| Annual Returns | Tech Firm Sample | Finance Firm Sample | |||
| 1 | 8.60% | 1 | 10.10% | ||
| 2 | 10.90% | 2 | 8.80% | ||
| 3 | 13.10% | 3 | 10.10% | ||
| 4 | 9.80% | 4 | 12.20% | ||
| 5 | 11.40% | 5 | 10.40% | ||
| 6 | 12.30% | 6 | 7.30% | ||
| 7 | 10.90% | 7 | 8.60% | ||
| 8 | 9.50% | 8 | 12.10% | ||
| 9 | 13.10% | 9 | 10.90% | ||
| 10 | 12.50% | 10 | 9.50% | ||
| 11 | 12.70% | 11 | 11.10% | ||
| 12 | 12.10% | 12 | 10.80% | ||
| 13 | 10.40% | ||||
| 14 |
12.50 |
||||
| Q8. Calculate the LCL and UCL for a 90% CI (α = 0.1) for the difference in mean returns between the two samples. |
| LCL ==> |
| UCL ==> |
| Based on this 90% interval, would you say the mean return for the Tech firms was higher than for Finance firms?(Yes/No) |
| Q9. Calculate the LCL and UCL for a 95% CI (α = 0.05) for the difference in mean returns between the two samples. |
| LCL ==> |
| UCL ==> |
| Based on this 95% interval, would you say the mean return for the Tech firms was higher than for Finance firms?(Yes/No) |
In: Statistics and Probability
| Annual Returns | Tech Firm Sample | Finance Firm Sample | |||
| 1 | 8.60% | 1 | 10.10% | ||
| 2 | 10.90% | 2 | 8.80% | ||
| 3 | 13.10% | 3 | 10.10% | ||
| 4 | 9.80% | 4 | 12.20% | ||
| 5 | 11.40% | 5 | 10.40% | ||
| 6 | 12.30% | 6 | 7.30% | ||
| 7 | 10.90% | 7 | 8.60% | ||
| 8 | 9.50% | 8 | 12.10% | ||
| 9 | 13.10% | 9 | 10.90% | ||
| 10 | 12.50% | 10 | 9.50% | ||
| 11 | 12.70% | 11 | 11.10% | ||
| 12 | 12.10% | 12 | 10.80% | ||
| 13 | 10.40% | ||||
| 14 |
12.50 |
||||
| Q8. Calculate the LCL and UCL for a 90% CI (α = 0.1) for the difference in mean returns between the two samples. |
| LCL ==> |
| UCL ==> |
| Based on this 90% interval, would you say the mean return for the Tech firms was higher than for Finance firms?(Yes/No) |
| Q9. Calculate the LCL and UCL for a 95% CI (α = 0.05) for the difference in mean returns between the two samples. |
| LCL ==> |
| UCL ==> |
| Based on this 95% interval, would you say the mean return for the Tech firms was higher than for Finance firms?(Yes/No) |
In: Statistics and Probability
There are two traffic lights on a commuter's route to and from work. Let X1 be the number of lights at which the commuter must stop on his way to work, and X2 be the number of lights at which he must stop when returning from work. Suppose that these two variables are independent, each with the pmf given in the accompanying table (so X1, X2 is a random sample of size n = 2).
| x1 | 0 | 1 | 2 |
|---|---|---|---|
| p(x1) | 0.3 | 0.4 | 0.3 |
μ = 1, σ2 = 0.6
Calculate σTo2.
σTo2 =
How does it relate to σ2, the population variance?
σTo2 = · σ2
(d)
Let X3 and X4 be the number of lights at which a stop is required when driving to and from work on a second day assumed independent of the first day. With To = the sum of all four Xi's, what now are the values of E(To) and V(To)?
E(To)=V(To)=
(e)
Referring back to (d), what are the values of
P(To = 8) and P(To ≥ 7)
[Hint: Don't even think of listing all possible outcomes!]
P(To = 8)
=
P(To ≥ 7)
=
In: Statistics and Probability
8. Madsen Motors's bonds have 23 years remaining to maturity. Interest is paid annually, they have a $1,000 par value, the coupon interest rate is 7%, and the yield to maturity is 9%. What is the bond's current market price? Round your answer to the nearest cent.
9. A bond has a $1,000 par value, 10 years to maturity, and a 7% annual coupon and sells for $985.
10. Nesmith Corporation's outstanding bonds have a $1,000 par value, a 6% semiannual coupon, 12 years to maturity, and a 10% YTM. What is the bond's price? Round your answer to the nearest cent.
11. A firm's bonds have a maturity of 10 years with a $1,000 face value, have an 8% semiannual coupon, are callable in 5 years at $1,049.23, and currently sell at a price of $1,095.02. What are their nominal yield to maturity and their nominal yield to call? Do not round intermediate calculations. Round your answers to two decimal places.
In: Finance
The following data lists the ages of a random selection of actresses when they won an award in the category of Best Actress, along with the ages of actors when they won in the category of Best Actor. The ages are matched according to the year that the awards were presented. Complete parts (a) and (b) below.
