Mathematics, 17.10.2019 19:00, garciagang0630

The table below shows the surface area y, in square feet, of a shrinking lake in x days: time (x) (days) 5 10 15 20 surface area (y) (square feet) 90 85 70 61 part a: what is the most likely value of the correlation coefficient of the data in the table? based on the correlation coefficient, describe the relationship between time and surface area of the lake. [choose the value of the correlation coefficient from −1, −0.98, −0.5, −0.02.] (4 points) part b: what is the value of the slope of the graph of surface area versus time between 15 and 20 days, and what does the slope represent? (3 points) part c: does the data in the table represent correlation or causation? explain your answer. (3 points)

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PART A:
Given the table below showing the surface area y, in square feet, of a shrinking lake in x days.

Time (x) (hours): 5 10 15 20
Surface area (y) (square inches): 90 85 70 61

We can find the correlation coeficient of the data using the table below:
$\begin{center}
\begin{tabular}
{|c|c|c|c|c|}
x & y & x^2 & y^2 & xy \\ [1ex]
5 & 90 & 25 & 8,100 & 450\\
10 & 85 & 100 & 7,225 & 850\\
15 & 70 & 225 & 4,900 & 1,050\\
20 & 61 & 400 & 3,721 & 1,220\\ [1ex]
\Sigma x=50 & \Sigma y=306 & \Sigma x^2=750 & \Sigma y^2=23,946 & \Sigma xy=3,570
\end{tabular}
\end{center}$

Recall that the correlation coefitient is given by the equation:
$r= \frac{n(\Sigma xy)-(\Sigma x)(\Sigma y)}{ \sqrt{(n\Sigma x^2-(\Sigma x)^2)(n\Sigma y^2-(\Sigma y)^2)} } \\ \\ = \frac{4(3,570)-(50)(306)}{ \sqrt{(4(750)-(50)^2)(4(23,946)-(306)^2)} } \\ \\ = \frac{14,280-15,300}{ \sqrt{(3,000-2,500)(95,784-93,636)}} = \frac{-1,020}{ \sqrt{500(2,148)}} \\ \\ = \frac{-1,020}{ \sqrt{1,074,000} } = \frac{-1,020}{1,036} =-0.98$

From the value of the correlation coeffeicient, it can be deduced that the surface area of the lake has a strong negative relationship with the time.

Recall the for the value of the correlation coeficient closer to +1, the relationship is strong positive, for the value closer to -1, the value is strong negative and for the values closer to zero, either way of zero is a weak positive if it is positive and weak negative if it is negative.

PART B:
Recall that the slope of a straight line passing through two points (x_1,y_1)

and

is given by
$m= \frac{y_2-y_1}{x_2-x_1}$

Thus, the slope of the graph of surface area versus time between 15 and 20 days, [i.e. the line passes through points (15, 70) and (20, 61)] is given by
$m= \frac{61-70}{20-15}= \frac{-9}{5} =-1.8$

The value of the slope means that the surface area y, in square feet, of the shrinking lake, shrinks by 1.8 square feet every day between the day 15 and day 20.

PART C:
We can say that the data above represent both correlation and causation.

Recall that correlation expresses the relationship between two variables while causation expresses that an event is as a result of another event.

From the information above, we have seen that there is a relationship (correlation) between the passing of days and the shrink in the surface area of the lake.

Also we can conclude that the shrink in the surface area of the lake is a function of the passing of days, i.e. the shrink in the surface area of the lake is as a result of the passing of days.

Therefore, the data in the table represent both correlation and causation.

answered: Guest

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answered: Guest

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The table below shows the surface area y, in square feet, of a shrinking lake in x days: time (x) (...