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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?Which of the following is not a plot of residuals typically used in multiple regression analysis?Select one:a. None of these b. Residuals versus correlation coefficients..c. Residuals versus X1.d. Residuals versus timee. Residuals versus X2.Which of the following assumptions is not necessary for unbiasedness of a slope coefficient in a multiple regression model? MLR1 MLR 4 Homoskedasticity Random sampling
- The following data is a regression model where the U.S. Department of Transportation has tried to relate the rate of fatal traffic accidents (per 1000 licenses) to the percentage of motorists under the age of 21. Data has been collected for 42 major cities in the United States. SUMMARY OUTPUT Regression Statistics Multiple R 0.83938748 R Square 0.70457134 Adjusted R Square 0.69718562 Standard Error 0.58935028 Observations 42 ANOVA df SS MS F Regression 1 33.13441764 33.1344 95.3964 Residual 40 13.89335048 0.34733 Total 41 47.02776812 Coefficients Standard Error t Stat P-value Intercept -1.5974138 0.371671454 -4.2979 0.00010 Percent Under 21 0.28705317 0.029389769 9.76711 3.79E-9…The accompanying data resulted from an experiment in which weld diameter and shear strength (in pounds) were determined for five different spot welds on steel. Below are the data collected and the regression equation. Diameter Strength 200.1 813.7 210.1 785.3 220.1 960.4 230.1 1118.0 240.0 1076.2 Strength = -941.6992 + 8.5988*Diameter The predicted y-hat value for a diameter of 201 is 864. if we observed a weld that had a diameter of 235 that had a strength 1000, what would be its residual?A researcher notes that, in a certain region, a disproportionate number of software millionaires were born around the year 1955. Is this a coincidence, or does birth year matter when gauging whether a software founder will besuccessful? The researcher investigated this question by analyzing the data shown in the accompanying table. Complete parts a through c below. a. Find the coefficient of determination for the simple linear regression model relating number (y) of software millionaire birthdays in a decade to total number (x) of births in the region. Interpret the result. The coefficient of determination is 1.___? (Round to three decimal places as needed.) This value indicates that 2.____ of the sample variation in the number of software millionaire birthdays is explained by the linear relationship with the total number of births in the region. (Round to one decimal place as needed.) b. Find the coefficient of determination for the simple linear regression model…
- which of the following regressions represents the strongest negative linear relationship between x and y? (Attached in picture provided)Question #6 Listed below are altitudes (thousands of feet) and outside air temperatures (°F) recorded during a flight. Find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. There is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions. For the prediction interval, use a 95% confidence level with the altitude of 6327 ft (or 6.327 thousand feet). Altitude 4 8 14 24 27 31 32 Temperature 55 37 20 −5 −27 −41 −57 a. Find the explained variation. ______________ (Round to two decimal places as needed.) b. Find the unexplained variation. _______________ (Round to five decimal places as needed.) c. Find the indicated prediction interval. _____________°F < y < ____________ °F (Round to four decimal places as needed.)Consider the following computer output of a multiple regression analysis relating annual salary to years of education and years of work experience. Regression Statistics Multiple R 0.73650.7365 R Square 0.54240.5424 Adjusted R Square 0.52250.5225 Standard Error 2124.60962124.6096 Observations 4949 ANOVA dfdf SSSS MSMS F� Significance F� Regression 22 246,127,958.1791246,127,958.1791 123,063,979.0896123,063,979.0896 27.262927.2629 1.6E-081.6E-08 Residual 4646 207,642,442.8821207,642,442.8821 4,513,966.14964,513,966.1496 Total 4848 453,770,401.0612453,770,401.0612 Coefficients Standard Error t� Stat P-value Lower 95%95% Upper 95%95% Intercept 14256.268814256.2688 2,513.30952,513.3095 5.67235.6723 0.0000008950.000000895 9197.23929197.2392 19,315.298419,315.2984 Education (Years) 2353.85412353.8541 336.0719336.0719 7.00407.0040 0.0000000090.000000009 1677.37651677.3765 3030.33173030.3317 Experience (Years) 832.8371832.8371…
- You estimated a regression with the following output. Source | SS df MS Number of obs = 389 -------------+---------------------------------- F(1, 387) = 52779.08 Model | 514021570 1 514021570 Prob > F = 0.0000 Residual | 3769038.12 387 9739.11659 R-squared = 0.9927 -------------+---------------------------------- Adj R-squared = 0.9927 Total | 517790608 388 1334511.88 Root MSE = 98.687 ------------------------------------------------------------------------------ Y | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- X | 20.14493 .0876869 229.74 0.000 19.97253 20.31733 _cons | 105.6206 9.345057 11.30 0.000 87.24716 123.994 ------------------------------------------------------------------------------ Which of the following is the estimated regression line? Group of answer choices Y = 9.35 + 0.09*X Y = 20.14 + 105.62*X Y = 105.62 + 20.14*X Y = 0.09 + 9.35*Xtable 7 autocorrelations of the residuals from estimating the regression ΔgPMt = 0.0006 − 0.33301 ΔgPMt −1 + εt 1Q:1992–4Q:2001 (40 Observations) regression Statistics R-squared Standard error Observations Durbin–watson intercept ΔgPMt −1 ΔgPMt −4 Coefficient −0.0001 −0.0608 0.8720 0.9155 0.0057 40 2.6464 Standard error 0.0009 0.0687 0.0678 t-Statistic −0.0610 −0.8850 12.8683 lag 1 2 3 4 5 autocorrelation −0.1106 −0.5981 −0.1525 0.8496 −0.1099 table 8 shows the output from a regression on changes in the gPM for home Depot, where we have changed the specification of the ar regression. table 8 Change in gross Profit Margin for home Depot 1Q:1992–4Q:2001 a. identify the change that was made to the regression model. b. Discuss the rationale for changing the regression specificationIf the standard error of the estimate for a regression model fitted to a large number of paired observations is 1.75, approximately 95% of the residuals would lie within ______. −3.50 and +3.50 −1.75 and +1.75 −0.95 and +0.95 −0.68 and +0.68 −0.97 and +0.97