Introduction to the Practice of Statistics
Introduction to the Practice of Statistics
9th Edition
ISBN: 9781319013387
Author: David S. Moore, George P. McCabe, Bruce A. Craig
Publisher: W. H. Freeman
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Chapter 2.4, Problem 74E

(a)

To determine

To find: The predicted count values.

(a)

Expert Solution
Check Mark

Answer to Problem 74E

Solution: The predicted count values obtained are

c^1=6.3324

c^2=5.8112

c^3=5.29

c^4=4.7688

Explanation of Solution

Calculation: The provided least-squares regression equation is

log count=6.593(0.2606×time)

The time values are provided in the data as 1, 3, 5, and 7.

Substituting the values of time in the above linear regression equation, the following results are obtained:

For time(t1)=1, the predicted log count (c^1) is calculated as

log count=6.593(0.2606×time)=6.593(0.2606×t1)=6.593(0.2606×1)c^1=6.3324

For time(t2)=3, the predicted count (c^2) is calculated as

log count=6.593(0.2606×time)=6.593(0.2606×t2)=6.593(0.2606×3)c^2=5.8112

For time(t3)=5, the predicted count (c^3) is calculated as

log count=6.593(0.2606×time)=6.593(0.2606×t3)=6.593(0.2606×5)c^3=5.29

For time(t4)=7, the predicted count (c^4) is calculated as

log count=6.593(0.2606×time)=6.593(0.2606×t4)=6.593(0.2606×7)c^4=4.7688

Hence, the predicted count values obtained are 6.3324, 5.8112, 5.29, and 4.7688.

(b)

Section 1

To determine

To find: The difference between the observed and the predicted counts.

(b)

Section 1

Expert Solution
Check Mark

Answer to Problem 74E

Solution: The differences obtained are

d1=0.02717

d2=0.0523

d3=0.02321

d4=0.00188

Explanation of Solution

Calculation: The observed counts are provided in the exercise as

c1=6.35957

c2=5.75890

c3=5.31321

c4=4.77068

The predicted counts obtained from part (a) are

c^1=6.3324

c^2=5.8112

c^3=5.29

c^4=4.7688

All of the above values are for the times 1, 3, 5, and 7.

The differences (dis where i=1,2,3, and 4) between the observed and the predicted counts are calculated as follows:

For time(t1)=1, the difference (d1) is calculated as

d1=c1c^1=6.359576.3324=0.02717

For time(t2)=3, the difference (d2) is calculated as

d2=c2c^2=5.758905.8112=0.0523

For time(t3)=5, the difference (d3) is calculated as

d3=c3c^3=5.313215.29=0.02321

For time(t4)=7, the difference (d4) is calculated as

d4=c4c^4=4.770684.7688=0.00188

Hence, the differences obtained are 0.02717, 0.0523, 0.02321, and 0.00188.

Section 2

To determine

The number of positive differences between the observed and the predicted counts.

Section 2

Expert Solution
Check Mark

Answer to Problem 74E

Solution: There are three difference values that are positive as 0.02717, 0.02321, and 0.00188.

Explanation of Solution

The differences between the observed and the predicted counts are obtained in the Section 1 of part (b) above as

d1=0.02717

d2=0.0523

d3=0.02321

d4=0.00188

Clearly, the positive difference values are 0.02717, 0.02321, and 0.00188.

Hence, three differences d1, d3, and d4 are positive.

To determine

The number of negative differences between the observed and the predicted counts.

Expert Solution
Check Mark

Answer to Problem 74E

Solution: The negative difference is

d2=0.0523

Explanation of Solution

The differences between the observed and the predicted counts are obtained in the Section 1 of part (b) above as:

d1=0.02717

d2=0.0523

d3=0.02321

d4=0.00188

Clearly, the negative difference value is 0.0523. Hence, the difference d2 is negative.

(c)

Section 1

To determine

To find: The squares of the differences obtained in part (b).

