Mean or Expected Value and Standard Deviation tài liệu, giáo án, bài giảng , luận văn, luận án, đồ án, bài tập lớn về tấ...
Designing value and valuing design Research Executive Summaries Series Vol. 2, No. 5 By Dr. Myfanwy Trueman and Professor Richard Pike Bradford University School of Management I SSN 1744 - 7038 (online) I SSN 1744 - 702X (print) Introduction Businesses recognise that good design can have a powerful impact on competitive advantage and profitability. It can differentiate products and services and enhance their value – while poor design can threaten the survival of an organisation. It has been argued that good design can improve communication and integration throughout the organisation, help to reduce complexity and cost, and enhance brand value. It can als o help companies to balance the needs of managers and shareholders with the cost, value and quality requirements of customers. However, good design rarely just happens, but stems from an effective development process. In the 1997 House of Lords debate on design, Lord Currie emphasised that design is a multi-skilled, multidisciplinary function: ‘Design involves not just designers and not just those working for design consultancies but also engineers, scientists and all those including senior management and – dare I mention it? – finance directors and accountants who contribute and influence the process of innovation and new product development (Lord Currie, House of Lords, 1997).’ There is growing recognition of the importance of people outside the formal design process in influencing the nature and form of new products. Nixon et al (1997) acknowledged the role of‘silent designers’ – including specialists from R&D, production, marketing and accounting, who can influence the nature and form of new products.These silent designers provide ‘essential information and strategic links in managing design parameters.’ Research has also shown that most of a product’s life cycle cost is ‘locked in’at the early design stage, and that an y c hanges to the product after this point incur very high costs. If this is the case,then to ensure that products are produced at lowest cost, accountants need to set parameters and become involved at the early stage in the product development process. The purpose of this study was to explore how management accounting can facilitate product development and effective design in or der to enhance br and v alue. It f ocused on the small business sector , where companies may not have the resources to investigate design potential in the same way as large organisations do, but can still benefit from a strong design input. The research posed two questions: • What evidence is there to show that accountants and designers are working together to reduce complexity and cost in product design? • Is the accountant’s role constructive, and how could it be improved? Project overview For this research 30 semi-structured interviews were held with designers, marketing staff and accountants from 16 small and medium companies in the UK. The companies came from a range of sectors, but one-third were known to have a strong design focus, one-third were technology-driven and one-third came from a random sample used in previous research) . A model developed from previous research was used as a framework for the interviews as a tool to analyse product development in the 16 companies. This model identifies four hierarchical dimensions of design; value, image, process and production (VIPP), where image, process and production dimensions feed into value, which is seen to be the key dynamic for success. An effective innovation strategy is likely to take a balanced approach to all these four dimensions. Related to these four dimensions are a number of design attributes (benefits), which also provided a basis for discussion in Mean or Expected Value and Standard Deviation Mean or Expected Value and Standard Deviation By: OpenStaxCollege The expected value is often referred to as the "long-term" average or mean This means that over the long term of doing an experiment over and over, you would expect this average You toss a coin and record the result What is the probability that the result is heads? If you flip a coin two times, does probability tell you that these flips will result in one heads and one tail? You might toss a fair coin ten times and record nine heads As you learned in [link], probability does not describe the short-term results of an experiment It gives information about what can be expected in the long term To demonstrate this, Karl Pearson once tossed a fair coin 24,000 times! He recorded the results of each toss, obtaining heads 12,012 times In his experiment, Pearson illustrated the Law of Large Numbers The Law of Large Numbers