Foundations of Quantitative Research Techniques and Statistics - latest Review Questions, Exams of Nursing

Download Foundations of Quantitative Research Techniques and Statistics - latest Review Questions and more Exams Nursing in PDF only on Docsity! Foundations of Quantitative Research Techniques and Statistics -   latest Update 2022/2023 -BMAL 590 - Liberty University   Review Questions 1. A company has developed a new smartphone whose average lifetime is unknown. In order to estimate this average, 200 smartphones are randomly selected from a large production line and tested; their average lifetime is found to be 5 years. The 200 smartphones represent a blank . o Sample 2. Which of the following is a measure of the reliability of a statistical inference? o A Significance Level 3. The process of using sample statistics to draw conclusions about population parameters is called blank . o doing inferential statistics 4. Which of the following statements involve descriptive statistics as opposed to inferential statistics? o The Alcohol, Tobacco and Firearms Department reported that Houston had 1,791 registered gun dealers in 1997. 5. A population of all college applicants exists who have taken the SAT exam in the United States in the last year. A parameter of the population are blank . o SAT Scores 6. Which of the following statements is true regarding the design of a good survey? o The questions should be kept as short as possible 7. Which method of data collection is involved when a researcher counts and records the number of students wearing backpacks on campus on a given day? o Direct observation 8. The manager of the customer service division of a major consumer electronics company is interested in determining whether the customers who have purchased a videocassette recorder over the past 12 months are satisfied with their products. If there are four different brands of videocassette recorders made by the company, the best sampling strategy would be to use a blank . o stratified random sample 9. Which of the following types of samples is almost always biased? o Self-selected samples 10. is an expected error based only on the observations limited to a sample taken from a population. o sampling error 11.Bayes's Law is used to compute blank . o posterior probabilities 12.The classical approach describes a probability blank . o in terms of the proportion of times that an event can be theoretically expected to occur 13.If a set of events includes all the possible outcomes of an experiment, these events are considered to be blank . o Exhaustive 14.Which of the following statements is not correct? o If event A does not occur, then its complement A' will also not occur. 15.The blank can determine the union of two events such as event A and event B. o addition rule 16. The that allows us to draw conclusions about the population based strictly on sample data without having any knowledge about the distribution of the underlying population is blank . o the central limit theorem 17. Each of the following are characteristics of the sampling distribution of the mean except blank . o if the original population is not normally distributed, the sampling distribution of the mean will also be approximately normal for large sample sizes 18. Suppose you are given 3 numbers that relate to the number of people in a university student sample. The three numbers are 10, 20, and 30. If the standard deviation is 10, the standard error equals blank o 5.77 19. You are tasked with finding the sample standard deviation. You are given 4 numbers. The numbers are 5, 10, 15, and 20. The sample standard deviation equals blank . o 6.455 20. Two methods exist to create a sampling distribution. One involves using parallel samples from a population and the other is to use the blank . o rules of probability 21. The hypothesis of most interest to the researcher is blank . o the alternative hypothesis 22. A Type I error occurs when we blank . 34. Assume a null hypothesis is found true. By dividing the sum of squares of all observations or SS(Total) by (n - 1), we can retrieve the blank . o sample variance 35. Which of the following is true about one-way analysis of variance? o n1 = n2 = … = nk is not required. 36. A tabular presentation that shows the outcome for each decision alternative under the various states of nature is called a blank . o payoff table 37. Which of the following statements is false regarding the expected monetary value (EMV)? o In general, the expected monetary values represent possible payoffs. 38.In the context of an investment decision, blank is the difference between what the profit for an act is and the potential profit given an optimal decision. o an opportunity loss 39. The branches in a decision tree are equivalent to o events and acts 40. Which of the following is not necessary to compute posterior probabilities? o EMV 41. Which of the following statements about decision analysis is false? o Decisions can never be made without the benefit of knowledge gained from sampling 42. can identify when two events are relational o Conditional probability 43. An approach of assigning probabilities, which assumes that all outcomes of the experiment are equally likely is referred to as the o Classical Approach 44. The expected value of perfect information is the same as o The expected opportunity loss for the best alternative 45. Which of the following is False? o The EMV decision is always different from the EOL decision 46. A payoff table lists monetary