描述统计学chap02
描述统计第二章PPT课件
特点
典型单位的选择具有主观性。因此只能作为 全面调查的补充
种类
划类选典式和解剖麻雀式
数据的收集方法
数据收集方法
询问调查
观察
访问调查
邮寄调查
电话调查
电脑辅助
座谈会
个别深访
实验
访问调查
(personal interview)
1. 调查者与被调查者通过 面对面地交谈而获得资料
2. 有标准式访问和非标准 式访问
5. 从互联网或图书馆查阅到的相关资料
Internet
http//WWW.
中 国中 人国 口市 统场 计统 年计 鉴年
鉴
中国商品交易市场统计年鉴 中国连锁餐饮企业统计年鉴中国连锁零售业统计年鉴 中国能源统计年鉴 全国农产品成本收益资料汇编 国际统计年鉴 中国对外经济贸易统计年鉴 中国基本单位统计年鉴 中国民政统计年鉴 中国高技术产业统计年鉴 中国农村统计年鉴 中国农村住户调查年鉴 中国农村住户调查年鉴中国乡镇统计资料 中国县（市）社会经济调查年鉴 中国西部农村统计资料 中国建制镇统计资料
2.1 数据来源(data sources)
2.1.1 数据的间接来源 2.1.2 数据的直接来源
二手数据的来源
1. 统计部门和政府部门公布的有关资料， 如各类统计年鉴
2. 各类经济信息中心、信息咨询机构、专 业调查机构等提供的数据
3. 各类专业期刊、报纸、书籍所提供的资 料
4. 各种会议，如博览会、展销会、交易会 及专业性、学术性研讨会上交流的有关 资料
电话调查
(telephone survey)
1. 调查者利用电话与被调查者 进行语言交流以获得信息
2. 时效快、成本低 3. 问题的数量不宜过多
统计学完整全套PPT课件
模型的参数估计
阐述非线性回归模型的参数估计方 法，如最小二乘法、极大似然法等 ，并探讨其计算过程和注意事项。
模型的检验与诊断
介绍非线性回归模型的检验方法， 如拟合优度检验、参数的显著性检 验等，以及模型的诊断方法，如残 差分析、异常值识别等。
方差
各数据与平均数之差的平方的 平均数
03
标准差
方差的平方根04四源自位数间距上四分位数与下四分位数之差
偏态与峰态分析
01
02
03
偏态系数
描述数据分布偏斜程度的 统计量
峰态系数
描述数据分布尖峭或扁平 程度的统计量
正态性检验
如Jarque-Bera检验等， 用于判断数据是否服从正 态分布
03
推论性统计方法
模型评估与优化
预测结果展示与应用
通过比较模型的预测结果与实际股票价格 的差异，评估模型的预测性能，并进行优 化和改进。
将模型的预测结果进行可视化展示，为投资 者提供决策参考。
THANKS
感谢观看
统计学完整全套PPT课件
目录
• 统计学基本概念与原理 • 描述性统计方法 • 推论性统计方法 • 非参数统计方法 • 回归分析及其应用 • 时间序列分析与预测
01
统计学基本概念与原理
Chapter
统计学的定义及作用
统计学定义
统计学是一门研究如何收集、整理、分析和解释数 据的科学，它使用数学方法对数据进行建模和预测 ，以揭示数据背后的规律和趋势。
游程检验
游程检验的基本原理
以上内容仅供参考，具体细节和扩展内 容需要根据实际需求和背景知识进行补 充和完善。
ch02统计学 江西财大
Slide 15
Frequency Distribution
The three steps 1. Determine the number of nonoverlapping classes. 2. Determine the width of each class 3. Determine the class limits Guidelines for Selecting Number of Classes • Use between 5 and 20 classes. • Data sets with a larger number of elements usually require a larger number of classes. • Smaller data sets usually require fewer classes.
