The U.S. Geological Survey compiled historical data about Old Faithful Geyser (Yellowstone National Park) from 1870 to 1987. Let x1 be a random variable that represents the time interval (in minutes) between Old Faithful eruptions for the years 1948 to 1952. Based on 9520 observations, the sample mean interval was x1 = 61.2 minutes. Let x2 be a random variable that represents the time interval in minutes between Old Faithful eruptions for the years 1983 to 1987. Based on 25,340 observations, the sample mean time interval was x2 = 71.2 minutes. Historical data suggest that σ1 = 8.35 minutes and σ2 = 12.41 minutes. Let μ1 be the population mean of x1 and let μ2 be the population mean of x2.
(a) Compute a 95% confidence interval for μ1 – μ2. (Use 2 decimal places.)
| lower limit | |
| upper limit |
(b) Comment on the meaning of the confidence interval in the context of this problem. Does the interval consist of positive numbers only? negative numbers only? a mix of positive and negative numbers? Does it appear (at the 95% confidence level) that a change in the interval length between eruptions has occurred? Many geologic experts believe that the distribution of eruption times of Old Faithful changed after the major earthquake that occurred in 1959.
Because the interval contains only positive numbers, we can say that the interval length between eruptions has gotten shorter.
Because the interval contains both positive and negative numbers, we can not say that the interval length between eruptions has gotten longer.
We can not make any conclusions using this confidence interval.
Because the interval contains only negative numbers, we can say that the interval length between eruptions has gotten longer.
In: Statistics and Probability
The U.S. Geological Survey compiled historical data about Old Faithful Geyser (Yellowstone National Park) from 1870 to 1987. Let x1 be a random variable that represents the time interval (in minutes) between Old Faithful eruptions for the years 1948 to 1952. Based on 9280 observations, the sample mean interval was x1 = 62.0 minutes. Let x2 be a random variable that represents the time interval in minutes between Old Faithful eruptions for the years 1983 to 1987. Based on 24,170 observations, the sample mean time interval was x2 = 69.6 minutes. Historical data suggest that σ1 = 8.35 minutes and σ2 = 12.76 minutes. Let μ1 be the population mean of x1 and let μ2 be the population mean of x2.
(a) Compute a 99% confidence interval for μ1 – μ2. (Use 2 decimal places.)
| lower limit | |
| upper limit |
(b) Comment on the meaning of the confidence interval in the context of this problem. Does the interval consist of positive numbers only? negative numbers only? a mix of positive and negative numbers? Does it appear (at the 99% confidence level) that a change in the interval length between eruptions has occurred? Many geologic experts believe that the distribution of eruption times of Old Faithful changed after the major earthquake that occurred in 1959.
Because the interval contains only positive numbers, we can say that the interval length between eruptions has gotten shorter.Because the interval contains both positive and negative numbers, we can not say that the interval length between eruptions has gotten longer. We can not make any conclusions using this confidence interval.Because the interval contains only negative numbers, we can say that the interval length between eruptions has gotten longer.
In: Statistics and Probability
Exercise 4
On January 1, 2017, Park Rapids Lumber Company issued $80 million in 20-year, 10% bonds payable. Interest is payable semiannually on June 30th and December 31st. Bond discounts and premiums are amortized straight-line at each interest payment date.
a. Record the journal entry when the bonds were issued on January 1, 2017, make the necessary the journal entry to record the payment of bond interest on June 30, 2017, under each of the following assumptions:
1. The bonds were issued at 98. Round your answers to the nearest dollar.
2. The bonds were issued at 101. Round your answers to the nearest dollar.
b. Compute the net bond liability at December 31, 2017, under assumptions 1 and 2 above. Round to the nearest dollar.
c. Under which of the above assumptions, 1 or 2 would the investor’s effective rate of interest be higher? Explain.
Exercise 5
Speed World Cycles sells high-performance motorcycles and Motocross racers. One of Speed World’s most popular models is the Kazomma 900 dirt bike. During the current year, Speed World purchased eight of these cycles at the following costs:
Purchase Date Units Purchased Unit Cost Total Cost
July 1 2 $4,950 $9,900
July 22 3 5,000 15,000
August 3 3 5,100 15,300
------ ------------
8 $40,200
On July 28, Speed World sold four Kazomma 900 dirt bikes to the Vince Wilson racing team. The remaining four bikes remained in inventory at September 30, the end of Speed World’s fiscal year.
