounding to four decimal places. Calculate the following: a = ______ b = ______ correlation = ______ n = ______ equation: y = ______ Is the correlation significant? Why or why not? (Answer in complete sentences.) Supply an answer for the following senarios: For a textbook with 400 pages, predict the cost. For a textbook with 600 pages, predict the cost. Obtain the graph on your calculator or computer. Sketch the regression line. Discussion Questions Answer each question in complete sentences. Does the line seem to fit the data? Why? What does the correlation imply about the relationship between the number of pages and the cost? Are there any outliers? If so, which point(s) is an outlier? Should the outlier, if it exists, be removed? Why or why not?

Math


Regression (Fuel Efficiency) Regression (Fuel Efficiency) Class Time: Names: Student Learning Outcomes The student will calculate and construct the line of best fit between two variables. The student will evaluate the relationship between two variables to determine if that relationship is significant. Collect the Data Find a reputable source that provides information on total fuel efficiency (in miles per gallon) and weight (in pounds) of new model cars with automatic transmissions. We will use this data to determine the relationship, if any, between the fuel efficiency of a car and its weight. Using your random number generator, randomly select 20 cars from the list and record their weights and fuel efficiency into . Weight Fuel Efficiency Which variable should be the dependent variable and which should be the independent variable? Why? By hand, do a scatterplot of “weight” vs. “fuel efficiency”. Plot the points on graph paper. Label both axes with words. Scale both axes accurately. Analyze the Data Enter your data into your calculator or computer. Write the linear equation, rounding to 4 decimal places. Calculate the following: a = ______ b = ______ correlation = ______ n = ______ equation: ŷ = ______ Obtain the graph of the regression line on your calculator. Sketch the regression line on the same axes as your scatter plot. Discussion Questions Is the correlation significant? Explain how you determined this in complete sentences. Is the relationship a positive one or a negative one? Explain how you can tell and what this means in terms of weight and fuel efficiency. In one or two complete sentences, what is the practical interpretation of the slope of the least squares line in terms of fuel efficiency and weight? For a car that weighs 4,000 pounds, predict its fuel efficiency. Include units. Can we predict the fuel efficiency of a car that weighs 10,000 pounds using the least squares line? Explain why or why not. Answer each question in complete sentences. Does the line seem to fit the data? Why or why not? What does the correlation imply about the relationship between fuel efficiency and weight of a car? Is this what you expected? Are there any outliers? If so, which point is an outlier?

Math


Introduction One-way ANOVA is used to measure information from several groups. (credit: modification of work “Magazine Stack” by thebittenword.com/ Flickr, CC BY 2.0) Chapter Objectives By the end of this chapter, the student should be able to: Interpret the F probability distribution as the number of groups and the sample size change. Discuss two uses for the F distribution: one-way ANOVA and the test of two variances. Conduct and interpret one-way ANOVA. Conduct and interpret hypothesis tests of two variances. Many statistical applications in psychology, social science, business administration, and the natural sciences involve several groups. For example, an environmentalist is interested in knowing if the average amount of pollution varies in several bodies of water. A sociologist is interested in knowing if the amount of income a person earns varies according to their upbringing. A consumer looking for a new car might compare the average gas mileage of several models. For hypothesis tests comparing averages between more than two groups, statisticians have developed a method called " Analysis of Variance " (abbreviated ANOVA). In this chapter, you will study the simplest form of ANOVA called single factor or one-way ANOVA . You will also study the F distribution, used for one-way ANOVA, and the test of two variances. This is just a very brief overview of one-way ANOVA. You will study this topic in much greater detail in future statistics courses. One-Way ANOVA, as it is presented here, relies heavily on a calculator or computer.

Math


One-Way ANOVA The purpose of a one-way ANOVA test is to determine the existence of a statistically significant difference among several group means. The test actually uses variances to help determine if the means are equal or not. In order to perform a one-way ANOVA test, there are five basic assumptions to be fulfilled: Each population from which a sample is taken is assumed to be normal. All samples are randomly selected and independent. The populations are assumed to have equal standard deviations (or variances) . The factor is a categorical variable. The response is a numerical variable. The Null and Alternative Hypotheses The null hypothesis is simply that all the group population means are the same. The alternative hypothesis is that at least one pair of means is different. For example, if there are k groups: H 0 : μ 1 = μ 2 = μ 3 = ... = μ k H a : At least two of the group means μ 1 , μ 2 , μ 3 , ..., μ k are not equal. That is, μ i ≠ μ j for some i ≠ j . The graphs, a set of box plots representing the distribution of values