states that, as the number of trials in a probability experiment increases, the difference between the theoretical probability of an event and the relative frequency approaches zero (the theoretical probability and the relative frequency get closer and closer together) When evaluating the long-term results of statistical experiments, we often want to know the “average” outcome This “long-term average” is known as the mean or expected value of the experiment and is denoted by the Greek letter μ In other words, after conducting many trials of an experiment, you would expect this average value NOTE To find the expected value or long term average, μ, simply multiply each value of the random variable by its probability and add the products A men's soccer team plays soccer zero, one, or two days a week The probability that they play zero days is 0.2, the probability that they play one day is 0.5, and the probability that they play two days is 0.3 Find the long-term average or expected value, μ, of the number of days per week the men's soccer team plays soccer 1/19 Mean or Expected Value and Standard Deviation To the problem, first let the random variable X = the number of days the men's soccer team plays soccer per week X takes on the values 0, 1, Construct a PDF table adding a column x*P(x) In this column, you will multiply each x value by its probability Expected Value TableThis table is called an expected value table The table helps you calculate the expected value or longterm average x P(x) x*P(x) 0.2 (0)(0.2) = 0.5 (1)(0.5) = 0.5 0.3 (2)(0.3) = 0.6 Add the last column x*P(x) to find the long term average or expected value: (0)(0.2) + (1)(0.5) + (2)(0.3) = + 0.5 + 0.6 = 1.1 The expected value is 1.1 The men's soccer team would, on the average, expect to play soccer 1.1 days per week The number 1.1 is the long-term average or expected value if the men's soccer team plays soccer week after week after week We say μ = 1.1 Find the expected value of the number of times a newborn baby's crying wakes its mother after midnight The expected value is the expected number of times per week a newborn baby's crying wakes its mother after midnight Calculate the standard deviation of the variable as well You expect a newborn to wake its mother after midnight 2.1 times per week, on the average x P(x) x*P(x) (x – μ)2 ⋅ P(x) P(x = 0) = 50 (0) ( 502 ) = (0 – 2.1)2 ⋅ 0.04 = 0.1764 P(x = 1) = ( 1150 ) (1) ( 1150 ) = 1150 (1 – 2.1)2 ⋅ 0.22 = 0.2662 P(x = 2) = 23 50 (2) ( 2350 ) = 4650 (2 – 2.1)2 ⋅ 0.46 = 0.0046 P(x = 3) = 27 50 (3) ( 509 ) = 2750 (3 – 2.1)2 ⋅ 0.18 = 0.1458 2/19 Mean or Expected Value and Standard Deviation x P(x) x*P(x) (x – μ)2 ⋅ P(x) P(x = 4) = 50 (4) ( 504 ) = 1650 (4 – 2.1)2 ⋅ 0.08 = 0.2888 P(x = 5) = 50 (5) ( 501 ) = 505 (5 – 2.1)2 ⋅ 0.02 = 0.1682 Add the values in the third column of the table to find the expected value of X: 105 μ = Expected Value = 50 = 2.1 Use μ to complete the table The fourth column of this table will provide the values you need to calculate the standard deviation For each value x, multiply the square of its deviation by its probability (Each deviation has the format x – μ) Add the values in the fourth column of the table: 0.1764 + 0.2662 + 0.0046 + 0.1458 + 0.2888 + 0.1682 = 1.05 The standard deviation of X is the square root of this sum: σ = √1.05 ≈ 1.0247 Try It A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-hour shift For a random sample of 50 patients, the following information was obtained What is the expected value? x P(x) P(x = 0) = 50 P(x = 1) = 50 P(x = 2) = 16 50 P(x = 3) = 14 50 P(x = 4) = 50 P(x = 5) = 50 The expected value is 2.24 4 16 14 (0) 50 + (1) 50 + (2) 50 + (3) 50 + (4) 50 + (5) 50 = + 50 + 32 50 42 + 50 + 24 50 + 10 50 = 116 50 = 2.24 3/19 Mean or Expected Value and Standard Deviation Suppose you play a game of chance in which five numbers are chosen from 0, 1, 2, 3, 4, 5, 6, 7, 8, A computer randomly selects five numbers from zero to nine with replacement You pay $2 to play ... [WHITE PAPER] INTERNATIONAL HEADQUARTERS: Eagle House, The Ring Bracknell, RG12 1HS TEL: +44 1344 746000 FAX: +44 1344 746001 THE IMPACT OF THE RECESSION ON VALUE AND PREMIUM GOODS [ ] By Tash Altay, Commercial Director, IRI, UK 2 Copyright © 2008 Information Resources, Inc. All rights reserved. Information Resources, Inc., IRI, the IRI logo and the names of IRI products and services referenced herein are either trademarks or re g istered trademarks of Information Resources , Inc. All other trademarks a r e the p ro p ert y of their res p ective owners. [ ] The Impact of the recession on Value and Premium Goods “What can retailers and manufacturers do to stay ahead of the competition and encourage greater consumption?” Nigel Howlett, IRI BACKGROUND Recessions, like any other part