Assume that Speed World uses a perpetual inventory system.
a. Compute the cost of goods sold relating to the sale on July 28 and the ending inventory of Kazomma 900 dirt bikes at September 30, using the following cost flow assumptions:
1. Average cost
2. FIFO
3. LIFO
Show the number of units and the unit costs of each layer comprising the cost of goods sold and ending inventory.
b. Using the cost figures computed in part a. answer the following questions:
1. Which of the three cost flow assumptions will result in Speed World Cycles reporting the highest net income for the current year? Would this always be the case? Explain.
2. Which of the three cost flow assumptions will minimize the income taxes owed by Speed World Cycles for the year? Would you expect this usually to be the case? Explain.
3. May Speed World Cycles use the cost flow assumption that results in the highest net income for the current year in its financial statements, but use the cost flow assumption that minimizes taxable income for the current year in its income tax return? Explain.
In: Accounting
The U.S. Geological Survey compiled historical data about Old Faithful Geyser (Yellowstone National Park) from 1870 to 1987. Let x1 be a random variable that represents the time interval (in minutes) between Old Faithful eruptions for the years 1948 to 1952. Based on 9580 observations, the sample mean interval was x1 = 61.8 minutes. Let x2 be a random variable that represents the time interval in minutes between Old Faithful eruptions for the years 1983 to 1987. Based on 23,000 observations, the sample mean time interval was x2 = 69.2 minutes. Historical data suggest that σ1 = 8.49 minutes and σ2 = 11.78 minutes. Let μ1 be the population mean of x1 and let μ2 be the population mean of x2.(a) Compute a 99% confidence interval for μ1 – μ2. (Use 2 decimal places.)
| lower limit | |
| upper limit |
(b) Comment on the meaning of the confidence interval in the context of this problem. Does the interval consist of positive numbers only? negative numbers only? a mix of positive and negative numbers? Does it appear (at the 99% confidence level) that a change in the interval length between eruptions has occurred? Many geologic experts believe that the distribution of eruption times of Old Faithful changed after the major earthquake that occurred in 1959.
Because the interval contains only positive numbers, we can say that the interval length between eruptions has gotten shorter.Because the interval contains both positive and negative numbers, we can not say that the interval length between eruptions has gotten longer. We can not make any conclusions using this confidence interval.Because the interval contains only negative numbers, we can say that the interval length between eruptions has gotten longer.
In: Statistics and Probability
Organism 3
Field Notes: Specimen collected from shaded area along stream in
South Cumberland State Park (Grundy County, TN)
Laboratory Analysis:
Body: Large leaves emerging from underground rhizome
Size: 63cm
Chromosomal Analysis: Plant body is diploid --chromosomes number of
44
Lignin test: Positive
Cuticle: Present
Leaves: Present -- large with branched veins. Underside has
sori(containing haploid spores)
Roots: Present-----branch from the inside
Stem:Present--- vascular tissue(xylem and phloem)present
Life History: Diploid sporophyte dominant generation. Haploid
spores germinate into heart-shaped, haploid, gametophyte. Water
required for fertilization due to flagellated sperm; no seed is
produced. Diploid zygote develops into sporophyte of life ---each
bearing ether megasporangia or microsporangia but not both.
Insects, especially beetles, appear important in
pollination
Question: Explain which domain, kingdom and phylum
you believe this plant should be classified in.
Communication: The local media features the work of your team on their nightly news. During a live interview the reporter asks you " Apparently this plant requires water for fertilization, can you explain, can you explain why"?
Response: ------------------------
In: Biology
The Fukushima Power plant is planning construction of a new plant to generate electricity four years hence and must decide now between a small, medium, or large-sized plant. The exact size needed is uncertain because future demands can only be estimated. Forecasters have estimated future demands and their likelihoods as follows:
|
Level of Demand |
Probability |
|
High |
0.30 |
|
Medium |
0.55 |
|
Low |
0.15 |
In the following, all the future costs and earnings have been adjusted to their present worth:
Conduct a decision-tree analysis on Excel to determine the size of the power-generating plant the company should build now. What size of the power-generating plant the company should build now? Briefly discuss your answer.
