Chapter 10, Problem 33PT3

(a)

To determine

To find: The equation of the least square regression line.

Answer to Problem 33PT3

The equation is $y=-73.64+5.7188x$

Explanation of Solution

Given information:

A nuclear power plant releases water into a nearby lake every afternoon at 4:51 P.m. Environmental researchers are concerned that fish are being driven away from the

area around the plant.

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 scatter plot 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.

Conclusion:

The equation is $y=-73.64+5.7188x$

(b)

To determine

To analyse: The slope of the regression line.

Answer to Problem 33PT3

The dots in the residual plot appear to be randomly scattered about zero, which indicates that the linear model is appropriate.

Explanation of Solution

Given information:

The slope of the regression line in context.

A linear model is appropriate for describing the relationship between temperature and distance to the nearest fish because the pattern in the scatter plot 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.

Conclusion:

The dots in the residual plot appear to be randomly scattered about zero, which indicates that the linear model is appropriate.

(c)

To determine

To analyse: The relationship between temperature and distance.

Answer to Problem 33PT3

The distance to the nearest fish increases by b=5.7188 m per⁰ C.

Explanation of Solution

Given information:

The linear model appropriate for describing the relationship between temperature and distance.

The regression line is as follows:

  $y=-73.64+5.7188x$

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

Thus, the value is b=5.7188.

So, on interpret that on average, the distance to the nearest fish increases by b=5.7188 m per⁰ C.

Conclusion:

The distance to the nearest fish increases by b=5.7188 m per⁰ C.

(d)

To determine

To analyse: The linear model is over predicted or under predicted.

Answer to Problem 33PT3

A linear model over predict the distance from the outflow pipe to the nearest fish found in the water for temperature of 29⁰.

Explanation of Solution

Given information:

The linear model in part (a) over predict or under predict the measured distance from the outflow pipe to the nearest fish found in the water for a temperature of 29°.

Formula used:

Substitution method is used.

Calculation:

The coefficient for constant is −73.64 and the coefficient for Temperature is 5.7188.

Thus, the regression equation is, Distance = −73.64 + 5.7188 Temperature.

Substitute Temperature = 29, in the regression equation.

That is,

  $\begin{array}{l}\text{Distance}=–73.64+5.7188\left(29\right)\\ \text{Distance}=-73.64+165.8452\\ \text{Distance}=92.2052\end{array}$

Thus, the predicted distance from the outflow pipe to the nearest fish found in the water for temperature of 29⁰ is 92.2052 m.

Moreover, from the scatter plot it is obtained that, the actual distance from the outflow pipe to the nearest fish found in the water for temperature of 29⁰ is about 80 m.

Thus, a linear model over predict the distance from the outflow pipe to the nearest fish found in the water for temperature of 29⁰.

Conclusion:

A linear model over predict the distance from the outflow pipe to the nearest fish found in the water for temperature of 29⁰.