ABC Corporation has current E&P of ($200,000). E&P at the beginning of the year was ($100,000). ABC Corp. makes a distribution of $350,000 to its sole shareholder on the least day of the year. The shareholder had a tax basis of $50,000. Determine the following (make sure that you show all calculations): How much of the distribution is treated like a dividend to the shareholder? After the distribution takes place, what is the shareholder's tax basis in the ABC Corp. stock after the distribution? What is the balance in E&P for ABC Corp. the day after the distribution? Discuss why E&P is measured. If E&P was not measured in the scenario presented above, what are the potential problems? Requirements: Back up your discussion with research from two scholarly sources (you may not use the course textbook to fulfill this requirement). Your responses to the questions above as well as your analysis should be 4-5 pages in length not counting the title and reference pages, which you must include. Your paper must be formatted according to the CSU-Global Guide to Writing and APA Requirements. I need help proof reading my paper That I have attached bellow and the original question above. If the corporation has a current E&P deficit and an accumulated E&P deficit, none of the distributions are treated as dividends. All distributions are treated as a return of capital until the shareholders stock basis is reduced to zero. Any additional amounts are treated a capital gain. In case of ABC Corp, since the distribution of 350000 is considered to be neither from Current E & P nor Accumulated E & P, but instead as return on capital, which will be tax free to the shareholder to the extent of the adjusted tax basis i-e 50000. Return on capital distributions in excess of tax basis i-e 300000 (350000-50000) will be treated as capital gains. Balance in accumulated E & P of ABC corp. will be NIL after day of distribution. Since Shareholder tax basis is 50000, will be considered as tax free return of capital . Excess received will be treated as capital gain. Since the beginning and current E&P is negative they will not consider any distribution as dividend.After the distribution the tax basis in the ABC corp. stock will still remain same and it is 50000 Balance in E&P = -100000 - 200000 + 350000 = - 650000 Earning and profit is required for calculating stockholder equity. if you do not find it then the balance sheet cannot be formed and it is published Distribution treated as dividend Beginning E&P......................................................100,000 Current E&P.........................................................200,000 Total E&P............................................................300,000 Distribution.........................................................(350,000) Return on capital....................................................(50,000) Distribution treated as dividend: Distribution..........................................................350,000 Return on capital...................................................(50,000) ........................................................................300,000 The taxability of ABC corporations $350000 distribution their sole shareholder is determined under section 316 and 301 (C). To the extent of ABC Corporations current balance of $200000 , the distribution is treated as dividend. Thus the Shareholder includes $200000 in taxable income under diction 301 (C)( 1).Of the remaining $150000 distribution $50000 is returned as the Share holders capital, and reduces Shareholders tax basis from $50000 to $0..The remaining $100000 distribution is treated as the sale of the stock, generating Capital Gain . Shareholders tax basis after stock distribution Taxable dividend..................................................................300,000 Nontaxable return of capital.......................................................50,000 Tax basis..............................................................................50,000 Tax basis Taxable dividend..................................................................300,000 Tax basis..............................................................................50,000 Taxable capital gain................................................................250,000 Balance in E&P Ending E&P = Beginning accumulated E&P +Current E&P -Distribution out of EP Beginning E&P..................................................................100,000 Current E&P.....................................................................200,000 Total E&P........................................................................300,000 Division distribution...........................................................(300,000) Balance Nil Why E&P is measured E&P is used to help in measuring the economic ability of a corporation to pay dividends. It assists corporations to know how much of the distribution to shareholders is out of the accumulated & current E&P. This is because any distribution which is not made from E&P is paid from stocks. It is therefore helpful in determining how much of the distribution should be a nontaxable return. The primary purpose of calculating E & P is to determine whether a distribution represents a taxable dividend, a non taxable return of shareholders capital, or capital gain to the recipient shareholder. Section 316 defines a dividend as an distribution of property made by a Corporation out of current or accumulated E & P. and section 301 (C) provides that distribution of constituting dividend must be included in gross income of the shareholder while amount distributed in excess of those considered as dividend are first treated as return of shareholders return of capital to the extent shareholders stock basis , with any remaining distribution treated as gain from the sale of stock resulting in capital gain. Potential problems if E&P was not measured in the scenario presented above Stock dividends could have been subjected to tax. It would have not been possible to know how much of the amount distributed could have been treated as nontaxable return. potential problem is that if E & P was not measured we would have treated $350000 as dividend but in the above scenario it is not dividend as E & P is negative and it is return on capital.