with the group means indicated by a horizontal line through the box, help in the understanding of the hypothesis test. In the first graph (blue box plots), H 0 : μ 1 = μ 2 = μ 3 and the three populations have the same distribution if the null hypothesis is true. The variance of the combined data is approximately the same as the variance of each of the populations. If the null hypothesis is false, then the variance of the combined data is larger which is caused by the different means as shown in the second graph (green box plots). (a) H 0 is true. All means are the same; the differences are due to random variation. (b) H 0 is not true. The means are not all the same; the differences are too large to be due to random variation. Chapter Review Analysis of variance extends the comparison of two groups to several, each a level of a categorical variable (factor). Samples from each group are independent, and must be randomly selected from normal populations with equal variances. We test the null hypothesis of equal means of the response in every group versus the alternative hypothesis of one or more group means being different from the others. A one-way ANOVA hypothesis test determines if several population means are equal. The distribution for the test is the F distribution with two different degrees of freedom. Assumptions: Each population from which a sample is taken is assumed to be normal. All samples are randomly selected and independent. The populations are assumed to have equal standard deviations (or variances). Use the following information to answer the next five exercises. There are five basic assumptions that must be fulfilled in order to perform a one-way ANOVA test. What are they? Write one assumption. Each population from which a sample is taken is assumed to be normal. Write another assumption. Write a third assumption. The populations are assumed to have equal standard deviations (or variances). Write a fourth assumption. Write the final assumption. The response is a numerical value. State the null hypothesis for a one-way ANOVA test if there are four groups. State the alternative hypothesis for a one-way ANOVA test if there are three groups. H a : At least two of the group means μ 1 , μ 2 , μ 3 are not equal. When do you use an ANOVA test? Homework Three different traffic routes are tested for mean driving time. The entries in the are the driving times in minutes on the three different routes. Route 1 Route 2 Route 3 30 27 16 32 29 41 27 28 22 35 36 31 State SS between , SS within , and the F statistic. SS between = 26 SS within = 441 F = 0.2653 Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses. Northeast South West Central East 16.3 16.9 16.4 16.2 17.1 16.1 16.5 16.5 16.6 17.2 16.4 16.4 16.6 16.5 16.6 16.5 16.2 16.1 16.4 16.8 x ¯ = s 2 = State the hypotheses. H 0 : ____________ H a : ____________ Analysis of Variance also referred to as ANOVA, is a method of testing whether or not the means of three or more populations are equal. The method is applicable if: all populations of interest are normally distributed. the populations have equal standard deviations. samples (not necessarily of the same size) are randomly and independently selected from each population. The test statistic for analysis of variance is the F -ratio. One-Way ANOVA a method of testing whether or not the means of three or more populations are equal; the method is applicable if: all populations of interest are normally distributed. the populations have equal standard deviations. samples (not necessarily of the same size) are randomly and independently selected from each population. there is one independent variable and one dependent variable. The test statistic for analysis of variance is the F -ratio. Variance mean of the squared deviations from the mean; the square of the standard deviation. For a set of data, a deviation can be represented as x – x ¯ where x is a value of the data and x ¯ is the sample mean. The sample variance is equal to the sum of the squares of the deviations divided by the difference of the sample size and one.

Math


The F Distribution and the F-Ratio The distribution used for the hypothesis test is a new one. It is called the F distribution , named after Sir Ronald Fisher, an English statistician. The F statistic is a ratio (a fraction). There are two sets of degrees of freedom; one for the numerator and one for the denominator. For example, if F follows an F distribution and the number of degrees of freedom for the numerator is four, and the number of degrees of freedom for the denominator is ten, then F ~ F 4,10 . NOTE The F distribution is derived from the Student's t-distribution. The values of the F distribution are squares of the corresponding values of the t -distribution. One-Way ANOVA expands the t -test for comparing more than two groups. The scope of that derivation is beyond the level of this course. It is preferable to use ANOVA when there are more than two groups instead of performing pairwise t -tests because performing multiple tests introduces the likelihood of making a Type 1 error. To calculate the F ratio , two estimates of the variance are made. Variance between samples : An estimate of σ 2 that is the variance of the sample means multiplied by n (when the sample sizes are the same.). If the samples are different sizes, the variance between samples is weighted to account for the different sample sizes. The variance is also called variation due to treatment or explained variation. Variance within samples : An estimate of σ 2 that is the average of the sample variances (also