Suppose a least-squares regression line is given by y= 4.302x -3.293. What is the mean value of the response variable if x = 20? HA 20 =U (Round to one decimal place as needed.)
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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?Suppose a least-squares regression line is given by y=4.302x−3.293. What is the mean value of the response variable if x=20? μy20=_______? (Round to one decimal place as needed.)The least-squares regression line relating two statistical variables is given as = 24 + 5x. Compute the residual if the actual (observed) value for y is 38 when x is 2. 4 38 2