M. Statistics for Management II Notes PDF

UNIT 1 1. Sampling and Sampling Sampling Distributions 1.1 Sampling Theory Theory Common Terminologies o Population: Refers to a set of all elements of i...

UNIT 1 1. Sampling and Sampling Sampling Distributions 1.1 Sampling Theory Theory Common Terminologies o Population: Refers to a set of all elements of interest under study or collection of all elements about which some inference is to be made. o Population: Refers to all items that have been chosen for study o Population: Refers to the totality of elements being studied ( inters grows of individual or object under consideration be noted by (N) Examples 1. The population of HIV carriers in the west shawa zone 2. The population of students in Finfine City 3. The incomes of all families in Ambo town 4. The number of miles per gallon achieved by all cars of a particular model Note: sometimes it is possible to practically examine every person or item in the population we wish to study which we call complete enumeration or census o Sample (n) refers to a part or subset of the population  Refers to a small portion of the population believed to represent the entire population  Sample- refers to a tool which inter something about population from which it selected Examples:- TV viewers in Ambo Town - A handful of bean - Numbers of HIV carriers in Ambo town. o Sampling: the process of selecting elements from the population or gathering information from parts of the population o Element: the basic unit (each unit) of the population o Census: gathering information from all elements in the population or refers to complete enumeration or the measurement of every individual or item in the population o Population parameter: refers to a measurement or the characteristics taken from the population Examples:- Mean of population Population Variance Population standard Deviation Population proportion Population range o Sample static: refers to a measurement or characteristic taken from the sample. Examples:- sample mean (X) Sample Variance (S2) Sample standard deviation ( S) Sample proportion 1 of 61 Sampling & sampling Distributions Tujuba A.  2.The The Needs for Sampling Some of the major reasons why sampling is necessary are:- o Contacting the entire (whole) population would often take more time (time-consuming) o The physical impossibility of checking all items in the population Examples: Population of birds population of snakes population of fish o The cost of studying all the items in a population is often prohibitive (too expensive or bidding) o Practical infeasibility of complete enumeration o Enough reliability of inference based on sampling o The adequacy of sample results o The quality of data collected o The destructive nature of the certain test Sampling Design & Sample Frame  Sampling Design: Design: refers to a definite plan for obtaining a sample from a given population  Sampling Design: Design refers to technique or the procedure the researcher would adopt in selecting items for the sample  Sample Design:the Design set of procedures for selecting the units from the population that are to be in the sample  Specifies for every possible sample the probability of being drawn. Characteristics of Good Sample Sample Design o Goal orientation:- should be oriented to the research objective. Should influence the choice of population o Measurability:- should enable the compotation of valid estimates of its sampling variability. Expressed in the form of standard errors in surveys o Practicality: sample design should be complete, correct, practical, and clear to implement sampling o Practicality -refers to the simplicity of the design (i.e. it should be capable of being understood and followed in the actual operation of the fieldwork o Economy:-Implies that the objectives of the survey shoo be achieved with minimum cost and offers o Independence:- Each element of the universe is independent. Each element has an equal chance of selection o Homogeneity:- there should not be any changes in the Universe where the survey is ducted (i.e. quality and nature of the units should remain the same) o Adequacy:- too much low units will not be desirable and too many high sampling units need more time and money for analysis. Sample Frame  Sampling Frame:the specific set of units from which the sample is actually drawn is the sampling frame  Sampling Frame: refers to a list of all the units of population Should be made up of update and free from errors of omission and duplication of sampling units 2 of 61 Sampling & sampling Distributions Tujuba A.  The frame can be perfect or defect Perfect frame:- identifies each element once and only once but it is seldom available in real life. Defect. Frame:- can be classified as o Incomplete frame o Inaccurate frame o Inadequate frame o Outdated frame Factors Influencing the Choice Between Between different designs include  Nature and Quality of the Frame  Availability of adequate information about units on the frame  Accuracy requirements and needs to measure accuracy  Cost/and operational concern When developing a sampling design the following factors are important: o Type of population (Universe) o Sampling unit (e.g. geographical, state, district, village, etc.) o Types of sample o Size of the sample o Parameters of interest o Budgetary constraints o Sampling procedures Sampling and Non- Non-Sampling Error Error The researcher must keep given the two causes of incorrect inferences in sampling 1. Systematic Bias 2. Sampling Error 1. Systematic Bias:- Bias:-results from errors in sampling procedure and it cannot be reduced or eliminated by increasing the sample size Systematic Bias: Bias: Refers to the faulty procedure of selecting representative and tendency problems to overlook elements that have peculiar characteristics  Systematic Bias is a result of one or more of the following factors  Inappropriate sampling frame  Defective Measuring devices  Nonresponse: Occur if you do not obtain full information from all the sampling unit  Indeterminacy Principle Principle: rinciple sometimes we find that individuals act differently when kept under observation than what they do when kept in non-observed situations Example: If somebody surveys the work to determine the average completion time of the job most workers slow than other times to minimize the average work completion time 3 of 61 Sampling & sampling Distributions Tujuba A.   Natural Bias in reporting of data:- data this often Causes a systematic bias in many inquiries. For Example: Example: there is always a downward bias in the income data find upward bias income data collected by some social organization. People understate their income if asked for tax purposes, but overstate the same if asked for social status or affluence. Generally in psychological surveys, people tend to give what they think is the correct answer rather than revealing their truth-telling.  Others: Errors arise from administrative procedures, collecting inconsistent reporting data (collecting data from faulty provider, sources..) 2. Sampling Error: refers to the difference between sample statics and its corresponding population parameter o Refers to the random variations in sample estimates around to true population parameters. o can be measured for a given sample design and size o The measurement of sampling error is usually called the precision of the sampling plan. o If we increase the sample size can cause high cost and can lake more time and effort (systematic bias) Example: suppose a population of five production employees had an efficiency rating of 97, 103, 96, 99, and 105, further suppose that a sample of two ratings of 97 and 105 is selected to estimate the population mean rating Hence the mean of sample would be x= = = 101 Another sample of two size selected 96 and 103, with X of x= = 99.5 Mean of All rating (population) would be µ= = 100 Therefore the sampling error for the first example is determined by X - µ = 101 - 100 = 1.00 Sampling error for the 2nd example is determined by X – µ = 99.5 – 100 = 0.5 Each of these differences 1.00 and 0.5 is the error mode in estimating the population means based on the sample mean and these sampling errors are due to chance. 1.3 Type of Samples Samples (Sampling (Sampling Techniques) In general, there are two types of sampling methods. These are 1. Probability (Random) samples 2. Non- probability (Nonrandom) samples 1. Probability (Random) Sampling Techniques Random Sampling: Sampling a sample selected in such a way when every item of the universe has an equal chance of inclusion in the sample 4 of 61 Sampling & sampling Distributions Tujuba A.  It is a lottery method in which individual units are picked up from the whole group by some mechanical process It is blind chance alone that determines whether one item or the other is selected Refers to sampling when all items (i.e. each element) in the population have a chance of being chosen in the sample and the probability of each element of the population included in the sample is known. Advantages of probability sampling  The only sampling method that provides essentially unbiased estimates having measurable precision  Permits the researcher to evaluate in quantitative terms the relative efficiency of alternative sampling techniques in a given situation.  