Actress (years) 31 25 29 31 35 25 25 42 30 32
Actor (years) 56 40 39 34 29 37 52 35 34 44
a. Use the sample data with a 0.01 significance level to test the claim that for the population of ages of Best Actresses and Best Actors, the differences have a mean less than 0 (indicating that the Best Actresses are generally younger than Best Actors).
In this example, μd is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the actress's age minus the actor's age. What are the null and alternative hypotheses for the hypothesis test?
H0: μd (1) _____ , _____ years
H1: μd (2) _____ , _____ years
(Type integers or decimals. Do not round.)
(1) >
<
≠
=
(2) <
=
≠
>
Identify the test statistic.
t= _____ (Round to two decimal places as needed.)
Identify the P-value.
P-value=_____ (Round to three decimal places as needed.)
What is the conclusion based on the hypothesis test?
Since the P-value is (3) _____ the significance level, (4) _____ the null hypothesis. There (5)_____ sufficient evidence to support the claim that actresses are generally younger when they won the award than actors.
(3) less than or equal to
greater than
(4) reject
fail to reject
(5) is
is not
b. Construct the confidence interval that could be used for the hypothesis test described in part (a). What feature of the confidence interval leads to the same conclusion reached in part (a)?
The confidence interval is _____ year(s)<μd< _____ year(s).
(Round to one decimal place as needed.)
What feature of the confidence interval leads to the same conclusion reached in part (a)?
Since the confidence interval contains (6) _____ (7) _____ the null hypothesis.
(6) zero,
only negative numbers,
only positive numbers,
(7) reject
fail to reject
In: Math
Demand for walnut fudge ice cream at the Sweet Cream Dairy can be approximated by a normal distribution with a mean of 17 gallons per week and a standard deviation of 3.2 gallons per week. The new manager desires a service level of 90 percent. Lead time is two days, and the dairy is open seven days a week. (Hint: Work in terms of weeks.)
a-1. If an ROP model is used, what ROP would be consistent with the desired service level? (Do not round intermediate calculations. Round your final answer to 2 decimal places.)
ROP ______ gallons
a-2. How many days of supply are on hand at the ROP, assuming average demand? (Do not round intermediate calculations. Round your final answer to 2 decimal places.)
Days _______
b-1. If a fixed-interval model is used instead of an ROP model, what order size would be needed for the 90 percent service level with an order interval of 7 days and a supply of 8 gallons on hand at the order time? (Do not round intermediate calculations. Round your final answer to the nearest whole number.)
Order size _______ gallons
b-2. What is the probability of experiencing a stockout before this order arrives? (Do not round intermediate calculations. Round your final answer to the nearest whole percent. Omit the "%" sign in your response.)
Probability _________ %
c. Suppose the manager is using the ROP model described in part a. One day after placing an order with the supplier, the manager receives a call from the supplier that the order will be delayed because of problems at the supplier’s plant. The supplier promises to have the order there in two days. After hanging up, the manager checks the supply of walnut fudge ice cream and finds that 2 gallons have been sold since the order was placed. Assuming the supplier’s promise is valid, what is the probability that the dairy will run out of this flavor before the shipment arrives? (Do not round intermediate calculations. Round your final answer to the nearest whole percent. Omit the "%" sign in your response.)
Risk probability _________ %
In: Advanced Math
Step 4:
What percent of the variation in corn yield is explained by these two variables? Give your answers to 2 decimal places and do not include units in your answers.
Percent explained by the model = %
Step 5:
Using the regression equation, find a point estimate for the corn yield for 2014 Assume that the soy bean yield for that year is 42.
Point Estimate = (Give your answer to 1 decimal place.)
