91Ó°ÊÓ

Given data:

The scatter plot does not have much curvature, a linear model between the two variables of the scatter plot would be appropriate. As a result, using a linear relationship between $\ln \left(count\right)$and time is reasonable.

Taking the exponential

Given data:

General equation of a least square regression line:

$\hat{y}=b_0+b_{1}x$

In the row "constant" and the column "Coef" of the computer's output, the calculated constant $b_0$is given.

$b_0=5.97316$

In the row "Time" and the column "Coef" of the computer output, the slope $b_1$is found.

$b_1=-0.218425$

Substituting the value of $b_0$and $b_1$:

$\hat{y}=b_0+b_{1}x$

$\hat{y}=5.97316-0.218425x$

Where $x$represents the current time and $y$is the ln (count)

Given data:

Substituting the value of $x$:

$\ln \hat{y}=5.97316-0.218425x$

$\ln \hat{y}=5.97316-0.218425\left(17\right)$

$\ln \hat{y}=2.259935$

Taking the exponential

$\hat{y}=e^{\ln \hat{y}}$

$=e^{2.259935}$

$=9.58247$