© 2003 Thomson/South-Western
Slide 1
2.1 Summarizing Qualitative Data
Frequency Distribution Relative Frequency Distribution Percent Frequency Distribution Bar Graph Pie Chart
© 2003 Thomson/South-Western
Slide 16
Frequency Distribution
© 2003 Thomson/South-Western
Slide 11
Example: Marada Inn
Pie Chart
Exc. Poor 5% 10% Above Average 45% Below Average 15% Average 25%
《统计学(第二版)》电子课件 第2章 数据的描述
《统计学》第2章数据的描述
2-19
抽样调查
抽样调查(sampling survey):是从研究对 象的总体中随机抽取一部分个体作为样 本进行调查，并根据调查结果来推断总 体数量特征的一种非全面调查方法。
抽样调查的特点：经济性好、实效性强、 适应面广、准确性高。
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《统计学》第2章数据的描述
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《统计学》第2章数据的描述
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【例2.2】
——条形图的绘制
图2.1 30名教师职称分布条形图
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《统计学》第2章数据的描述
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【例2.3】
——饼图的绘制
（数据文件为）根据表资料, 用SPSS绘制 饼图。
解：打开数据文件example2.1.sav;
选择→“图形”→点击“旧对话框 （L）”→“饼图（E）”→在“图表中的 数据为”中选“个案组摘要（G）→点击 “定义”→ 在“分区的表征”中选中“个 案数（N）”→将“职称”选入“定义分区 （B）”→点击“确定”，可得图。
100.00
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组距分组中的几个基本概念
组限:每个组两端的数值。分为上限和 下限。
组距：一个组的上限与下限两端的距 离。
全距：所有变量值中最大值与最小值 之差 。
组中值：每个组的上限与下限的中点 值。
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数据的计量尺度 数据的类型
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《统计学》第2章数据的描述
2-4
数据的计量尺度
按照对现象计量程度的不同，可以将数据 计量尺度分为四种，即：定类尺度、定序 尺度、定距尺度、定比尺度。
描述统计学chap05
Business Statistics B siness StatisticsChapter 5Some Important Discrete Probability Distributions Probability DistributionsChapter GoalsChapter GoalsAfter completing this chapter, you should be Aft l ti thi h t h ld b able to:Interpret the mean and standard deviation for a I t t th d t d d d i ti f discrete probability distributionExplain covariance and its application in finance E l i i d it li ti i fiUse the binomial probability distribution to find probabilitiesb bilitiDescribe when to apply the binomial distribution Use Poisson discrete probability distributions to find probabilitiesDefinitionsRandom VariablesRandom VariablesA random variable represents a possible numerical value from an uncertain event. numerical value from an uncertain event.Discrete random variables produce outcomes random variables produce outcomes that come from a counting process (e.g. number of courses you are taking this semester).of courses you are taking this semester).random variables produce Continuous random variables produce outcomes that come from a measurement (e.g. your annual salary, or your weight).your annual salary,or your weight).R d V Definitions i blRandom Variables RandomVariablesC Discrete ContinuousRandom Variable Random Variable Ch. 5Ch. 6Can only assume a countable number of values CProbability Distribution for aDiscrete Random VariableDiscrete Random VariableA probability distribution (or probability mass function )(pdf) A b bilit di t ib ti(b bilit f ti)(df for a discrete random variable is a mutually exclusive listing of all possible numerical outcomes for that random variable of all possible numerical outcomes for that random variable such that a particular