Question

1. data for the global co₂ emissions from aviation for the years 2010 - 2019 is given in the table. because the data follows an exponential pattern, the emissions data was adjusted using a log transformation. let y = global co₂ emissions from aviation (in millions of tons) and log(y)=log(global co₂).

year	y	log(y)
2011	710	2.85125835
2012	745.5	2.87244765
2013	783.8	2.89420526
2014	821.9	2.91481898
2015	863.1	2.93606112
2016	906.2	2.95722406
2017	953.5	2.9793207
2018	999.04	2.99958288
2019	1,048.9	3.02073409

the computer output for a linear regression model for log(global co₂ emissions from aviation) is given below.

predictor	coef	se coef	t	p
years since 2010	0.021	0	563.781	< 0.001

a. give the equation of the least - squares regression line. define any variables used.

b. an exponential regression model for the original data was calculated as ŷ = 676.3153(1.05)^x where x is the number of years since 2010 and y is the global co₂ emissions from aviation. calculate log(676.3153) and log(1.05).

c. how are the values of the parameters in the linear regression related to the values of the parameters in the original exponential regression model?

Explanation:

Step1: Find the least - squares regression line equation for part a

Let $x$ be the number of years since 2010 and $\log(y)$ be the logarithm of global CO₂ emissions from aviation. The general form of a linear regression line is $\hat{\log(y)}=a + bx$. From the computer - output, $a = 2.83$ (the coefficient of the constant) and $b=0.021$ (the coefficient of years since 2010). So the equation is $\hat{\log(y)}=2.83 + 0.021x$.

Step2: Calculate logarithms for part b

Using the property of logarithms, $\log(676.3153)\approx2.8301$ (using a calculator: $\log_{10}(676.3153)$) and $\log(1.05)\approx0.0212$ (using a calculator: $\log_{10}(1.05)$).

Step3: Analyze parameter relationships for part c

The exponential regression model is $\hat{y}=676.3153(1.05)^x$. Taking the logarithm of both sides gives $\log(\hat{y})=\log(676.3153)+x\log(1.05)$. Comparing this with the linear regression model $\hat{\log(y)} = 2.83+0.021x$, we can see that the intercept of the linear regression model ($2.83$) is approximately equal to $\log(676.3153)$ and the slope of the linear regression model ($0.021$) is approximately equal to $\log(1.05)$.

Answer:

a. $\hat{\log(y)}=2.83 + 0.021x$, where $x$ is the number of years since 2010 and $\hat{\log(y)}$ is the predicted logarithm of global CO₂ emissions from aviation.
b. $\log(676.3153)\approx2.8301$, $\log(1.05)\approx0.0212$
c. The intercept of the linear regression model is approximately equal to $\log$ of the initial - value parameter in the exponential regression model, and the slope of the linear regression model is approximately equal to $\log$ of the growth - factor parameter in the exponential regression model.