...Statistics Final Review 1.|Using the z-table (Table E), find the critical value (or values) for a = .08 two-tailed test.| |A) 1.4051 B) 1.7507 C) 1.4051 D) 1.7507| 2.|A report states that 38% of home owners had a vegetable garden. How large a sample is needed to estimate the true proportion of home owners who have vegetable gardens to within 5% with 90% confidence?| |A) 64 B) 128 C) 255 D) 510| 3.|If the equation for the regression line is y = –8x + 5, then the intercept of this line is| |A) 10 B) 5 C) –3 D) –8| 4.|For the samples summarized below, test the hypothesis at =.05 that the two variances are equal.|Variance|Number of data values|Sample 1|25|7|Sample 2|9|17|| A)|Accept the hypothesis because the test value 7.72 is greater than the critical value 3.34.| B)|Reject the hypothesis because the test value 2.78 is less than the critical value 3.16.| C)|Reject the hypothesis because the test value 7.72 is greater than the critical value 3.16.| D)|Accept the hypothesis because the test value 2.78 is less than the critical value 3.34.| 5.|For the conjecture "The average age of students in this class is 23", the null hypothesis is:| A)|We accept the hypothesis that the average age of students in this class is 23| B)|The average age of students in this class is 23| C)|The average age of students in this class is not 23| D)|We reject the hypothesis that he average age of students in this class is 23| 6.|Compute...
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...probability theory to understand how the process of taking a random sample will blur the information in a population. But first, we need to understand why and how the information is blurred. Sampling Variability Although the average social networking hours for all US residents is a fixed number, the average of a sample of 100 residents depends on precisely which sample is taken. In other words, the sample mean is subject to “sampling variability”. The problem is that by reporting x alone, we don’t take account of the variability caused by the sampling procedure. If we had polled different residents, we might have gotten a different average social networking hours. In general, the characteristics of the observed distribution (mean, median, variance, range, IQR, etc.), change from sample to sample, and may never exactly match the population quantities. To visualize properties of sampling distributions, we will use the sampling lab, and the very nice website at: http://onlinestatbook.com/stat_sim/sampling_dist/index.html Statistical Inference: A body of...
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...1. Results Interpretation: What do each of the following variance results tell us about the Clinic (i.e., what's the good news and what's the bad news)? Analyze in the aggregate and for each product line, as appropriate. If an analysis applies to more than one product line, you may combine product lines in your discussion for convenience. Use of bullet points are fine and, to the extent possible, encouraged. Profit (total) variance = actual profit – static profit Revenue= Member months x premium We see that the overall profit is -$329,366 than what was expected . What caused this variation? When it comes to revenue, ideally HMOs would like to see enrollment high and utilization low along with costs down making profit high. The main contributors to profit variances are: Revenue issues, Cost issues, or BOTH. Revenue – Cost= Profit. HMOs make money through member months (enrollment) x premiums. • The main issue with total profit variance is insufficient revenue from PC Commercial and increased costs from the other three product lines due to high utilization from which created negative variances causing decreased profits. • The static budget expected margin was 12.7% The actual profit was significantly less than static budget with the actual total margin cost being at 2.7% Revenue Variance: • The PMPM rate increased in PC comm. the biggest provider, which impacted its revenue through decreased enrollment. Its enrollee amounts were less than the aggregates...
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...Revere Street Working Paper Series Financial Economics 272-18 Mean-Variance Analysis versus Full-Scale Optimization Out of Sample First Version: November 11, 2005 This Draft: December 13, 2005 Timothy Adler Windham Capital Management, LLC 5 Revere Street Cambridge, MA 02138 617 234-9459 tadler@windhamcapital.com Abstract For three decades, mean-variance analysis has served as the standard procedure for constructing portfolios. Recently, investors have experimented with a new optimization procedure, called full-scale optimization, to address certain limitations of mean-variance optimization. Specifically, mean-variance optimization assumes that returns are normally distributed or that investor preferences are well approximated by mean and variance. Full-scale optimization relies on sophisticated search algorithms to identify the optimal portfolio given any set of return distributions and based on any description of investor preferences. Full-scale optimization yields the truly optimal portfolio in sample, whereas the mean-variance solution is an approximation to the insample truth. Both approaches to portfolio formation, however, suffer from estimation error. Mean-variance analysis requires investors to estimate the means and variances of all assets and the covariances of all asset pairs. To the extent the out-of-sample experience of these parameters departs from the in-sample parameter values, the mean-variance approximation will be even less accurate. Full-scale optimization requires...