probability of occurrence is associated with each outcome.Number of Classes ProbabilityTaken20.230.40440.2450.16016Experiment: Toss 2 Coins. Let X = # heads.4 possible outcomes4possibleS MSummary MeasuresE t d V l()f di tExpected Value (or mean)of a discrete distribution (Weighted Average)∑===µN1iii)X(PXE(X)E l T2iX P(X) Example:Toss 2 coins,X= # of heads,t t d l f X0 .251 .50compute expected value of X:E(X) = (0 x .25) + (1 x .50) + (2 x .25)=102 .25= 1.0(continued) Variance(continued)σThe Covariance The CovarianceThe covariance measures the strength of thelinear relationship between two variables ea e a o s p be ee o a ab es The covariance :)Y X (P )]Y (E Y )][(X (E X [σNi i i i XY ∑−−=1i =where:X = discrete variable X=the i th outcome of X X i = the i outcome of X Y = discrete variable Y Y i = the i th outcome of Y )=probability of occurrence of the condition affectingP(X i Y i ) probability of occurrence of the condition affecting the i th outcome of X and the i th outcome of YComputing the Mean for I RInvestment Returns Return per $1,000 for two types of investmentsP(X i Y i )Economic condition Passive Fund XAggressive Fund YInvestment.2Recession -$ 25-$200.5Stable Economy + 50+ 60.3Expanding Economy+ 100+ 350E(X) = µX = (-25)(.2) +(50)(.5) + (100)(.3) = 50E(Y)(200)(2)(60)(5)(350)(3)95E(Y) = µY = (-200)(.2) +(60)(.5) + (350)(.3) = 95Computing the Standard DeviationP(X.2.5.3Computing the Covariance f I Rfor Investment Returns P(X i Y i )Economic condition Passive Fund XAggressive Fund YInvestment.2Recession -$ 25-$200.5Stable Economy + 50+ 60.3Expanding Economy+ 100+ 35095)(.5)50)(60(5095)(.2)200-50)((-25σY X ,−−−−+−−=825095)(.3)50)(350(100=+Interpreting the Results forI RInvestment ReturnsThe aggressive fund has a higher expectedreturn,but much more risk return, but much more risk=95>µ=50µY = 95 > µX = 50but19321>4330σY = 193.21 > σX = 43.30 The Covariance of 8250 indicates that the twoinvestments are positively related and will vary in the same direction in the same directionPortfolio ExamplePortfolio ExampleReturnsProbability Stock X Stock Y0.1-$100$5001$1000.301500.380-200.3150-100You are trying to develop a strategy for investing in two different stocks. The anticipated annual return for a different stocks The anticipated annual return for a $1,000 investment in each stock has the above probability distributions.probability distributions.Compute the1) Expected return for stock X2) Expected return for stock Y3) Standard deviation for stock X3)Standard deviation for stock X4) Standard deviation for stock Y5) Covariance of stock X and stock Y5)C i f t k X d t k Y6) Do you think you will invest in stock X or stock Y?Explain.Suppose you wanted to create a portfolio that consists of stock X and stock Y. Compare the portfolio expected return and portfolio risk for each of the following proportions invested in stock X:7) 0.1 8) 0.77)018)079)0.3 10) 0.9 11) 0.5On the basis of the results of 7)—11), which portfolio would you recommend? Explain.Probability Distributions Probability Distributions Probability Distributions Continuous Probability