Requires relatively little universe Knowledge the only two things to be known are I. Way of identifying each universe element uniquely II. The total number of universe elements. Disadvantages of Random sampling o pensive than non-probability sampling o The method of selection can be complex and time-consuming Types of probability sampling The various probability sampling methods are as follows 1. Simple Random Sampling 2. Systematic Random Sampling 3. Stratified Sampling 4. Cluster Sampling 5. Multi- Multi-Stage Sampling 6. Area Sampling 1. Simple Random sampling Simple random sampling: random sampling in which each unit of the population have an equal chance of being included in the sample (also known as the lottery method)  Simplest and most popular techniques of sampling  Each element of the population has an equal chance of being included in the sample and all choices are independent of each other  Each possible sample combination have an equal chance of being chosen or every element has the same chance of being selected for the given sample size Example 1: If we want to take a sample of 25 persons out of a population of 150, the procedure is to write the names of all the 150 persons on separate slips of paper, fold these slips, mix them thoroughly and then make a blindfold selection of 25 slips 5 of 61 Sampling & sampling Distributions Tujuba A.  Example 2: Suppose Belay, Nur, G/Tsadik, and Chala are on office staff population. All four employees wants the same vacation period, but only two can be away at the same time The chips lettered B, N, G and C are shaken in a container and then the manager draw out two chips. Therefore, the possible size two samples from B, N, G, and C populations are B can appear in three of the six samples, so P(B) = = similarly P(C) = = P(N) = P(G) = P(C) = P(B) = ½ Note simple random sampling depends on whether the population is finite or infinite For a simple random sample of size from a finite population of size N, the sample selected has the same chance of being selected i.e. n/N The number of different simple random samples of the size that can be selected from a finite N population of size N is ! N ! ! = N Example: for the above example the number of samples can be ! N = ! ! ! = ! ! ! = = =6 Using software like MINITAB and excel we can randomly select 2. Systematic Random Sampling The mechanics of taking a systematic sample can be performed through the method suppose the population consists of ordered N units (number 1 to N) and a sample of size n is selected from the population in such a way that N/n = K (rounded to the nearest integer). Here, k is called sampling interval the event integer) Her, k is called sampling interval TN every kth unit is selected in the sample. The items or individuals of a population are arranged in the same way Examples: Alphabetically in file drawer by data received or by other methods the random starting point is selected and every ith or Kth member of the population is selected ! "# $% ! & $' & ( &$) * &+ ,- &$. Sampling interval 6 of 61 Sampling & sampling Distributions Tujuba A.  e.g. If the 1st item selected randomly is the 10th and if we want every 20th item to be the sample, The sample consists of 10th, 30th, 50th, 70th, 90th, etc. elements of the sample Note The first (e.g. 10th) element should be chosen by using a random process since the Note: first item is chosen at a random all items have the same likelihood of being selected for the sample, therefore it is a probability sampling  Under certain conditions, it may produce a biased or unrepresentative sample 3.S Stratified Random Sampling Stratified sample in one in which random selection is done not from the universe as whole but different parts or strata of a universe.  Concerned with dividing a population into non over loping groups and choosing an element from each stratum by random (selecting samples from each stratum) Note: stratified random sample gives more precise results than simple random if the difference between strata is greater than different within strata Factors to be considered in stratified sampling a. Base of stratification b. Number of strata c. Sample size within strata Example 2 If a Researcher wants to deal with the income inequality situation in Ambo town the researcher can divide the households into different groups as follows:  Civil servant  Merchant  Petty traders and local drink sellers Note: In stratified sampling, the population is divided into well-defined groups, whereas each group (strata) has homogeneity within itself but wider heterogeneity (or variation) among the groups. In the case of cluster sampling, the situation is reversed for stratified sampling (i.e. The different cluster's age homogenous but elements in each cluster age heterogeneous) 4.Cluster 4.Cluster Random sampling In this method the universe (population) is divided into some recognizable subgroups, called clusters, After this simple random sample of these clusters is drawn and all elements belonging to the selected Chester longitude are the sample.  This method is always employed to reduce the cost of sampling for a population scattered over large geographic areas.  Involves groupings of the heterogeneous population elements in the subdivision or groups or cluster and selecting a random sample of the group  Closter sampling can be of different stages 7 of 61 Sampling & sampling Distributions Tujuba A.  *one one stage: stage when the sample is selected from clusters or groups of the population * Two stages: when random sampling is made from selected samples of the cluster ( Note: all elements in the selected sample of clusters cloud be included in the sample  Cluster sampling is mainly used because it is very economy (The goal of sampling is to achieve a desired degree of passion at the lowest cost.  In cluster sampling, only a fraction (or a small fraction) of clusters in a population is sampled Features of Cluster Sampling o The population is divided into many subgroups and each group have a few elements o Viewed (concerned) selecting heterogeneous samples within the groups and homogeneous between the subgroups. o Only a few subgroups are selected (randomly) selected and elements within subgroups are covered in the study. o Cluster samples work best for relatively homogeneous populations (i.e. population having a uniform composition throughout) Example: N = 200 (η1 = 50) (η3 = 50) E.g. please refer to the mo. module for the example (η2 = 50 ) (η4 = 50) 5. Multi- ulti-stage Sampling Sampling When different sampling techniques are combined at the various stages to select the representative from the Universe it is known as multi-stage sampling  Combines different sampling methods  More complex in nature when compared to another method Example: consider the idea of sampling Oromia Region residents for a face-to-face interview. To sample in the real ground we have to do the same cluster sampling for the first stage of the process. This can be performed like township, census tract … we can make stratified sampling process in the cluster, etc. 2. Non- Non-probability (Non (Non-Random) Random) Sampling Non- Non-Probability Sampling: Sampling refers to sampling Methodology in which personal knowledge and opinion play a major role in identifying which elements of the population are to be included in the population to be included in the sample is not known. *Non-probability Sampling can be classified as 1. Convenience (Accidental) sampling 2. Purposive sampling sampling 3. Quota sampling 4. Snowball sampling Haphazard or Convenience Sampling: 1. Accidental Or Haphazard ing select anyone conveniently. It can produce ineffective, highly unrepresentative samples and is not recommended. Such samples are cheap and quick; however, the bias and systematic errors that easily occur make them worse than no sample at all. The person on the street interview conducted by television program is an example of a haphazard sample. 8 of 61 Sampling & sampling Distributions Tujuba A.  Sampling: is an improvement over haphazard, but it, too, is a weak type of sampling. In 2. Quota Sampling quota sampling, a researcher first identifies categories of people (e.g. male and female), then decides how many to get in each category. Thus, the number of people in various categories of the sample is fixed. The researcher interviews the first 5 males that he/she encounters. Similarly, the first 5 females encountered are interviewed 3. Purposive or Judgmental Sampling: - is an acceptable kind of sampling for a special institution. It uses the judgment of an expert in selecting cases or it selects cases with a specific purpose in mind. The researcher never knows whether the case selected represents the population. It is used in exploratory research or field research. Purposive sampling can be appropriate in some situations. E.g., in a study of labor problems, you may want to talk only with those who have experienced on-the-job discrimination, Companies often try out new product ideas on their employees. The rationale is that one would expect the firm’s employees to be more favorably disposed toward a new product idea than the public. If the product does not pass this group, it does not have prospects for success in the general market. 4. Snowball Sampling (also called or Reputation) Sampling: is a called Network, Chain Referral, or method for identifying and sampling (or selecting) the cases in a network. The network could be scientists around the world investigating the same problem, the elites of a medium-sized city, the members of an organized crime family, persons who sit on the boards of directors of major corporations, etc. Each person or unit is connected with another through a direct or indirect linkage. This does not mean that each person directly knows, interacts with, or is influenced by every other person in the network. 