Compare the projects based on the present worth.

Question

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Answer 1

Question

Chapter 5, Problem 15P

To determine

Compare the projects based on the present worth.

Expert Solution & Answer

Explanation of Solution

Project A: Cost (C) is $80,000, salvage value (S) is $15,000, operating cost (OC) is $30,000 in year 1 and it increases (IO) by $4,000 each year, time period (n) is 3 years, and interest rate (i) is 12% per year.

Project A: Cost (C) is $120,000, salvage value (S) is $40,000, operating cost (OC) is $8,000 in year 1 and it increases (IO) by $6,500 each year, time period (n) is 3 years, and interest rate (i) is 12% per year.

The present worth of project A (PW) can be calculated as follows:

PWA=−C−(OC((1+i)n−1i(1+i)n)+IO×1i((1+i)n−1i(1+i)n−n(1+i)n))+SV(1+i)n=−80,000−(30,000((1+0.12)3−10.12(1+0.12)3)+4,000×10.12((1+0.12)3−10.12(1+0.12)3−3(1+0.12)3))+15,000(1+0.12)3=−80,000−(30,000(1.404928−10.12(1.404928))+4,000×8.3333(1.404928−10.12(1.404928)−31.404928))+15,0001.404928=−80,000−(30,000(0.4049280.16859136)+4,000×8.3333(0.4049280.16859136−2.135341))+10,676.7037=−80,000−(30,000(2.401831)+4,000×8.3333(2.401831−2.135341))+10,676.7037=−80,000−(30,000(2.401831)+4,000×8.3333(0.26649))+10,676.7037=−80,000−(72,054.93+8,882.96)+10,676.7037=−150,258.19

The present worth of project A is -$142,444.19.

The present worth of project B (PW) can be calculated as follows:

PWA=−C−(OC((1+i)n−1i(1+i)n)+IO×1i((1+i)n−1i(1+i)n−n(1+i)n))+SV(1+i)n=−120,000−(8,000((1+0.12)3−10.12(1+0.12)3)+6,500×10.12((1+0.12)3−10.12(1+0.12)3−3(1+0.12)3))+40,000(1+0.12)3=−120,000−(8,000(1.404928−10.12(1.404928))+6,500×8.3333(1.404928−10.12(1.404928)−31.404928))+40,0001.404928=−120,000−(8,000(0.4049280.16859136)+6,500×8.3333(0.4049280.16859136−2.135341))+28,471.2099=−120,000−(8,000(2.401831)+6,500×8.3333(2.401831−2.135341))+28,471.2099=−120,000−(8,000(2.401831)+6,500×8.3333(0.26649))+28,471.2099=−120,000−(19,214.648+14,434.82)+28,471.2099=−112,475.62125,178.26

The present worth of project B is -$125,178.56. Since the present worth of project B is greater than project A, select project B.

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Suppose that a random sample of 216 twenty-year-old men is selected from a population and that their heights and weights are recorded. A regression of weight on height yields Weight = (-107.3628) + 4.2552 x Height, R2 = 0.875, SER = 11.0160 (2.3220) (0.3348) where Weight is measured in pounds and Height is measured in inches. A man has a late growth spurt and grows 1.6200 inches over the course of a year. Construct a confidence interval of 90% for the person's weight gain. The 90% confidence interval for the person's weight gain is ( ☐ ☐) (in pounds). (Round your responses to two decimal places.)

Suppose that (Y, X) satisfy the assumptions specified here. A random sample of n = 498 is drawn and yields Ŷ= 6.47 + 5.66X, R2 = 0.83, SER = 5.3 (3.7) (3.4) Where the numbers in parentheses are the standard errors of the estimated coefficients B₁ = 6.47 and B₁ = 5.66 respectively. Suppose you wanted to test that B₁ is zero at the 5% level. That is, Ho: B₁ = 0 vs. H₁: B₁ #0 Report the t-statistic and p-value for this test. Definition The t-statistic is (Round your response to two decimal places) ☑ The Least Squares Assumptions Y=Bo+B₁X+u, i = 1,..., n, where 1. The error term u; has conditional mean zero given X;: E (u;|X;) = 0; 2. (Y;, X¡), i = 1,..., n, are independent and identically distributed (i.i.d.) draws from i their joint distribution; and 3. Large outliers are unlikely: X; and Y, have nonzero finite fourth moments.