(c)

Section 1

Expert Solution
Check Mark

Answer to Problem 74E

Solution: The squares of the differences are

d12=0.000738

d22=0.002735

d32=0.000539

d42=3.61×106(approximately 0)

Explanation of Solution

Calculation: The differences obtained in the part (b) above are

d1=0.02717

d2=0.0523

d3=0.02321

d4=0.00188

The squares of the differences obtained are calculated as follows:

The square of the difference (d1) is calculated as:

d12=d1×d1=0.02717×0.02717=0.000738

The square of the difference (d2) is calculated as:

d22=d2×d2=(0.0523)×(0.0523)=0.002735

The square of the difference (d3) is calculated as:

d32=d3×d3=0.02321×0.02321=0.000539

The square of the difference (d4) is calculated as:

d42=d4×d4=0.0019×0.0019=3.61×1060

Hence, the squares of the differences obtained are 0.000738, 0.002735, 0.000539 and 3.61×106(approximately equal to 0).

Section 2

To determine

To find: The sum of the squares of the differences obtained in Section 1 above.

Section 2

Expert Solution
Check Mark

Answer to Problem 74E

Solution: The sum of the squares of the differences obtained is 0.004011_.

Explanation of Solution

Calculation: The squares of the differences are obtained in Section 1 above as

d12=0.000738

d22=0.002735

d32=0.000538

d42=3.61×106(approximately 0).

The sum of the differences (Σdn2) where n=1,2,3, and 4 is calculated as follows:

Σdn2n=14=d12+d22+d32+d42=(0.000738)+(0.002735)+(0.000538)+(0)=0.004011

Hence, the sum of the differences obtained is 0.004011.

(d)

Section 1

To determine

To find: The predicted count values for the new regression equation.

(d)

Section 1

Expert Solution
Check Mark

Answer to Problem 74E

Solution: The predicted count values obtained are:

c^1=6.8

c^2=6.4

c^3=6

c^4=5.6

Explanation of Solution

Calculation: The provided least-squares regression equation is

log count=7(0.2×time)

The time values are provided in the data as 1, 3, 5, and 7.

Substituting the values of time in the above linear regression equation, the following results are obtained:

For time(t1) as 1, the predicted count (c^5) is calculated as

log count=7(0.2×time)=7(0.2×t5)=7(0.2×1)c^5=6.8

For time(t2) as 3, the predicted count (c^6) is calculated as

log count=7(0.2×time)=7(0.2×t6)=7(0.2×3)c^6=6.4

For time(t3) as 5, the predicted count (c^7) is calculated as

log count=7(0.2×time)=7(0.2×t7)=7(0.2×5)c^7=6

For time(t4) as 7, the predicted count (c^8) is calculated as

log count=7(0.2×time)=7(0.2×t8)=7(0.2×7)c^8=5.6

Hence, the predicted count values obtained are 6.8, 6.4, 6, and 5.6 for the times 1, 3, 5 and 7 respectively.

Section 2:

To determine

To find: The difference between the observed and the predicted counts.

Section 2:

Expert Solution
Check Mark

Answer to Problem 74E

Solution: The differences obtained are 0.44043, 0.6411, 0.68679, and 0.82932.

Explanation of Solution

Calculation: The observed counts are provided in the exercise as:

c1=6.35957

c2=5.75890

c3=5.31321

c4=4.77068

The predicted counts obtained from section 1 above are:

c^5=6.8

c^6=6.4

c^7=6

c^8=5.6

In both the cases, the time is 1, 3, 5 and 7 respectively.

The differences (dis where i are 5, 6, 7, and 8) between the observed and the predicted counts are calculated as follows:

For time(t1) as 1, the difference (d5) is calculated as

d5=c1c^5=6.359576.8=0.44043

For time(t2) as 3, the difference (d6) is calculated as

d6=c2c^6=5.758906.4=0.6411

For time(t3) as 5, the difference (d7) is calculated as

d7=c3c^7=5.313216=0.68679

For time(t4) as 7, the difference (d8) is calculated as

d8=c4c^8=4.770685.6=0.82932

Hence, the differences obtained are 0.44043, 0.6411, 0.68679, and 0.82932.