1.4 Sampling Distributions Sampling distribution: refers to the probability distribution of a given statistic based on a random sample of a certain size n. The distribution of the statistic for all possible samples of a given size Sampling Distribution: is a probability distribution for all possible values of a sample statistic (such as sample mean.) Note: The normal probability distribution is used to determine probabilities for normally distributed individual measurements given the mean and standard division. Symbolically, the variable is the measurement X, with the population mean µ and population standard deviation δ. in contrast to such distribution of individuals measurement, the sampling distribution is the probability distribution for possible values of the sample statistic Standard Deviation of the sampling distribution of a statistic is referred to as the Standard Error of that quantity. 9 of 61 Sampling & sampling Distributions Tujuba A.  The sampling distribution is based on (depends on) the underlying distribution of population, statistics being considered and the sample size used. Sampling Distribution of Sample Mean Sampling distribution of mean: is the probability distribution of all possible sample means of size η that could be taken from a population of size N. NB: sampling distribution of the mean is not sample distribution, Which is the distribution of the measured valves of X in one random sample rather, the sampling distribution of the mean is the probability distribution for X, the sample mean.  Sampling distribution of the mean is described by two parameters (or E(x)  The expected valve (X) = X or mean of sampling disturb of mean  standard deviation of the mean δX, the standard error of the mean. Properties of the sampling distribution of the Means 1. The mean of the sampling distribution of the means is equal to the population mean Algebraically µ = µX = X (or E(X) 2. The standard deviation of the sampling distribution the means (standard Error) equals to population standard deviation divided by the square root of the sample size δX= this is true if and only if η0.05N / 3 4 - called (referred) population correction factor As a role of thumb, we can omit the multiplier when η > 0.05N. Iη application, n usually less than 5% of N. so the sample formula to compute the error of the mean is / 3 4 Example : Let N=10000 N=100 / 3 4 =/ =/ = √0.9900990099 n< 0.05N or 0.01 , 0.05 = 0.995037 = 0.995≈ 1 As we increase the size of the sample, the spread of the distribution of the sample mean becomes smaller 3. The sampling distribution of means is approximately normal for a sufficiently large sample size (η≥ 30) Exercise: Assume there is a population that contains only five numbers (elements): 0,3,6,3 and 18 which is designated by letters A, B, C, D, and E respectively If a sample size of 3 is selected: Required: a. Compute for sample number can drown from the given population of size 3. b. What is the population mean of the given data? c. What is sampling distribution of the sample means for the given sample size of 3 d. Compute mean of sampling distribution of the mean Exercise 2: Naol Trading has seven production employees thus engaged by hourly earnings. The hourly earnings of each employee are as follow Employee Hourly Earning in Birr Belay Birr 7 Tola 7 Samuel 8 Ayele 8 Wako 7 Nur 8 Gada 9 11 of 61 Sampling & sampling Distributions Tujuba A.  Required i) What is population mean of the given population ii) What is the sampling distribution of the sample mean of the size of 2? iii) What is the mean of the sampling distribution iv) What observation can be made about the population and sampling distribution Exercise 3: A population consists of the following ages 10, 20, 30, 40, and 50, A random sample of three is to be selected from this population Required a. Compute for size three sample can drown from the given population b. What is the population mean of the given data? c. What is the sampling distribution of the sample means for the given sample size of 3 d. What is the mean of a sampling distribution Central Central Limit Theorem (CLT) The concept was first developed by Abraham de Moivre in 1733, but it was not formalized till 1930, when Hungarian mathematician George Polya named the Central Limit Theorem. In probability theory, the central limit theorem (CLT) states that the distribution of a sample variable approximates to normal distribution (i.e. Bell curve”) as the sample size becomes larger, assuming that all samples are identical in size, and regardless of the population's actual distribution shape. Central Limit Theorem is a statistical premise that, given a sufficiently large sample size from a population with a finite level of variance, the mean of all sampled variables from the same population will be approximately equal to the mean of the whole population. Furthermore, these samples approximate a normal distribution, with their variances being approximately equal to the variance of the population as the sample size gets larger, according to the law of large numbers. The central limit theorem is often used in conjunction with the law of large numbers, which states that the average of the sample means and standard deviations will come closer to equaling the population mean and standard deviation as the sample size grows, which is extremely useful in accurately predicting the characteristics of populations. To summarize  The central limit theorem (CLT) states that the distribution of sample means approximates a normal distribution as the sample size gets larger, regardless of the population's distribution.  Sample sizes equal to or greater than 30 are often considered sufficient for the CLT to hold.  A key aspect of CLT is that the average of the sample means and standard deviations will equal the population mean and standard deviation.  A sufficiently large sample size can predict the characteristics of a population more accurately  If the population is normally distributed the distribution of sample means is normal regardless of the sample size 12 of 61 Sampling & sampling Distributions Tujuba A.   If the population from which samples are taken is not normal, the distribution of sample means will be approximately normal if the sample size (n) is sufficiently large (N≥30) the larger the sample size used the closer the sampling distribution is to the normal curve  The significance of CLT is that it permits to use of sample statist to make inferences about population parameters without know knowing anything about the shape of the frequency distribution of that population rather than what we can get from the sample  Permits to use normal distribution curve for analyzing distributions whose shape is unknown.  Creates potential for applying the normal distribution to many problems when the sample is sufficiently large Exercise 1: the distribution of annual earnings of all bank tell with the five years of experience is skewed negatively This distribution has a mean of mean birr 15,000 and standard deviation of birr 2000 if we draw a random sample of 30 tellers what is the probability that their earning will average more than birr 15,750 annually? Exercise 2: the hourly wage of workers in a given industry has a mean wage rate of birr 5 per hour and a standard deviation of birr 0.6 Required: What is the prob. that the mean wage of a random sample of 50 workers will be between birr 5.1 and 5.2? Exercise 3: The average income tax paid by a family is Birr 2200 with a standard deviation of Birr 800 If a sample of 100 families is selected, what is the prob. That the sample means will be the average income tax paid by the family is more than birr 2000 and Birr 2400? 8) 1.5 Sampling Distribution of proportions ((7 In some (certain) circumstances (situation) in statistics, it is important to know a proportion of certain characteristics in a population Examples: Examples: A quality central engineer might want to know what proportion of products on assembly lines are defective A labor economist might want to know what proportion of the labor force is unemployed o The mean is computed by averaging a set of values, and o The sample proportion is computed by dividing the frequency that a given characteristic occurs in a sample by the number of items in the sample i.e 9̅ = Where 9̅ = sample proportion ; = number of items in a sample that possess the character = number of items in the sample 13 of 61 Sampling & sampling Distributions Tujuba A.  like other prob.distribution,sampling distribution of proportion can be described by two parameters  The mean of the sample proportions, ε(9̅  The standard deviation of the proportions (δ9̅ ?@>A@ BAACA CD =ℎB 9AC9>A=FC?< 8 Properties of sampling Distribution of 7 1As a sampling distribution of the mean does, the population proportion (p) always equals the means of the sample proportion (9̅ ) F. B. p=ε(9̅ 2. The standard error of the proportions is equaled to ,G δ9̅ = / Where, p = population proportion q = 1-p n = sample size. / ,G or δ9̅ = / , where/ = finite population correction factor Note : finite population correction factor is not necessary if n < 0.05 N Central Limit Theorem (CLT) and sampling Distributions of how does a researcher use the sample proportion in analysis (Answer by applying the central limit theorem Central Limit Theorem(CLT): states that normal distribution approximates the shape of the distribution of sample proportions if np and n are greater than five (5) consequently we solve problems involving sample proportion by using a normal distribution whose mean and standard deviations are H9̅ = ,G δ9̅ = / ,̅ , ,̅ Z9̅ = NB: The sampling distribution of proportion (9HHH can be approximated by a normal distribution whenever the sample size is large (i.e np and ηq > 5) Exercise1 Exercise1: Suppose that 60% of electrical contractors in a region use a particular brand of wire. Of taking a random sample of size 120 from that electrical contractor and find the probability that 0.5 or less use that brand of wire? Exercise 2: If 10% of a population of parts is defective what is the probability of randomly selecting 80 parts and finding that 12 or more are defective? 