Answer 2

Question

Chapter 5, Problem 15P

To determine

Compare the projects based on the present worth.

Expert Solution & Answer

Explanation of Solution

Project A: Cost (C) is $80,000, salvage value (S) is $15,000, operating cost (OC) is $30,000 in year 1 and it increases (IO) by $4,000 each year, time period (n) is 3 years, and interest rate (i) is 12% per year.

Project A: Cost (C) is $120,000, salvage value (S) is $40,000, operating cost (OC) is $8,000 in year 1 and it increases (IO) by $6,500 each year, time period (n) is 3 years, and interest rate (i) is 12% per year.

The present worth of project A (PW) can be calculated as follows:

PWA=−C−(OC((1+i)n−1i(1+i)n)+IO×1i((1+i)n−1i(1+i)n−n(1+i)n))+SV(1+i)n=−80,000−(30,000((1+0.12)3−10.12(1+0.12)3)+4,000×10.12((1+0.12)3−10.12(1+0.12)3−3(1+0.12)3))+15,000(1+0.12)3=−80,000−(30,000(1.404928−10.12(1.404928))+4,000×8.3333(1.404928−10.12(1.404928)−31.404928))+15,0001.404928=−80,000−(30,000(0.4049280.16859136)+4,000×8.3333(0.4049280.16859136−2.135341))+10,676.7037=−80,000−(30,000(2.401831)+4,000×8.3333(2.401831−2.135341))+10,676.7037=−80,000−(30,000(2.401831)+4,000×8.3333(0.26649))+10,676.7037=−80,000−(72,054.93+8,882.96)+10,676.7037=−150,258.19

The present worth of project A is -$142,444.19.

The present worth of project B (PW) can be calculated as follows:

PWA=−C−(OC((1+i)n−1i(1+i)n)+IO×1i((1+i)n−1i(1+i)n−n(1+i)n))+SV(1+i)n=−120,000−(8,000((1+0.12)3−10.12(1+0.12)3)+6,500×10.12((1+0.12)3−10.12(1+0.12)3−3(1+0.12)3))+40,000(1+0.12)3=−120,000−(8,000(1.404928−10.12(1.404928))+6,500×8.3333(1.404928−10.12(1.404928)−31.404928))+40,0001.404928=−120,000−(8,000(0.4049280.16859136)+6,500×8.3333(0.4049280.16859136−2.135341))+28,471.2099=−120,000−(8,000(2.401831)+6,500×8.3333(2.401831−2.135341))+28,471.2099=−120,000−(8,000(2.401831)+6,500×8.3333(0.26649))+28,471.2099=−120,000−(19,214.648+14,434.82)+28,471.2099=−112,475.62125,178.26

The present worth of project B is -$125,178.56. Since the present worth of project B is greater than project A, select project B.

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Answer 3

Question

Chapter 5, Problem 15P

To determine

Compare the projects based on the present worth.

Expert Solution & Answer

Explanation of Solution

Project A: Cost (C) is $80,000, salvage value (S) is $15,000, operating cost (OC) is $30,000 in year 1 and it increases (IO) by $4,000 each year, time period (n) is 3 years, and interest rate (i) is 12% per year.

Project A: Cost (C) is $120,000, salvage value (S) is $40,000, operating cost (OC) is $8,000 in year 1 and it increases (IO) by $6,500 each year, time period (n) is 3 years, and interest rate (i) is 12% per year.

The present worth of project A (PW) can be calculated as follows:

PWA=−C−(OC((1+i)n−1i(1+i)n)+IO×1i((1+i)n−1i(1+i)n−n(1+i)n))+SV(1+i)n=−80,000−(30,000((1+0.12)3−10.12(1+0.12)3)+4,000×10.12((1+0.12)3−10.12(1+0.12)3−3(1+0.12)3))+15,000(1+0.12)3=−80,000−(30,000(1.404928−10.12(1.404928))+4,000×8.3333(1.404928−10.12(1.404928)−31.404928))+15,0001.404928=−80,000−(30,000(0.4049280.16859136)+4,000×8.3333(0.4049280.16859136−2.135341))+10,676.7037=−80,000−(30,000(2.401831)+4,000×8.3333(2.401831−2.135341))+10,676.7037=−80,000−(30,000(2.401831)+4,000×8.3333(0.26649))+10,676.7037=−80,000−(72,054.93+8,882.96)+10,676.7037=−150,258.19

The present worth of project A is -$142,444.19.