Discrete Probability Ch. 5Ch. 6ProbabilityDistributionsProbability Distributions Binomial NormalHypergeometric UniformPoisson ExponentialBinomial Probability Distribution Binomial Probability DistributionA fixed number of observations n A fixed number of observations, ne.g., 15 tosses of a coin; ten light bulbs taken from a warehouse Each observation is categorized as to whether or not the g“event of interest” occurrede.g., head or tail in each toss of a coin; defective or not defective light b lb bulb Since these two categories are mutually exclusive and collectively exhaustive(Generally called “success ” and “failure ”) When the probability of the event of interest is represented as π, then the probability of the event of interest not occurring is 1 -πConstant probability for the event of interest occurring ()for Constant probability for the event of interest occurring (π) for each observationProbability of getting a tail is the same each time we toss the coin y g gBinomial Probability Distribution Binomial Probability Distribution(continued)Observations are independentThe outcome of one observation does not affect theoutcome of the otherp g pTwo sampling methods deliver independenceInfinite population without replacementFinite population with replacementPossible Applications for theppBinomial DistributionA manufacturing plant labels items as either defective or acceptableeither defective or acceptableA firm bidding for contracts will either get a contract or notA marketing research firm receives survey A marketing research firm receives survey responses of “yes I will buy” or “no I will not” New job applicants either accept the offer Ne job applicants either accept the offer or reject itThe Binomial DistributionCounting TechniquesCounting TechniquesSuppose the event of interest is obtaining heads on the toss of a fair coin.You are to toss the coin three times. toss of a fair coin. You are to toss the coin three times. In how many ways can you get two heads?Possible ways: HHT, HTH, THH, so there are threeP ibl HHT HTH THH th thways you can getting two heads.This situation is fairly simple. We need to be able to count the number of ways for more complicated situations.31Binomial Distribution Formula Binomial Distribution FormulaP(X)n X !π(1-π)X n X !=−P(X)=probability of events of interest in X !n X ()!−P(X) = probability of X events of interest in ntrials, with the probability of an “event ofinterest” being πfor each trialExample:Flip a coin four times, let x = # heads:X = number of “events of interest” in sample, (X = 0, 1, 2, ..., n )l i (b f t i l,n = 4π= 0.5n = sample size (number of trials or observations)π= probability of “event of interest” 1 -π= (1 -0.5) = 0.5X = 0, 1, 2, 3, 4p yThe Binomial Distribution E Example lSuppose the probability of purchasing a defective computer p is 0.02. What is the probability p y of purchasing 2 defective computers in a group of 10? X = 2, n = 10, and π = .02n! P(X = 2) = π X (1 − π ) n − X X!(n − X)! 10! = (.02) 2 (1 − .02)10 − 2 2!