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...PROC FREQ [DATA=dataset] [ORDER=FREQ|INTERNAL|DATA]; [TABLES vars ... [/ [MISSING] [NOCUM] ;] DATA; INPUT dept$ pos$; CARDS; CM CLK CM MGR BA CLK BA CLK CM MGR EE MGR EE CLK CM CLK CM MGR BA CLK BA CLK CM MGR EE MGR EE CLK RUN; PROC FREQ ORDER=FREQ; TABLES dept*pos / MISSING NOCUM; RUN; Frequency‚ Percent ‚ Row Pct ‚ Col Pct ‚CLK ‚MGR ‚ Total ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆ CM ‚ 2 ‚ 4 ‚ 6 ‚ 14.29 ‚ 28.57 ‚ 42.86 ‚ 33.33 ‚ 66.67 ‚ ‚ 25.00 ‚ 66.67 ‚ ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆ BA ‚ 4 ‚ 0 ‚ 4 ‚ 28.57 ‚ 0.00 ‚ 28.57 ‚ 100.00 ‚ 0.00 ‚ ‚ 50.00 ‚ 0.00 ‚ ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆ EE ‚ 2 ‚ 2 ‚ 4 ‚ 14.29 ‚ 14.29 ‚ 28.57 ‚ 50.00 ‚ 50.00 ‚ ‚ 25.00 ‚ 33.33 ‚ ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆ Total 8 6 14 57.14 42.86 100.00 Or we can use he WEIGHT statement specifies a numeric variable with a value that represents the frequency of the observation. DATA; INPUT dept$ pos$ weight; CARDS; CM CLK 1 CM MGR 2 BA CLK 2 EE MGR 1 ...
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...[Type the company name] | [Type the document title] | [Type the document subtitle] | | | [Pick the date] | [Type the abstract of the document here. The abstract is typically a short summary of the contents of the document. Type the abstract of the document here. The abstract is typically a short summary of the contents of the document.] | Contents Executive Summary 3 Introduction 3 Analysis and methods section 3 Do charges incurred by a patient depend on the type of insurance the patient has? If so how? 5 Do charges incurred by patients depend on which doctor treats them? If so how? 5 Bibliography 7 Appendix 8 Executive Summary Hospital are required to bill for individual items or services provided to a patient. Patients admitted to hospitals are charged for their room, supplies, drugs, labs, x-rays, operating room time and other care. It is important to know that hospitals submit a bill to the insurance company for all the services provided to the patient and the payor determines the amount owed to the hospital based upon the insurance company’s contract with the hospital for specific services [ (Henry Ford Health) ]. Introduction The purpose of this case study is to understand the relationship between hospital charges for certain physicians and insurance carriers. Cost finding and cost analysis are the techniques of allocating data that we were provided as part of the case study. The information that we are going to use to analyze...
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...Hypothesis Testing- Practice Questions Q: 1 A local school board member wants to know if sophomore students at Lincoln High School have approximately the same reading level as the state average for tenth graders. The state average is 150 words per minute with a standard deviation of 15. The level of the test is to be set at .05. A random sample size of 100 tenth graders has been drawn, and the resulting average is 157 words per minute. Q: 2 A can of peeled whole tomatoes is supposed to contain an average of 19 ounces of tomatoes (excluding juice). The actual weight is normally distributed random variable whose standard deviation is known to be 1.9 ounces. A random sample of 25 cans yield 18 ounces as the average weight. At 1% level of significance, test the hypothesis that the mean weight is below 19 ounces. Q: 3 A U.S. dime is supposed to weight 2.268 grams with a SD of 0.026 grams. A random sample of 15 circulated dimes showed a mean weight of 2.256 grams. Test the hypothesis, under the assumption that data is normally distributed, the mean weight of all the circulated dimes lower than the specification. Q: 4 A company currently pays its production employees a mean wage of $15.00 per hour with a standard deviation of $2.40. The company is planning to build a new factory and is considering several locations. The availability of labor at a rate less than $15.00 per hour is a major factor in the location decision. For one location, a sample of 40 workers showed a current mean...