The present worth of project B (PW) can be calculated as follows:

PWA=−C−(OC((1+i)n−1i(1+i)n)+IO×1i((1+i)n−1i(1+i)n−n(1+i)n))+SV(1+i)n=−120,000−(8,000((1+0.12)3−10.12(1+0.12)3)+6,500×10.12((1+0.12)3−10.12(1+0.12)3−3(1+0.12)3))+40,000(1+0.12)3=−120,000−(8,000(1.404928−10.12(1.404928))+6,500×8.3333(1.404928−10.12(1.404928)−31.404928))+40,0001.404928=−120,000−(8,000(0.4049280.16859136)+6,500×8.3333(0.4049280.16859136−2.135341))+28,471.2099=−120,000−(8,000(2.401831)+6,500×8.3333(2.401831−2.135341))+28,471.2099=−120,000−(8,000(2.401831)+6,500×8.3333(0.26649))+28,471.2099=−120,000−(19,214.648+14,434.82)+28,471.2099=−112,475.62125,178.26

The present worth of project B is -$125,178.56. Since the present worth of project B is greater than project A, select project B.

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Answer 4

Question

Chapter 5, Problem 15P

To determine

Compare the projects based on the present worth.

Expert Solution & Answer

Explanation of Solution

Project A: Cost (C) is $80,000, salvage value (S) is $15,000, operating cost (OC) is $30,000 in year 1 and it increases (IO) by $4,000 each year, time period (n) is 3 years, and interest rate (i) is 12% per year.

Project A: Cost (C) is $120,000, salvage value (S) is $40,000, operating cost (OC) is $8,000 in year 1 and it increases (IO) by $6,500 each year, time period (n) is 3 years, and interest rate (i) is 12% per year.

The present worth of project A (PW) can be calculated as follows:

PWA=−C−(OC((1+i)n−1i(1+i)n)+IO×1i((1+i)n−1i(1+i)n−n(1+i)n))+SV(1+i)n=−80,000−(30,000((1+0.12)3−10.12(1+0.12)3)+4,000×10.12((1+0.12)3−10.12(1+0.12)3−3(1+0.12)3))+15,000(1+0.12)3=−80,000−(30,000(1.404928−10.12(1.404928))+4,000×8.3333(1.404928−10.12(1.404928)−31.404928))+15,0001.404928=−80,000−(30,000(0.4049280.16859136)+4,000×8.3333(0.4049280.16859136−2.135341))+10,676.7037=−80,000−(30,000(2.401831)+4,000×8.3333(2.401831−2.135341))+10,676.7037=−80,000−(30,000(2.401831)+4,000×8.3333(0.26649))+10,676.7037=−80,000−(72,054.93+8,882.96)+10,676.7037=−150,258.19

The present worth of project A is -$142,444.19.

The present worth of project B (PW) can be calculated as follows:

PWA=−C−(OC((1+i)n−1i(1+i)n)+IO×1i((1+i)n−1i(1+i)n−n(1+i)n))+SV(1+i)n=−120,000−(8,000((1+0.12)3−10.12(1+0.12)3)+6,500×10.12((1+0.12)3−10.12(1+0.12)3−3(1+0.12)3))+40,000(1+0.12)3=−120,000−(8,000(1.404928−10.12(1.404928))+6,500×8.3333(1.404928−10.12(1.404928)−31.404928))+40,0001.404928=−120,000−(8,000(0.4049280.16859136)+6,500×8.3333(0.4049280.16859136−2.135341))+28,471.2099=−120,000−(8,000(2.401831)+6,500×8.3333(2.401831−2.135341))+28,471.2099=−120,000−(8,000(2.401831)+6,500×8.3333(0.26649))+28,471.2099=−120,000−(19,214.648+14,434.82)+28,471.2099=−112,475.62125,178.26

The present worth of project B is -$125,178.56. Since the present worth of project B is greater than project A, select project B.