known as a pooled variance). When the sample sizes are different, the variance within samples is weighted. The variance is also called the variation due to error or unexplained variation. SS between = the sum of squares that represents the variation among the different samples SS within = the sum of squares that represents the variation within samples that is due to chance. To find a "sum of squares" means to add together squared quantities that, in some cases, may be weighted. We used sum of squares to calculate the sample variance and the sample standard deviation in Descriptive Statistics . MS means " mean square ." MS between is the variance between groups, and MS within is the variance within groups. Calculation of Sum of Squares and Mean Square k = the number of different groups n j = the size of the j th group s j = the sum of the values in the j th group n = total number of all the values combined (total sample size: ∑ n j ) x = one value: ∑ x = ∑ s j Sum of squares of all values from every group combined: ∑ x 2 Total sum of squares: S S total = ∑ x 2 ( ∑ x ) 2 n Explained variation: sum of squares representing variation among the different samples: SS between = ∑ [ ( s j ) 2 n j ] − ( ∑ s j ) 2 n Unexplained variation: sum of squares representing variation within samples due to chance: S S within = S S total – S S between df 's for different groups ( df 's for the numerator): d f between = k – 1 Equation for errors within samples ( df 's for the denominator): df within = n – k Mean square (variance estimate) explained by the different groups: MS between = S S between d f between Mean square (variance estimate) that is due to chance (unexplained): MS within = S S within d f within MS between and MS within can be written as follows: M S between = S S between d f between = S S between k − 1 M S w i t h i n = S S w i t h i n d f w i t h i n = S S w i t h i n n − k The one-way ANOVA test depends on the fact that MS between can be influenced by population differences among means of the several groups. Since MS within compares values of each group to its own group mean, the fact that group means might be different does not affect MS within . The null hypothesis says that all groups are samples from populations having the same normal distribution. The alternate hypothesis says that at least two of the sample groups come from populations with different normal distributions. If the null hypothesis is true, MS between and MS within should both estimate the same value. NOTE The null hypothesis says that all the group population means are equal. The hypothesis of equal means implies that the populations have the same normal distribution, because it is assumed that the populations are normal and that they have equal variances. F -Ratio or F Statistic F = M S between M S within If MS between and MS within estimate the same value (following the belief that H 0 is true), then the F -ratio should be approximately equal to one. Mostly, just sampling errors would contribute to variations away from one. As it turns out, MS between consists of the population variance plus a variance produced from the differences between the samples. MS within is an estimate of the population variance. Since variances are always positive, if the null hypothesis is false, MS between will generally be larger than MS within .Then the F -ratio will be larger than one. However, if the population effect is small, it is not unlikely that MS within will be larger in a given sample. The foregoing calculations were done with groups of different sizes. If the groups are the same size, the calculations simplify somewhat and the F -ratio can be written as: F -Ratio Formula when the groups are the same size F = n ⋅ s x ¯ 2 s 2 pooled where ... n = the sample size df numerator = k – 1 df denominator = n – k s 2 pooled = the mean of the sample variances (pooled variance) s x ¯ 2 = the variance of the sample means Data are typically put into a table for easy viewing. One-Way ANOVA results are often displayed in this manner by computer software. Source of Variation Sum of Squares ( SS ) Degrees of Freedom ( df ) Mean Square ( MS ) F Factor (Between) SS (Factor) k – 1 MS (Factor) = SS (Factor)/( k – 1) F = MS (Factor)/ MS (Error) Error (Within) SS (Error) n – k MS (Error) = SS (Error)/( n – k ) Total SS (Total) n – 1 Three different diet plans are to be tested for mean weight loss. The entries in the table are the weight losses for the different plans. The one-way ANOVA results are shown in . Plan 1: n 1 = 4 Plan 2: n 2 = 3 Plan 3: n 3 = 3 5 3.5 8 4.5 7 4 4 3.5 3 4.5 s 1 = 16.5, s 2 =15, s 3 = 15.5 Following are the calculations needed to fill in the one-way ANOVA table. The table is used to conduct a hypothesis test. S S ( b e t w e e n ) = ∑ [ ( s j ) 2 n j ] − ( ∑ s j ) 2 n = s 1 2 4 + s 2 2 3 + s 3 2 3 − ( s 1 + s 2 + s 3 ) 2 10 where n 1 = 4, n 2 = 3, n 3 = 3 and n = n 1 + n 2 + n 3 = 10 = ( 16.5 ) 2 4 + ( 15 ) 2 3 + ( 15.5 ) 2 3 − ( 16.5 + 15 + 15.5 ) 2 10 S S ( b e t w e e n ) = 2.2458 S ( t o t a l ) = ∑ x 2 − ( ∑ x ) 2 n = ( 5 2 + 4.5 2 + 4 2 + 3 2 + 3.5 2 + 7 2 + 4.5 2 + 8 2 + 4 2 + 3.5 2 ) − ( 5 + 4.5 + 4 + 3 + 3.5 + 7 + 4.5 + 8 + 4 + 3.5 ) 2 10 = 244 − 47 2 10 = 244 − 220.9 S S ( t o t a l ) = 23.1 S S ( w i t h i n ) = S S ( t o t a l ) − S S ( b e t w e e n ) = 23.1 − 2.2458 S S ( w i t h i n ) = 20.8542 One-Way ANOVA Table: The formulas for SS (Total) , SS (Factor) = SS (Between) and SS (Error) = SS (Within) as shown previously. The same information is provided by the TI calculator hypothesis test function ANOVA in STAT TESTS (syntax is