(a) Answer The equation of the least-squares regression line is 92.29 – 0.05762 (Wildebeest) Explanation Given: Calculation: We think that “wildebeest” will help explain “percent burned”. So, “wildebeest” is the explanatory variable and “percent burned” is the response variable. From the out put we see that, Y- Intercept = 92.29 Slope= -0.05762 Equation of least squares regression line is: Percent burned = 92.29 – 0.05762 (Wildebeest) Conclusion: Hence,the equation of the least-squares regression line is 92.29 – 0.05762 (Wildebeest) (b) To determine To explain: the slope of the regression line Answer Slope is negative, we can say that every wildebeest abundance (in 1000s) increases percent of the grass burned decreases by 0.05762. Explanation The slope b of the regression line Y ^ = a + b X is the rate at which the predicted response changes along the line as explanatory variable X changes. Specifically, b is the predicted change in Y when X increases by 1 unit. (c) To determine To find: The correlation Answer The value of Correlation (r) is − 0.8037 Explanation Calculation: From the output we have, R − s q = 64.6 % r 2 = 0.646 r = ± 0.8037 The graph 1 shows a clear direction: The overall pattern moves from upper left to lower right. That is, if wildebeest abundance increases, percent of the grass burned decreases. We call this a negative association between the two variables. Conclusion: Hence, the value of Correlation (r) is − 0.8037 This means a strong relationship between variables. (d) To determine To find: whether a linear model is appropriatefor describing the relationship Answer A line does seem to be an appropriate model for these data. Explanation A line does seem to be an appropriate model for these data. The residual plot shows a fairly random scatter of points about the horizontal line representing error = 0. The residuals are between +30 to -30 except one point. On average, predictions of percent of the grass burned would be off by about s = 15.9880 for the data. That may be too much prediction error.

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 3, Problem 13PT

(a)

To determine

To give: the equation of the least-squares regression line.

(a)

Expert Solution

Answer to Problem 13PT

The equation of the least-squares regression line is 92.29–0.05762 (Wildebeest)

Explanation of Solution

Given:

The Practice of Statistics for AP - 4th Edition, Chapter 3, Problem 13PT , additional homework tip 1

The Practice of Statistics for AP - 4th Edition, Chapter 3, Problem 13PT , additional homework tip 2

Calculation:

We think that “wildebeest” will help explain “percent burned”. So, “wildebeest” is the explanatory variable and “percent burned” is the response variable. From the out put we see that,

Y- Intercept = 92.29

Slope= -0.05762

Equation of least squares regression line is:

Percent burned =92.29–0.05762 (Wildebeest)

Conclusion:

Hence,the equation of the least-squares regression line is 92.29–0.05762 (Wildebeest)

(b)

To determine

To explain: the slope of the regression line

(b)

Expert Solution

Answer to Problem 13PT

Slope is negative, we can say that every wildebeest abundance (in 1000s) increases percent of the grass burned decreases by 0.05762.

Explanation of Solution

The slope b of the regression line Y^=a+bX is the rate at which the predicted response changes along the line as explanatory variable X changes.

Specifically, b is the predicted change in Y when X increases by 1 unit.

(c)

To determine

To find: The correlation

(c)

Expert Solution

Answer to Problem 13PT

The value of Correlation (r) is −0.8037

Explanation of Solution

Calculation:

From the output we have,

R−sq=64.6%r2=0.646r=±0.8037

The graph 1 shows a clear direction: The overall pattern moves from upper left to lower right.

That is, if wildebeest abundance increases, percent of the grass burned decreases. We call this a negative association between the two variables.

Conclusion:

Hence, the value of Correlation (r) is −0.8037

This means a strong relationship between variables.

(d)

To determine

To find: whether a linear model is appropriatefor describing the relationship

(d)

Expert Solution

Answer to Problem 13PT

A line does seem to be an appropriate model for these data.

Explanation of Solution

A line does seem to be an appropriate model for these data. The residual plot shows a fairly random scatter of points about the horizontal line representing error = 0. The residuals are between +30 to -30 except one point.

On average, predictions of percent of the grass burned would be off by abouts=15.9880 for the data.

That may be too much prediction error.