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Answer 5

Chapter 5, Problem 15P

To determine

Compare the projects based on the present worth.

Expert Solution & Answer

Explanation of Solution

Project A: Cost (C) is $80,000, salvage value (S) is $15,000, operating cost (OC) is $30,000 in year 1 and it increases (IO) by $4,000 each year, time period (n) is 3 years, and interest rate (i) is 12% per year.

Project A: Cost (C) is $120,000, salvage value (S) is $40,000, operating cost (OC) is $8,000 in year 1 and it increases (IO) by $6,500 each year, time period (n) is 3 years, and interest rate (i) is 12% per year.

The present worth of project A (PW) can be calculated as follows:

PWA=−C−(OC((1+i)n−1i(1+i)n)+IO×1i((1+i)n−1i(1+i)n−n(1+i)n))+SV(1+i)n=−80,000−(30,000((1+0.12)3−10.12(1+0.12)3)+4,000×10.12((1+0.12)3−10.12(1+0.12)3−3(1+0.12)3))+15,000(1+0.12)3=−80,000−(30,000(1.404928−10.12(1.404928))+4,000×8.3333(1.404928−10.12(1.404928)−31.404928))+15,0001.404928=−80,000−(30,000(0.4049280.16859136)+4,000×8.3333(0.4049280.16859136−2.135341))+10,676.7037=−80,000−(30,000(2.401831)+4,000×8.3333(2.401831−2.135341))+10,676.7037=−80,000−(30,000(2.401831)+4,000×8.3333(0.26649))+10,676.7037=−80,000−(72,054.93+8,882.96)+10,676.7037=−150,258.19

The present worth of project A is -$142,444.19.

The present worth of project B (PW) can be calculated as follows:

PWA=−C−(OC((1+i)n−1i(1+i)n)+IO×1i((1+i)n−1i(1+i)n−n(1+i)n))+SV(1+i)n=−120,000−(8,000((1+0.12)3−10.12(1+0.12)3)+6,500×10.12((1+0.12)3−10.12(1+0.12)3−3(1+0.12)3))+40,000(1+0.12)3=−120,000−(8,000(1.404928−10.12(1.404928))+6,500×8.3333(1.404928−10.12(1.404928)−31.404928))+40,0001.404928=−120,000−(8,000(0.4049280.16859136)+6,500×8.3333(0.4049280.16859136−2.135341))+28,471.2099=−120,000−(8,000(2.401831)+6,500×8.3333(2.401831−2.135341))+28,471.2099=−120,000−(8,000(2.401831)+6,500×8.3333(0.26649))+28,471.2099=−120,000−(19,214.648+14,434.82)+28,471.2099=−112,475.62125,178.26

The present worth of project B is -$125,178.56. Since the present worth of project B is greater than project A, select project B.

Answer 6

Chapter 5, Problem 15P

To determine

Compare the projects based on the present worth.

Expert Solution & Answer

Explanation of Solution

Project A: Cost (C) is $80,000, salvage value (S) is $15,000, operating cost (OC) is $30,000 in year 1 and it increases (IO) by $4,000 each year, time period (n) is 3 years, and interest rate (i) is 12% per year.

Project A: Cost (C) is $120,000, salvage value (S) is $40,000, operating cost (OC) is $8,000 in year 1 and it increases (IO) by $6,500 each year, time period (n) is 3 years, and interest rate (i) is 12% per year.

The present worth of project A (PW) can be calculated as follows:

PWA=−C−(OC((1+i)n−1i(1+i)n)+IO×1i((1+i)n−1i(1+i)n−n(1+i)n))+SV(1+i)n=−80,000−(30,000((1+0.12)3−10.12(1+0.12)3)+4,000×10.12((1+0.12)3−10.12(1+0.12)3−3(1+0.12)3))+15,000(1+0.12)3=−80,000−(30,000(1.404928−10.12(1.404928))+4,000×8.3333(1.404928−10.12(1.404928)−31.404928))+15,0001.404928=−80,000−(30,000(0.4049280.16859136)+4,000×8.3333(0.4049280.16859136−2.135341))+10,676.7037=−80,000−(30,000(2.401831)+4,000×8.3333(2.401831−2.135341))+10,676.7037=−80,000−(30,000(2.401831)+4,000×8.3333(0.26649))+10,676.7037=−80,000−(72,054.93+8,882.96)+10,676.7037=−150,258.19

The present worth of project A is -$142,444.19.

