(a) Answer The equation is y = − 73.64 + 5.7188 x Explanation 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 + 5.7188 x Where x= Temperature of the water released. Conclusion: The equation is y = − 73.64 + 5.7188 x (b) To determine To analyse: The slope of the regression line. Answer The dots in the residual plot appear to be randomly scattered about zero, which indicates that the linear model is appropriate. Explanation 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 The distance to the nearest fish increases by b=5.7188 m per 0 C. Explanation Given information: The linear model appropriate for describing the relationship between temperature and distance. The regression line is as follows: y = − 73.64 + 5.7188 x 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 per 0 C. Conclusion: The distance to the nearest fish increases by b=5.7188 m per 0 C. (d) To determine To analyse: The linear model is over predicted or under predicted. Answer A linear model over predict the distance from the outflow pipe to the nearest fish found in the water for temperature of 29 0 . Explanation 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, Distance = – 73.64 + 5.7188 ( 29 ) Distance = − 73.64 + 165.8452 Distance = 92.2052 Thus, the predicted distance from the outflow pipe to the nearest fish found in the water for temperature of 29 0 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 29 0 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 29 0 . Conclusion: A linear model over predict the distance from the outflow pipe to the nearest fish found in the water for temperature of 29 0 .

The Practice of Statistics for AP - 4th Edition

4th Edition

ISBN: 9781429245593

Author: Starnes, Daren S., Yates, Daniel S., Moore, David S.

Publisher: Macmillan Higher Education

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Question

Chapter 10, Problem 33PT3

(a)

To determine

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

(a)

Expert Solution

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.

(b)

Expert Solution

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.

(c)

Expert Solution

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.

(d)

Expert Solution

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,

Distance=–73.64+5.7188(29)Distance=−73.64+165.8452 Distance=92.2052

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.

Chapter 10, Problem 32PT3

Chapter 10, Problem 34PT3

Chapter 10 Solutions

The Practice of Statistics for AP - 4th Edition

Ch. 10.1 - Prob. 1.1CYU Ch. 10.1 - Prob. 1.2CYU Ch. 10.1 - Prob. 1.3CYU Ch. 10.1 - Prob. 1.4CYU Ch. 10.1 - Prob. 2.1CYU Ch. 10.1 - Prob. 3.1CYU Ch. 10.1 - Prob. 3.2CYU Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E

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