Although the normal distribution is a reasonable input distribution in many situations, it does have two potential drawbacks: (1) It allows negative values, even though they may be extremely improbable, and (2) It is a symmetric distribution. Many situations are modeled better with a distribution that allows only positive values and is skewed to the right. Two of these that have been used in many real applications are the gamma and lognormal distributions. @RISK enables you to generate observations from each of these distributions. The @RISK function for the gamma distribution is RISKGAMMA, and it takes two arguments, as in =RISKGAMMA(3,10). The first argument, which must be positive, determines the shape. The smaller it is, the more skewed the distribution is to the right; the larger it is, the more symmetric the distribution is. The second argument determines the scale, in the sense that the product of it and the first argument equals the mean of the distribution. Also, the product of the second argument and the square root of the first argument is the standard deviation of the distribution. The @RISK function for the lognormal distribution is RISKLOGNORM. It has two arguments, as in =RISKLOGNORM(40,10). These arguments are the mean and standard deviation of the distribution. Rework Example 15.2 for the following demand distributions. Do the simulated outputs have any different qualitative properties with these skewed distributions than with the triangular distribution used in the example? a. Gamma distribution with parameters 2 and 85 b. Gamma distribution with parameters 5 and 35 c. Lognormal distribution with mean 170 and standard deviation 60
How important is the assumption "The sampled population is normally distributed" to the use of Student's t-distribution? Using a computer, simulate drawing 100 samples of size 10 from each of three different types of population distributions, namely, a normal, a uniform, and an exponential. First generate 1000 data values from the population and construct a histogram to see what the population looks like. Then generate 100 samples of size 10 from the same population; each row represents a sample. Calculate the mean and standard deviation for each of the 100 samples. Calculate t_ for each of the 100 samples. Construct histograms of the 100 sample means and the 100 t_ values. (Additional details can be found in the Student Solutions Manual.) For the samples from the normal population: a. Does the distribution appear to be normal? Find percentages for intervals and compare them with the normal distribution. b. Does the distribution of t_ appear to have a t-distribution with df = 9? Find percentages for intervals and compare them with the t-distribution. For the samples from the rectangular or uniform population: c. Does the distribution appear to be normal? Find percentages for intervals and compare them with the normal distribution. d. Does the distribution of t_ appear to have a t-distribution with df = 9? Find percentages for intervals and compare them with the t-distribution. For the samples from the skewed (exponential) population: e. Does the distribution appear to be normal? Find percentages for intervals and compare them with the normal distribution. f. Does the distribution of t_ appear to have a t-distribution with df = 9? Find percentages for intervals and compare them with the t-distribution. In summary: g. In each of the preceding three situations, the sampling distribution for appears to be slightly different from the distribution of t_. Explain why. h. Does the normality condition appear to be necessary in order for the calculated test statistic t_ to have a Student's t-distribution? Explain
The exponential distribution is an example of a skewed distribution. It is defined by the density function with parameter l. The mean of an exponential distribution The exponential distribution is used to model the failure rate of electrical components. For many electrical components, a plot of the failure rate over time appears to be shaped like a bathtub. The exponential distribution is used widely in reliability engineering for the constant hazard rate portion of the bathtub curve, where components spend most of their useful life. The portion of the curve prior to the constant rate is referred to as the burn-in period, and the portion after the constant rate is referred to as the wear-out period. Another interesting property of the exponential distribution is the memoryless property. This says that the time to failure of a component does not depend on how long the component has been in use. For example, the probability that a component will fail in the next hour is the same whether the component has been in use for 10, 50, or 100 hours, and so on. In our discussion of summary statistics, we have seen that the mean is used as a measure of the center for data sets whose distribution is (roughly) symmetric with no outliers. In the presence of outliers or skewness, we used the median as the measure of the center because it is resistant to extreme values. However, for skewed distributions, it is often difficult to determine which measure of center is more useful. In some cases, the mean, though not resistant, is easier to find than the median or is more intuitive to the reader within the context of the problem. Here we look at the sampling distributions of the mean and median for two different populations, one normal (symmetric) and one exponential (skewed). 1. Using statistical software, generate 250 samples (rows) of size n = 80 (columns) from a normal distribution with mean 50 and standard deviation 10. 