The present worth of project B (PW) can be calculated as follows:

PWA=−C−(OC((1+i)n−1i(1+i)n)+IO×1i((1+i)n−1i(1+i)n−n(1+i)n))+SV(1+i)n=−120,000−(8,000((1+0.12)3−10.12(1+0.12)3)+6,500×10.12((1+0.12)3−10.12(1+0.12)3−3(1+0.12)3))+40,000(1+0.12)3=−120,000−(8,000(1.404928−10.12(1.404928))+6,500×8.3333(1.404928−10.12(1.404928)−31.404928))+40,0001.404928=−120,000−(8,000(0.4049280.16859136)+6,500×8.3333(0.4049280.16859136−2.135341))+28,471.2099=−120,000−(8,000(2.401831)+6,500×8.3333(2.401831−2.135341))+28,471.2099=−120,000−(8,000(2.401831)+6,500×8.3333(0.26649))+28,471.2099=−120,000−(19,214.648+14,434.82)+28,471.2099=−112,475.62125,178.26

The present worth of project B is -$125,178.56. Since the present worth of project B is greater than project A, select project B.

Answer 7

Chapter 5, Problem 15P

To determine

Compare the projects based on the present worth.

Answer 8

Expert Solution & Answer

Explanation of Solution

Project A: Cost (C) is $80,000, salvage value (S) is $15,000, operating cost (OC) is $30,000 in year 1 and it increases (IO) by $4,000 each year, time period (n) is 3 years, and interest rate (i) is 12% per year.

Project A: Cost (C) is $120,000, salvage value (S) is $40,000, operating cost (OC) is $8,000 in year 1 and it increases (IO) by $6,500 each year, time period (n) is 3 years, and interest rate (i) is 12% per year.

The present worth of project A (PW) can be calculated as follows:

PWA=−C−(OC((1+i)n−1i(1+i)n)+IO×1i((1+i)n−1i(1+i)n−n(1+i)n))+SV(1+i)n=−80,000−(30,000((1+0.12)3−10.12(1+0.12)3)+4,000×10.12((1+0.12)3−10.12(1+0.12)3−3(1+0.12)3))+15,000(1+0.12)3=−80,000−(30,000(1.404928−10.12(1.404928))+4,000×8.3333(1.404928−10.12(1.404928)−31.404928))+15,0001.404928=−80,000−(30,000(0.4049280.16859136)+4,000×8.3333(0.4049280.16859136−2.135341))+10,676.7037=−80,000−(30,000(2.401831)+4,000×8.3333(2.401831−2.135341))+10,676.7037=−80,000−(30,000(2.401831)+4,000×8.3333(0.26649))+10,676.7037=−80,000−(72,054.93+8,882.96)+10,676.7037=−150,258.19

The present worth of project A is -$142,444.19.

The present worth of project B (PW) can be calculated as follows:

PWA=−C−(OC((1+i)n−1i(1+i)n)+IO×1i((1+i)n−1i(1+i)n−n(1+i)n))+SV(1+i)n=−120,000−(8,000((1+0.12)3−10.12(1+0.12)3)+6,500×10.12((1+0.12)3−10.12(1+0.12)3−3(1+0.12)3))+40,000(1+0.12)3=−120,000−(8,000(1.404928−10.12(1.404928))+6,500×8.3333(1.404928−10.12(1.404928)−31.404928))+40,0001.404928=−120,000−(8,000(0.4049280.16859136)+6,500×8.3333(0.4049280.16859136−2.135341))+28,471.2099=−120,000−(8,000(2.401831)+6,500×8.3333(2.401831−2.135341))+28,471.2099=−120,000−(8,000(2.401831)+6,500×8.3333(0.26649))+28,471.2099=−120,000−(19,214.648+14,434.82)+28,471.2099=−112,475.62125,178.26