ANOVA(L1, L2, L3) where L1, L2, L3 have the data from Plan 1, Plan 2, Plan 3 respectively). Source of Variation Sum of Squares ( SS ) Degrees of Freedom ( df ) Mean Square ( MS ) F Factor (Between) SS (Factor) = SS (Between) = 2.2458 k – 1 = 3 groups – 1 = 2 MS (Factor) = SS (Factor)/( k – 1) = 2.2458/2 = 1.1229 F = MS (Factor)/ MS (Error) = 1.1229/2.9792 = 0.3769 Error (Within) SS (Error) = SS (Within) = 20.8542 n – k = 10 total data – 3 groups = 7 MS (Error) = SS (Error)/( n – k ) = 20.8542/7 = 2.9792 Total SS (Total) = 2.2458 + 20.8542 = 23.1 n – 1 = 10 total data – 1 = 9 Try It As part of an experiment to see how different types of soil cover would affect slicing tomato production, Marist College students grew tomato plants under different soil cover conditions. Groups of three plants each had one of the following treatments bare soil a commercial ground cover black plastic straw compost All plants grew under the same conditions and were the same variety. Students recorded the weight (in grams) of tomatoes produced by each of the n = 15 plants: Bare: n 1 = 3 Ground Cover: n 2 = 3 Plastic: n 3 = 3 Straw: n 4 = 3 Compost: n 5 = 3 2,625 5,348 6,583 7,285 6,277 2,997 5,682 8,560 6,897 7,818 4,915 5,482 3,830 9,230 8,677 Create the one-way ANOVA table. The one-way ANOVA hypothesis test is always right-tailed because larger F -values are way out in the right tail of the F -distribution curve and tend to make us reject H 0 . Notation The notation for the F distribution is F ~ F df ( num ), df ( denom ) where df ( num ) = df between and df ( denom ) = df within The mean for the F distribution is μ = d f ( d e n o m ) d f ( d e n o m ) – 2 References Tomato Data, Marist College School of Science (unpublished student research) Chapter Review Analysis of variance compares the means of a response variable for several groups. ANOVA compares the variation within each group to the variation of the mean of each group. The ratio of these two is the F statistic from an F distribution with (number of groups – 1) as the numerator degrees of freedom and (number of observations – number of groups) as the denominator degrees of freedom. These statistics are summarized in the ANOVA table. Formula Review S S between = ∑  [ ( s j ) 2 n j ] − ( ∑  s j ) 2 n S S total = ∑  x 2 − ( ∑  x ) 2 n S S within = S S total − S S between df between = df ( num ) = k – 1 df within = df(denom) = n – k MS between = S S between d f between MS within = S S within d f within F = M S between M S within F ratio when the groups are the same size: F = n s x ¯ 2 s 2 p o o l e d Mean of the F distribution: µ = d f ( n u m ) d f ( d e n o m ) − 2 where: k = the number of groups n j = the size of the j th group s j = the sum of the values in the j th group n = the total number of all values (observations) combined x = one value (one observation) from the data s x ¯ 2 = the variance of the sample means s 2 p o o l e d = the mean of the sample variances (pooled variance) Use the following information to answer the next eight exercises. Groups of men from three different areas of the country are to be tested for mean weight. The entries in are the weights for the different groups. Group 1 Group 2 Group 3 216 202 170 198 213 165 240 284 182 187 228 197 176 210 201 What is the Sum of Squares Factor? 4,939.2 What is the Sum of Squares Error? What is the df for the numerator? 2 What is the df for the denominator? What is the Mean Square Factor? 2,469.6 What is the Mean Square Error? What is the F statistic? 3.7416 Use the following information to answer the next eight exercises. Girls from four different soccer teams are to be tested for mean goals scored per game. The entries in the table are the goals per game for the different teams. The one-way ANOVA results are shown in . Team 1 Team 2 Team 3 Team 4 1 2 0 3 2 3 1 4 0 2 1 4 3 4 0 3 2 4 0 2 What is SS between ? What is the df for the numerator? 3 What is MS between ? What is SS within ? 13.2 What is the df for the denominator? What is MS within ? 0.825 What is the F statistic? Judging by the F statistic, do you think it is likely or unlikely that you will reject the null hypothesis? Because a one-way ANOVA test is always right-tailed, a high F statistic corresponds to a low p -value, so it is likely that we will reject the null hypothesis. Homework Use the following information to answer the next three exercises. Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses. Northeast South West Central East 16.3 16.9 16.4 16.2 17.1 16.1 16.5 16.5 16.6 17.2 16.4 16.4 16.6 16.5 16.6 16.5 16.2 16.1 16.4 16.8 x ¯ = s 2 = H 0 : µ 1 = µ 2 = µ 3 = µ 4 = µ 5 Hα : At least any two of the group means µ 1 , µ 2 , …, µ 5 are not equal. degrees of freedom – numerator: df ( num ) = _________ degrees of freedom – denominator: df ( denom ) = ________ df ( denom ) = 15 F statistic = ________

Math


