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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?I have some doubts regarding linear regression. if any 2 variables in X1, X2 AND Y have a positive correlation, then in the linear regression Y = b0 + b1X1 +b2X2 +e, will the sign of b1 and b2 both be positive? will the residuals that we get from linear regression will always be uncorrelated given X?Suppose that the sales of a company (Y) is regressed on advertising expenditure (x) and labor cost (z), and the estimated regression equation is Y = 5 + 0.5x + 0.7z + u (where u is the error term). Here, sales, advertising expenditure and labor cost are measured in million Tk. Standard error for the coefficient of x is 0.04, standard error for the coefficient of z is 0.01, and the sample size is 20. Can we conclude that advertising expenditure is a statistically significant variable?
- There is a linear regression: Yi = B0+ B1(Xi^2)+ ei present, where Xi is squared. ei ∼ N(0,σ2). How would I derive LSE for B0 and B1 and their variance?If there is a positive correlation between X and Y, then the regression equation Y = bX + a will have _____.Consider a simple linear regression model with predictor variable x and response variable y, where the regression line is represented by the equation y = β0 + β1x. If β0 = -5 and β1 = 3, what is the predicted value of y for a given x = 4?
- If other factors are held constant, if the Pearson correlation between X and Y is r = 0.50, then the regression equation will produce more accurate predictions than would be obtained if r = 0.70. True or false?1. Suppose that the sales of a company (Y) is regressed on advertising expenditure (x) and labor cost (z), and the estimated regression equation is Y = 5 + 0.5x + 0.7z + u (where u is the error term). Here, sales, advertising expenditure and labor cost are measured in million Tk. Standard error for the coefficient of x is 0.4, standard error for the coefficient of z is 0.01, and the sample size is 20. Based on this information, find out whether labor cost is a statistically significant variable using an appropriate statistical test.η =8, sum x=28, sum y=384, Sigma xy=480, Sigma x^ 2 =134 Find the estimated regression line
- A fitted linear regression model is (y=10+2x ). If x = 0 and the corresponding observed value of y = 9, the residual at this observation is:he following estimated regression model was developed relating yearly income (y in $1000s) of 30 individuals with their age (x1) and their gender (x2) (0 if male and 1 if female).ŷ = 30 + 0.7x1 + 3x2Also provided are SST = 1200 and SSE = 384. The yearly income of a 24-year-old female individual isThe following estimated regression model was developed relating yearly income (Y in $1,000s) of 30 individuals with their age in years (X1) and their gender (X2) (0 if male and 1 if female). The yearly income of a 24-year-old female individual is