ID Year CornYield SoyBeanYield 1 1957 48.3 23.2 2 1958 52.8 24.2 3 1959 53.1 23.5 4 1960 54.7 23.5 5 1961 62.4 25.1 6 1962 64.7 24.2 7 1963 67.9 24.4 8 1964 62.9 22.8 9 1965 74.1 24.5 10 1966 73.1 25.4 11 1967 80.1 24.5 12 1968 79.5 26.7 13 1969 85.9 27.4 14 1970 72.4 26.7 15 1971 88.1 27.5 16 1972 97 27.8 17 1973 91.3 27.8 18 1974 71.9 23.7 19 1975 86.4 28.9 20 1976 88 26.1 21 1977 90.8 30.6 22 1978 101 29.4 23 1979 109.5 32.1 24 1980 91 26.5 25 1981 108.9 30.1 26 1982 113.2 31.5 27 1983 81.1 26.2 28 1984 106.7 28.1 29 1985 118 34.1 30 1986 119.4 33.3 31 1987 119.8 33.9 32 1988 84.6 27.0 33 1989 116.3 32.3 34 1990 118.5 34.1 35 1991 108.6 34.2 36 1992 131.5 37.6 37 1993 100.7 32.6 38 1994 138.6 41.4 39 1995 113.5 35.3 40 1996 127.1 37.6 41 1997 126.7 38.9 42 1998 134.4 38.9 43 1999 133.8 36.6 44 2000 136.9 38.1 45 2001 138.2 39.6 46 2002 129.3 38.0 47 2003 142.2 33.9 48 2004 160.3 42.2 49 2005 147.9 43.1 50 2006 149.1 42.9 51 2007 150.7 41.7 52 2008 153.9 39.7 53 2009 164.7 44.0 54 2010 152.8 43.5 55 2011 147.2 41.9 56 2012 123.4 39.8 57 2013 158.8 43.3
In: Statistics and Probability
Jennifer Nguyen, a Humber College Healthcare Management program graduate who always had only perfect marks in statistics, was hired by the famous Healthy Lifemedical insurance company. Jennifer is assigned to conduct statistical analysis of medical and financial data. As Jennifer is on probation, please help her to complete the following six tasks. In problems 2-6, state hypotheses H0and H1and provide detailed conclusions (based on P-values or critical values/test statistics) together with the Exceloutput. For your convenience the data are given in the Major Assignment Data file. You can also find useful information on the Blackboard in Excel Instructions folder. Jennifer’s manager Dr. Jonathan Steinberg, who has degrees and publications in both mathematical statistics and medical science, asked her to find estimates of the average dental claim reimbursement for 2019. As Healthy Lifehas many thousands of clients it is virtually impossible to calculate the population mean. Using the Excel Random Number Generator function, Jennifer found a random sample of 52 dental claims submitted to Healthy Life. The amounts covered by insurance you can see in the Major Assignment Data file. Please help Jennifer Nguyen to construct90%,95%, and 99%confidence intervals for the true average reimbursement. Make sure that t-distribution is applicable: build a histogram with the bin values, for example, $100, $200, $300, $400, and $500, and check whether it is approximately symmetric and bell-shaped. Then, use Descriptive Statistics function from Data Analysis. Constructing confidence intervals, please round values to two decimal places.
| Problem 1. | ||||
| Dental Claim Number | Amount Covered | Bin | ||
| 1 | $192.75 | 100 | ||
| 2 | $192.75 | 200 | ||
| 3 | $350.25 | 300 | ||
| 4 | $200.00 | 400 | ||
| 5 | $225.00 | 500 | ||
| 6 | $95.00 | |||
| 7 | $375.50 | |||
| 8 | $380.00 | |||
| 9 | $192.75 | |||
| 10 | $400.00 | |||
| 11 | $230.00 | |||
| 12 | $245.00 | |||
| 13 | $150.00 | |||
| 14 | $250.00 | |||
| 15 | $250.00 | |||
| 16 | $340.00 | |||
| 17 | $225.50 | |||
| 18 | $156.25 | |||
| 19 | $300.00 | |||
| 20 | $350.00 | |||
| 21 | $435.00 | |||
| 22 | $192.75 | |||
| 23 | $192.75 | |||
| 24 | $250.00 | |||
| 25 | $225.00 | |||
| 26 | $230.00 | |||
| 27 | $245.00 | |||
| 28 | $250.00 | |||
| 29 | $250.00 | |||
| 30 | $250.00 | |||
| 31 | $350.00 | |||
| 32 | $98.00 | |||
| 33 | $405.00 | |||
| 34 | $295.00 | |||
| 35 | $205.00 | |||
| 36 | $230.00 | |||
| 37 | $245.00 | |||
| 38 | $750.00 | |||
| 39 | $250.00 | |||
| 40 | $250.00 | |||
| 41 | $340.00 | |||
| 42 | $225.50 | |||
| 43 | $192.75 | |||
| 44 | $192.75 | |||
| 45 | $250.00 | |||
| 46 | $225.00 | |||
| 47 | $350.00 | |||
| 48 | $250.00 | |||
| 49 | $250.00 | |||
| 50 | $340.00 | |||
| 51 | $195.00 | |||
| 52 | $385.00 | |||
In: Statistics and Probability
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136 |
97 |
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99 |
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What is the Range? Blank 1
What is the Class Width? Blank 2
Fill in the table after creating it on paper. rf and crf are rounded to two decimal places. To input the Class Limits and Class Boundaries, you will need to put the lower class limit/boundary in the first blank and the upper class limit/boundary in the blank below where you input the lower. It is the way the Blackboard tool works.
|
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f |
cf |
rf |
crf |
|
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In: Statistics and Probability
Assignment Purpose
Write a well commented java program that demonstrates the use and re-use of methods with input validation.
Instructions
“I dn'ot gvie a dman for a man taht can olny sepll a wrod one way.” (Mrak Taiwn)
“We aer all moratls, Hamrbee is an immoratl, and tehre is no question abuot it.” (Kevin Unknown)
Hint: First generate two random integers in range of the length of the string. Then use substring method to access characters at those locations. Rest is left to your imagination.
Submission
A .java and a .html (generated with Javadoc) file
In: Computer Science