of the economic cycle, bring with them challenges as well as opportunities for businesses. Consumer confidence across Europe is currently at a particularly low ebb, as macro-economic factors force shoppers to reappraise their consumption habits. Issues such as rising fuel costs, food and commodity price rises, the tightening of credit and decreasing property values have all encouraged a more cautious approach to spending. On the CPG side inflation is hurting turnover, as price rises in raw materials force firms to push up in-store prices. In some cases, the price rises, when combined with other factors, are forcing customers to trade down to cheaper brands or private label products, while other goods which are seen as unnecessary to their needs may even be sidelined altogether. However, at the other end of the scale, certain premium products are appearing more resilient to the recession, as consumers sacrifice items in order to satisfy their need – whether societal, psychological or emotional – for these goods. So what can retailers and manufacturers do to stay ahead of the competition and encourage greater consumption? How do they pick the right marketing and promotional strategies in such challenging business conditions? IRI has used its powerful Infoscan point- of-sale data gathering technology to compile a wide-ranging study into shopping habits over a range of categories, during the past two years. The body of evidence provided in this paper is a result of this data gathering and analysis tool. The vast data warehouses of retailers across the UK, France, Germany, Spain and Italy have been trawled to provide us with a comprehensive picture of shopping habits in these regions. Categories are broken down into three tiers – value, regular and premium – products assigned a tier according to their average price per volume. A picture then begins to emerge of where the opportunities lie, and where the challenges are. IRI has also polled consumers themselves in these five regions about their preferences and perceptions of value to gain a vital snapshot into their ever- volatile shopping behaviour. In conclusion, IRI makes recommendations so that both CPG retailers and manufacturers can approach the coming months and quarters with the confidence that they will make the right decisions to not only survive but thrive during this downturn. 3 Copyright © 2008 Information Resources, Inc. All rights reserved. Information Resources, Inc., IRI, the IRI logo and the names of IRI products and services referenced herein are INEQUALITIES INVOLVING THE MEAN AND THE STANDARD DEVIATION OF NONNEGATIVE REAL NUMBERS OSCAR ROJO Received 22 December 2005; Revised 18 August 2006; Accepted 21 Septe mber 2006 Let m(y) = n j =1 y j /n and s(y) = m(y 2 ) −m 2 (y) be the mean and the standard deviation of the components of the vector y = (y 1 , y 2 , , y n−1 , y n ), where y q = (y q 1 , y q 2 , , y q n −1 , y q n ) with q a positive integer. Here, we prove that if y ≥ 0,thenm(y 2 p )+(1/ √ n −1)s(y 2 p ) ≤ m(y 2 p+1 )+(1/ √ n −1)s(y 2 p+1 )forp = 0,1,2, The equality holds if and only if the (n − 1) largest components of y are equal. It follows that (l 2 p (y)) ∞ p=0 , l 2 p (y) = (m(y 2 p )+(1/ √ n −1)s(y 2 p )) 2 −p , is a strictly increasing sequence converging to y 1 ,the largest component of y,exceptifthe(n −1) largest components of y are equal. In this case, l 2 p (y) = y 1 for all p. Copyright © 2006 Oscar Rojo. This is an op en access article distributed under the Cre- ative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 1. Introduction Let m(x) = n j =1 x j n , s(x) = m x 2 − m 2 (x) (1.1) be the mean and the standard deviation of the components of x = (x 1 ,x 2 , , x n−1 ,x n ), where x q = (x q 1 ,x q 2 , , x q n −1 ,x q n ) for a positive integer q. The following theorem is due to Wolkowicz and Styan [3, Theorem 2.1.]. Theorem 1.1. Let x 1 ≥ x 2 ≥···≥x n−1 ≥ x n . (1.2) Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2006, Article ID 43465, Pages 1–15 DOI 10.1155/JIA/2006/43465 2 Inequalities on the mean and standard deviation Then m(x)+ 1 √ n −1 s(x) ≤ x 1 , (1.3) x 1 ≤ m(x)+ √ n −1s(x). (1.4) Equality holds in ( 1.3)ifandonlyifx 1 = x 2 =···=x n−1 . Equality holds in (1.4)ifandonly if x 2 = x 3 =···=x n . Let x 1 ,x 2 , ,x n−1 ,x n be complex numbers such that x 1 is a positive real number and x 1 ≥ x 2 ≥···≥ x n−1 ≥ x n . (1.5) Then, x p 1 ≥ x 2 p ≥···≥ x n−1 p ≥ x n p (1.6) for any positive integer p.WeapplyTheorem 1.1 to (1.6)toobtain m | x| p + 1 √ n −1 s | x| p ≤ x p 1 , x p 1 ≤ m | x| p + √ n −1s | x| p , (1.7) where |x|=(|x 1 |,|x 2 |, , |x n−1 |,|x n |). Then, l p (x) = m | x| p + 1 √ n −1 s | x| p 1/p (1.8) isasequenceoflowerboundsforx 1 and u p (x) = m | x| p + √ n −1s | x| p 1/p (1.9) isasequenceofupperboundsforx 