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...gasoline, reported last evening that the mean price nationwide is $1.50 per gallon for self-serve regular unleaded. A random sample of 35 stations in the Milwaukee, WI, area revealed that the mean price was $1.52 per gallon and that the standard deviation was $0.05 per gallon. At the .05 significance level, can we conclude that the price of gasoline is higher in the Milwaukee area? Calculate the p-value and interpret. PROBLEM 2: (Weight 20 points). Suppose Babsie generated the following probability distribution: X p(x) 5 7 10 12 15 .25 .30 .25 .05 .15 a. b. c. d. Is this probability distribution discrete or continuous? Explain your reasoning. Calculate the expected value of X. Show your work!! Calculate the variance of X. Show your work!!! Calculate the standard deviation. Show your work!! PROBLEM 3: (Weight 20 points). Babsie is a public affairs specialist at Park University. A press release issued by Babsie based on some research claims that Park University students study at least as much as the national average for students at four year universities. Across the nation, 73 percent of all students at four year universities study at least four hours per week. Seventy percent of one hundred randomly selected Park University students surveyed claimed to study more than four hours per week. Should the University retract its previous statement? Explain why or why not. Answer using a 95 percent confidence interval. PROBLEM 4: (Weight...
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...Total Plan Risk: Integrating Assets into a Consistent Risk Framework Dan diBartolomeo Northfield Information Services, Inc. FactSet PMW Conference, Atlanta , November 2002 Do We Want to Measure Risk or Manage It? § Measuring risk is an exercise in forecasting § Managing risk requires decision making § Managing risk well requires rational decision making based on an understanding of utility theory What Risks are of Concern to Us? § Asset/liability mismatch risks § Asset class volatility § Style and active management risks How about Multiple Portfolios? § The firm-wide (plan-wide) risk problem w Multiple portfolios with multiple benchmarks w Across countries, across asset classes w Mixture of liquid, and illiquid assets, derivatives w Need to integrate liabilities Approach Number #1 § Build factor risk model for each portfolio separately and aggregate the risks § Arises from the existing stock of models § Advantage is that you are probably using the same models at the portfolio level so you have internal consistency § Problems w Not intuitive, as you can’t add exposures w Lots of factors may lead to covariance matrix which is not positive definite w Use high frequency data or an EM algorithm w Inclusion of liabilities or illiquid assets Approach #2 § Proxy each asset class with indices and then use full covariance. Adapted from trading desk systems § Advantage is simplicity. Works well for asset classes where instruments within the class are homogeneous...
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...attached the following reports for your review/records: • Critical Tasks • Over Allocated Resources • Over Budget Tasks • Resource Usage Solution After the test subjects have been finalized the next task will be to prepare the product. This is a critical task that has been assigned to an over-allocated resource. This is a constraint that will lead to other risks in the scheduling, and since we are already behind schedule, we will base our remaining duration estimates on a worse-case scenario using the Program Evaluation and Review Technique (PERT). This technique accounts for three different determined estimate criteria (Most Pessimistic (P), Most Optimistic (O), or Most Likely (M) – explained below) where a standard deviation variance will be used to weigh out the probability that the duration of a task will be efficient. The PERT criteria are defined below: • Most Pessimistic time - The longest determined...
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...allow for variation in the underlying volume. Why not simply send engineers in search of the problem whenever e-mail use exceeds a rate of say 1000 messages? Because the number of messages per day is 7680, there is a high chance that engineers will get false alarms. So the 1000 messages limit is not feasible and even if it is, it requires aggressive monitoring. Explain why it is important to monitor both the mean and the variance of the volume of email on this system It is important to monitor both the mean and the variance because a change in mean implies a change in the number of users using the email system. This is important for diagnostic purposes and identifying the usage patterns of the system. Depending on the shift in mean, the upper control limits and the lower control limits can be modified. A shift in the variance depicts how the number of users vary every hour. By looking at this, we can find out which part of the day there are more number of users and how much deviation is a safe limit. So it is necessary to monitor both the mean and the variance. Because the computer support team is well-staffed, there is minimal cost in having someone check for a problem. On the other hand, failing to identify a problem could be serious because it would allow the problem to grow in magnitude. What value do you recommend for α, the chance of a Type I error? For a value lying between the Upper Control Limit & Lower Control Limit, there is no error. If the system goes out of...
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...SAMPLING DISTRIBUTIONS Simple Random Sample A simple random sample X 1 , , X n , taken from a population represented by a random variable X with mean and standard deviation , has the following characteristics. Each X i , i 1,, n , is a random variable that has the same distribution as X, and thus the same mean and standard deviation . The X i ’s are independent random variables implying the following identity: V X 1 X n V X 1 V X n 2 2 n 2 The sample mean, X where X X1 X n , is a continuous random variable n with mean X and standard deviation X of the mean. Sampling Distribution of the Mean , known as the standard error n Result 1: If X is Normal, then X is Normal and Z of the sample n. X n , regardless of the size Result 2: If n is large enough, n 30 , then X is approximately normal and X Z , regardless of the distribution of the population random variable X. n Sampling Distribution of the Proportion For a binomial random variable X with parameters n (sample size) and (probability of X success or population proportion), the sample proportion is given by p . n The sample proportion p is a continuous random variable with mean p and standard deviation p 1 n . Result 3: If n is large enough, n 5 and n1 5 , then p is approximately p normal and Z . 1 n Why? We want to use the...
