Bonnie and Clyde are the only two shareholders in Getaway Corporation. Bonnie owns 60 shares with a basis of $6,600, and Clyde owns the remaining 40 shares with a basis of $15,000. At year-end, Getaway is considering different alternatives for redeeming some shares of stock. Evaluate whether each of the following stock redemption transactions will qualify for sale and exchange treatment.
a. Getaway redeems 10 of Bonnie’s shares for $5,000. Getaway has $26,000 of E&P at year-end and Bonnie is unrelated to Clyde.
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b. Getaway redeems 29 of Bonnie’s shares for $10,000. Getaway has $26,000 of E&P at year-end and Bonnie is unrelated to Clyde.
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c. Getaway redeems 8 of Clyde’s shares for $5,500. Getaway has $26,000 of E&P at year-end and Clyde is unrelated to Bonnie.
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In: Accounting
Suppose that, prior to the passage of the Truth in Lending Simplification Act and Regulation Z, the demand for consumer loans was given by Qdpre-TILSA = 14 -90P (in billions of dollars) and the supply of consumer loans by credit unions and other lending institutions was QSpre-TILSA = 6 + 70P (in billions of dollars). The TILSA now requires lenders to provide consumers with complete information about the rights and responsibilities of entering into a lending relationship with the institution, and as a result, the demand for loans increased to Qdpost-TILSA = 26 -90P (in billions of dollars). However, the TILSA also imposed “compliance costs” on lending institutions, and this reduced the supply of consumer loans to QSpost-TILSA = 2 + 70P (in billions of dollars). Based on this information, compare the equilibrium price and quantity of consumer loans before and after the Truth in Lending Simplification Act.(Note: Q is measured in billions of dollars and P is the interest rate). Instruction: Enter your responses for the equilibrium price in percentage terms, and round all responses to one decimal place. Equilibrium price (interest rate) before TILSA: percent Equilibrium quantity (in billions of dollars) before TILSA: $ billion Equilibrium price (interest rate) after TILSA: percent Equilibrium quantity (in billions of dollars) after TILSA: $ billion
In: Economics
Suppose that, prior to the passage of the Truth in Lending Simplification Act and Regulation Z, the demand for consumer loans was given by Qdpre-TILSA = 14 -90P (in billions of dollars) and the supply of consumer loans by credit unions and other lending institutions was QSpre-TILSA = 6 + 70P (in billions of dollars). The TILSA now requires lenders to provide consumers with complete information about the rights and responsibilities of entering into a lending relationship with the institution, and as a result, the demand for loans increased to Qdpost-TILSA = 26 -90P (in billions of dollars). However, the TILSA also imposed “compliance costs” on lending institutions, and this reduced the supply of consumer loans to QSpost-TILSA = 2 + 70P (in billions of dollars).
Based on this information, compare the equilibrium price and
quantity of consumer loans before and after the Truth in Lending
Simplification Act.(Note: Q is measured in
billions of dollars and P is the interest rate).
Instruction: Enter your responses for the
equilibrium price in percentage terms, and round all responses to
one decimal place.
Equilibrium price (interest rate) before TILSA: percent
Equilibrium quantity
(in billions of dollars) before TILSA:
$ billion
Equilibrium price (interest rate) after TILSA: percent
Equilibrium quantity (in billions of dollars) after TILSA: $ billion
In: Economics
Suppose that, prior to the passage of the Truth in Lending Simplification Act and Regulation Z, the demand for consumer loans was given by Qdpre-TILSA = 12 -100P (in billions of dollars) and the supply of consumer loans by credit unions and other lending institutions was QSpre-TILSA = 5 + 150P (in billions of dollars). The TILSA now requires lenders to provide consumers with complete information about the rights and responsibilities of entering into a lending relationship with the institution, and as a result, the demand for loans increased to Qdpost-TILSA = 18 -100P (in billions of dollars). However, the TILSA also imposed “compliance costs” on lending institutions, and this reduced the supply of consumer loans to QSpost-TILSA = 3 + 150P (in billions of dollars).
Based on this information, compare the equilibrium price and
quantity of consumer loans before and after the Truth in Lending
Simplification Act.(Note: Q is measured in
billions of dollars and P is the interest rate).
Instruction: Enter your responses for the
equilibrium price in percentage terms, and round all responses to
one decimal place.
Equilibrium price (interest rate) before TILSA: ____ percent
Equilibrium quantity (in billions of dollars) before TILSA: $
___ billion
Equilibrium price (interest rate) after TILSA: _____percent
Equilibrium quantity (in billions of dollars) after TILSA: $ _____billion
In: Accounting
Do students reduce study time in classes where they achieve a higher midterm score? In a Journal of Economic Education article (Winter 2005), Gregory Krohn and Catherine O’Connor studied student effort and performance in a class over a semester. In an intermediate macroeconomics course, they found that “students respond to higher midterm scores by reducing the number of hours they subsequently allocate to studying for the course.” Suppose that a random sample of n = 8 students who performed well on the midterm exam was taken and weekly study times before and after the exam were compared. The resulting data are given in Table 10.6. Assume that the population of all possible paired differences is normally distributed.