Facts About the F Distribution Here are some facts about the F distribution. The curve is not symmetrical but skewed to the right. There is a different curve for each set of df s. The F statistic is greater than or equal to zero. As the degrees of freedom for the numerator and for the denominator get larger, the curve approximates the normal. Other uses for the F distribution include comparing two variances and two-way Analysis of Variance. Two-Way Analysis is beyond the scope of this chapter. Let’s return to the slicing tomato exercise in . The means of the tomato yields under the five mulching conditions are represented by μ 1 , μ 2 , μ 3 , μ 4 , μ 5 . We will conduct a hypothesis test to determine if all means are the same or at least one is different. Using a significance level of 5%, test the null hypothesis that there is no difference in mean yields among the five groups against the alternative hypothesis that at least one mean is different from the rest. The null and alternative hypotheses are: H 0 : μ 1 = μ 2 = μ 3 = μ 4 = μ 5 H a : μ i ≠ μ j some i ≠ j The one-way ANOVA results are shown in Source of Variation Sum of Squares ( SS ) Degrees of Freedom ( df ) Mean Square ( MS ) F Factor (Between) 36,648,561 5 – 1 = 4 36,648,561 4 = 9,162,140 9,162,140 2,044,672 .6 = 4 .4810 Error (Within) 20,446,726 15 – 5 = 10 20,446,726 10 = 2,044,672 .6 Total 57,095,287 15 – 1 = 14 Distribution for the test: F 4,10 df ( num ) = 5 – 1 = 4 df ( denom ) = 15 – 5 = 10 Test statistic: F = 4.4810 Probability Statement: p -value = P ( F > 4.481) = 0.0248. Compare α and the p -value: α = 0.05, p -value = 0.0248 Make a decision: Since α > p -value, we reject H 0 . Conclusion: At the 5% significance level, we have reasonably strong evidence that differences in mean yields for slicing tomato plants grown under different mulching conditions are unlikely to be due to chance alone. We may conclude that at least some of mulches led to different mean yields. To find these results on the calculator: Press STAT. Press 1:EDIT . Put the data into the lists L 1 , L 2 , L 3 , L 4 , L 5 . Press STAT, and arrow over to TESTS, and arrow down to ANOVA. Press ENTER, and then enter L 1 , L 2 , L 3 , L 4 , L 5 ). Press ENTER. You will see that the values in the foregoing ANOVA table are easily produced by the calculator, including the test statistic and the p -value of the test. The calculator displays: F = 4.4810 p = 0.0248 ( p -value) Factor df = 4 SS = 36648560.9 MS = 9162140.23 Error df = 10 SS = 20446726 MS = 2044672.6 Try It There are multiple variants of the virus that causes COVID-19. The length of hospital stays for patients afflicted with various strains of COVID-19 is shown in . Delta Strain Omicron Strain Alpha Strain Gamma Strain Beta Strain 13.9 11.7 18.2 16.9 9.3 14.9 15.1 14.6 12.8 15.8 16.8 9.9 10.1 11.2 16.4 Test whether the mean length of hospital stay is the same or different for the various strains of COVID-19. Construct the ANOVA table, find the p -value, and state your conclusion. Use a 5% significance level. We test for equality of mean length of hospital stay: H 0 : μ 1 = μ 2 = μ 3 = μ 4 = μ 5 H a : μ i ≠ μ j , some i ≠ j The one-way ANOVA table results are shown: Source of Variation Sum of Squares (SS) Degrees of Freedom (df) Mean Square (MS) F P-value Factor (Between) 14.056 5 – 1 = 4 3.514 0.353095 0.8362 Error (Within) 99.52 15 – 5 = 10 9.952 Total 113.576 15 – 1 = 14 Distribution for the test: F 4 , 10 Probability Statement: p - v a l u e = P ( F > 0 . 3531 ) = 0 . 8362 Compare  and the p -value: α = 0 . 05 , p - v a l u e = 08362 , α > p - v a l u e Make a decision: Since  > p -value, we do not reject H 0 . Conclusion: At the 5% significance level, there is insufficient evidence from these data that different strains of COVID-19 virus will cause a significant difference in the length of a hospital stay. Four sororities took a random sample of sisters regarding their grade means for the past term. The results are shown in . MEAN GRADES FOR FOUR SORORITIES Sorority 1 Sorority 2 Sorority 3 Sorority 4 2.17 2.63 2.63 3.79 1.85 1.77 3.78 3.45 2.83 3.25 4.00 3.08 1.69 1.86 2.55 2.26 3.33 2.21 2.45 3.18 Using a significance level of 1%, is there a difference in mean grades among the sororities? Let μ 1 , μ 2 , μ 3 , μ 4 be the population means of the sororities. Remember that the null hypothesis claims that the sorority groups are from the same normal distribution. The alternate hypothesis says that at least two of the sorority groups come from populations with different normal distributions. Notice that the four sample sizes are each five. NOTE This is an example of a balanced design , because each factor (i.e., sorority) has the same number of observations. H 0 : μ 1 = μ 2 = μ 3 = μ 4 H a : Not all of the means μ 1 , μ 2 , μ 3 , μ 4 are equal. Distribution for the test: F 3,16 where k = 4 groups and n = 20 samples in total df ( num )= k – 1 = 4 – 1 = 3 df ( denom ) = n – k = 20 – 4 = 16 Calculate the test statistic: F = 2.23 Graph: Probability statement: p -value = P ( F > 2.23) = 0.1241 Compare α and the p -value: α = 0.01 p -value = 0.1241 α < p -value Make a decision: Since α < p -value, you cannot reject H 0 . Conclusion: There is not sufficient evidence to conclude that there is a difference among the mean grades for the sororities. Put the data into lists L 1 , L 2 , L 3 , and L 4 . Press STAT and arrow over to TESTS . Arrow down to F:ANOVA . Press ENTER and Enter ( L1,L2,L3,L4 ). The calculator displays the F statistic, the p -value and the values for the one-way ANOVA table: F = 2.2303 p = 0.1241 ( p -value) Factor df = 3 SS = 2.88732 MS = 0.96244 Error df = 16 SS = 6.9044 MS = 0.431525 Try It Four sports teams took a random sample of players regarding their GPAs for the last year. The results are shown in . GPAs FOR FOUR SPORTS TEAMS Basketball Baseball Hockey Lacrosse 3.6 2.1 4.0 2.0 2.9 2.6 2.0 3.6 2.5 3.9 2.6 3.9 3.3 3.1 3.2 2.7 3.8 3.4 3.2 2.5 Use a significance level of 5%, and determine if there is a difference in GPA among the teams. A fourth grade class is studying the environment. One of the assignments is to grow bean plants in different soils. Tommy chose to grow his bean plants in soil found outside his classroom mixed with dryer lint. Tara chose to grow her bean plants in potting soil bought at the local nursery. Nick chose to grow his bean plants