1 . We recall that the p-norm and the infinity-norm of a vector x = (x 1 ,x 2 , , x n )are x p = n i=1 x i p 1/p ,1≤ p<∞, x ∞ = max i x i . (1.10) It is well known that lim p→∞ x p =x ∞ . Oscar Rojo 3 Then, l p (x) = ⎛ ⎜ ⎝ x p p n + 1 n(n −1) x 2p 2p − x 2p p n ⎞ ⎟ ⎠ 1/p , u p (x) = ⎛ ⎜ ⎝ x p p n + n −1 n x 2p 2p − x 2p p n ⎞ ⎟ ⎠ 1/p . (1.11) In [2, Theorem 11], we proved that if y 1 ≥ y 2 ≥ y 3 ≥···≥y n ≥ 0, then m y 2 p + √ n −1s y 2 p ≥ m y 2 p+1 + √ n −1s y 2 p+1 (1.12) for p = 0,1,2, The equality holds if and only if y 2 = y 3 =···=y n . Using this inequal- ity, we proved in [2, Theorems 14 and 15] that if y 2 = y 3 =···=y n ,thenu p (y) = y 1 for all p,andify i <y j for some 2 ≤ j<i≤ n,then(u 2 p (y)) ∞ p=0 is a strictly decreasing sequence converging to y 1 . The main purpose of this paper is to prove that if y 1 ≥ y 2 ≥ y 3 ≥···≥y n ≥ 0, then m y 2 p + 1 √ n −1 s y 2 p ≤ m y 2 p+1 + 1 √ n −1 s y 2 p+1 (1.13) for p = 0,1,2, The equality holds if and only Genet. Sel. Evol. 36 (2004) 455–479 455 c INRA, EDP Sciences, 2004 DOI: 10.1051/gse:2004011 Original article A simulation study on the accuracy of position and effect estimates of linked QTL and their asymptotic standard deviations using multiple interval mapping in an F 2 scheme Manfred M a∗ ,YuefuL b , Gertraude F a a Research Unit Genetics and Biometry, Research Institute for the Biology of Farm Animals, Dummerstorf, Germany b Centre of the Genetic Improvement of Livestock, University of Guelph, Ontario, Canada (Received 4 August 2003; accepted 22 March 2004) Abstract – Approaches like multiple interval mapping using a multiple-QTL model for simul- taneously mapping QTL can aid the identification of multiple QTL, improve the precision of estimating QTL positions and effects, and are able to identify patterns and individual elements of QTL epistasis. Because of the statistical problems in analytically deriving the standard errors and the distributional form of the estimates and because the use of resampling techniques is not feasible for several linked QTL, there is the need to perform large-scale simulation studies in order to evaluate the accuracy of multiple interval mapping for linked QTL and to assess con- fidence intervals based on the standard statistical theory. From our simulation study it can be concluded that in comparison with a monogenetic background a reliable and accurate estima- tion of QTL positions and QTL effects of multiple QTL in a linkage group requires much more information from the data. The reduction of the marker interval size from 10 cM to 5 cM led to a higher power in QTL detection and to a remarkable improvement of the QTL position as well as the QTL effect estimates. This is different from the findings for (single) interval mapping. The empirical standard deviations of the genetic effect estimates were generally large and they were the largest for the epistatic effects. These of the dominance effects were larger than those of the additive effects. The asymptotic standard deviation of the position estimates was not a good criterion for the accuracy of the position estimates and confidence intervals based on the standard statistical theory had a clearly smaller empirical coverage probability as compared to the nominal probability. Furthermore the asymptotic standard deviation of the additive, domi- nance and epistatic effects did not reflect the empirical standard deviations of the estimates very well, when the relative QTL variance was smaller/equal to 0.5. The implications of the above findings are discussed. mapping / QTL / simulation / asymptotic standard error / confidence interval ∗ Corresponding author: mmayer@fbn-dummerstorf.de 456 M. Mayer et al. 1. INTRODUCTION In their landmark paper Lander and Botstein [15] proposed a method that uses two adjacent markers to test for the existence of a quantitative trait locus (QTL) in the interval by performing a likelihood ratio test at many positions in the interval and to estimate the position and the effect of the QTL. This approach was termed interval mapping. It is well known however, that the ex- istence of other QTL in the linkage group can distort the identification and quantification of QTL [10,11,15,31]. Therefore, QTL mapping combining in- terval mapping with multiple marker regression analysis was proposed [11,30]. The method of Jansen [11] is known as multiple QTL mapping and Zeng [31] named his approach composite interval mapping. Liu and Zeng [19] extended the composite interval mapping approach to mapping QTL from various cross designs of multiple inbred lines. In the literature, numerous studies on the power of data designs and map- ping strategies for single QTL models like interval mapping and composite interval mapping can be found. But these