(10 − 2)! = (45)(.0004)(.8508) = .015315-31Example Suppose that 20% of all copies of a particular textbook fail a certain binding strength test test. Let X denote the number among 15 randomly selected copies that fail the test test. Find the probability that at most 8 fail the test Find the probability that exactly 8 fail the test Find the probability that at least 8 fail the test5-32The Binomial Distribution Shape The shape of the bi binomial i l di distribution t ib ti depends on the values of π and n Here, n = 5 and π = .1P(X) .6 .4 .2 0 0 P(X) .6 .4 .2 0 0 1 2 3 4 5 X5-33n = 5 π = 0.112345Xn = 5 π = 0.5Here, , n = 5 and π = .5The Binomial Distribution Using Binomial Tablesn = 10 x 0 1 2 3 4 5 6 7 8 9 10 … … … … … … … … … … … … … π=.20 0.1074 0 2684 0.2684 0.3020 0.2013 0.0881 0 0264 0.0264 0.0055 0.0008 0.0001 0 0000 0.0000 0.0000 π=.80 π=.25 0.0563 0 1877 0.1877 0.2816 0.2503 0.1460 0 0584 0.0584 0.0162 0.0031 0.0004 0 0000 0.0000 0.0000 π=.75 π=.30 0.0282 0 1211 0.1211 0.2335 0.2668 0.2001 0 1029 0.1029 0.0368 0.0090 0.0014 0 0001 0.0001 0.0000 π=.70 π=.35 0.0135 0 0725 0.0725 0.1757 0.2522 0.2377 0 1536 0.1536 0.0689 0.0212 0.0043 0 0005 0.0005 0.0000 π=.65 π=.40 0.0060 0 0403 0.0403 0.1209 0.2150 0.2508 0 2007 0.2007 0.1115 0.0425 0.0106 0 0016 0.0016 0.0001 π=.60 π=.45 0.0025 0 0207 0.0207 0.0763 0.1665 0.2384 0 2340 0.2340 0.1596 0.0746 0.0229 0 0042 0.0042 0.0003 π=.55 π=.50 0.0010 0 0098 0.0098 0.0439 0.1172 0.2051 0 2461 0.2461 0.2051 0.1172 0.0439 0 0098 0.0098 0.0010 π=.50 10 9 8 7 6 5 4 3 2 1 0 xExamples:n = 10, π = .35, x = 3: n = 10, π = .75, x = 2: P(x = 3|n =10, π = .35) = .2522 P(x = 2|n =10, π = .75) = .00045-34Binomial Distribution Ch Characteristics t i ti Meanµ = E(x) ( ) = nπσ = nπ ( (1 - π )σ = nπ ( (1 - π )2Variance and Standard DeviationWheren = sample size π = probability of the event of interest for any trial (1 – π) = probability of no event of interest for any trial5-35The Binomial Distribution CharacteristicsExamplesµ = nπ = (5)(.1) (5)( 1) = 0.5 05σ = nπ (1 - π ) = (5)(.1)(1 − .1) = 0.6708P(X) .6 .4 .2 0 0 P(X) .6 .4 .2 0 0 1 1n = 5 π = 0.12345Xµ = nπ = (5)(.5) = 2.5σ = nπ (1 - π ) = (5)(.5)(1 − .5) = 1.118 1 118n = 5 π = 0.52345X5-36Using Excel For The Binomial Distribution5-37Using PHStat Select PHStat / Probability & Prob. Distributions / Binomial…5-38Using PHStat(continued) Enter desired values in dialog box Here: n = 10 π = .35 Output for X = 0 to X = 10 will be generated by PHStat Optional check boxes for additional output5-39PHStat OutputP(X = 3 | n = 10, π = .35) = .2522P(X > 5 | n = 10, 10 π = .35) 35) = .0949 09495-40=Binomdist (numbers_s, trials,probability_s, cumulative)b bilit l ti)The Hypergeometric Distribution The Hypergeometric Distribution“n” trials in a sample taken from a finiteof size Npopulation of size NSample taken without replacementOutcomes of trials are dependentConcerned ith finding the probabilit of“X”Concerned with finding the probability of “X” successes in the sample where there are “A” successes in the population■Example:ExampleFive individuals from an animal population thought to be near extinction in a certain region thought to be near