in soil from his mother's garden. No chemicals were used on the plants, only water. They were grown inside the classroom next to a large window. Each child grew five plants. At the end of the growing period, each plant was measured, producing the data (in inches) in . Tommy's Plants Tara's Plants Nick's Plants 24 25 23 21 31 27 23 23 22 30 20 30 23 28 20 Does it appear that the three media in which the bean plants were grown produce the same mean height? Test at a 3% level of significance. This time, we will perform the calculations that lead to the F' statistic. Notice that each group has the same number of plants, so we will use the formula F' = n ⋅ s x ¯ 2 s 2 pooled . First, calculate the sample mean and sample variance of each group. Tommy's Plants Tara's Plants Nick's Plants Sample Mean 24.2 25.4 24.4 Sample Variance 11.7 18.3 16.3 Next, calculate the variance of the three group means (Calculate the variance of 24.2, 25.4, and 24.4). Variance of the group means = 0.413 = s x ¯ 2 Then MS between = n s x ¯ 2 = (5)(0.413) where n = 5 is the sample size (number of plants each child grew). Calculate the mean of the three sample variances (Calculate the mean of 11.7, 18.3, and 16.3). Mean of the sample variances = 15.433 = s 2 pooled Then MS within = s 2 pooled = 15.433. The F statistic (or F ratio) is F = M S between M S within = n s x ¯ 2 s 2 p o o l e d = ( 5 ) ( 0.413 ) 15.433 = 0.134 The dfs for the numerator = the number of groups – 1 = 3 – 1 = 2. The dfs for the denominator = the total number of samples – the number of groups = 15 – 3 = 12 The distribution for the test is F 2,12 and the F statistic is F = 0.134 The p -value is P ( F > 0.134) = 0.8759. Decision: Since α = 0.03 and the p -value = 0.8759, do not reject H 0 . (Why?) Conclusion: With a 3% level of significance, from the sample data, the evidence is not sufficient to conclude that the mean heights of the bean plants are different. To calculate the p -value: *Press 2nd DISTR *Arrow down to Fcdf (and press ENTER . *Enter 0.134, E99 , 2, 12) *Press ENTER The p -value is 0.8759. Try It Another fourth grader also grew bean plants, but this time in a jelly-like mass. The heights were (in inches) 24, 28, 25, 30, and 32. Do a one-way ANOVA test on the four groups. Are the heights of the bean plants different? Use the same method as shown in . From the class, create four groups of the same size as follows: men under 22, men at least 22, women under 22, women at least 22. Have each member of each group record the number of states in the United States they have visited. Run an ANOVA test to determine if the average number of states visited in the four groups are the same. Test at a 1% level of significance. Use one of the solution sheets in . References Data from a fourth grade classroom in 1994 in a private K – 12 school in San Jose, CA. Hand, D.J., F. Daly, A.D. Lunn, K.J. McConway, and E. Ostrowski. A Handbook of Small Datasets: Data for Fruitfly Fecundity. London: Chapman & Hall, 1994. Hand, D.J., F. Daly, A.D. Lunn, K.J. McConway, and E. Ostrowski. A Handbook of Small Datasets. London: Chapman & Hall, 1994, pg. 50. Hand, D.J., F. Daly, A.D. Lunn, K.J. McConway, and E. Ostrowski. A Handbook of Small Datasets. London: Chapman & Hall, 1994, pg. 118. “MLB Standings – 2012.” Available online at http://espn.go.com/mlb/standings/_/year/2012. Mackowiak, P. A., Wasserman, S. S., and Levine, M. M. (1992), "A Critical Appraisal of 98.6 Degrees F, the Upper Limit of the Normal Body Temperature, and Other Legacies of Carl Reinhold August Wunderlich," Journal of the American Medical Association , 268, 1578-1580. Chapter Review The graph of the F distribution is always positive and skewed right, though the shape can be mounded or exponential depending on the combination of numerator and denominator degrees of freedom. The F statistic is the ratio of a measure of the variation in the group means to a similar measure of the variation within the groups. If the null hypothesis is correct, then the numerator should be small compared to the denominator. A small F statistic will result, and the area under the F curve to the right will be large, representing a large p -value. When the null hypothesis of equal group means is incorrect, then the numerator should be large compared to the denominator, giving a large F statistic and a small area (small p -value) to the right of the statistic under the F curve. When the data have unequal group sizes (unbalanced data), then techniques from need to be used for hand calculations. In the case of balanced data (the groups are the same size) however, simplified calculations based on group means and variances may be used. In practice, of course, software is usually employed in the analysis. As in any analysis, graphs of various sorts should be used in conjunction with numerical techniques. Always look of your data! An F statistic can have what values? What happens to the curves as the degrees of freedom for the numerator and the denominator get larger? The curves approximate the normal distribution. Use the following information to answer the next seven exercise. Four basketball teams took a random sample of players regarding how high each player can jump (in inches). The results are shown in . Team 1 Team 2 Team 3 Team 4 Team 5 36 32 48 38 41 42 35 50 44 39 51 38 39 46 40 What is the df(num) ? What is the df(denom) ? ten What are the Sum of Squares and Mean Squares Factors? What are the Sum of Squares and Mean Squares Errors? SS = 237.33; MS = 23.73 What is the F statistic? What is the p -value? 