In: Statistics and Probability
Craft Pro Machining produces machine tools for the construction industry. The following details about overhead costs were taken from its company records.
| Production Activity | Indirect Labor | Indirect Materials | Other Overhead | ||||||||
| Grinding | $ | 390,000 | |||||||||
| Polishing | $ | 125,000 | |||||||||
| Product modification | 450,000 | ||||||||||
| Providing power | $ | 240,000 | |||||||||
| System calibration | 500,000 | ||||||||||
Additional information on the drivers for its production activities
follows.
| Grinding | 17,000 | machine hours |
| Polishing | 17,000 | machine hours |
| Product modification | 1,300 | engineering hours |
| Providing power | 20,000 | direct labor hours |
| System calibration | 1,100 | batches |
| Job 3175 | Job 4286 | |||
| Number of units | 140 | units | 1,750 | units |
| Machine hours | 500 | MH | 5,000 | MH |
| Engineering hours | 33 | eng. hours | 24 | eng. hours |
| Batches | 15 | batches | 45 | batches |
| Direct labor hours | 410 | DLH | 3,690 | DLH |
Required:
1. Classify each activity as unit level, batch
level, product level, or facility level.
Required:
2, 3 & 4. Compute the activity overhead rates using
ABC. Combine the grinding and polishing activities into a single
cost pool. Determine overhead costs to assign to the above jobs
using ABC. What is the overhead cost per unit for Job 3175? What is
the overhead cost per unit for Job 4286? (Round your
activity rate and average overhead cost per unit to 2 decimal
places. Round "overhead assigned" to the nearest whole
dollar.)
5. If the company uses a plantwide overhead rate based on direct labor hours, what is the overhead cost for each unit of Job 3175? Of Job 4286? (Do not round intermediate calculations. Round "OH Cost per unit" answers to 2 decimal places.)
In: Accounting
BUDGETED PROFIT AND LOSS ACCOUNT
Assume that you are working as a Budegt Control Manger in Cinimax Ltd. It has recently a fully equipped theatre and 3 cinema houses at a cost of £ 30 million. The theatre has a capacity of 800 seats nd each cinema has a capacity of 600 seats. Information and projections for the first year of operations are as follows:
1. Fixed administration and maintenance cost of the entire facility is £ 4.5 million per year.
2. The average cost of master print of a Hollywood film is £ 4 million while the cost of master print of a Bollywood film is £ 6.5 million.
3. Two cinema houses are dedicated for Hollywood films which show the same film at the same time while one cinema house will show Bollywood films.
4. Each Bollywood film is displayed for 6 weeks and the average occupancy level is 70%. Each Hollywood film is displayed for 4 weeks and the average occupancy level is 65%. On weekdays, there are 2 shows while on weekends (Sat and Sun), 3 shows are displayed. Ticket price has been fixed at £ 350.
5. Variable cost per show is £ 35,000 and setup cost of each film is £ 500,000.
6. No films would be shown during 8 weeks of the year.
7. Theatre is rented to production houses at £ 60,000 per day. Each play requires setup time of 2 days while rehearsal time needs 1 day. Each play is staged 45 times. One show is staged on weekdays whereas two shows are staged on weekends.
8. There is an interval of 2 days whenever a new play is to be staged.
9. The construction costs of theatre and cinema houses are to be depreciated over a period of 15 years.
10. Assume 52 weeks in a year and 30 days in a month.
Required: Prepare budgeted profit and loss account for the first year.
In: Accounting
Nina's Construction Bonds carry a 4% coupon, mature on 4/1/2030 and pay coupons on 10/1 and 4/1 each year using a 30/360 accrual convention.
Given this information, what is the accrued interest if
you owned $30,000 par value as of the end of the day
March 15th, 2024?
In: Finance
Nina's Construction Bonds carry a 4% coupon, mature on 4/1/2030 and pay coupons on 10/1 and 4/1 each year using a 30/360 accrual convention.
Given this information, what is the accrued interest if
you owned $30,000 par value as of the end of the day
March 15th, 2024?
In: Finance