extinction in a certain regionhave been caught, tagged, and released tomixed into the population. After they have hadp p yan opportunity to mix, a random sample of 10 ofthese animals is selected . Let X be the numberof tagged animals in the second sample. If theref t d i l i th d l If thare actually 25 animals of this type in the region,what is the probability thatwhat is the probability that1) X=2?2) X2?2)X≤Hypergeometric Distributionin PHStati PHSSelect:PHStat / Probability & Prob. Distributions / Hypergeometric …Hypergeometric Distributionin PHStati PHSt t(continued)Complete dialog box entries and get output…N 10n 3N=10n=3A = 4X = 2P(X = 2) = 0.3The Poisson DistributionThe Poisson DistributionApply the Poisson Distribution when:You wish to count the number of times an event occurs in a given area of opportunityThe probability that an event occurs in one area of t it i th f ll f t it opportunity is the same for all areas of opportunity The number of events that occur in one area of opportunity is independent of the number of events opportunity is independent of the number of events that occur in the other areas of opportunityThe probability that two or more events occur in an The probability that two or more events occur in an area of opportunity approaches zero as the area of opportunity becomes smallerThe average number of events per unit is λ(lambda)。
统计学基础
目录
01. 描述统计 02. 自由主题 03. 推断统计学
描述统计
01
描述统计
集中量数 差异量数 相对量数
描述统计
集中量数
A
平均 数
B
众数
C
中位 数
描述统计
差异量数
A
标准 差
B
方差
C
其他
描述统计
相对量数
01
Z分数
02
相对差 异量数
03
百分等 级
自由主题
02
自由主题
推断统计学
03
推断统计学
参数估
01
计
回归分
04
析
参数检
02
验
卡方检
05
验
方差分
03
析
非参数
06
检验
推断统计学
基本问题
推断统计学
参数估计
总体均数 的估计
子主题
总体均数的估计
标准误 方法
正态 近似正态
参数估计
一．子主题
推断统计学
参数检验
0 1
差异及差异显
著性检验
0 2
假设与假设检
验
0 3
显著性水平
0 4
B
双因 素
C
多因 素
推断统计学
回归分析
概念
一元
多元
回归分析
概念 2
拟合优 度检验
03
独立检 验
卡方检验
意义 拟合优度检验 独立检验
推断统计学
非参数检验
A
意义
D
子主 题
B
相关 样本
E
第2章 描述统计学
样本方差
样本标准差 性质1（简化计算）
性质2（简化计算）设
则
2.3 数据的汇总: 差异特征
例2.3.7 世界各地商业航班空难事故数据。计 算这些年事故数的样本方差。
2.3 数据的汇总: 分布特征
分位数：对于0p 1, 样本100p%分位数是 100p %的值小于等于它而100(1-p) %的值大于 等于它。如果有两个值满足以上条件，则样本 100p %分位数是这两个值的算术平均值。特别 地，中位数是50%分位数。 计算方法：先对数据按从小到大排序。如果np 不是整数，则第[np]+1个数据是100p%分位数。 如果np是一个整数，那么100p%分位数取第[np] 和第[np]+1个值的平均值。
线图、柱形图、折线图
2.2 数据的描述
例：42位具有电气工程学士学位毕业生起始年 薪(千美元)的频数表
2.2 数据的描述
柱形图（用Excel）
2.2 数据的描述
名词： 相对频数=频数/总的样本数（每个数据出现次 数的百分比） 相对频数图：反映每个数据及其相对频数，横 坐标为不同数据，纵坐标为相对频数
茎叶图自然形成直方图(旋转90度)
例
某个会计事务所完成20个样本 客户年终审计所要求的时间 数（以天为单位）
12 14 19 18 15 15 18 17 20 27 22 23 22 21 33 28 14 18 16 13
组数：选择组数为5 组距：组宽=（最大数据值-最小数据值）/组数=4.2，确定组宽为5
≤ 34 20 1.00
2.3 数据的汇总：中心特征
样本数据： 样本均值
统计chap002
Self-Review 2–1(cont.)
(a) Is the data qualitative or quantitative? Why? (b) What is the table called? What does it show? (c) Develop a bar chart to depict the information. (d) Develop a pie chart using the relative frequencies.
A relative frequency captures the relationship between a class total and the total number of observations.
2-7
Graphic Presentation of Qualitative Data
2-4
class frequency
To summarize this qualitative data, we classify the vehicles as either domestic (coded as 1) or foreign (coded as 0) and count the number in each class. We use vehicle type to develop a frequency table
