You are given: (i) Losses follow an exponential distribution with mean 0 . (ii) A random sample of losses is distributed as follows: Loss Range Number of Losses (0 – 100] 32 (100 – 200] 21 (200 – 400] 27 (400 – 750] 16 (750 – 1000] (1000 – 1500] Total 100 Estimate 0 by matching at the 80™ percentile. (A) 249 (B) 253 (C) 257 (D) 260 (E) 263
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![You are given:
(i)
Losses follow an exponential distribution with mean 0 .
(ii)
A random sample of losses is distributed as follows:
Loss Range
Number of Losses
(0 – 100]
32
(100 – 200]
21
(200 – 400]
27
(400 – 750]
16
(750 – 1000]
2
(1000 – 1500]
Total
100
Estimate 0 by matching at the 80" percentile.
(A)
249
(В)
253
(C)
257
(D)
260
(E)
263](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0b4d35de-88fd-40a5-a122-8015485b71a5%2Faacc21fd-a7cc-4137-915d-1ce3f86778d6%2Fw0zvhhs_processed.jpeg&w=3840&q=75)
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- In the book Business Research Methods (5th ed.). Donald R. Cooper and C. William Emory discuss studying the relationship between on-the-job accidents and smoking. Cooper and Emory describe the study as follows: Suppose a manager implementing a smoke-free workplace policy is interested in whether smoking affects worker accidents. Since the company has complete reports of on-the-job accidents, she draws a sample of names of workers who were involved in accidents during the last year. A similar sample from among workers who had no reported accidents in the last year is drawn. She interviews members of both groups to determine if they are smokers or not. The sample results are given in the following table. On-the-Job No 7 Smoker Heavy Moderate Nonsmoker 16 Column total 34 Heavy Moderate Yes Nonsmoker Total 12 6 Expected counts are below observed counts No Accident Total 19 7 9.21 9 Row 1 Row 2 Row 3 Column 1 Column 2 16 32 Accident 12 9.79 6 Accident Row Total 7.27 16 15.52 32 7.73 16…From the data in the table it shows the 2019 daily return data for the S&P 500 Index (stocks) and the US 30 year Bond (bonds). How do I test the hypothesis that mean daily return on stocks exceeds the mean daily return on bonds? Please show the steps. S&P500 dailt returns data: Date Price Daily Return 2019-12-31 3230.78 0.294602% 2019-12-30 3221.29 -0.578082% 2019-12-27 3240.02 0.003398% 2019-12-26 3239.91 0.512817% 2019-12-24 3223.38 -0.019545% 2019-12-23 3224.01 0.086614% 2019-12-20 3221.22 0.494478% 2019-12-19 3205.37 0.445929% 2019-12-18 3191.14 -0.043230% 2019-12-17 3192.52 0.033529% US 30 Year Bond Date Price Daily Return 31-Dec-19 2.389 2.32% 30-Dec-19 2.335 0.78% 29-Dec-19 2.317 -0.03% 27-Dec-19 2.318 -0.16% 26-Dec-19 2.321 -0.71% 25-Dec-19 2.338 0.15% 24-Dec-19 2.334 -0.83% 23-Dec-19 2.354 0.31% 22-Dec-19 2.346 0.12% 20-Dec-19 2.344 -0.43%Group A5- 5 86- 1 3 77- 0 1 1 2 5 6 7 8 8 88- 1 2 3 3 3 4 4 5 6 89- 0 2 710- 0 0 Key: 5|5 represents 55 Group B5- 76- 4 87- 1 2 2 5 68- 0 2 4 5 5 5 7 79- 1 3 4 5 6 7 7 8 8 9 910- 0 0 0Key: 5|7 represents 57 1. Calculate the mean, median, range, and interquartile range for each data set. 2. Using the appropriate measures, compare the center and variability of the two groups of students. Explain your reasoning based on the shapes of the distributions.