14 of 61 Sampling & sampling Distributions Tujuba A.  Unit 2 2. Statistical Estimation Concepts of Statistical Estimation Sampling distribution of the mean shows how far sample means could be from a known population mean. The sampling distribution the proportion shows how far sample proportions could be from a known population proportion while the aim of estimation is to determine how far unknown population mean could from the mean of a simple random sample selected from that population or how far population proportion could be from a sample proportion. Or it is the concerns of statistical inference in which an unknown population parameter is derived from information contained in random sample selected from the population. Estimation: Refers to the process of using sample statistic as estimates of population parameter Estimation: is any procedure where sample information is used to estimate (predict) the numerical value of some population measure (parameter) Estimator: any sample statistic that is used to estimate a population parameter (sample mean, sample standard deviation, sample variance, sample proportion) Statistical Inference: branch of statistics which is concerned with using probability concepts to deal with uncertainty in decision making o It is based on estimation and hypothesis testing and any inference drawn about the population (i.e based on sample statistic or function of sample information Important Properties of Good Estimators 1. Un-biasedness 2. Efficiency 3. Consistency 4. Sufficiency 1. Un-biasedness : The estimator is said to be unbiased when the mean of sampling estimator is equals to population parameter 2. Efficiency: Refers to the size of standard error of statistics. The most efficient estimator is the one with smallest variance Example: f there are two estimator for I with var (I1 is to be more efficient than the second one (I2 , if var (I1 < var (I2 3. Consistency: related to the relationship between the sample size/the larger the sample size and the variance. A statistic is a consistent estimator of population parameter if sample size is large/increases, it becomes almost more certain, the value of the statistic comes very close to the value of the population parameter. Variance approaches to zero (0) as sample size (n) increases 4. Sufficiency (or Sufficient Statistic): Refers to an estimator that utilize/include all the information is the better the estimator or the sample contains all information about the parameter is the good estimator 15 of 61 Sampling & sampling Distributions Tujuba A.  Example: the sample mean is a sufficient estimator of the population mean that means no other estimator of the population mean from same sample data (e.g sample median) can add any further information about the parameter Estimate: is a specific numerical value of statistical data value estimator Example: ;̅ = 9, 2, 5 ;̅ ,pH, s,2 s ………………….. Estimators µ , p, δ 2, δ …………… Items being estimated 1 ,0.5, 9,2,3 ………. Estimates Type of Estimates We can make two types of estimates about a population the population parameter 1. Point Estimate 2. Interval Estimate Point Estimate Point Estimate: is a single number that is used to estimate an unknown population parameter/characteristics. Refers to a single value taken from the sample and used as an estimator of the corresponding population parameter. The most important point estimates (given that they are single values) are: Sample mean (;̅ ) for population mean( µ ) Sample proportion (pH) for population proportion (p) Sample variance (s2) for population variance ( δ 2) Sample standard deviation (s) for population standard deviation ( δ ) Example: Price earnings ratios for a random sample of stocks traded on xy city stock exchange on Dec 28,2018 were 10, 16, 5, 10, 12, 8, 4, 6, 5, 4 Required: Find the point estimates of the population mean, variance, standard deviation and proportion of stock in the population for which the price earning ratio exceeded 8.5 Solution: To find the first there of these sample quantities, we show the calculations in tabular from No xi xi2 1 10 100 2 10 256 3 5 25 4 10 100 5 12 144 6 8 64 7 4 16 8 6 36 9 5 25 10 4 16 Sum 80 782 N = 10  xi2 = 782  xi = 80 16 of 61 Sampling & sampling Distributions Tujuba A.  ;̅ =  xi  Sample mean (x) is = 80 = 8 this refers to point estimate of the population mean N 10  A point estimate of population variance can be computed by S2x =  ( xi2 − nx2) or  ( xi2 − nx2) n-1 n-1 2 = 782 – 10(8) 10-1 = 782 – 640 9 = 142 9 = 15.78  Point Estimate of population standard deviation Sx = σ 2 = 15.78 = 3.97 = 4  The number of stock for which the price warning ratio exceeds 8.5 is 4 hence point estimate of the population proportion is: pHx = x = 4 = 0.4 n 10 An Interval Estimate An interval Estimate: Refers to the range of values used to estimate a population parameter. It estimates describes the range of values within which a parameter might lie. Can be stated differently an interval estimate is a range of values within which the analyst can declare with some confidence that the population parameter with fall Example: suppose we have the sample 10, 20, 30, 40 and 50 selected randomly from population whose mean µ is unknown The sample mean (x) =  xi = 10+20+30+40+50 = 30 i.e 30 is a point estimate of µ. N 5  On the other hand if we state that the mean or µ, is between x + 10, the range of values from 20(30-10) to 40 (30+10) is an interval estimate  Point estimators of population parameter do not convey as much information as interval estimators, because point estimation produces a single value for unknown population parameter, or the estimate may or may not be close to the parameter, or estimate may be incorrect  An interval estimate refers to a range of values that conveys the fact that estimation is an uncertain process. 17 of 61 Sampling & sampling Distributions Tujuba A.   Interval estimate is also called confidence interval, for this reasons interval estimators are considered as more desirable than point estimators. Interval Estimation for Population Mean(µ) As a result of Central Limit Theorem(CLT) the following z formula for sample means can be used when sample size are large regardless of the shape of the population distribution or for smallest size if the population is normally distributed Z=; −µ σ or z = ;̅ − µ where δ;̅ = n δ δ;̅ Since sample mean can be greater than or less than the population mean z can be positive or negative. Thus the preceding expression takes the form M = ;̅ ± Z O n Therefore, the value of population mean M lies between this range (this expression yields O ≤ µ ≤ ; +Z the confidence interval for the population mean F. B ; 8 −Z O n n The confident interval for population mean is affected by  The population distribution (i.e. whether the population is normally distributed or not)  The standard deviation (i.e. whether µ known or not  The sample size (i.e. whether the sample size (n) is large or not) Condition 1:Confidence Interval Estimate of for(µ) when Distribution is Normal and δ is Known A confidence Interval Estimate for µ is an interval estimate together with a statement of how confident we are that the interval estimate is correct. When the population distributing is normal and at the sometime δ is known, we can estimate µ (regardless of the sample size) using the following formula µ = ;̅ ± α σ z2 n ;̅ = Sample mean Where δ = Population standard deviation Z= Value from the standard normal table reflecting confidence level O = 9C9QR>=FC? ?@>A@ @BSF>=FC? ∝ = =ℎB 9C9QR>=FC? CD F?UCAABU= =BVB?= F. B ∝ = 1 − U F. B ∝ = 1 − U 18 of 61 Sampling & sampling Distributions Tujuba A.  n = sample size M = Q?W?CX? 9C9QR>=FC? VB>? c = confidence level  Form this formula we can understand that an interval in the estimate is constructed by adding and subtracting the error term to and from the point estimate. That is, the point estimate is found at the center of the confidence interval Steps: To Determine Interval Population Mean 1. Compute the standard error of the mean ( ;̅ ) α 2. Compute from the confidence coefficient 2 α 3. Find the z value for the from the table 2 4. Construct the confidence interval 5. Interpret the result Exercise 1: The Vice President of Operations for Ethiopian Tele Communication Corporation (ETC) is in the process of developing a strategic plan. He believes that the ability to estimate the length of the average phone call on the system. He takes a random sample of 60 calls from the company records and finds that the mean sample length for a call is 4.26 minutes; past history for these types of calls has shown that the population standard deviation for calls length is 1.1 minutes. Assuming that the population is normally distributed ,and 95% confidence level estimate the population mean Exercise2: Time Magazine reports information on the time required for caffeine from products such as coffee and soft drinks to leave the body after consumption. Assume that the 99% confidence interval estimate of the population mean time for adults is 5.6 Hrs to 6.4 Hrs. Required (a) What is the point estimate of the mean time for caffeine to leave the body after consumption (b) If the population standard deviation is 2 Hrs, how large sample was to provide the interval estimate? Condition 2:Confidence interval estimate of µ for normal distribution (when δ unknown and n is large) If the population standard deviation unknown, it has to be estimated from the sample (i.e when δ is unknown we use sample standard deviation Ε( xi − x) 2 S=Y ?