91影视

A linear model is appropriate for describing the relationship between temperature and distance to the nearest fish because the pattern in the scatterplot does not contain a lot of curvature and the pattern in the residual plot does not contain a lot of curvature either. Moreover, the dots in the residual plot appear to be randomly scattered about zero, which indicates that the linear model is appropriate.

It is given that researchers want to study about is the temperature of the water discharged by the plants causes harm to the fishes of the water body or not. Thus, a scatterplot was been made on the temperature and distance to the nearest fish. Then, computer output from a least square regression analysis on these data and residual plot information is given in the question.

From that we know that, the coefficients for the regression line is listed under the heading "Coef" and the constant is the intercept. Also, temperature is the slope. Thus, now we can calculate the regression line as:

$y=-73.64+5.7188x$

Where $x=$Temperature of the water released.

It is given that researchers want to study about is the temperature of the water discharged by the plants causes harm to the fishes of the water body or not. Thus, a scatterplot was been made on the temperature and distance to the nearest fish. Then, computer output from a least square regression analysis on these data and residual plot information is given in the question.

From part (b), we know that the regression line is as follows:

$y=-73.64+5.7188x$

The slope as we know, is coefficient of $x$in the least square regression equation and represents the average increase or decrease of $y$per unit of $x$.

Thus, we have $b=5.7188$

So, we can interpret that on average, the distance to the nearest fish increases by$b=5.7188mper掳C$

It is given that the point is $(29,78)$, which implies that,

$x=29andy=78$

Now, we know from part (b), that the regression line is as follows:

$y=-73.64+5.7188x$

Let us now find out the predicted value, which can be calculated by evaluating the least square regression line at $x=29$. Thus, we have,

$$\begin{gathered} \hat{y}=-73.64+5.7188x \\ =-73.64+5.7188\left(29\right) \\ =92.2052 \end{gathered}$$

Now, as we know residual is the difference between the actual value and the predicted value. Thus it is calculated as:

Residual $=y-\hat{y}$

$$\begin{gathered} =78-92.2052 \\ =-14.2052 \end{gathered}$$

Thus, we conclude that, the predicted distance is$14.2052$ meters higher than the actual distance, when the temperature is$29掳C$.