- b) r=10:38. (a) Construct the cost of living index number from the table given below. Index for 1998 550 Expenditure Group 1. Food Clothing 3. Fuel & Lighting 46% 10% 2. 215 7% 220 12% 4. House Rent 150 25% 5. Miscellaneous 275 (b) How is the overall index number altered: (i) if all the group index numbers are changed in the same ratio ; (ii) if all the weights and group indices are changed in the same ratio ; (iii) if all the tion. group index numbers are increased by 10 and all the weights doubled ? reYou take a simple random sample of size 25 from a very large population in which the true proportion is p = 0.1. Which statement below best describes what you know about the sampling distribution of p ? %3D Hp-hat 0.1; we cannot use the formula Op-hat = (0.1)(0.9) : the distribution is not approximately normal, 25 Hp-hat = 0.1; op-hat = (0.1)(0.9) : the distribution is skewed. 25 Hp-hat = 0.1; Op-hat = (0.10.9) ; the distribution is approximately normal. 25 Hn-hat = 0.1; we cannot use the formula op-hat = V (0.1)(0.9) 25 Op-hat ; the distribution is approximately normal. * Previous IMG 0218.jpeg Next Activa PAUSE PRT F12 F11 F10 F9 M3 PO UP F8 M2 HOME F7 M1 F6 INS F5 F4 M6 PO DN M4 DEL END
- Benford's Law claims that numbers chosen from very large data files tend to have "1" as the first nonzero digit disproportionately often. In fact, research has shown that if you randomly draw a number from a very large data file, the probability of getting a number with "1" as the leading digit is about 0.301. Suppose you are an auditor for a very large corporation. The revenue report involves millions of numbers in a large computer file. Let us say you took a random sample of n = 250 numerical entries from the file and r = 60 of the entries had a first nonzero digit of 1. Let p represent the population proportion of all numbers in the corporate file that have a first nonzero digit of 1. Test the claim that p is less than 0.301 by using α = 0.01. What does the area of the sampling distribution corresponding to your P-value look like? a. The area in the right tail of the standard normal curve. b. The area not including the right tail of the standard normal curve.…Age is a variable that is important in most if not all epidemiological studies of health because our health is so strongly related to age. The following table is a two-dimensional table that summarizes the distribution of age in a certain population every five years starting with 1970 (the first row) to 2000 (the last row). Each row contains information for one year. Reading across the row (over the column), the table shows the number of individuals in each age group. If you choose this option, please do the following: (a) using the data provided, produce a new two-dimensional table that shows, for each year, the percentage of that year’s total population in each age interval (Tip! In each row, the sum of your percentages should be 100%); and (b) write a brief report (a paragraph or so, no more!) on what these summaries suggest regarding the possibility (or lack thereof) of changes over time in the age distribution of this population. Interval of Age, years…(25) If coefficient of variation-70 and mean -10 then variance is (a) 49 (b) 7 (c) 100
- A researcher wants to investigate the effects of environmental factors on IQ scores. For an initial study, she takes a sample of 400 people who grew up as the only child. She finds that 48.5% of them have an IQ score over 100. It is known that 50% of the general population has an IQ score exceeding 100. (a) Find the mean of p, where p is the proportion of people with IQ scores over 100 in a random sample of 400 people. (b) Find the standard deviation of p. (c) Compute an approximation for P(p is greater than or equal to 0.485), which is the probability that there will be 48.5% or more individuals with IQ scores over 100 in a random sample of 400. Round answer to 4 decimal places.The values under n, N, and (n/N,)x 100% for the following table are not real, they are presented here for reasons of clarity. Table 1: Age distribution. Age Group Number of Patients Size of Age Group in Catchment A rea (N,) Proportion (n/ N,)x100% (n) 0- 4 5 - 14 15 - 44 45 - 59 2 60 925 4675 5762 19.75 15.63 58.08 901 4213 7253 23.93 13.84 153 635 76 549 i. Explain why the use of the proportions (n,/ N, )x100%, i= 1, 2,. .. 5, in the construction of the multiple bar chart is comparatively better than the use of the number of patients, (n,). ii. Assuming you are the bio-statis tician, use the multiple bar chart to write a report to management Be guided by the purpose for which multiple bar charts are used.
