Chapter 3.6, Problem 29E

a.

To determine

To find: an exponential model and a power model for the data and identify the coefficient of determination for each model using the regression feature of a graphing utility.

Answer to Problem 29E

Exponential model: $\mathrm{S}=4.114{(1.1960)}^{\mathrm{t}},{\mathrm{r}}^2\approx \mathrm{0.9986.}$

Power model: $\mathrm{S}=3.153{\mathrm{t}}^{0.766},{\mathrm{r}}^2\approx 0.9514$

Explanation of Solution

Given information: The table shows the yearly sales estimates S (in billions of dollars).

for the online grocery shopping market for the years 2012 through 2018.

Year	Sales, S (in billions)
2012	6.0
2013	6.9
2014	8.4
2015	10.1
2016	12.0
2017	14.2
2018	17.5

Calculation:

Let t represent the year, with t = 2 corresponding to 2012 ,then the above data table can be written as below.

Year	Sales, S (in billions)
2	6.0
3	6.9
4	8.4
5	10.1
6	12.0
7	14.2
8	17.5

Using the graphing utility, the exponential model of the above table data is:

  $\mathrm{S}=4.114{(1.1960)}^{\mathrm{t}},{\mathrm{r}}^2\approx \mathrm{0.9986.}$

Using the graphing utility, the power model of the above table data is:

  $\mathrm{S}=3.153{\mathrm{t}}^{0.766},{\mathrm{r}}^2\approx 0.9514$

b.

To determine

To graph: each model with the data using the graphing utility.

Answer to Problem 29E

Explanation of Solution

Given information: Exponential model: $\mathrm{S}=4.114{(1.1960)}^{\mathrm{t}},{\mathrm{r}}^2\approx \mathrm{0.9986.}$

Power model: $\mathrm{S}=3.153{\mathrm{t}}^{0.766},{\mathrm{r}}^2\approx 0.9514$

Calculation:

The graph of Exponential model: $\mathrm{S}=4.114{(1.1960)}^{\mathrm{t}},{\mathrm{r}}^2\approx \mathrm{0.9986.}$ is shown below.

  

The graph of Power model: $\mathrm{S}=3.153{\mathrm{t}}^{0.766},{\mathrm{r}}^2\approx 0.9514$ is shown below.

  

c.

To determine

To find: which model fits the data better.

Answer to Problem 29E

The exponential model fits the data better because ${\mathrm{r}}^2\approx 0.9986$ is close to 1 than ${\mathrm{r}}^2\approx 0.9514$ .

Explanation of Solution

Given information:

Calculation:

The exponential model fits the data better because ${\mathrm{r}}^2\approx 0.9986$ is close to 1 than ${\mathrm{r}}^2\approx 0.9514$ .