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- Olympic Pole Vault The graph in Figure 7 indicates that in recent years the winning Olympic men’s pole vault height has fallen below the value predicted by the regression line in Example 2. This might have occurred because when the pole vault was a new event there was much room for improvement in vaulters’ performances, whereas now even the best training can produce only incremental advances. Let’s see whether concentrating on more recent results gives a better predictor of future records. (a) Use the data in Table 2 (page 176) to complete the table of winning pole vault heights shown in the margin. (Note that we are using x=0 to correspond to the year 1972, where this restricted data set begins.) (b) Find the regression line for the data in part ‚(a). (c) Plot the data and the regression line on the same axes. Does the regression line seem to provide a good model for the data? (d) What does the regression line predict as the winning pole vault height for the 2012 Olympics? Compare this predicted value to the actual 2012 winning height of 5.97 m, as described on page 177. Has this new regression line provided a better prediction than the line in Example 2?The following fictitious table shows kryptonite price, in dollar per gram, t years after 2006. t= Years since 2006 0 1 2 3 4 5 6 7 8 9 10 K= Price 56 51 50 55 58 52 45 43 44 48 51 Make a quartic model of these data. Round the regression parameters to two decimal places.Which of the following statements is NOT true? a.If the correlation coefficent, rr, is negative, then the slope of the regression line must always be negative. b.For variables x and y, the value of the correlation coefficent will not change c.if the units of x and y are changed, (i.e., if both x and y are changed from say inches to meters) d.If r≠0r≠0 , then a significant linear correlation will exist between the two variables e.The value of the correlation coefficient must be within the interval: −1≤r≤1
- If other factors are held constant and the Pearson correlation value between X and Y is r = 0.80, then the regression equation will tend to produce more accurate predictions than would be obtained if the Pearson correlation value was r = 0.60. True or FalseA professor obtains SAT scores and freshman grade point averages (GPAs) for a group of n = 15 college students. The SAT scores have a mean of M = 580 with SS = 22,400, and the GPAs have a mean of 3.10 with SS = 1.26, and SP = 84. A) Find the regression equation for predicting GPA from SAT scores. B) What percentage of the variance in GPAs is accounted for by the regression equation (i.e., compute the correlation, r, then find r2)? C) Does the regression equation account for a significant portion of the variance in GPA? Use a = .05 to evaluate the F-ratio.2. The following data, adapted from Montgomery, Peck, and Vining (2001), present the number of certified mental defectives per 10,000 of estimated population in the United Kingdom ( y) and the number of radio receiver licenses issued (x) by the BBC (in millions) for the years 1924 through 1937. Fit a regression model relating y and x. Comment on the model. Specifically, does the existence of a strong correlation imply a cause-and-effect relationship?
- A set of n = 15 pairs of X and Y values has a correlation of r = +0.80 with SSY = 75, and the regression equation for predicting Y is computed. Find the standard error of estimate for the regression equation. How big would the standard error be if the sample size were n = 30.If a sample of 25 pairs of data yields a correlation coefficient, r, of 0.390 and the scatterplot displays a linear trend, can you use the regression equation to make predictions, assuming your x-values are within the domain of the data set? Choose your answer from the multiple choice answers below A.) Yes, because rcrit = 0.396 and the regression coefficient, r, is less than this value. B.) Yes, because rcrit = 0.381 and the regression coefficient, r, is greater than this value. C.) No, because rcrit = 0.381 and the regression coefficient, r, is greater than this value. D.) No, because rcrit = 0.396 and the regression coefficient, r, is less than this value.A sample of n = 15 pairs of X and Y scores produces a Pearson correlation of r = 0.45, SSY = 90. a) If the regression equation was found for these scores, how much of the Y variability would be predicted by the regression equation (SSregression) and how much would not be predicted (SSresidual)? b) Does the regression equation predict a significant portion of the variability for the Y scores? (Equivalently, is the Pearson correlation significant?)
- A professor obtains SAT scores and freshman grade point averages (GPAs) for a group of n=15 college students. The SAT scores have a mean of M=580 with SS = 22,400, and the GPAs have a mean of 3.10 with SS = 1.26, and SP = 84. Find the regression equation for predicting GPA from SAT scores. What percentage of the variance in GPAs is accounted for by the regression equation? (Compute the correlation, r, then find r2.) C) Does the regression equation account for a significant portion of the variance in GPA? Use alpha = 0.05 to evaluate the F-ratio. A) b=0.00375; a=0.925; Y (hat) = 0.00375X+0.925 B) r= 0.5 r^2= 0.25Suppose that you run a correlation and find the correlation coefficient is 0.421 and the regression equation is ˆy=2.2x+6.96y^=2.2x+6.96. The center of your x-data was 5.2 and the center of your y-data was 18.6.If the critical value is .396, use the appropriate method to predict the yy value when xx is 6.8A random sample of nonindustrialized countries was selected, and the life expectancy in years is listed for both men and women. Men 61.7 42.8 72.6 57.9 55.4 62.1 Women 50.4 74.2 75.3 75.4 49.1 65.2 The correlation coefficient for the data is =r−0.003 and =α0.05. Should regression analysis be done? Now Find the equation of the regression line. Round the coefficients to at least three decimal places. =y′+abx =a =b Now Find women's life expectancy in a country where men's life expectancy =60 years. Round your answer to at least three decimal places. Women's life expectancy is ____ years.