−1 Then the standard error of the mean(δ;̅ ) is estimated from standard error of the mean S;̅ = s √? 19 of 61 Sampling & sampling Distributions Tujuba A. Therefore the confidence interval to estimate µ when population standard deviation is unknown, when population is normal and n is large µ =;̅ ± Z α & 2 √ Exercise: Suppose that the car Renting Firm in wants to estimate the average number of miles traveled by each of its car rented. A random sample of 110 rented cars revealed that the sample mean travel distance per day is 85.5 mile, with sample standard deviation of 19.3 miles. Assuming normal Distributions Construct 99% confidence interval to estimate population mean Condition 3:Confidence interval for µ , when δ unknown, n small and population is normal If sample size is small (n < 30) we can develop an interval estimate of a population mean only if the population has a normal probability distribution. If the sample stands standard deviation (s) is used as an estimator of the population standard deviation (δ ) and if the population has a normal distribution, interval estimation of the population mean can be based up on a probability distribution known as t-distribution * t-distribution is a family of probability distribution. Characteristics of t-Distribution  The t-distribution is symmetric about its mean and ranges from - ∞ to ∞  The t-distribution is bell-shaped (uni-modal) and has approximately the same appearance as the standard normal distribution (z – distribution)  The t-distribution depends on a parameter v (Greek NU) called the degrees of freedom of distributions v = n-1 where n is sample size. The degrees of freedom (v) refers to the number of values we can choose freely The degrees of freedom for a t-statistic are degrees of freedom associated with sum of squares used to obtain an estimate of the variance. The variance estimate depends up on not only on the sample size but also how many parameters must be estimate with the sample degrees of freedom = number of – number of parameters that observation must be estimated beforehand. Here we calculate sample variance by using n observations and estimating one parameter (the mean). Thus there are (n-1) degrees of freedom %  The variance of the t-distribution is = % for v>2  The variance of t-distribution always exceeds 1  As v increases, the variance of t-distribution appropriates to 1 and the shape approaches to standard normal distribution  The variance of two t-distribution exceeds 1.0 while the variance of the z – distribution equals to 1, the t-distribution is slightly flatter in the middle than the z – distribution and has thicker tail 20 of 61 Sampling & sampling Distributions Tujuba A.   The t – distribution is a family of distributions with a different density function corresponding to each different value of the parameter (v). In proper statically language we would sy t-distribution for each sample size that is there is a separate t- distribution for each sample size. In prier statistical language, we would say, there is a different t-distribution for each of the possible degrees of freedom.  The t-formula for sample when δ is unknown, the sample size is small and population is normally distributed is t= & = ;− µ µ δ n This formula is essentially the same as the z Formula but the distribution table values are not the confidence interval to estimate µ becomes δ M; ±z α ,v n 2 Where x = sample mean ἀ = 1- c V = n – 1 (degrees of freedom) S = sample standard deviation n = sample size µ = unknown population mean Exceptional α & 2 √ z. – Called error of estimation (e) which results from the sampling process and it is (the difference between x and µ Steps 1. Calculate for the degrees of freedom (v = n-1) and the sample standard error of the mean α 2. Compute for (compute for incorrect proportion) 2 α 3. Look up t , v value from t-distribution 2 4. Construct the confidence interval 5. Interpret the result 21 of 61 Sampling & sampling Distributions Tujuba A. Interval Estimation of the Population Proportion Sample proportion (p) is unbiased estimator of population proportion(p), and if the sample size is large, the sampling distribution of p is normal with p-p , , [\] =;= 4 Z= /δ ,  Here p is unknown and we want to estimate P using p and z becomes 9^ p-p [\] Z= , O9 is substituted by sp = / 4 ? P can by formulated from the above formula as: √,G P = p- z since z can assume both position and negative values, p becomes 9^ P = p ± z/ ? 9^ Since z represents the confidence level, we can formulate for the population proportion as P=p±z ,/ α ? or p = p± z , sp 2 Where p = sample proration ∝ = 1 –c q=1–p n = sample size p = unknown population proration Exercise 1:: A study of 87 randomly selected companies with telemarketing operation was completed. The study revealed that 39% of the sampled companies had used telemarketing to assist them in order processing using the information estimate the population proportion of lacing 95% confidence mater interval level. (Answer P = 0.2875 ≤ p ≤ 0.4925) telemarketing companies who use their telemarking operation to assist them in order processing by Exercise 2: A fast food restaurant took a randomly sample of 400 customers to determine the proportion of customers who are female. A confidence confidence interval of 0.73 to 0.87 was reported Required (a) Find the number of females and the sample proportion(Answer proportion( = 320 Females)) (b) Find the level of confidence of this interval Determination of Sample Size The basic reason for sampling is it is too cost to gather data from the whole population but collecting sample data is also cost and the larger sample size, the higher the cost. To hold cost down we want to use the small sample size while in contrary we want the large sample size to provide good approximation approximation (estimates of population parameter therefore how large should the sample be? The answer depends on their factors 22 of 61 Sampling & sampling Distributions Tujuba A.  (1) How precise (narrow) do we want a confidence interval to be? (2) How confident do we want to be that the interval estimate is correct (3) How variable variable is the population being sampled? Sample size to estimate population mean (`) ( ) Driving for sample size & M=x±z Confidence interval for M α √. 2 From the above expression α & √ z. Called error of estimation (e) That is the difference between x and M results 2 From the sampling process α & √ e=z. 2 Squaring both sides results α & √ e2 = z. 2 When solving for n α & √ e2 = z. 2. 1.96(Formulate Decision Rule) Step4: Collect the sample data, and compute the value of the test statistic OR Formulate a decision rule). A test statistic is a random variable whose value is used to determine whether we reject the Ho. Example sample Z=2.00 Step5: Compare the value of the test statistic to the critical value (s) and make the decision which mean either reject Ho or accept Ho/don not reject Ho) Don not reject Ho or reject Ho Decision and/or Reject Ho and accept Ha Type I and Type II Errors There are four possible outcomes of any hypothesis test, two of which are correct and two of which are in correct. The incorrect ones are. 1. Type I Error 2. Type II Error Type I Error o Type I error is committed when a true null hypothesis rejected or Rejecting the null hypothesis (Ho) when it is true o The possibility of committing a type I error is represented by alpha (ἀ) or the level of significance. Alpha (ἀ) is sometimes referred as the amount of risk taken in an experiment. o Alpha (ἀ) represents the proportion of the area of curve occupied by the rejection region. The most widely used values of ἀ are 0.001, 0.01,0.05,0.10 o The larger the area of rejection the greater the risk of committing type I error 26 of 61 Sampling & sampling Distributions Tujuba A.  Type II Error Type II error is committed by failing to reject a false null hypothesis. Which means accepting a null hypothesis when it is false? The probability of committing type II error is represented by Beta(ß) Researcher Null Hypothesis Accept (Ho) Reject(Ho) (Ho) Ho is true Correct Decision Type I error Ho is false Type II Error Correct Decision One –Tailed and Two-tailed Tests There are three possible Ho along with their corresponding Ha. 1. Ho: M = y 2. Ho: M ≤ y 3. Ho: M ≥ y Ha: M ≠ y Ha: M > y Ha: M y Leads to two- tailed test leads to one tailed leads to one tailed test ( left – tailed test) (right – tailed test) Note: Hypothesis testing about population proportion is based on the difference between sample proportion (p) and hypnotized valve p. CLT : applied to sample proportion (p) when the values are normally distributed with a mean of P and standard deviation of O p ([9^/? when np and nq are greater than or equal to 5.  