2. Compute the mean for the first 10 values in each row and store the values in C81. You might label this column "n10mn" for normal distribution, sample size 10, mean. 3. Compute the median for the first 10 values in each row and store the values in C82. You might label this column "n10med" for normal distribution, sample size 10, median. 4. Repeat parts 2 and 3 for samples of size 20 (C1-C20), 40 (C1-C40), and 80 (C1-C80), storing the results in consecutive columns C83, C84, C85, and so on. 5. Compute the summary statistics for your data in columns C81-C88. 6. How do the averages of your sample means compare to the actual population mean for the different sample sizes? How do the averages of your sample medians compare to the actual population median for the different sample sizes? 7. How does the standard deviation of sample means compare to the standard deviation of sample medians for different sample sizes? 8. Based on your results in parts 6 and 7, which measure of center seems more appropriate? Explain. 9. Construct histograms for each column of summary statistics. Describe the effect, if any, of increasing the sample size on the shape of the distribution of sample means and sample medians. 10. Using statistical software, generate 250 samples of size n = 80 from an exponential distribution with mean 50. This could represent the distribution of failure times for a component with a failure rate of Î» = 0.02. Note: The threshold should be set to 0. Store the results in columns C101-C180. 11. Compute the mean for the first 10 values in each row and store the values in C181. You might label this column "e10mn" for exponential distribution, sample size 10, mean. 12. Compute the median for the first 10 values in each row and store the values in C182. You might label this column "e10med" for exponential distribution, sample size 10, median. 13. Repeat parts 11 and 12 for samples of size 20 (C101-C120), 40 (C101-C140), and 80 (C101-C180), storing the results in consecutive columns C183, C184, C185, and so on. 14. Compute the summary statistics for your data in columns C181-C188. 15. How do the averages of your sample means compare to the actual population mean for the different sample sizes? How do the averages of your sample medians compare to the actual population median for the different sample sizes? 16. How does the standard deviation of sample means compare to the standard deviation of sample medians for different sample sizes? 17. Based on your results in parts 15 and 16, which measure of center seems more appropriate? Explain. 18. Construct histograms for each column of summary statistics. Describe the effect, if any, of increasing the sample size on the shape of the distribution of sample means and sample medians. 19. Can the Central Limit Theorem be used to explain any of the results in part 18? Why or why not?
Multiple choice questions: 1. A sampling distribution is the probability distribution of a. A population parameter b. A sample statistic c. Any random variable 2. Non-sampling errors are a. The errors that occur because the sample size is too large in relation to the population size b. The errors made while collecting, recording, and tabulating data c. The errors that occur because an untrained person conducts the survey 3. A sampling error is a. The difference between the value of a sample statistic based on a random sample and the value of the corresponding population parameter b. The error made while collecting, recording, and tabulating data c. The error that occurs because the sample is too small 4. The mean of the sampling distribution of is always equal to a. µ b. µ – 5 c. σ/√n 5. The condition for the standard deviation of the sample mean to be is that a. np > 5 b. n/N |< .05 c. n > 30 6. The standard deviation of the sampling distribution of the sample mean decreases when a. x increases b. n increases c. n decreases 7. When samples are selected from a normally distributed population, the sampling distribution of the sample mean has a normal distribution a. If n > 30 b. If n/N < .05 c. All the time 8. When samples are selected from a non-normally distributed population, the sampling distribution of the sample mean has an approximately normal distribution a. If n < 30 b. If n/N < .05 c. Always 9. In a sample of 200 customers of a mail-order company, 174 are found to be satisfied with the service they receive from the company. The proportion of customers in this sample who are satisfied with the company’s service is a. .87 b. .174 c. .148 10. The mean of the sampling distribution of is always equal to a. p b. µ c. p̂ 11. The condition for the standard deviation of the sampling distribution of the sample proportion to be √pq/n is a. np > 5 and nq > 5 b. n > 30 c. n/N < .05 12. The sampling distribution of is (approximately) normal if a. np > 5 and nq > 5 b. n > 30 c. n/N < .05
Hydropump, Inc., produces and sells high-quality pumps to business customers. Its marketing research shows a growing market for a similar type of pump aimed at final consumers— for use with Jacuzzi-style tubs in home remodeling jobs. Hydropump will have to develop new channels of distribution to reach this target market because most consumers rely on a retailer for advice about the combination of tub, pump, heater, and related plumbing fixtures they need. Hydropump’s marketing manager, Robert Black, is trying to decide between intensive and selective distribution. With intensive distribution, he would try to sell through all the plumbing supply, bathroom fixture, and hot-tub retailers who will carry the pump. He estimates that about 5,600 suitable retailers would be willing to carry a new pump. With selective distribution, he would focus on about 280 of the best hot-tub dealers (2 or 3 in the 100 largest metropolitan areas). Intensive distribution would require Hydropump to do more mass selling—primarily advertising in home renovation magazines—to help stimulate consumer familiarity with the brand and convince retailers that Hydropump equipment will sell. The price to the retailer might have to be lower too (to permit a bigger markup) so they will be motivated to sell Hydropump rather than some other brand offering a smaller markup. With intensive distribution, each Hydropump sales rep could probably handle about 300 retailers effectively. With selective distribution, each sales rep could handle only about 70 retailers because more merchandising help would be necessary. Managing the smaller sales force and fewer retailers, with the selective approach, would require less manager overhead cost. Going to all suitable and available retailers would make the pump available through about 20 times as many retailers and have the potential of reaching more customers. However, many customers shop at more than one retailer before making a final choice—so selective distribution would reach almost as many potential customers. Further, if Hydropump is using selective distribution, it would get more in-store sales attention for its pump and a larger share of pump purchases at each retailer. Black has decided to use a spreadsheet to analyze the benefits and costs of intensive versus selective distribution. a. Based on the initial spreadsheet, which approach seems to be the most sensible for Hydropump? Why? b. A consultant points out that even selective distribution needs national promotion. If Black has to increase advertising and spend a total of $100,000 on mass selling to be able to recruit the retailers he wants for selective distribution, would selective or intensive distribution be more profitable? c. With intensive distribution, how large a share (percent) of the retailers’ total unit sales would Hydropump have to capture to sell enough pumps to earn $200,000 profit?
Entomologists study the distribution of insects across agricultural fields. A study of fire ant hills across pasture lands is conducted by dividing pastures into 50- meter squares and count-ing the number of fire ant hills in each square. The null hypothesis of a Poisson distribution for the counts is equivalent to a random distribution of the fire ant hills over the pasture. Rejection of the hypothesis of randomness may occur due to one of two possible alternatives. The distribution of fire ant hills may be uniformâ€” that is, the same number of hills per 50- meter squareâ€” or the distribution of fire ants may be clustered across the pasture. A random distribution would have the variance in counts equal to the mean count, Ïƒ2 = Î¼. If the distribution is more uniform than random, then the distribution is said to be underdispersed, Ïƒ2, Î¼. If the distribution is more clustered than random, then the distribution is said to be overdispersed, Ïƒ2. Î¼. The number of fire ant hills was recorded on one hundred 50- meter squares. In the data set, yi is the number of fire ant hills per square, and ni denotes the number of 50- meter squares with yi ant hills. a. Estimate the mean and variance of the number of fire ant hills per 50- meter square; that is, compute y and Ïƒ2 using the formulas from Chapter 3. b. Do the fire ant hills appear to be randomly distributed across the pastures? Use a chi-square test of the adequacy of the Poisson distribution to fit the data using a = .05. c. If you reject the Poisson distribution as a model for the distribution of fire ant hills, does it appear that fire ant hills are more clustered or uniformly distributed across the pastures?
a. Calculate the chi-square test statistic for the simulated table in Question 20. Since the table cell counts (at least for the concave malignant group) were assumed to be randomly generated, this test statistic can be thought of as one random sample drawn from a chi- square distribution with 1 degree of freedom. b. Use software to generate 10,000 possible counts for the concave malignant group, using the cancer cell data. Calculate the chi-square test statistic for each of the 10,000 simulations. Create a histogram of the 10,000 chi-square test statistics. Describe the shape of this simulated chi-square distribution with 1 degree of freedom. Since 37 observations is still a fairly small sample size, the simulated distribution does not look completely like the theoretical chi- square distribution. Larger sample sizes will cause the simulated distribution to look much more like the true chi- square distribution. c. Use statistical software to randomly generate 10,000 values from a chi-square distribution with 1 degree of freedom. Compare this distribution to the one generated in Part B. Do they look the same? d. Repeat Part B for a larger sample size. Assume that there are 160 round and 210 concave nuclei. In addition, assume that there are 240 malignant and 130 benign cells. As in Part B, generate 10,000 possible counts for the concave malignant group for this new table. Calculate the chi-square test statistic for each of the 10,000 simulations. Create a histogram of the 10,000 chi-square test statistics. Describe the shape of this simulated chi-square distribution with 1 degree of freedom. The theoretical chi-square distribution is continuous. As shown in Part B and C, when the number of counts in each cell is small, the test statistic does not accurately follow a chi-square distribution. While the chi-square test works best with large sample sizes, most statisticians agree that the cell counts in Part B are sufficient to be modeled by the chi- square distribution.