2-12
Example(cont.)
1. What type of measurement scale is used for ease of navigation?
2. Draw a bar chart for the survey results. 3. Draw a pie chart for the survey results.
统计学第二章描述优秀课件
散点图
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james
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均值
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no
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差异（ 离散）
score 6 12 18 24 30 36 42 48 54 60
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59 .0
自由度
▪ 计算样本方差时应除以n-1，而不是n
s2 x x 2 n 1 ▪ 这里n-1叫自由度（degree of freedom）, 表示样本可自由取值的数目
自由度？
如果某班只有1位学 生，身高为172
如果从某班抽取1位学生 调查其身高为172
四分位数
▪ 观测值按大小顺序排列后，均分为四部分， 处于分界点上的数
• 2/4位置:中位数 • 1/4位置:下四分位数 • 3/4位置:上四分位数
四分位数
▪ 詹姆斯:
Ql 2,2Qu32
▪ 杜兰特:
Ql 2,7Qu36
月薪
从某公司随机抽取13位职工，调查他们的月薪如下：
2000 2600 3500 1800 2500 4800 2800 3000 2200 3300 5200 4600 4000
杜兰特
31 32 25 43 42 29 30 37 18 28 25 25 38 27 28 26 54 33 30 38 31 33 27 51 37 31 36 34 36 24 25 36 27 35 28 26 37 29 29 20 15 26 23 35 42 26 33 24 33 33 28 15 38 30 28 33 30 17 27 33 39 30 28 29 38 41 48 32 32 37 27 36 28 42 43 32 21 30 25 23 40 33 31 27 36 36 48 28 24 33 36 42 29 34 41 46 24 31 19 13 42
《统计学第二章》课件
THANKS
感谢观看
多元线性回归分析
总结词
多元线性回归分析是研究多个因变量与 多个自变量之间的线性关系的统计方法 。
VS
详细描述
多元线性回归分析通过建立多元线性回归 方程来描述多个因变量与多个自变量之间 的平均变化关系。这种方法可以同时考虑 多个自变量对因变量的影响，并通过对回 归方程的参数进行估计和检验来评估关系 的强度和方向。多元线性回归分析在经济 学、社会学和生物医学等领域有广泛应用 。
离散型随机变量的概率分布
1 2
离散型随机变量
随机变量只取有限个或可数个值。
离散型随机变量的概率分布
描述离散型随机变量取各个可能值的概率。
3
离散型随机变量的期望值和方差
描述离散型随机变量的数学期望和离散程度的量 。
连续型随机变量的概率分布
连续型随机变量
01
随机变量可以取任何实数值。
连续型随机变量的概率分布
提出原假设和备择假设、构造检验统计量、确定临界值、做出决 策。
单样本假设检验的示例
检验某班级学生的平均成绩是否达到预期水平。
单样本假设检验的适用场景
只有一个总体需要检验的情况。
双样本假设检验
双样本假设检验的基本步骤
提出原假设和备择假设、构造检验统计量、确定临界值、 做出决策。
双样本假设检验的示例
比较两个不同班级学生的平均成绩是否存在显著差异。
双样本假设检验的适用场景
需要对两个总体进行比较的情况。
06
CATALOGUE
回归分析与方差分析
一元线性回归分析
总结词
一元线性回归分析是研究一个因变量与一个自变量之间的线性关系的统计方法。
详细描述
统计学02-课件
在 = 0.05的水平上拒绝H0
结论：
说明该机器的性能不好
2 未知小样本均值的检验
(P 值的计算与应用)
第1步：进入Excel表格界面，选择“插入”下拉菜 单 第2步：选择“函数”点击，并在函数分类中点击 “统
计” ，然后，在函数名的菜单中选择字符 “TDIST”，确定 第3步：在弹出的X栏中录入计算出的t值3.16 在自由度(Deg-freedom)栏中录入9 在Tails栏中录入2，表明是双侧检验(单测 检验则在该栏内录入1) 8 - 42 P值的结果为0.01155<0.025，拒绝H0
第4步：将Z的绝对值2.83录入，得到的函数值为
0.997672537
P值=2(1－0.997672537)=0.004654
8 - 34
P值远远小于/2，故拒绝H0
2 已知均值的检验
(小样本例题分析)
【例】根据过去大量资料，
某厂生产的灯泡的使用寿命 服 从 正 态 分 布 N~(1020 ， 1002) 。 现 从 最 近 生 产 的 一 批 产 品 中 随 机 抽 取 16 只 ， 测 得 样 本 平 均 寿 命 为 1080 小 时 。 试 在 0.05 的 显 著 性 水 平 下 判 断这批产品的使用寿命是否 有显著提高？(＝0.05)
8.2 一个总体参数的检验
8.2.1 检验统计量的确定 8.2.2 总体均值的检验 8.2.3 总体比例的检验 8.2.4 总体方差的检验
8 - 27
一个总体参数的检验
一个总体
均值
比例
方差
Z 检验
t 检验
Z 检验
（单尾和双尾） （单尾和双尾） （单尾和双尾）
2检验
（单尾和双尾）
概率论与统计学基本知识chapt2english
Dot Plot
30
35
40
45
Stem-and-Leaf Display 茎叶图
Stem-and-leaf display combines graphic technique and sorting technique. It is very popular for summarizing numerical data.
1. Bar graph shows the amount of data that belongs to each class as proportionally sized rectangular areas
2. Pie chart shows the amount of data that belongs to each class as a proportional part of a circle
The researchers want to determine whether one type of aphasia occurs more often than any other, and, if so, how often.
