The table below gives the estimated world population (in billions) for various years.Years198019901997200020052011Population 440051005852608064507000What is the Linear Regression Equation?ỹ = 48.19x – 546 785y = -163 766x + 84.91ŷ 84.91x 163 766y = 546 785x + 48.19

x y X - Mx Y - My (X - Mx)2 (X - Mx)(Y - My) 198019901997200020052011…

Answered: The table below gives the estimated…

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter4: Equations Of Linear Functions

Section4.6: Regression And Median-fit Lines

Problem 13PPS

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Life Expectancy The following table shows the average life expectancy, in years, of a child born in the given year42 Life expectancy 2005 77.6 2007 78.1 2009 78.5 2011 78.7 2013 78.8 a. Find the equation of the regression line, and explain the meaning of its slope. b. Plot the data points and the regression line. c. Explain in practical terms the meaning of the slope of the regression line. d. Based on the trend of the regression line, what do you predict as the life expectancy of a child born in 2019? e. Based on the trend of the regression line, what do you predict as the life expectancy of a child born in 1580?2300
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.
Cellular Phone Subscribers The table shows the numbers of cellular phone subscribers y in millions in the United States from 2008 through 2013. Source: CTIA- The Wireless Association Year200820092010201120122013Number,y270286296316326336 (a) Find the least squares regression line for the data. Let x represent the year, with x=8 corresponding to 2008. (b) Use the linear regression capabilities of a graphing utility to find a linear model for the data. How does this model compare with the model obtained in part a? (c) Use the linear model to create a table of estimated values for y. Compare the estimated values with the actual data.
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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The table below gives the estimated world population (in billions) for various years. 1980 1990 1997 2000 2005 2011 Years Population 4400 5100 5852 6080 6450 7000 What is the Linear Regression Equation? Oy = 48.19x - 546 785 y = -163 766x + 84.91 O y = 84.91x - 163 766 y = -546 785x + 48.19