Table 10.6
| Weekly Study Time Data for Students Who Perform Well on the MidTerm | ||||||||
| Students | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| Before | 14 | 12 | 14 | 13 | 15 | 12 | 18 | 17 |
| After | 6 | 7 | 4 | 9 | 10 | 4 | 8 | 3 |
Paired T-Test and CI: Study Before, Study After
| Paired T for Study Before - Study After | ||||
| N | Mean | StDev | SE Mean | |
| StudyBefore | 8 | 14.3750 | 2.1998 | .7778 |
| StudyAfter | 8 | 6.3750 | 2.5600 | .9051 |
| Difference | 8 | 8.00000 | 3.33809 | 1.18019 |
95% CI for mean difference: (5.20929, 10.79071)
T-Test of mean difference = 0 (vs not = 0): T-Value = 6.78, P-Value = .0003
(a) Set up the null and alternative hypotheses to test whether there is a difference in the true mean study time before and after the midterm exam.
H0: µd = versus Ha: µd ≠
(b) Above we present the MINITAB output for the paired differences test. Use the output and critical values to test the hypotheses at the .10, .05, and .01 level of significance. Has the true mean study time changed? (Round your answer to 2 decimal places.)
t = We have (Click to select)strongvery strongextremely strongno evidence.
(c) Use the p-value to test the hypotheses at the .10, .05, and .01 level of significance. How much evidence is there against the null hypothesis?
There is (Click to select)extermly strong evidenceno evidencestrong evidencevery strong evidence against the null hypothesis.
In: Statistics and Probability
1-Look up the requested values using book tables. You may check with a calculator, but enter the table value.
Let T denote a t-distribution variable with 14 degrees of freedom. Find:
P(T>1.761)=?
Note: Enter X.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 27 is entered as 27.00, 3.5 is entered as 3.50, 0.3750 is entered as 0.38
2-Look up the requested values using book tables. You may check with a calculator, but enter the table value.
Find the normal distribution P VALUE used for a Greater Than one-sided alternative hypothesis test with a test statistic z=2.32.
Note: Enter X.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 27 is entered as 27.00, 3.5 is entered as 3.50, 0.3750 is entered as 0.3
3-Look up the requested values using book tables. You may check with a calculator, but enter the table value.
Find the standard normal distribution Critical VALUE used for a Less Than one-sided alternative hypothesis test with a 1% significance level.
Note: Enter X.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 27 is entered as 27.00, 3.5 is entered as 3.50, 0.3750 is entered as 0.3
4-Look up the requested values using book tables. You may check with a calculator, but enter the table value.
Let T denote a t-distribution variable with 25 degrees of freedom. Find:
P(T<-0.684)=?
Note: Enter X.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 27 is entered as 27.00, 3.5 is entered as 3.50, 0.3750 is entered as 0.38
5-Use a CALCULATOR to compute the below probability.
Suppose a basketball player hits and average of 60% of his free throws. In a game with 15 independent free throws, what is the probability he makes exactly 12 baskets?
Note: Enter X.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 27 is entered as 27.00, 3.5 is entered as 3.50, 0.3750 is entered as 0.3
In: Statistics and Probability
Do students reduce study time in classes where they achieve a higher midterm score? In a Journal of Economic Education article (Winter 2005), Gregory Krohn and Catherine O’Connor studied student effort and performance in a class over a semester. In an intermediate macroeconomics course, they found that “students respond to higher midterm scores by reducing the number of hours they subsequently allocate to studying for the course.” Suppose that a random sample of n = 8 students who performed well on the midterm exam was taken and weekly study times before and after the exam were compared. The resulting data are given in Table 10.6. Assume that the population of all possible paired differences is normally distributed.
Table 10.6
| Weekly Study Time Data for Students Who Perform Well on the MidTerm | ||||||||
| Students | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| Before | 17 | 11 | 16 | 18 | 15 | 18 | 17 | 13 |
| After | 9 | 9 | 8 | 11 | 10 | 7 | 10 | 11 |
Paired T-Test and CI: Study Before, Study After
| Paired T for Study Before - Study After | ||||
| N | Mean | StDev | SE Mean | |
| StudyBefore | 8 | 15.6250 | 2.5036 | .8851 |
| StudyAfter | 8 | 9.3750 | 1.4079 | .4978 |
| Difference | 8 | 6.25000 | 3.10530 | 1.09789 |
95% CI for mean difference: (3.65391, 8.84609)
T-Test of mean difference = 0 (vs not = 0): T-Value = 5.69, P-Value = .0007
(a) Set up the null and alternative hypotheses to test whether there is a difference in the true mean study time before and after the midterm exam.
H0: µd = versus Ha: µd ≠
(b) Above we present the MINITAB output for the paired differences test. Use the output and critical values to test the hypotheses at the .10, .05, and .01 level of significance. Has the true mean study time changed? (Round your answer to 2 decimal places.)
t = We have (Click to select)noextremely strongvery strongstrong evidence.
(c) Use the p-value to test the hypotheses at the .10, .05, and .01 level of significance. How much evidence is there against the null hypothesis?