0.1614 At the 5% significance level, is there a difference in the mean jump heights among the teams? Use the following information to answer the next seven exercises. A video game developer is testing a new game on three different groups. Each group represents a different target market for the game. The developer collects scores from a random sample from each group. The results are shown in Group A Group B Group C 101 151 101 108 149 109 98 160 198 107 112 186 111 126 160 What is the df(num) ? two What is the df(denom) ? What are the SS between and MS between ? SS = 5,700.4; MS = 2,850.2 What are the SS within and MS within ? What is the F Statistic? 3.6101 What is the p -value? At the 10% significance level, are the scores among the different groups different? Yes, there is enough evidence to show that the scores among the groups are statistically significant at the 10% level. Use the following information to answer the next three exercises. Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses. Northeast South West Central East 16.3 16.9 16.4 16.2 17.1 16.1 16.5 16.5 16.6 17.2 16.4 16.4 16.6 16.5 16.6 16.5 16.2 16.1 16.4 16.8 x ¯ = s 2 = Enter the data into your calculator or computer. p -value = ______ State the decisions and conclusions (in complete sentences) for the following preconceived levels of α . α = 0.05 a. Decision: ____________________________ b. Conclusion: ____________________________ α = 0.01 a. Decision: ____________________________ b. Conclusion: ____________________________ Homework DIRECTIONS Use a solution sheet to conduct the following hypothesis tests. The solution sheet can be found in . Three students, Linda, Tuan, and Javier, are given five laboratory rats each for a nutritional experiment. Each rat's weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a specified time period, each rat is weighed again, and the net gain in grams is recorded. Using a significance level of 10%, test the hypothesis that the three formulas produce the same mean weight gain. Weights of Student Lab Rats Linda's rats Tuan's rats Javier's rats 43.5 47.0 51.2 39.4 40.5 40.9 41.3 38.9 37.9 46.0 46.3 45.0 38.2 44.2 48.6 H 0 : µ L = µ T = µ J H a : at least any two of the means are different df ( num ) = 2; df ( denom ) = 12 F distribution 0.67 0.5305 Answers may vary. Decision: Do not reject null hypothesis; Conclusion: There is insufficient evidence to conclude that the means are different. A grassroots group opposed to a proposed increase in the gas tax claimed that the increase would hurt working-class people the most, since they commute the farthest to work. Suppose that the group randomly surveyed 24 individuals and asked them their daily one-way commuting mileage. The results are in . Using a 5% significance level, test the hypothesis that the three mean commuting mileages are the same. working-class professional (middle incomes) professional (wealthy) 17.8 16.5 8.5 26.7 17.4 6.3 49.4 22.0 4.6 9.4 7.4 12.6 65.4 9.4 11.0 47.1 2.1 28.6 19.5 6.4 15.4 51.2 13.9 9.3 Use the following information to answer the next two exercises. lists the number of pages in four different types of magazines. home decorating news health computer 172 87 82 104 286 94 153 136 163 123 87 98 205 106 103 207 197 101 96 146 Using a significance level of 5%, test the hypothesis that the four magazine types have the same mean length. Eliminate one magazine type that you now feel has a mean length different from the others. Redo the hypothesis test, testing that the remaining three means are statistically the same. Use a new solution sheet. Based on this test, are the mean lengths for the remaining three magazines statistically the same? H a : µ c = µ n = µ h At least any two of the magazines have different mean lengths. df ( num ) = 2, df ( denom ) = 12 F distribtuion F = 15.28 p -value = 0.0005 Answers may vary. Alpha: 0.05 Decision: Reject the Null Hypothesis. Reason for decision: p -value < alpha Conclusion: There is sufficient evidence to conclude that the mean lengths of the magazines are different. A researcher wants to know if the mean times (in minutes) that people watch their favorite news station are the same. Suppose that shows the results of a study. CNN FOX Local 45 15 72 12 43 37 18 68 56 38 50 60 23 31 51 35 22 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. Are the means for the final exams the same for all statistics class delivery types? shows the scores on final exams from several randomly selected classes that used the different delivery types. Online Hybrid Face-to-Face 72 83 80 84 73 78 77 84 84 80 81 81 81 86 79 82 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. H 0 : μ o = μ h = μ f At least two of the means are different. df ( n ) = 2, df ( d ) = 13 F 2,13 0.64 0.5437 Answers may vary. Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for decision: p -value > alpha Conclusion: The mean scores of different class delivery are not different. Are the mean number of times a month a person eats out the same for White, Black, Hispanic/Latino and Asian people? Suppose that shows the results of a study. White Black Hispanic/Latino Asian 6 4 7 8 8 1 3 3 2 5 5 5 4 2 4 1 6 6 7 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. Are the mean numbers of daily visitors to a ski resort the same for the three types of snow conditions? Suppose that shows the results of a study. Powder Machine Made Hard Packed 1,210 2,107 2,846 1,080 1,149 1,638 1,537 862 2,019 941 1,870 1,178 1,528 2,233 1,382 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. H 0 : μ p = μ m = μ h At least any two of the means are different. df ( n ) = 2, df ( d ) = 12 F 2,12 3.13 0.0807 Answers may vary. Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for decision: p -value > alpha Conclusion: There is not sufficient evidence to conclude that the mean numbers of daily visitors are different. Sanjay made identical paper airplanes out of three different weights of paper, light, medium and heavy. He made four airplanes from each of the weights, and launched them himself across the room. Here are the distances (in meters) that his planes flew. Paper Type/Trial Trial 1 Trial 2 Trial 3 Trial 4 Heavy 5.1 meters 3.1 meters 4.7 meters 5.3 meters Medium 4 meters 3.5 meters 4.5 meters 6.1 meters Light 3.1 meters 3.3 meters 2.1 meters 1.9 meters Take a look at the data in the graph. Look at the spread of data for each group (light, medium, heavy). Does it seem reasonable to assume a normal distribution with the same variance for each group? Yes or No. Why is this a balanced design? Calculate the sample mean and sample standard deviation for each group. Does the weight of the paper have an effect on how far the plane will travel? Use a 1% level of significance. Complete the test using the method shown in the bean plant example in . variance of the group means __________ MS between = ___________ mean of the three sample variances ___________ MS within = _____________ F statistic = ____________ df(num) = __________, df(denom) = ___________ number of groups _______ number of observations _______ p -value = __________ ( P ( F > _______) = __________) Graph the p -value. decision: _______________________ conclusion: _______________________________________________________________ DDT is a pesticide that has been banned from use in the United States and most other areas of the world. It is quite effective, but persisted in the environment and over time became seen as harmful to higher-level organisms. Famously, egg shells of eagles and other raptors were believed to be thinner and prone to breakage in the nest because of ingestion of DDT in the food chain of the birds. An experiment was conducted on the number of eggs (fecundity) laid by female fruit flies. There are three groups of flies. One group was bred to be resistant to DDT (the RS group). Another was bred to be especially susceptible to DDT (SS). Finally there was a control line of non-selected or typical fruitflies (NS). Here are the data: RS SS NS RS SS NS 12.8 38.4 35.4 22.4 23.1 22.6 21.6 32.9 27.4 27.5 29.4 40.4 14.8 48.5 19.3 20.3 16 34.4 23.1 20.9 41.8 38.7 20.1 30.4 34.6 11.6 20.3 26.4 23.3 14.9 19.7 22.3 37.6 23.7 22.9 51.8 22.6 30.2 36.9 26.1 22.5 33.8 29.6 33.4 37.3 29.5 15.1 37.9 16.4 26.7 28.2 38.6 31 29.5 20.3 39 23.4 44.4 16.9 42.4 29.3 12.8 33.7 23.2 16.1 36.6 14.9 14.6 29.2 23.6 10.8 47.4 27.3 12.2 41.7 The values are the average number of eggs laid daily for each of 75 flies (25 in each group) over the first 14 days of their lives. Using a 1% level of significance, are the mean rates of egg selection for the three strains of fruitfly different? If so, in what way? Specifically, the researchers were interested in whether or not the selectively bred strains were different from the nonselected line, and whether the two selected lines were different from each other. Here is a chart of the three groups: The data appear normally distributed from the chart and of similar spread. There do not appear to be any serious outliers, so we may proceed with our ANOVA calculations, to see if we have good evidence of a difference between the three groups. H 0 : μ 1 = μ 2 = μ 3 ; H a : μ i ≠ μ j some i ≠ j . Define μ 1 , μ 2 , μ 3 , as the population mean number of eggs laid by the three groups of fruit flies. F statistic = 8.6657; p -value = 0.0004 Decision: Since the p -value is less than the level of significance of 0.01, we reject the null hypothesis. Conclusion: We have good evidence that the average number of eggs laid during the first 14 days of life for these three strains of fruitflies are different. Interestingly, if you perform a two sample t -test to compare the RS and NS groups they are significantly different ( p = 0.0013). Similarly, SS and NS are significantly different ( p = 0.0006). However, the two selected groups, RS and SS are not significantly different ( p = 0.5176). Thus we appear to have good evidence that selection either for resistance or for susceptibility involves a reduced rate of egg production (for these specific strains) as compared to flies that were not selected for resistance or susceptibility to DDT. Here, genetic selection has apparently involved a loss of fecundity. The data shown is the recorded body temperatures of 130 subjects as estimated from available histograms. Traditionally we are taught that the normal human body temperature is 98.6 F. This is not quite correct for everyone. Are the mean temperatures among the four groups different? Calculate 95% confidence intervals for the mean body temperature in each group and comment about the confidence intervals. FL FH ML MH FL FH ML MH 96.4 96.8 96.3 96.9 98.4 98.6 98.1 98.6 96.7 97.7 96.7 97 98.7 98.6 98.1 98.6 97.2 97.8 97.1 97.1 98.7 98.6 98.2 98.7 97.2 97.9 97.2 97.1 98.7 98.7 98.2 98.8 97.4 98 97.3 97.4 98.7 98.7 98.2 98.8 97.6 98 97.4 97.5 98.8 98.8 98.2 98.8 97.7 98 97.4 97.6 98.8 98.8 98.3 98.9 97.8 98 97.4 97.7 98.8 98.8 98.4 99 97.8 98.1 97.5 97.8 98.8 98.9 98.4 99 97.9 98.3 97.6 97.9 99.2 99 98.5 99 97.9 98.3 97.6 98 99.3 99 98.5 99.2 98 98.3 97.8 98 99.1 98.6 99.5 98.2 98.4 97.8 98 99.1 98.6 98.2 98.4 97.8 98.3 99.2 98.7 98.2 98.4 97.9 98.4 99.4 99.1 98.2 98.4 98 98.4 99.9 99.3 98.2 98.5 98 98.6 100 99.4 98.2 98.6 98 98.6 100.8

Math