Chapter 12.2, Problem 43E

(a)

To determine

To Explain: the reasonable to use the exponential model to explain the relation between count for the bacteria and time.

Answer to Problem 43E

Scatter plot is not having strong curvature

Explanation of Solution

Given:

  

  

It is noticed that the given scatter plot is not having strong curvature and therefore it would be appropriate to utilize a linear model between the two variables of the scatter plot. Therefore it is appropriate to utilize a linear model between ln(count) and time.

General linear model to expect ln(count) and time

  $\ln \left(count\right)=a+b\left(time\right)$

Taking the exponential

  $\begin{array}{c}count=e^{\ln \left(count\right)}\\ =e^{a+b\left(time\right)}\\ =e^a{e}^{b\left(time\right)}\end{array}$

(b)

To determine

To find: the equation of the least-squares regression line.

Answer to Problem 43E

  $\ln \widehat{y}=5.97316-0.218425x$

Explanation of Solution

Given:

  

Calculation:

General equation for the least square regression line

  $\widehat{y}=b_0+b_{1}x$

The calculated of the constant $b_0$ is mention in the row “constant” and in the column “Coef” of the output of the computer

  $b_0$ =5.97316

The calculated of the slope $b_1$ is mention in the row “Time” and in the column “Coef” of the output of the computer

  $b_1$ =-0.218425

Putting the value of $b_0$ and $b_1$

  $\begin{array}{l}\widehat{y}=b_0+b_{1}x\\ \widehat{y}=5.97316-0.218425x\end{array}$

Where x shows the time and y shows the ln (count)

  $\ln \widehat{y}=5.97316-0.218425x$

Where $x$ is representing the time and y is representing the count.

(c)

To determine

To Calculate: the number of bacteria after 17 minutes.

Answer to Problem 43E

9.58247 hundred bacteria or 958.247 bacteria

Explanation of Solution

Given:

  

Calculation:

From the part (b)

  $\ln \widehat{y}=5.97316-0.218425x$

Putting the value of $x$

  $\begin{array}{l}\ln \widehat{y}=5.97316-0.218425x\\ \ln \widehat{y}=5.97316-0.218425\left(17\right)\\ \ln \widehat{y}=2.259935\end{array}$

Taking the exponential

  $\begin{array}{l}\widehat{y}=e^{\ln \widehat{y}}=e^{2.259935}\\ =9.58247\end{array}$

Therefore the expected number of bacteria after 17 minutes is 9.58247 hundred bacteria or 958.247 bacteria.