mapping methods often provide only point estimates of QTL positions and effects. To get an idea of the preci- sion of a mapping study, it is important to compute the standard deviations of the Workforce Project time, days FIGURE 19 Workforce requirements based on project schedule Statistics, Probability, and Their Applications Basic Statistics If the value assumed by a variable on a given occasion cannot be predicted because it is influenced by chance, the variable is known as a random, or stochastic, variable The number of times the variable assumes a given value is called the frequency of that value A number that may be considered representative of all values of the variable is called an average There are several types of averages, such as arithmetic mean, geometric mean, harmonic mean, median, mode, etc Each is used for a specific purpose The standard deviation is a measure of the dispersion, or scatter, of the values of the random variable; it is approximately equal to the amount by which a given value of the variable may be expected to differ from the arithmetic mean, in either direction The variance is the square of the standard deviation Where no mention of frequency appears, it is understood that all frequencies are Notational System Here X= random variable; X1 = ith value assumed by X\ft - frequency of X1; n = sum of frequencies; X = arithmeticmean of values of X\ Xmed = median of these values; dt = deviation of Xb from X = Xi-X\ A = an assumed arithmetic mean; dA>i = deviation of X1 from A=X1-A', s = standard deviation; s2 = variance In the following material, the subscript / will be omitted DETERMINATIONOFARITHMETICMEAN, MEDIAN, AND STANDARD DEVIATION Column of Table 22 presents the number of units of a commodity that were sold monthly by a firm for consecutive months Find the arithmetic mean, median, and standard deviation of the number of units sold monthly TABLE 22 (2) (1) Month Total Number of units soldX 32 49 51 44 37 41 _47 301 (3) d = X-41 -11 -6 -2 _4 O (4) d2 121 36 64 36 _16 278 (5) dA=X-40 -8 11 -3 21 (6) d2A 64 81 121 16 _49 341 Calculation Procedure: Compute the arithmetic mean Let X= number of units_sold monthly Find the sum of the values of X9 which is 301 Then setX = QX)In, or X = 301/7 = 43 Find the median Consider that all values of X are arranged in ascending order of magnitude If n is odd, the value that occupies the central position in this array is called the median If n is even, the median is taken as the arithmetic mean of the two values that occupy the central positions In either case, the total frequency of values below the median equals the total frequency of values above the median The median is useful as an average because the arithmetic mean can be strongly influenced by an extreme value at one end of the array and thereby offer a misleading view of the data In the present instance, the array is 32, 37, 41, 44, 47, 49, 51 The fourth value in the array is 44; then Xmed = 44 Compute the standard deviation Compute the deviations of the X values from X, and record the results in column of Table 22 The sum of the deviations must be O Now square the deviations, and record the results in column Find the sum of the squared deviations, which is 278 Set the variance s2 = &d2)/n = 278/7 Then set the standard deviation s = V278/7 = 6.30 Compute the arithmetic mean by using an assumed arithmetic mean SQtA = 40 Compute the deviations of the Jf values from A9 recordjthe results in column of Table 22, and find the sum of the deviations, which is 21 Set X = A + (2dA)/n, or X = 40 + 21/7 = 43 Compute the standard deviation by using the arithmetic mean assumed in step Square the deviations from A9 record the results in column of Table 22, and find their sum, which is 341 Set s2 = (ZdJ)In - [&dA)/n]2 = 341/7 - (21/7)2 = 341/7 - = 278/7 Thens = V278/7 = 6.30 Related Calculations: Note that the equation applied in step does not contain the true mean X This equation serves to emphasize that the standard deviation is purely a measure of dispersion and thus is independent of the arithmetic mean For example, if all values of Xincrease by a constant /z, then X increases by h, but s remains constant ... − μ)2P(x) Complete the expected value table x P(x) x*P(x) 0.2 0.2 0.4 0.2 Find the expected value from the expected value table 9/19 Mean or Expected Value and Standard Deviation x P(x) x*P(x)... monetary value of X to determine the expected value 13/19 Mean or Expected Value and Standard Deviation Card Event X net gain/loss P(X) Face Card and Heads ( 1252 )( 12 ) = ( 526 ) Face Card and Tails... back and $256 What is your expected profit of playing the game over the long term? 4/19 Mean or Expected Value and Standard Deviation Let X = the amount of money you profit The x-values are –$1 and