Section 3

To determine

To find: The number of positive differences between the observed and the predicted counts.

Section 3

Expert Solution
Check Mark

Answer to Problem 74E

Solution: None of the differences are positive.

Explanation of Solution

The differences between the observed and the predicted counts are obtained in the Section 2 above as 0.44043, 0.6411, 0.68679, and 0.82932.

Clearly, none of the difference values are positive. Hence, zero difference values are positive.

To determine

To find: The number of negative differences between the observed and the predicted counts.

Expert Solution
Check Mark

Answer to Problem 74E

Solution: There are four negative differences as: 0.44043, 0.6411, 0.68679, and 0.82932.

Explanation of Solution

The differences between the observed and the predicted counts are obtained in the Section 1 above as:

d5=0.44043

d6=0.6411

d7=0.68679

d8=0.82932

Clearly, all of the values are negative. Hence, the differences d5, d6, d7, and d8 are negative.

Section 4

To determine

To find: The squares of the differences obtained in section 2 of part (d).

Section 4

Expert Solution
Check Mark

Answer to Problem 74E

Solution: The squares of the differences are 0.19397859, 0.41100921, 0.47168050, and 0.68777166.

Explanation of Solution

Calculation: The differences obtained in the Section 2 of part (d) above are

d5=0.44043

d6=0.6411

d7=0.68679

d8=0.82932

The squares of the differences obtained are calculated as follows:

The square of the difference (d5) is calculated as

d52=d5×d5=(0.44043)×(0.44043)=0.19397859

The square of the difference (d6) is calculated as

d62=d6×d6=(0.6411)×(0.6411)=0.41100921

The square of the difference (d7) is calculated as

d72=d7×d7=(0.68679)×(0.68679)=0.47168050

The square of the difference (d8) is calculated as

d82=d8×d8=(0.82932)×(0.82932)=0.68777166

Hence, the squares of the differences obtained are 0.19397859, 0.41100921, 0.47168050, and 0.68777166.

Section 5

To determine

To find: The sum of the squares of the differences obtained in Section 3 above.

Section 5

Expert Solution
Check Mark

Answer to Problem 74E

Solution: The sum of the differences is 1.76443996_.

Explanation of Solution

Calculation: The squares of the differences are obtained in Section 4 above as

d52=0.19397859

d62=0.41100921

d72=0.47168050

d82=0.68777166

The sum of the differences (Σdn2) where nis 5, 6, 7, and 8 is calculated as follows:

Σdn2n=58=d52+d62+d72+d82=(0.19397859)+(0.41100921)+(0.47168050)+(0.68777166)=1.76443996

Hence, the sum of the differences is 1.76443996.

(e)

To determine

To explain: The least-squares inference based on the calculations performed.

(e)

Expert Solution
Check Mark

Answer to Problem 74E

Solution: The following least-squares regression is a better measure of the relationship between the log count and the time.

log count=6.593(0.2606×time)

Explanation of Solution

In a linear least-squares equation, the graph of the residuals must have an unbiased pattern, that is, the scatter plot should be scattered above and below the x-axis. The residual plots must be both positive and negative. In the calculations performed above, it is observed that for the regression line

log count=6.593(0.2606×time)

The differences (residuals) are both positive and negative values whereas for the regression line

log count=7(0.2×time)

all the differences (residuals) are extremely low negative values. Thus, the residual plots for the first regression line will lie both below and above the x-axis. Also, its predicted regression line will lie much near to the observed regression line. Whereas, the residual plots for the second regression line will lie only below the x-axis, indicating it to be not a much accurate measure. Also, with such low negative differences, the predicted regression line will lie far from the observed regression line. Hence, the first line better predicts the relationship between the variables log count and the time because it gives a better approximate value of the response variable.

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Chapter 2 Solutions

Introduction to the Practice of Statistics

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