There is (Click to select)no evidencevery strong evidenceextermly strong evidencestrong evidence against the null hypothesis.
In: Statistics and Probability
Do students reduce study time in classes where they achieve a higher midterm score? In a Journal of Economic Education article (Winter 2005), Gregory Krohn and Catherine O’Connor studied student effort and performance in a class over a semester. In an intermediate macroeconomics course, they found that “students respond to higher midterm scores by reducing the number of hours they subsequently allocate to studying for the course.” Suppose that a random sample of n = 8 students who performed well on the midterm exam was taken and weekly study times before and after the exam were compared. The resulting data are given in Table 10.6. Assume that the population of all possible paired differences is normally distributed.
Table 10.6
| Weekly Study Time Data for Students Who Perform Well on the MidTerm | ||||||||
| Students | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| Before | 15 | 19 | 12 | 17 | 16 | 15 | 11 | 16 |
| After | 11 | 18 | 9 | 10 | 8 | 9 | 11 | 10 |
Paired T-Test and CI: Study Before, Study After
| Paired T for Study Before - Study After | ||||
| N | Mean | StDev | SE Mean | |
| StudyBefore | 8 | 15.1250 | 2.5877 | .9149 |
| StudyAfter | 8 | 10.7500 | 3.1053 | 1.0979 |
| Difference | 8 | 4.37500 | 2.87539 | 1.01660 |
95% CI for mean difference: (1.97112, 6.77888)
T-Test of mean difference = 0 (vs not = 0): T-Value = 4.30, P-Value = .0036
(a) Set up the null and alternative hypotheses to test whether there is a difference in the true mean study time before and after the midterm exam.
H0: µd = versus Ha: µd ?
(b) Above we present the MINITAB output for the paired differences test. Use the output and critical values to test the hypotheses at the .10, .05, and .01 level of significance. Has the true mean study time changed? (Round your answer to 2 decimal places.)
t = We have (Click to select)novery strongextremely strongstrong evidence.
(c) Use the p-value to test the hypotheses at the .10, .05, and .01 level of significance. How much evidence is there against the null hypothesis?
There is (Click to select)very strong evidenceextermly strong evidencestrong evidenceno evidence against the null hypothesis.
In: Statistics and Probability
Do students reduce study time in classes where they achieve a higher midterm score? In a Journal of Economic Education article (Winter 2005), Gregory Krohn and Catherine O’Connor studied student effort and performance in a class over a semester. In an intermediate macroeconomics course, they found that “students respond to higher midterm scores by reducing the number of hours they subsequently allocate to studying for the course.” Suppose that a random sample of n = 8 students who performed well on the midterm exam was taken and weekly study times before and after the exam were compared. The resulting data are given in Table 10.6. Assume that the population of all possible paired differences is normally distributed.
Table 10.6
| Weekly Study Time Data for Students Who Perform Well on the MidTerm | ||||||||
| Students | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| Before | 18 | 15 | 11 | 17 | 16 | 15 | 12 | 19 |
| After | 5 | 8 | 6 | 6 | 5 | 9 | 14 | 5 |
Paired T-Test and CI: Study Before, Study After
| Paired T for Study Before - Study After | ||||
| N | Mean | StDev | SE Mean | |
| StudyBefore | 8 | 15.3750 | 2.7742 | .9808 |
| StudyAfter | 8 | 7.2500 | 3.1053 | 1.0979 |
| Difference | 8 | 8.12500 | 5.24915 | 1.85585 |
95% CI for mean difference: (3.73660, 12.51340)
T-Test of mean difference = 0 (vs not = 0): T-Value = 4.38, P-Value = .0032
(a) Set up the null and alternative hypotheses to test whether there is a difference in the true mean study time before and after the midterm exam.
H0: µd = versus Ha: µd ?
(b) Above we present the MINITAB output for the paired differences test. Use the output and critical values to test the hypotheses at the .10, .05, and .01 level of significance. Has the true mean study time changed?(Round your answer to 2 decimal places.)
t = We have (Click to select)strongvery strongextremely strongno evidence.
(c) Use the p-value to test the hypotheses at the .10, .05, and .01 level of significance. How much evidence is there against the null hypothesis?
There is (Click to select)no evidencevery strong evidencestrong evidenceextermly strong evidence against the null hypothesis.
In: Math
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(v) GHL purchased a factory site in Malaysia on 1 April 2019 with intention for industrial use. Land prices in the area had increased significantly in the years immediately prior to 31 March 2020. Nearby sites had been acquired and converted into residential use. It is felt that, should the GHL’s |
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site also be converted into residential use, the factory site would have a market value of $27 mil- lion. $1.5 million of costs are estimated to be required to demolish the factory and to obtain plan- ning permission for the conversion. GHL was not intending to convert the site at 1 April 2019 and had not sought planning permission at that date. The current replacement cost and carrying amount of the factory site are correctly calculated as $25.1 million and $28 million respectively as at 31 March 2020 before revaluation. Fanny did not reflect the change in fair value of the factory site even the factory site is measured using the revaluation model under HKAS 16. |
Discuss the approach described in HKFRS 13 ‘Fair Value Measurement’ to measure the non- financial asset.
In: Accounting