The following estimated regression equation is based on 10 observations was presented. ŷ = 29.1270 + 0.5906x1 + 0.4980x2 . Here SST=6,724.125 , SSR = 6,216.375 , sb1 = 0.0813, sb2 = 0.0567. Compute MSR & MSE to 3 decimals, then compute F using the appropriate F test (round answer to 3 decimals). Use α = 0.05.
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The following estimated regression equation is based on 10 observations was presented. ŷ = 29.1270 + 0.5906x1 + 0.4980x2 . Here SST=6,724.125 , SSR = 6,216.375 , sb1 = 0.0813, sb2 = 0.0567.
Compute MSR & MSE to 3 decimals, then compute F using the appropriate F test (round answer to 3 decimals). Use α = 0.05.
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- Find the equation of the regression line for the following data set. x 1 2 3 y 0 3 4For the following exercises, use Table 4 which shows the percent of unemployed persons 25 years or older who are college graduates in a particular city, by year. Based on the set of data given in Table 5, calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient. Round to three decimal places of accuracyOlympic 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 estimated regression equation based on 30 observations was presented. ŷ = 17.6 + 3.8x1 − 2.3x2 + 7.6x3 + 2.7x4 The values of SST and SSR are 1,807 and 1,757, respectively. (a) Compute R2. (Round your answer to three decimal places.) R2 = (b) Compute Ra2. (Round your answer to three decimal places.) Ra2 = (c) Comment on the goodness of fit. (For purposes of this exercise, consider a proportion large if it is at least 0.55.) The estimated regression equation did not provide a good fit as a large proportion of the variability in y has been explained by the estimated regression equation. The estimated regression equation provided a good fit as a large proportion of the variability in y has been explained by the estimated regression equation. The estimated regression equation provided a good fit as a small proportion of the variability in y has been explained by the estimated regression equation. The estimated regression equation did not provide a good…The following estimated regression equation based on 30 observations was presented. ŷ = 17.6 + 3.8x1 − 2.3x2 + 7.6x3 + 2.7x4 The values of SST and SSR are 1,808 and 1,780, respectively. (a) Compute R2. (b) Compute Ra2. (c) Comment on the goodness of fit.The following estimated regression equation based on 30 observations was presented. ŷ = 17.6 + 3.8x1 − 2.3x2 + 7.6x3 + 2.7x4 The values of SST and SSR are 1,801 and 1,758, respectively. (a)Compute R2. (Round your answer to three decimal places.) R2 = (b)Compute Ra2.(Round your answer to three decimal places.) Ra2 = (c) Comment on the goodness of fit. (For purposes of this exercise, consider a proportion large if it is at least 0.55.) The estimated regression equation provided a good fit as a small proportion of the variability in y has been explained by the estimated regression equation.The estimated regression equation did not provide a good fit as a large proportion of the variability in y has been explained by the estimated regression equation. The estimated regression equation did not provide a good fit as a small proportion of the variability in y has been explained by the estimated regression equation.The estimated regression equation provided a good fit as a…
- In a regression analysis involving 25 observations, the following estimated regression equation was developed.ŷ = 10 – 18x1 + 3x2 + 14x3Also, the following standard errors and the sum of squares were obtained.Sb1 = 3 Sb1 = 6 Sb1 = 7SST = 4,800 SSE = 1,296If we are interested in testing for the significance of the relationship among the variables (i.e., significance of the model), the critical value of F at α = .05 is _____.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.The following estimated regression equation based on 10 observations was presented. ŷ = 29.1260 + 0.5306x1 + 0.4680x2 The values of SST and SSR are 6,728.125 and 6,215.375, respectively. (a) Find SSE. SSE = (b) Compute R2. (Round your answer to three decimal places.) R2 = (c) Compute Ra2. (Round your answer to three decimal places.) Ra2 = (d) Comment on the goodness of fit. (For purposes of this exercise, consider a proportion large if it is at least 0.55.) The estimated regression equation provided a good fit as a small proportion of the variability in y has been explained by the estimated regression equation.The estimated regression equation did not provide a good fit as a small proportion of the variability in y has been explained by the estimated regression equation. The estimated regression equation provided a good fit as a large proportion of the variability in y has been explained by the estimated regression equation.
- The following results are from data concerning the amount withdrawn from an ATM machine based on the amount of time spent at the ATM machine (SECONDS) and the gender, FEMALE (dummy variable = 1 for females and = 0 for males) and an interaction term, SECONDS*FEMALE Based on the regression results, what is the predicted amount withdrawn for a male who spends 20 at the ATM machine? (please express your answer using 1 decimal places)For the 2011 season, suppose the average number of passing yards per attempt for a certain NFL team was 6.1. Use the estimated regression equation developed in part (c) to predict the percentage of games won by that NFL team. (Note: For the 2011 season, suppose this NFL team's record was 7 wins and 9 losses. Round your answer to the nearest integer.)The following estimated regression equation based on 10 observations was presented. ŷ = 25.1270 + 0.5309x1 + 0.4920x2 Here, SST = 6,738.125, SSR = 6,221.375, sb1 = 0.0818, and sb2 = 0.0563. Perform a t test for the significance of β1. Use α = 0.05. State the null and alternative hypotheses. a. H0: β1 > 0 Ha: β1 ≤ 0 b. H0: β1 ≠ 0 Ha: β1 = 0 c. H0: β1 < 0 Ha: β1 ≥ 0 d. H0: β1 = 0 Ha: β1 > 0 e. H0: β1 = 0 Ha: β1 ≠ 0 5. Find the value of the test statistic. (Round your answer to two decimal places.) 6. Find the p-value. (Round your answer to three decimal places.) p-value = 7. State your conclusion. a. Do not reject H0. There is sufficient evidence to conclude that β1 is significant. b. Reject H0. There is insufficient evidence to conclude that β1 is significant. c. Reject H0. There is sufficient evidence to conclude that β1 is significant. d. Do not reject H0. There is insufficient evidence to conclude that…