Chapter 3, Problem 12PT

Chapter 4.1, Problem 1.1CYU

Chapter 3 Solutions

The Practice of Statistics for AP - 4th Edition

Ch. 3.1 - Prob. 1.1CYU Ch. 3.1 - Prob. 1.2CYU Ch. 3.1 - Prob. 2.1CYU Ch. 3.1 - Prob. 2.2CYU Ch. 3.1 - Prob. 2.3CYU Ch. 3.1 - Prob. 2.4CYU Ch. 3.1 - Prob. 2.5CYU Ch. 3.1 - Prob. 3.1CYU Ch. 3.1 - Prob. 3.2CYU Ch. 3.1 - Prob. 1E

Ch. 3.1 - Prob. 2E Ch. 3.1 - Prob. 3E Ch. 3.1 - Prob. 4E Ch. 3.1 - Prob. 5E Ch. 3.1 - Prob. 6E Ch. 3.1 - Prob. 7E Ch. 3.1 - Prob. 8E Ch. 3.1 - Prob. 9E Ch. 3.1 - Prob. 10E Ch. 3.1 - Prob. 11E Ch. 3.1 - Prob. 12E Ch. 3.1 - Prob. 13E Ch. 3.1 - Prob. 14E Ch. 3.1 - Prob. 15E Ch. 3.1 - Prob. 16E Ch. 3.1 - Prob. 17E Ch. 3.1 - Prob. 18E Ch. 3.1 - Prob. 19E Ch. 3.1 - Prob. 20E Ch. 3.1 - Prob. 21E Ch. 3.1 - Prob. 22E Ch. 3.1 - Prob. 23E Ch. 3.1 - Prob. 24E Ch. 3.1 - Prob. 25E Ch. 3.1 - Prob. 26E Ch. 3.1 - Prob. 27E Ch. 3.1 - Prob. 28E Ch. 3.1 - Prob. 29E Ch. 3.1 - Prob. 30E Ch. 3.1 - Prob. 31E Ch. 3.1 - Prob. 32E Ch. 3.1 - Prob. 33E Ch. 3.1 - Prob. 34E Ch. 3.2 - Prob. 1.3CYU Ch. 3.2 - Prob. 1.4CYU Ch. 3.2 - Prob. 2.1CYU Ch. 3.2 - Prob. 3.1CYU Ch. 3.2 - Prob. 3.2CYU Ch. 3.2 - Prob. 3.3CYU Ch. 3.2 - Prob. 4.1CYU Ch. 3.2 - Prob. 4.2CYU Ch. 3.2 - Prob. 5.1CYU Ch. 3.2 - Prob. 5.2CYU Ch. 3.2 - Prob. 35E Ch. 3.2 - Prob. 36E Ch. 3.2 - Prob. 37E Ch. 3.2 - Prob. 38E Ch. 3.2 - Prob. 39E Ch. 3.2 - Prob. 40E Ch. 3.2 - Prob. 41E Ch. 3.2 - Prob. 42E Ch. 3.2 - Prob. 43E Ch. 3.2 - Prob. 44E Ch. 3.2 - Prob. 45E Ch. 3.2 - Prob. 46E Ch. 3.2 - Prob. 47E Ch. 3.2 - Prob. 48E Ch. 3.2 - Prob. 49E Ch. 3.2 - Prob. 50E Ch. 3.2 - Prob. 51E Ch. 3.2 - Prob. 52E Ch. 3.2 - Prob. 53E Ch. 3.2 - Prob. 54E Ch. 3.2 - Prob. 55E Ch. 3.2 - Prob. 56E Ch. 3.2 - Prob. 57E Ch. 3.2 - Prob. 58E Ch. 3.2 - Prob. 59E Ch. 3.2 - Prob. 60E Ch. 3.2 - Prob. 61E Ch. 3.2 - Prob. 62E Ch. 3.2 - Prob. 63E Ch. 3.2 - Prob. 64E Ch. 3.2 - Prob. 65E Ch. 3.2 - Prob. 66E Ch. 3.2 - Prob. 67E Ch. 3.2 - Prob. 68E Ch. 3.2 - Prob. 69E Ch. 3.2 - Prob. 70E Ch. 3.2 - Prob. 71E Ch. 3.2 - Prob. 72E Ch. 3.2 - Prob. 73E Ch. 3.2 - Prob. 74E Ch. 3.2 - Prob. 75E Ch. 3.2 - Prob. 76E Ch. 3.2 - Prob. 77E Ch. 3.2 - Prob. 78E Ch. 3.2 - Prob. 79E Ch. 3.2 - Prob. 80E Ch. 3.2 - Prob. 81E Ch. 3 - Prob. 1CRE Ch. 3 - Prob. 2CRE Ch. 3 - Prob. 3CRE Ch. 3 - Prob. 4CRE Ch. 3 - Prob. 5CRE Ch. 3 - Prob. 6CRE Ch. 3 - Prob. 7CRE Ch. 3 - Prob. 1PT Ch. 3 - Prob. 2PT Ch. 3 - Prob. 3PT Ch. 3 - Prob. 4PT Ch. 3 - Prob. 5PT Ch. 3 - Prob. 6PT Ch. 3 - Prob. 7PT Ch. 3 - Prob. 8PT Ch. 3 - Prob. 9PT Ch. 3 - Prob. 10PT Ch. 3 - Prob. 11PT Ch. 3 - Prob. 12PT Ch. 3 - Prob. 13PT

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