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...Q1. a) The T-value is 3.5 and the P-value is 0.001. The chance of an association between Expend and Number of this magnitude occurring just by chance is around 1 in 1000. So we very strongly suspect that there really is an underlying association. b) The average or expected monthly expenditure on extras this year is $12.26 for customers who did not take any options last year. For every option the customer took last year you can add another $2.79 to this. c) The prediction equation is not very accurate. The model error measures the typical difference between the fitted values and the observed values. This is a rough estimator of the prediction error. The sample size was n=83 so there is a typical prediction error of [pic] which is the large in comparison with the typical size of expenditures in this data set (roughly $10-$30 judging by the given mean and standard deviation). The percentage of variation in expenditure explained by optional purchases last year is only 12%. d) We predict $12.26+5x$2.79=$26.21 e) This is a simple regression (single input variable) and chapter 15 gave an explicit formula for the slope estimate. The equation for slope is [pic] and we know this equals 2.787. We also know that the standard deviation of expenditure is 14.064 and of number is 1.828. Solving for the correlation gives the answer 0.362. For simple regression, you can calculate the slope from the correlation or visa-versa. For multiple regression, it is the partial correlations...
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...TABLE OF CONTENTS Question 1: ……………………………………………..…………………..………......1 - 3 1.1 1.2 1.2.1 1.2.2 1.2.3 1.3 Question 2: ……………………………………………………………..…………...….4 - 5 2.1 2.2 2.3 2.4 Question 3: ………………………………………………………………..………………..6 Question 4: …………………………………………………………..................…..…7 - 8 Question 5: ……………………………..……………………………………………..9 - 11 Bibliography: ………………………………………………………...…….……………..12 1 QUESTION 1 1.1 Class Interval 20 up to 50 50 up to 80 80 up to 110 110 up to 140 140 up to 170 Frequency 06 12 14 04 04 40 Less than type Cumulative Frequency 00+06 = 06 06+12 = 18 18+14 = 32 32+04 = 36 36+04 = 40 1.2 Draw to scale 1.2.1 Histogram 1.2.2 Frequency Polygon 1.2.3 “Less-than” Ogive REFER TO THE GRAPH ON THE NEXT PAGE FOR ANSWERS 2 3 1.3 Use the Ogive to determine 1.3.1. the 65th percentile To determine the 65th percentile we calculate :65% of 40 = 26. The 65% percentile occurs at position 26. From the graph it gives the value of 82 1.3.2 The inter-quartile range is a measure of dispersion and is equal to the difference between the third and first quartiles. Half of the inter-quartile range is called semi inter-quartile range or Quartile deviation. Symbolically it is defined as; Q.D = (Q3 - Q1)/ 2 where Q1 and Q3 are the first and third quartiles of the data. !""#$%&'(%#)*$ !"#$% ! !""!!"# ! ! !"#$%&'( !!!"#$%&'( ! ! ������������������������������������������������ ������������������������������������������������������...
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...Introduction Purpose of this Project,to find out the releationship between population and the square of countries is tried to be understood. This Project includes 129 observations and 30 samples.We tested with 95% cofidence interval Step 1:State null and alternate hypotheses Step 2:Select a level of significance Step 3:Identify the test statistic Step 4:Formulate a decision rule Step 5:Take a sample,arrive at a decision Step 1 Hypothesis Test: Ho:The average population of countries is equal or less than 39.786.098 H1:The average population of countries is more than 39.786.098 Step 2The significance level is 0.05 Step 3Because the sample is 30,z distribution has been used Step 4Reject Ho if z>-1.96 or z<-1.96 or if p<0.05 Step 5z<-1.65 Ho is rejected.We can conclude the average population of countries is more than 39.786.098 POPULATION: China | India | Indonesia | Brazil | Bangladesh | Nigeria | Japan | Ethiopia | Egypt | Germany | Turkey | Iran | Congo, Democratic Republic of the | France | Italy | Myanmar (Burma) | Korea, South | Colombia | Argentina | Kenya | Algeria | Canada | Morocco | Iraq | Afghanistan | Nepal | Malaysia | Ghana | Korea, North | Mozambique | Madagascar | Australia | Cote d'Ivoire | Cameroon | Chile | Netherlands | Burkina Faso | Niger | Malawi | Kazakhstan | Ecuador | Cambodia | Mali | Guatemala | Angola | Cuba | Greece | Chad | ...
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