If np and nq are greater than or equal to 5, a z – test is used to test hypothesis about p. Exercise 1: A magazine claims that 25% of its readers are college students. A random sample of 200 is taken and it found that 42 of these readers are college students, use a 10% level of significance and test the magazine’s claim. Exercise 2: An Economist states that no mare than 35% new lobo force is unemployed you don’t know whether the economist estimates is too high or too low, Thus you want to test the economist’s claim using a 5% level of significance you obtain a random sample of 400 people in the labor force of whom 125 are unemployed, would you reject the economist’s claim. Exercise 3: A survey of the morning beverage market has shown that the primary breakfast beverage for at least 60% of a given country town and city dwellers is tea. The country and tea authority believes that the figure is higher for the capital city, To test this idea one of the employees of the country’s coffee and tea authority contracts a random sample of 500 residents in the capital city and asks which primary beverage they consumed for breakfast that day. Suppose 325 replied that tea was the primary beverage. Using a 1% level of significance, test the idea that the lea figure is higher for capital city. 32 of 61 Sampling & sampling Distributions Tujuba A.  Chapter - 4 Chi- Chi-Square Distributions(X Distributions(X2) This test is applied when there is a categorical valuable from a single population Used to determine whether sample data are consistent with a hypothesized distribution. Chi-Square (X2 ) Distributions is a continuous distribution ordinarily derived as the sampling distribution of a sum of square of independent standard formal variables Characteristics of Chi- Chi-Square Distribution It is a continuous distribution The X2 distribution has a single parameter, the degrees of freedom (V) DR the 2 mathematical expression of X distributions contains only on parameter the no of degrees of freedom (v) The computed value (the variable of X2 ) cannot be negative because chi-square do not extend to the left of zero (i.e. the Difference between to and for is squared) When V exceeds 2(two) chi-square curve have one mode and are skewed to the light (positively skewed) When V is very large (i.e. n is large) the chi-square distribution is almost the same as formal distribution having a year equal to v and standard deviation equal to √2v i. Significance level ἀ will be right tail areas of chi-square distribution the symbol X22,v will mean the value of X2 such that the destruction with degrees of freedom v has a right tail area of alpha ii. The mean of chi-square distribution is v while the variance of X2 is 2v there are the mean and variance depends on the degrees of freedom iii. It is skewed destruction and only non-negative values of the variable X2 are possible. The skew nesses decreases as v increases and when v increases without limit it apparition to loyal destruction (Extends indefinitely in the positive direction) iv. The area under the curve is 1.0 X2 Distribution has the Following Areas of Application Application 1. Chi Chi--square test for independence 2. Goodness of fit tests (binominal, normal and poison destruction) 33 of 61 Sampling & sampling Distributions Tujuba A.  3. Test for the Equality of Proportions 1. Test for the Independence between Two Variables 2 A X test of independence is used to analyze the frequencies of two variables with multiple categories to determine whether the two variables are independent. The X2 distribution uses sample data to test for the independence of two variables.  Sample data are given in a two way table called contingency table (the test is sometimes referred to as contingency analysis) or contingency table test Examples: o Whether the sport favorite is independent of nationality o Whether the bear preference is independent of sex(gender) o Whether the type of financial investment is independent of geographic region o Whether employee absenteeism is independent of the job satisfaction (classification) Steps to test X2 distribution 1. State the Hypothesis (Null Hypothesis and Alternative Hypothesis. Hypothesis  The hypothesis are stated in such a way that they are mutually exclusive (i.e. it one is true the other is false and vice versa) 2. Formulate the Analysis Plan (Decision Rule)  Describes how to use sample data to accept or Reject the null hypothesis  The plan should specify: i. Significance level (e.g. 0.01,0.05,0.10 or any value between 0 and 1 ii.Test method (use test for independence chi-equate goodness fit test) 3. Analyze Sample Data Data (Compile the Test Statistic)  Using sample data find: a. Degrees of freedom (df or V) which refers to number of levels of the categorical variable minus one (df = (R - 1)(C - 1) b. Expected Frequency counts(fe) or Expected Observation Eij = ith Row Total x jth column total N (the total sample size) c. Test statistics: the test statistic is an X2 random variable defined by the following equation X2= ∑(oij-eij)2 Eij Where Oij= observed frequency count for the ith level of categorical variable Eij= expected frequency count for the ith level of categorical variable 34 of 61 Sampling & sampling Distributions Tujuba A.  d. P-value: refers to the probability of observing a sample statistic as extreme as the test statistic since the test statistic is a chi-square use the X2 distribution calculator to assess probability associated with test statistic use the degrees of freedom computed above 4. Interpret the result If the sample findings are unlikely to the null hypothesis, the researches rejects the Ho. Typically this involves comparing the p value to significance level, and rejecting the Ho when P-value is less than significance level. Exercise 1: A company planning a TV advertising campaign wants to determine which TV shows its target audience withers and thereby to know who thus the choice of TV program of an individual watch is independent of individual income. The table supporting this is shown below. Use 5 % level of significance and Ho. Income Basket ball Movie News Total Types of show Low 143 70 37 250 Medium 90 67 43 200 High 17 13 20 50 Total 250 150 100 500 Exercise 2: A human resource manager of EAGLE Inc. was interested in knowing the voluntary absence behavior of the firms employees was independent of material status. The employee ticks contained data on marital status and on the voluntary absenteeism behavior for a sample of 500 employees is shown below Marital Status Married Divorced Widowed Single Total Often absent 36 16 14 34 100 Absence Behavior Seldom absent 64 34 20 82 200 Never absent 50 50 16 84 200 Total 150 100 50 200 500 Required: Test the hypothesis that absence is independent of marital status of the significance level of 1% Exercise 3: A personnel Administration (Administrator) of xyz company provided the following data as ana example of selection among 40 male and 40 female applicants for 12 open positions. Statics Selected Not selected Total Applicants Male 7 33 40 Female 5 35 40 Total 12 68 80 Required (a) Test the X2 independence whether the way of determining the decision to hire of males and 5 females should have a selection bias in favor of males 35 of 61 Sampling & sampling Distributions Tujuba A.   Conduct a test of independence using a 10% level of significance what is your conclusion? (b) Using the same test, would the decision to hire 8 males and 4 females suggest to concern for a selection hire? (c) How many males could be hired for the 12 open positions before the produce would concern for a selection bias? 2. Chi- Chi-Square Goodness Goodness of Fit Test  Goodness Of Fit Test: Test used to determine whether or not the outcomes multinomial experiment fit a distribution with specified proportions of response in certain categories  It is non parametric test that is used to find out how the observed value of a given phenomena is significantly different from the expected values  The term X2 goodness of fit test is used to compare the observed sample distribution with the expected probability distribution  X2 goodness of fit test determines how well theoretical destruction (such as normal, Binomial or poison) fits the empirical distribution.  X2 goodness of fit test is divided into intervals therefore the number of points that falls into the intervals is compared with the expected numbers of points in each interval. Exercise 1: Assume each day a sales person calls on a prospective customers, and she records wither or not visit for sale. For a period of 100 days her record is as follows. Number of sales 0 1 2 3 4 5 Frequency 15 21 40 14 6 4 A marketing research thinks that a call results in a sale about 35% of the time. So he wants to test if this sampling of sales person’s effort fits a theoretical binomial distribution for the trials. This binomial distribution has the following probabilities and leads to the following expected values for 100 days of Sales (x) P(x) records. Ei = np 0 0.1161 1 0.3124 Where p = probability n= sample size 2 0.3364 3 0.1812 Use 5% level of significance to test for goodness fit 4 0.0487 5 0.0053 Exercise 2: the criminal investigation authority federal bureau of investigation (FBI) compiles data on crime and crime rates, and publishes the information. A violent crime is classified by the authority FBI as murders, forcible rape, robbery and aggravated assault. The table below provides (shows) the relative frequency distribution for a reported violent crimes in 2019. A simple random sample of violent crime report from best is yielded suppose we want to decide whether last year’s distribution of the violent crime is changed from 2019 Distribution use 5% level of significance significance. ficance. Types of violent crimes Relative Frequency Types of the violent Observed Frequency Murder 0.011 crime Forcible rape 0.063 Murder 3 Robbery 0.286 Forcible rape 37 Robbery 154 Aggravated assault 0.640 Aggravated assault 306 Sample result for 500 randomly selected violent crime report from last year. 36 of 61 Sampling & sampling Distributions Tujuba A.  