categories • All observations have the same probability of success It is okay to ignore dependence if the trials make up less than 10% of the population P hat is a statistic – p(hat)= X/n Bernoulli Distribution- simulates one sample and see the individual successes and failures Binominal Distribution- simulates several samples and see the number of successes and failures in each sample Sampling error- the difference between sample proportions and the true proportion. This is the sample variability
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金融计量第三次作业(本科+硕士) 要求: A. 提交截止时间:2014 年 4 月 23 日; B. 将所有数据处理工作写入 dofile,名称为 W3_姓名_学号.do; C. 所有分析过程和结果写入 word 文档,按照《经济学(季刊)》的格式排版,名称为 W3_ 姓名_学号.docx。注:这是我评分的主要依据。 D. 将上述两份文档(无需压缩,以便于我直接打开查看)发送至:ef_lianyujun@163.com。 E. 邮件标题为:本科_W3_姓名_学号(本科生) ;硕士_W3_姓名_学号(硕士生) 。不按照 我的要求填写邮件标题可能导致作业漏登。 F. 请勿发送数据文档。 你的任务: 目的:研究中国民营企业高管薪酬的影响因素。 1. 请查阅国内外相关文献,对高管薪酬影响因素方面的文献做简要评述(500-1000 字) 。 评述的主要内容包括: 文献中主要从哪些角度研究了高管薪酬问题?(2) 影响高管薪 (1) 酬的主要因素有哪些?(3) 现有文献中有关高管薪酬决定因素的研究中,已经比较一致 的观点有哪些?尚存争议的观点有哪些? 2. 统计分析各个一级行业(SIC)在各个年度上的平均
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Question 1: Hypothesis Test for US Public Transportation Year | Percent of Public Transportation | 1997 | 4.6 | 1999 | 4.9 | 2001 | 4.7 | 2003 | 4.4 | 2005 | 4.7 | 2006 | 4.8 | 2007 | 4.9 | 2008 | 5.0 | 2009 | 5.0 | 1/ Less than 5% of US citizens use public transportation according to several websites of U.S. Departments of Transportation Survey. = 4.778 SD = 0.199 n = 9 The test statistic: t= = = 71.276 We have enough evidence to reject H0 Thus, we
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Probability review (week 2) 1 Bernoulli, Binomial, Poisson and normal distributions. In this excercise we deal with Bernoulli, binomial, Poisson and normal random variables (RVs). A Bernoulli RV X models experiments, such as a coin toss, where success happens with probability p and failure with probability 1 − p. Success is indicated by X = 1 and failure by X = 0. Therefore, the probability mass function (pmf) of X is P {X = 0} = 1 − p, P {X = 1} = p (1) A binomial random variable (RV)
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Multi-regression Analysis Summer 2013 EC315: Quantitative Research Methods Professor Scott Sowder Introduction One day I was sitting in class with my classmates. Our GPA, the number of classes were are taking, ages, IQ and the amount of time we spend studying were all different. I became curious and wanted to know what effect the different variables had on the student’s GPA, if any. So I decided to a survey of 30 students with varies GPAs, IQs, ages, number of classes being taken
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Comandos SPSS * Análise Fatorial: ANALYZE DESCRIPTIVE EXPLORE * Normality plots with tests Analyze Dimension reduction Factor Analysis Descriptives * Initial solution * Coefficients * KMO and Bartlett´s test of sphericity * Anti-image Extraction * Correlation matrix * Unrotated factor solution * Scree plot * Fixed number of factors * Maximum Iterations for Convergence Rotation * Direct Oblimin * Loading Plot Scores * Save as
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years of education? _____________ d. What is the median beginning salary for minority males with 16 years of education? ____________ Note: Questions 3, 4, and 5 below do not involve SPSS. You will need to consult the “Table of Areas Under the Normal Curve” handout. 3. Assume grades on an exam are normally
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Researching medical devices before purchase takes place daily throughout the healthcare world. Not only does the proper device need to be found to match the need, but multiple manufacturers usually make the device needed. Each manufacturer tries to show the unique and what they consider the best way to tackle the need. In the large capital equipment world, many times a trial is set up between the manufacturers and the end users are surveyed. This survey takes on different levels of importance
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Summary This article discusses about twins separated and the genetic of personality differences and concentrated on studies that have been made in 1979 by the Minnesota Study of Twins Reared Apart ( MISTRA). In the research, the researcher focused on two aspects of cognitive and personality. However, in this article the discussion limited to discuss about the personality and behavioral differences only. The study conducted by Minnesota is to support the existence of an important genetic
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that 140 patients in his sample show a complete recovery. Use alpha=5%. :___p=0.3______:___p>0.3_______ Given a=0.05, the critical value is |Z(0.05)|=1.645 (check standard normal table) phat=140/300=0.47. The test statistic is: Z=(phat-p)/√[p*(1-p)/n]=(0.47-0.3)/sqrt(0.3*0.7/300)=6.43. The p-value is P(Z>6.43)=0 (check standard normal table). Since the p-value is less than 0.05, we reject Ho. Can the biochemist conclude that his new drug is more effective than the existing drug in the treatment
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