Consider the data.14Yi76.11The estimated regression equation for these data is ŷ= 0.80 + 2.40x.(a) Compute SSE, SST, and SSR using equations SSE = E(y; - ŷ;)², sST = E(y; - y)², and SSR = E(ŷ; - 7)².SSE =SST =SSR =(b) Compute the coefficient of determination r.p2 =Comment on the goodness of fit. (For purposes of this exercise, consider a proportion large if it is at least 0.55.)The least squares line provided a good fit as a large proportion of the variability in y has been explained by the least squares line.The least squares line did not provide a good fit as a large proportion of the variability in y has been explained by the least squares line.The least squares line did not provide a good fit as a small proportion of the variability in y has been explained by the least squares line.The least squares line provided a good fit as a small proportion of the variability in y has been explained by the least squares line.(c) Compute the sample correlation coefficient. (Round your answer to three decimal places.)13O O O

We have used the excel data analysis tool to run the regression analysis.

The estimated regression equatión for these data is ý = 0.80 + 2.40x. (a) Compute SSE, SST, and SSR using equations SSE = E(y, - ŷ)?, sST = E(y, - y)?, and SSR = E(ŷ, - )?. SSE = SST = SSR = (b) Compute the coefficient of determination r2. Comment on the goodness of fit. (For purposes of this exercise, consider a proportion large if it is at least 0.55.) O The least squares line provided a good fit as a large proportion of the variability in y has been explained by the least squares lir O The least squares line did not provide a good fit as a large proportion of the variability in y has been explained by the least squa O The least squares line did not provide a good fit as a small proportion of the variability in y has been explained by the least squ O The least squares line provided a good fit as a small proportion of the variability in y has been explained by the least squares lin (c) Compute the sample correlation coefficient. (Round your answer to three decimal places.)

The estimated regression equatión for these data is ý = 0.80 + 2.40x. (a) Compute SSE, SST, and SSR using equations SSE = E(y, - ŷ)?, sST = E(y, - y)?, and SSR = E(ŷ, - )?. SSE = SST = SSR = (b) Compute the coefficient of determination r2. Comment on the goodness of fit. (For purposes of this exercise, consider a proportion large if it is at least 0.55.) O The least squares line provided a good fit as a large proportion of the variability in y has been explained by the least squares lir O The least squares line did not provide a good fit as a large proportion of the variability in y has been explained by the least squa O The least squares line did not provide a good fit as a small proportion of the variability in y has been explained by the least squ O The least squares line provided a good fit as a small proportion of the variability in y has been explained by the least squares lin (c) Compute the sample correlation coefficient. (Round your answer to three decimal places.)

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)

6th Edition

ISBN:9781337111348

Author:Bruce Crauder, Benny Evans, Alan Noell

Publisher:Bruce Crauder, Benny Evans, Alan Noell

Chapter3: Straight Lines And Linear Functions

Section3.CR: Chapter Review Exercises

Problem 15CR: Life Expectancy The following table shows the average life expectancy, in years, of a child born in...

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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.
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?
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Assume that there is a positive linear correlation between the variable R (return rate in percent of a financial investment) and the variable t (age in years of the investment) given by the regression equation R = 2.3t + 4.8. Without further information, can we assume there is a cause-and-effect relationship between the return rate and the age of the investment? If the investment continues to grow at a constant rate, what is the expected return rate when the investment is 7 years old? If the investment continues to grow at a constant rate, how old is the investment when the return rate is 30%?