Explanation Given: The data set is x 1 3 7 y 8 4 1.5 Formula used: The regression line with values m and b is given by, y = m x + b The chain rule of differentiation states that, d d x ( f ( x ) ) n = n ( f ( x ) ) n − 1 f ′ ( x ) Calculation: Consider the data set, x 1 3 7 y 8 4 1.5 For an exponential function of the form y = a ⋅ b x logarithm is taken on both sides, ln y = ln ( a ⋅ b x ) Use the property, ln ( m ⋅ n ) = ln m + ln n , ln y = ln a + ln b x Apply the property, ln ( m n ) = n ln m , ln y = ln a + x ln b Consider ln y to be y 0 , ln a to be a 0 and ln b to be b 0 , thus, the equation becomes, y 0 = b 0 x + a 0 The table for the new data set is x 1 3 7 y 8 4 1.5 y 0 ln 8 ln 4 ln 1.5 The data points are ( 1 , l n 8 ) , ( 3 , l n 4 ) and ( 7 , ln 1.5 ) . According to the least-squares assumption, the line which best fits is the line for which the sum of the squares of the y -deviations is minimum. For that, the y -deviations are y 1 − ln 8 , y 2 − ln 4 , y 3 − ln 1.5 The sum of the squares of y -deviations has to be minimised. Thus, consider the sum of the squares is S , S = ( y 1 − ln 8 ) 2 + ( y 2 − ln 4 ) 2 + ( y 3 − ln 1.5 ) 2 …… (1) The points ( 1 , y 1 ) , ( 3 , y 2 ) and ( 7 , y 3 ) must be solutions of the equation y = m x + b , thus, y 1 = m ( 1 ) + b y 2 = m ( 3 ) + b y 3 = m ( 7 ) + b Substitute the values of y 1 , y 2 and y 3 in the equation (1), S = ( m + b − ln 8 ) 2 + ( 3 m + b − ln 4 ) 2 + ( 7 m + b − ln 1.5 ) 2 Differentiate the both sides with respect to b , d S d b = d ( ( m + b − ln 8 ) 2 + ( 3 m + b − ln 4 ) 2 + ( 7 m + b − ln 1.5 ) 2 ) d b Apply the chain rule of differentiation, d S d b = d d b ( m + b − ln 8 ) 2 + d d b ( 3 m + b − ln 4 ) 2 + d d b ( 7 m + b − ln 1.5 ) 2 = 2 ( m + b − ln 8 ) ( 0 + b 1 − 1 − 0 ) + 2 ( 3 m + b − ln 4 ) ( 0 + b 1 − 1 − 0 ) + 2 ( 7 m + b − ln 1.5 ) ( 0 + b 1 − 1 − 0 ) = 2 m + 2 b − 2 ln 8 + 6 m + 2 b − 2 ln 4 + 14 m + 2 b − 2 ln 1.5 = 22 m + 6 b − 7.7424 Differentiate the both sides with respect to m , d S d m = d ( ( m + b − ln 8 ) 2 + ( 3 m + b − ln 4 ) 2 + ( 7 m + b − ln 1.5 ) 2 ) d m Apply the chain rule of differentiation, d S d m = d d m ( m + b − ln 8 ) 2 + d d m ( 3 m + b − ln 4 ) 2 + d d m ( 7 m + b − ln 1.5 ) 2 = 2 ( m + b − ln 8 ) ( m 1 − 1 + 0 − 0 ) + 2 ( 3 m + b − ln 4 ) ( 3 m 1 − 1 + 0 − 0 ) + 2 ( 7 m + b − ln 1

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Chapter 6.4, Problem 5E

To determine

To calculate: The exponential regression curve for the data set represented by the table,

x 1 3 7

y 8 4 1.5

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Chapter 6.4, Problem 4E

Chapter 6.4, Problem 6E

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Five students took an aptitude test prior to their algebra course. What linear regression equation best predicts algebra performance, based on aptitude scores? If a student made an 80 on the aptitude test, what grade would we expect her to make in algebra? How well does the regression equation fit the data? Student aptitude test algebra grade 1 95 85 2 85 95 3 80 70 4 70 65 5 60 70

XLMiner Analysis ToolPak A В D G K M N P 1 Salesperson Years of Experience Annual Sales ($1000s) 78 SUMMARY OUTPUT 2 3 Anova: Single Factor Regression Statistics Multiple R R Square Adjusted R Square Standard Error 3 3 96 4 3 5 88 0.907961759 Anova: Two-Factor With Replication 4 4 107 0.824394555 6 5 6 105 0.802443875 Anova: Two-Factor Without Replication 7 7 116 8.072755139 8 7 10 123 Observations 10 Correlation 9 8 117 Covariance 10 9 10 122 ANOVA 11 10 14 139 df SS MS F Significance F Regression Descriptive Statistics 12 1 2447.544996 2447.544996 37.55667406 0.000280582 13 Residual 8 521.3550043 65.16937553 Total 9 2968.9 Exponential Smoothing 14 15 F-Test Two-Sample for Variances 16 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95% Upper 95% Intercept Years of Experience 4.575705731 0.746645624 6.128350028 0.000280582 2.853937836 6.297473627 2.853937836 6.297473627 17 76.61248931 5.883832818 13.02084741 1.14813E-06 63.04434651 90.18063211 63.04434651…

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The exponential regression curve for the data set represented by the table, x 1 3 7 y 8 4 1.5

Solution Summary: The author explains the exponential regression curve for the given data set, x 1 3 7 y 8 4 1.5. The chain rule of differentiation states that, mathrmln(m