Hydropump, Inc., produces and sells high-quality pumps to business customers. Its marketing research shows a growing market for a similar type of pump aimed at final consumers—for use with Jacuzzi-style tubs in home remodeling jobs. Hydropump will have to develop new channels of distribution to reach this target market because most consumers rely on a retailer for advice about the combination of tub, pump, heater, and related plumbing fixtures they need. Hydropump’s marketing manager, Robert Black, is trying to decide between intensive and selective distribution. With intensive distribution, he would try to sell through all the plumbing supply, bathroom fixture, and hot-tub retailers who will carry the pump. He estimates that about 5,600 suitable retailers would be willing to carry a new pump. With selective distribution, he would focus on about 280 of the best hot-tub dealers (2 or 3 in the 100 largest metropolitan areas). Intensive distribution would require Hydropump to do more mass selling—primarily advertising in home renovation magazines—to help stimulate consumer familiarity with the brand and convince retailers that Hydropump equipment will sell. The price to the retailer might have to be lower too (to permit a bigger markup) so they will be motivated to sell Hydropump rather than some other brand offering a smaller markup. With intensive distribution, each Hydropump sales rep could probably handle about 300 retailers effectively. With selective distribution, each sales rep could handle only about 70 retailers because more merchandising help would be necessary. Managing the smaller sales force and fewer retailers, with the selective approach, would require less manager overhead cost. Going to all suitable and available retailers would make the pump available through about 20 times as many retailers and have the potential of reaching more customers. However, many customers shop at more than one retailer before making a final choice—so selective distribution would reach almost as many potential customers. Further, if Hydropump is using selective distribution, it would get more in-store sales attention for its pump and a larger share of pump purchases at each retailer. Black has decided to use a spreadsheet to analyze the benefits and costs of intensive versus selective distribution. a. Based on the initial spreadsheet, which approach seems to be the most sensible for Hydropump? Why? b. A consultant points out that even selective distribution needs national promotion. If Black has to increase advertising and spend a total of $100,000 on mass selling to be able to recruit the retailers he wants for selective distribution, would selective or intensive distribution be more profitable? c. With intensive distribution, how large a share (percent) of the retailers’ total unit sales would Hydropump have to capture to sell enough pumps to earn $200,000 profit?
The dataset HollywoodMovies2011 contains information on all 136 movies to come out of Hollywood in 2011. We see in Exercise B.13 that 27 of the movies were comedies. If we take 1000 samples of size n = 25 from this dataset and record the proportion of movies in the sample that are comedies, we get the sampling distribution shown in Figure B.6. In each case below, fill in the blank with the best option provided. (a) The standard error of this sampling distribution is approximately ......................0.02 0.07 0.13 0.17 0.20 (b) If we create a new sampling distribution using samples of size n = 40, we expect the center of the new distribution to be the................center of the distribution shown in Figure B.6. smaller than about the same as larger than (c) If we create a new sampling distribution using samples of size n = 40, we expect the standard error of the new distribution to be...................the standard error of the distribution shown in Figure B.6. smaller than about the same as larger than. (d) If we create a new sampling distribution using 5000 samples of size n = 25, we expect the center of the new distribution to be the....................center of the distribution shown in Figure B.6. smaller than about the same as larger than (e) If we create a new sampling distribution using 5000 samples of size n = 25, we expect the...............standard error of the new distribution to be the standard error of the distribution shown in Figure B.6. Exercise B.6 Data 2.7 on page 93 introduces the dataset HollywoodMovies2011, which contains information on all 136 movies to come out of Hollywood in 2011. Twenty-seven of those movies were comedies.
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