91Ó°ÊÓ

$\begin{array}{ccccc}\text{Time}\mathit{t}& \text{Count}\mathit{y}& & \text{Time}\mathit{t}& \text{Count}\mathit{y}\\ 1& 355& & 9& 56\\ 2& 211& & 10& 38\\ 3& 197& & 11& 36\\ 4& 166& & 12& 32\\ 5& 142& & 13& 21\\ 6& 106& & 14& 19\\ 7& 104& & 15& 15\\ 8& 60& & & \\ & & & & \end{array}$

The scatter plot is:

As a result, we can see from the scatterplot that,

Because the scatterplot slopes downhill, the direction is negative.

Because the points are not in a straight line, the shape is curved.

Strength: Strong because all points in the same pattern are relatively near together.

We also see an anomaly in the scatterplot's upper left corner.

The values for the bacteria count and the time are now provided in the question. As a result, we can argue that using an exponential model to describe the relationship between the count of bacteria and time would be plausible because the related scatterplot has an approximately linear pattern with no apparent strong outliers, as seen in the scatterplot in part $1$. (a).

Minitab output from a linear regression analysis on the transformed data is shown below.

$\begin{array}{lcccc}\text{Predictor}& \text{Coef}& \text{SE Coef}& T& \text{P}\\ \text{Constant}& 5.97316& 0.05978& 99.92& 0.000\\ \text{Time}& -0.218425& 0.006575& -33.22& 0.000\end{array}$

$\mathrm{S}=0.110016\mathrm{R}-\mathrm{Sq}=98.8\%\mathrm{R}-\mathrm{Sq}\left(\mathrm{adj}\right)=98.7\%$

Now $t$denotes the current time and $y$denotes the number of microorganisms. As a result, the transformation is $Iny$, where is the count's natural logarithm.

As we already know, the regression line's general equation is as follows:

$\ln \hat{y}=a+bt$

The slope and constant in the computer output are as follows:

$a=5.97316$

$b=-0.218425$

As a result, the regression line looks like this:

localid="1652849978599" $$\begin{gathered} \ln \hat{y}=a+bt \\ =5.97316-0.218425t \end{gathered}$$

Minitab output from a linear regression analysis on the transformed data is shown below.

$\begin{array}{lcccc}\text{Predictor}& \text{Coef}& \text{SE Coef}& T& \text{P}\\ \text{Constant}& 5.97316& 0.05978& 99.92& 0.000\\ \text{Time}& -0.218425& 0.006575& -33.22& 0.000\end{array}$

$\mathrm{S}=0.110016\mathrm{R}-\mathrm{Sq}=98.8\%\mathrm{R}-\mathrm{Sq}\left(\mathrm{adj}\right)=98.7\%$

The regression line is

$$\begin{gathered} \ln \hat{y}=a+bt \\ =5.97316-0.218425t \end{gathered}$$

Calculation of equation is

$$\begin{gathered} \ln \hat{y}=a+bt \\ =5.97316-0.218425t \\ =5.97316-0.218425\left(17\right) \\ =2.259935 \\ =e^{2.259935} \\ =9.58247 \end{gathered}$$