Unit- Unit- 5 5. Analysis of Variance (ANOVA) 5.1. INTRODUCTION You use analysis of variance (ANOVA) to test hypotheses about the equality of two or more population means. ANOVA is based on experiments performed on subjects that are independent of each other; in other words, they are not related to each other. For example, suppose that a department store chain wants to compare the mean sales of household appliances at its stores in New York, Boston, and Philadelphia. Because these stores are in different geographical locations, the sales at one store don’t influence sales at the other stores (they’re independent of each other.) And because the sales at these stores are independent, ANOVA can be used to test the hypothesis that mean sales are equal at all three stores. ANOVA is a statistical method for determining the existence of differences among several population means. While the aim of ANOVA is to defect differences among several population means, the technique requires the analysis of different forms of variance associated which the random samples under study hence the name of analysis of variance. ASSUMPTIONS OF THE ANALYSIS OF VARIANCE The following are the assumptions implicit in the use of the analysis of variance 1. The data are quantitative in nature and are normally distributed. In case of nominal or ordinal data, analysis of variance should not be carryout. 2. The samples drawn from the population are on a random basis. 3. The variances of the population form which samples have been drawn are equal. 5.2. THE F-DISTRIBUTION Analysis-of-variance procedures rely on a distribution called the F-distribution, named in honor of Sir Ronald Fisher. 37 of 61 Sampling & sampling Distributions Tujuba A. As is the case with t-and chi square distributions, the shape of a particulate F- distribution curve depends on the number of degrees of freedom. A major difference between t- and chi- square distributions and F- distribution is that the former two distributions have only one number of degrees of freedom whereas the latter has two numbers of degrees of freedom. This is because it has degrees of freedom for the numerator as also for the denominator. This two numbers of the degrees of freedom are the parameters of the F- distribution. The two degrees of freedom taken together give a particular F- distribution curve. It may also be noted that the F- distribution is also a continuous distribution like normal, t- and chi- square distribution. It takes only positive values. The F- distribution curve is skewed to the right. But as the number of degrees of freedom increases the skewness decreases. A variable is said to have an F-distribution if its distribution has the shape of a special type of right-skewed curve, called an F-curve. There are infinitely many F-distributions, and we identify an F-distribution (and F-curve) by its number of degrees of freedom, just as we did for t-distributions and chi-square distributions. An F-distribution, however, has two numbers of degrees of freedom instead of one. The figure below depicts two different F-curves; one has df = (10, 2), and the other has df = (9, 50). The first number of degrees of freedom for an F-curve is called the degrees of freedom for the numerator, and the second is called the degrees of freedom for the denominator. Thus, for the F-curve in Fig. below with df = (10, 2), we have 38 of 61 Sampling & sampling Distributions Tujuba A. Basic Properties of F-Curves Property 1: The total area under an F-curve equals 1. Property 2: An F-curve starts at 0 on the horizontal axis and extends indefinitely to the right, approaching, but never touching, the horizontal axis as it does so. Property 3: An F-curve is right skewed. USING THE F-TABLE Percentages (and probabilities) for a variable having an F-distribution equal areas under its associated F-curve. To perform an ANOVA test, we need to know how to find the F- value having a specified area to its right. The symbol Fα denotes the F-value having area α to its right. The degrees of freedom for the denominator (dfd) are displayed in the outside columns of the table, the values of α in the next columns, and the degrees of freedom for the numerator (dfn) along the top. 39 of 61 Sampling & sampling Distributions Tujuba A. Finding the F-Value Having a Specified Area to Its Right For an F-curve with df = (4, 12), find F0.05; that is, find the F-value having area 0.05 to its right, as shown in Figure below. Solution We use Table VI to find the F-value. In this case, α = 0.05, the degrees of freedom for the numerator is 4, and the degrees of freedom for the denominator is 12. We first go down the dfd column to “12.” Next, we concentrate on the row for α labeled 0.05. Then, going across that row to the column labeled “4,” we reach 3.26. This number is the F-value having area 0.05 to its right, as shown in Figure below. In other words, for an F-curve with df = (4, 12), F0.05 = 3.26. 5.3. THE USUAL FIVE-STEP HYPOTHESIS-TESTING PROCEDURE Example Instructor Debele had the 22 students in his 3rd period Principles of Marketing rate his performance as Excellent, Good, Fair, or Poor. An undergraduate student collected the ratings and assured the students that Instructor Debele would not receive them until after course grades had been sent to the Registrar's office. The rating (i.e., the treatment) a student gave the professor was matched with his or her course grade, which could range from 0 to 100. The sample information is reported below. Is there a difference in the mean score of the students in each of the four rating categories? Use the.01 significance level. 40 of 61 Sampling & sampling Distributions Tujuba A.  Course Grades Excellent Good Fair Poor 94 75 70 68 90 68 73 70 85 77 76 72 80 83 78 65 88 80 74 68 65 65 Step 1: State the null hypothesis and the alternate hypothesis. The null hypothesis is that the mean scores are the same for the four ratings. Ho: µ1= µ2 =µ3 =µ4 The alternate hypothesis is that the mean scores are not all the same for the four ratings. H1: The mean scores are not all equal. We can also think of the alternate hypothesis as "at least two mean scores are not equal." If the null hypothesis is not rejected, we conclude that there is no difference in the mean course grades based on the instructor ratings. If Ho is rejected, we conclude that there is a difference in at least one pair of mean ratings, but at this point we do not know which pair or how many pairs differ. Step 2: Select the level of significance. We selected the.01 significance level. Step 3: Determine the test statistic. The test statistic follows the F distribution. Step 4: Formulate the decision rule. To determine the decision rule, we need the critical value. The critical value for the F statistic is found in F distribution table. The critical values for the.05 significance level are found on the first page and the.01 significance level on the second page. To use this table we need to know the degrees of freedom in the numerator and the denominator. The degree of freedom in the numerator equals the number of treatments, designated as k, minus 1. The degree of freedom in the denominator is the total number of observations, n, minus the number of treatments. For this problem there are four treatments and a total of 22 observations. Degrees of freedom in the numerator = k-1=4-1=3 Degrees of freedom in the denominator =n-k =22–4=18 Refer to F distribution table and the.01 significance level. Move horizontally across the top of the page to 3 degrees of freedom in the numerator. Then move down that column to 41 of 61 Sampling & sampling Distributions Tujuba A.  the row with 18 degrees of freedom. The value at this intersection is 5.09. So the decision rule is to reject Ho if the computed value of F exceeds 5.09. Step 5: Select the sample, perform the calculations, and make a decision. It is convenient to summarize the calculations of the F statistic in an ANOVA table. The format for an ANOVA table is as follows. Statistical software packages also use this format.  