Describing Qualitative data
Qualitative data are nonnumerical in nature, thus the value of a qualitative variable can only be classified into categories called classes. We can summarise such data numerically in two ways: (1) by counting
MPG
Histogram
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Organizing and Presenting Data Graphically
Data in raw form are usually not easy to use for decision making
Some type of organization is needed
The class boundaries (or class midpoints) are shown on the horizontal axis the vertical axis is either frequency, relative frequency, or percentage Bars of the appropriate heights are used to represent the number of observations within each class
6 7
1224 becomes
12
2
2-10
Using other stem units
(continued)
Using the 100’s digit as the stem:
The completed stem-and-leaf display:
Data:
613, 632, 658, 717, 722, 750, 776, 827, 841, 859, 863, 891, 894, 906, 928, 933, 955, 982, 1034, 1047,1056, 1140, 1169, 1224 Stem 6 7 Leaves 136 2258
Histogram : Daily High Tem perature 7 6 5 4 3 2 1 0 5 15 0 25 35 45 55 0 More
2-19
6 5 4 3 2
(No gaps between bars)
Frequency
Class Midpoints
Histograms in Excel
containing class groupings (categories or ranges within which the data falls) ...
and the corresponding frequencies with which data falls within each grouping or category
1 Select Tools/Data Analysis
2-20
Histograms in Excel
(continued)
2 Choose Histogram
(
Input data range and bin range (bin range is a cell
2-18
Histogram Example
Class 10 but less than 20 20 but less than 30 30 but less than 40 40 but less than 50 50 but less than 60 Class Midpoint Frequency 15 25 35 45 55 3 6 5 4 2
2-14
Frequency Distribution Example
Example: A manufacturer of insulation randomly selects 20 winter days and records the daily high temperature 24, 35, 17, 21, 24, 37, 26, 46, 58, 30,
Compute class midpoints: 15, 25, 35, 45,
55
Count observations & assign to classes
2-16
Frequency Distribution Example
(continued)
Data in ordered array:
12, 13, 17, 21, 24, 24, 26, 27, 27, 30, 32, 35, 37, 38, 41, 43, 44, 46, 53, 58
2-9
Using other stem units
Using the 100’s digit as the stem:
Round off the 10’s digit to form the leaves
Stem
Leaf
1 8
613 would become 776 would become ...
8
9 10 11 12
346699
13368 356 47 2
2-11
Tabulating Numerical Data: Frequency Distributions
What is a Frequency Distribution?
A frequency distribution is a list or a table …
2-4
The Ordered Array
A sorted list of data:
Shows range (min to max) Provides some signals about variability within the range May help identify outliers (unusual observations)
If the data set is large, the ordered array is less useful
2-5
The Ordered Array
(continued)
Data in raw form (as collected):
24, 26, 24, 21, 27, 27, 30, 41, 32, 38
range Width of interval num berof desired class groupings
Use at least 5 but no more than 15 groupings Class boundaries never overlap Round up the interval width to get desirable endpoints
Data in ordered array from smallest to largest: 21, 24, 24, 26, 27, 27, 30, 32, 38, 41
2-6
Stem-and-Leaf Diagram
A simple way to see distribution details in a data set METHOD: Separate the sorted data series into leading digits (the stem) and the trailing digits (the leaves)
Table Graph
Techniques reviewed here:
Ordered Array Stem-and-Leaf Display Frequency Distributions and Histograms Bar charts and pie charts Contingency tables
2-12
W?
A frequency distribution is a way to summarize data The distribution condenses the raw data into a more useful form...
2 3
1 8
2-8
Example
(continued)
Data in ordered array:
21, 24, 24, 26, 27, 27, 30, 32, 38, 41
Completed stem-and-leaf diagram:
Stem Leaves
2 3 4
1 4 4 6 7 7 0 2 8 1
and allows for a quick visual interpretation of the data
2-13
Class Intervals and Class Boundaries
Each class grouping has the same width Determine the width of each interval by
32, 13, 12, 38, 41, 43, 44, 27, 53, 27
2-15
Frequency Distribution Example
(continued)
Sort raw data in ascending order:
12, 13, 17, 21, 24, 24, 26, 27, 27, 30, 32, 35, 37, 38, 41, 43, 44, 46, 53, 58
2-3
Tables and Charts for Numerical Data
Numerical Data
Ordered Array
Frequency Distributions and Cumulative Distributions
Histogram Polygon Ogive
Stem-and-Leaf Display
Find range: 58 - 12 = 46 Select number of classes: 5 (usually between 5 and 15) Compute class interval (width): 10 (46/5 then round up)