Sum of Squares With in groups (SSW)  Total sum of squares (SST)  Error mean square (MSE)  Treatment mean square (MSTR) COURSE GRADES Excellent Good Fair Poor (94-87.25)2=45.56 (75-78.2)2=10.24 (70-73.86)2=14.89 (68-69)2=1 (90-87.25)2=7.56 (68-78.2)2=104.04 (73-73.86)2=0.74 (70-69)2=1 Marks (85-87.25)2=5.06 (77-78.2)2=1.44 (76-73.86)2=4.58 (72-69)2=9 (80-87.25)2=52.56 (83-78.2)2=23.04 (78-73.86)2=17.14 (65-69)2=16 (88-78.2)2=96.04 (80-73.86)2=37.69 (74-69)2=25 (68-73.86)2=34.34 (65-69)2=16 (65-73.86)2=78.49 Column 349 391 510 414 1664 total N 4 5 7 6 22 X=  xi 349 =87.25 391 =78.20 510 =73.86 414 =69 1664 = 75.64 n 4 5 7 6 22 Sum 110.74 234.80 187.87 68 601.41 Add up the sums to get the sum of squares within groups (SSW): 110.74+234.8+187.87+68=601.41 FINDING TOTAL SUM OF SQUARES (SST) Course Grades Excellent Good Fair Poor Total (94-75.64)2=337.09 (75-75.64)2=0.41 (70-75.64)2=31.81 (68-75.64)2=58.37 (90-75.64)2=206.21 (68-75.64)2=58.37 (73-75.64)2=6.97 (70-75.64)2=31.81 Marks (85-75.64)2=87.61 (77-75.64)2=1.85 (76-75.64)2=0.13 (72-75.64)2=13.25 (80-75.64)2=19.01 (83-75.64)2=54.17 (78-75.64)2=5.57 (65-75.64)2=113.21 (88-75.64)2=152.77 (80-75.64)2=19.01 (74-75.64)2=2.69 (68-75.64)2=58.37 (65-75.64)2=113.21 (65-75.64)2=113.21 Sum 649.91 267.57 235.07 332.54 1485.09 Total Sum of Squares is =674.27+267.57+259.42+332.54=1485.09 42 of 61 Sampling & sampling Distributions Tujuba A. Calculating Sum of Squares between groups or treatments (SSTR) SSTR = n1 ( X 1 − X ) 2 + n 2 ( X 2 − X ) 2 + n3 ( X 3 − X ) 2 + n4 ( X 4 − X ) 2 =4(87-75.6)2+5(78-75.6)2+7(73-75.6)2+6(69-75.6)2 =4(11.4)2+5(2.4)2+7(-2.6)2+6(-6.6)2 =4(129.96) +5(5.76) +7(6.76) +6(43.56) =519.84+28.8+47.32+261.36 =857.32 Getting the error mean square (MSE) SSW MSE / MSW = n−t = 601.4 22 − 4 601.4 = = 33.41 18 Getting the treatment mean square (MSTR) SSTR MSTR = t −1 857.32 = = 286 4 −1 MSTR 286 F= = MSW 33.41 F = 8.56 The computed value of F is 8.56, which is greater than the critical value of 5.09, so the null hypothesis is rejected. We conclude the population means are not all equal. The mean scores are not the same in each of the four ratings groups. It is likely that the grades students earned in the course are related to the opinion they have of the overall competency and classroom performance of Instructor Debele. At this point we can only conclude there is a difference in the treatment means. We cannot determine which treatment groups differ or how many treatment groups differ. Exercise:1 Suppose a manufacturer is considering releasing one of three new types of batteries and wants to determine whether one of these batteries has a longer mean lifetime than the others. If so, it will manufacture this battery exclusively. Otherwise, it will randomly pick one 43 of 61 Sampling & sampling Distributions Tujuba A. of the three to be manufactured. The proposed names for the three battery types are Electrical, Ready forever, and Voltage now. In this experiment, battery lifetime is referred to as the dependent variable. The hypothesis tested is that the mean battery lifetime is the same for Electrical, Ready forever, and Voltage now. The three battery types are treatments. If the mean lifetimes of the Electrical, Ready forever, and Voltage now batteries are different, this may be due to two different sources. These are known as variation between groups (battery types) and variation within groups (variation among batteries of the same type). The process used to test whether the mean battery lifetimes are equal for each type is known as one-way ANOVA. If the manufacturer wants to compare the mean lifetimes according to type and determine if there can be substantial differences within each type, a more complex version of ANOVA is used. This is known as two-way ANOVA. Writing the null and alternative hypotheses The null hypothesis is a statement that’s assumed to be true unless you find strong contrary evidence. For testing the hypothesis that three population means are equal, you write: H0:µ1=µ2=µ3 H1:µ1≠µ2≠µ3 In this expression, H0 represents the null hypothesis µ1, µ2, µ3 represent the means of population 1, 2 and 3 The alternative hypothesis is a statement that you accept in the event that the null hypothesis is rejected (for example, there’s strong evidence against it). When testing the hypothesis that three population means are equal, the alternative hypothesis is simply that the three population mean are not equal. This alternative hypothesis can be expressed in different ways, such as: H1: The three means are not all equal 44 of 61 Sampling & sampling Distributions Tujuba A.  H1: At least one of the three means is different from the others In these examples, H1 represents the alternative hypothesis. Choosing the level of significance To test a hypothesis, you have to choose a level of significance. The level of significance designated with α, equals the probability of incorrectly rejecting the null hypothesis when it’s actually true. This is called a Type I error. A Type II error occurs when you fail to reject the null hypothesis when it’s not true. For many business applications, the level of significance is chosen as 0.05 (5 percent). Other frequent choices include 0.001, 0.01, and 0.10. Computing the test statistic The test statistic is a numerical value that is used to determine if the null hypothesis should be rejected. The form of the test statistic depends on the type of hypothesis being tested. If the test statistic has an extremely large positive or negative value, this may be a sign that the null hypothesis is incorrect and should be rejected. Constructing the test statistic for ANOVA is quite complex, compared with other types of hypothesis tests. Referring to the battery example, assume that the manufacturer randomly chooses a sample of four Electrical batteries, four Ready forever batteries, and four Voltagenow batteries and then tests their lifetimes. Battery lifetimes (in hundreds of hours) Electrical Ready forever Voltagenow Battery 1 2.4 1.9 2.0 Battery 2 1.7 2.1 2.3 Battery 3 3.2 1.8 2.1 Battery 4 1.9 1.6 2.2 45 of 61 Sampling & sampling Distributions Tujuba A. Each element in this table can be represented as a variable with two indexes, one for the row and one for the column. In general, this is written as Xij. The subscript i represents the row index and j, represents the column index. For example, X23 represents the element found in the second row and third column. The first step in constructing the test statistic is to calculate the following three measures: Error sum of squares (SSE) Treatment sum of squares (SSTR) Total sum of squares (SST) The calculations are detailed in the following sections. Finding the error sum of squares (SSE) The error sum of squares (abbreviated SSE) is obtained by first computing the mean lifetime of each battery type. For each battery of a specified type, the mean is subtracted from each individual battery’s lifetime and then squared. The sum of these squared terms for all battery types equals the SSE. SSE is a measure of sampling error. This refers to the fact that the values computed from a sample will be somewhat different from one sample to the next. To compute the SSE for this example, the first step is to find the mean for each column. So, for example, you find the mean of column 1, with this formula: n  Xi1 X1 = i =1 n1 Here’s what each term means: ✓ X 1 is the mean of column 1 (the bar indicates that this is a mean). The subscripts indicate that this average is computed from all elements within column 1. ✓ X i1 is the value of X in row i and column 1. ✓ n1 is the number of elements in column 1. n  Xi1 X1 = i =1 n1 46 of 61 Sampling & sampling Distributions Tujuba A.  2.4 + 1. 7 + 3.2 + 1.9 = 4 9.2 = = 2.3 4 In other words, you sum the lifetimes of the four Electrical batteries and divide by 4. The mean lifetime of the Electrical batteries in this sample is 2.3. Similarly, you find the mean of column 2 (the Ready forever batteries) as 1.9 + 2.1 + 1.8 + 1.6 = 4 7.4 = 4 = 1.85 And column 3 (the Voltage now batteries) as 2.0 + 2.3 + 2.1 + 2. 2 = 4 8.6 = = 2.15 4 The next step is to subtract the mean of each column from each element within that column, then square the result. I set up the calculations in Table below. Battery lifetimes (in hundreds of hours) Electrical Ready forever Voltagenow Battery 1 (2.4-2.3)2=0.01 (1.9-1.85)2=0.0025 (2.0-2.15)2=0.0225 Battery 2 (1.7-2.3)2 (2.1-1.58)2 (2.3-2.15)2 Battery 3 (3.2-2.3)2 (1.8-1.85)2 (2.1-2.15)2 Battery 4 (1.9-2.3)2 (1.6-1.85)2 (2.2-2.15)2 Sum 1.34 0.13 0.05 For example, because 2.3 is the mean of column 1, you subtract 2.3 from each element in column 1. You square the result in each row, and the sum of these squared values is 1.34. Repeat the process for columns 2 and 3 to get sums of 0.13 and 0.05, respectively. Add up the sums to get the error sum of squares (SSE): 1.34 + 0.13 + 0.05 = 1.52. 47 of 61 Sampling & sampling Distributions Tujuba A. The error sum of squares shows how much variation there is among the lifetimes of the batteries of a given type. The smaller the SSE, the more uniform the lifetimes of the different battery types. Calculating the treatment sum of squares (SSTR) After you find the SSE, your next step is to compute the treatment sum of squares (SSTR). This is a measure of how much variation there is among the mean lifetimes of the battery types. With a low SSTR, the mean lifetimes of the different battery types are similar to each other. First, you need to calculate the overall average for the sample, known as the overall mean or grand mean. In the battery example from the previous sections, you have 12 total observations (four batteries chosen from each of three battery types; the data are in Table 13-1). You may obtain the overall mean by adding up the 12 sample values and dividing by 12: 2.4 + 1.7 + 3.2 + 1.9 + 1.9 + 2.1 + 1.8 + 1.6 + 2.0 + 2.3 + 2.1 + 2.2 25.2 = = 2.1 12 12 You then compute the SSTR with the following steps for each column: 1. Compute the squared difference between the column mean and the overall mean. 2. Multiply the result by the number of elements in the column. So in this example, SSTR equals SSTR = n1 ( X 1 − X ) 2 + n 2 ( X 2 − X ) 2 + n3 ( X 3 − X ) 2 + n4 ( X 4 − X ) 2 SSTR=4(2.3-2.1)2+4(1.85-2.1)2+4(2.15-2.1)2=0.42 The calculations are based on the following results obtained in previous sections:  There are four observations in each column.  The overall mean is 2.1.  The column means are 2.3 for column 1, 1.85 for column 2 and 2.15 for column 3. Computing the total sum of squares (SST) The total sum of squares (SST) equals the sum of the SSTR and the SSE (see the preceding sections). So using the battery example, you get 48 of 61 Sampling & sampling Distributions Tujuba A. SST=SSTR+SSE =0.42+1.52 =1.94 When you compute SSE, SSTR, and SST, you’re ready to proceed to the next step in computing the test statistic. The test statistic is computed from the mean (average) of SSE and SSTR; these are known as: Error mean square with in groups (MSW) Treatment mean square