values for each possible combination of the o Event (state of nature) and act (alternative) 47. When a person receives an email questionnaire and places it in their deleted items witout responding, they are contributing to o Non-response error 48. The Notion represents the probability of B when A has occurred o P(B/A) 49. If A nd B are mutually exclusive events with P(A) = 0.30 and P(B) = 0.40, then P(A or B) is o 0.7 50. Which of the following statements is true regarding the design of a good survey? o The questions should be kept as short as possible 51. Which of he following statements is false regarding he expected monetary value (EMV) o In general, the expected monetary values represent possible payoff 52. The primary interest of designing a randomized block experiment is to o reduce the within- treatments variation to more easily detect differences among the treatment means 53. Which of the following is not a goal of descriptive statistics? o Estimating characteristics of the population (this is inferential) 54.Consider a probability tree for selecting to puppies without replacement. The joint probability of P(F)=2/7 and the second selection is P(F/F) = 3/10 is o P(FF) = (2/7)(3/10) 55. The power of a test is measured by its capability of o Rejecting a null hypothesis that is false 56. A Sample of 500 Athletes is taken from a population of 11,000 Olympic athletes to measure work ethic. As a result, we o Can predict an outcome with some level of certainty 57. Assume EVPI=$50,000 and EMV=$35000. If perfect information exists the value of EPPI is o 85,000 58. A summary measure that is computed from a sample to describe a characteristic of the population is called o Inferential statistics 59. In one-way analysis of variance, between-treatments variation is measured by the o SST 60. A survey will be conducted to compare the United Way…3 corporations… ANOVA model will be o One-way analysis variance 61. When is the tukey multiple comparison method used o To test for difference in pairwise means 62.Which of the following statements is false o A confidence level expresses the degree of certainty that an INTERVAL will include the actual value of the SAMPLE STATISTIC 63. Though not the most efficient method rolling four dice enough times will result in theoretical probabilities being similar to o The relative frequency 64. The standard error is o The standard deviation of the sampling distribution 65.For a given level of significance, if the sample size increases. The probability of a type II error will o Decrease 66. A study is under way to determine the average height of all 32,000 adult pine trees in a certain national forest. The heights of 500 randomly selected adult pine trees are measured and analyzed. The sample in this study is o the 500 adult pine trees selected at random selected at random from this forest. 67. The average sales per customer at a home improvement store during the past year is $75 with a standard deviation of $12. The probability that the average sales per customer from a sample of 36 customers, taken at random from this population, exceeds $78 is o .0668 68. Which of the following is a rule violation in hypothesis testing o We accept the null hypothesis 69. 69. What Is Statistics? "Statistics is a way to get information from data." The Mode = The most occuring number. eg. Find the Mode of the list: 2 4 6 8 12 12 15 4 2 6 8 7 7 9 12 Mode = 12 eg. Find the Mode of the list: 2 8 6 4 10 12 2 4 8 8 4 Mode = 4 and 8 The mean of the data set 2, 7, 9 is 6. The median of the data set 1, 3, 3, 6, 7, 8, 9 is 6. The median of the data set 1, 2, 3, 4, 5, 6, 8, 9 is 4 plus 5 divided by 2 = 4.5. The mode of the dataset 2, 4, 6, 8, 12, 12, 15, 4, 6, 8, 7, 7, 9, 12 is 12. The modes of the dataset 2, 8, 6, 4, 10, 12, 2, 4, 8, 8, 4 are 4 and 8. A graphic with Frequency on the Y axis shows three different smooth curves. The first is a negatively skewed distribution which has a tail to the left of the peak; the mode is at the peak of the distribution with the median to the left and the mean to the left of the median. The second curve is a normal distribution which represents a perfectly symmetrical distribution where the mean, median, and mode are at the peak of the curve. The final curve is a positively skewed distribution with a tail to the right of the peak; the mode of the distribution is at the peak with the median to the right of the median and the mean to the right of the median. Next Page Descriptive Statistics Over the past several years, colleges and universities have signed exclusivity agreements with a variety of private companies. These agreements bind the university to sell that company's products exclusively on the campus. Many of the agreements involve food and beverage firms. A large university with a total enrollment of about 50,000 students has offered Pepsi-Cola an exclusivity agreement that would give Pepsi exclusive rights to sell its products at all university facilities for the next year with an option for future years. In return, the university would receive 35% of the on-campus revenues and an additional lump sum of $200,000 per year. Pepsi has been given 2 weeks to respond. The market for soft drinks is measured in terms of 12-ounce cans. Pepsi currently sells an average of 22,000 cans per week (over the 40 weeks of the year that the university operates). The cans sell for an average of 75 cents each. The costs including labor amount to 20 cents per can. Pepsi is unsure of its market share but suspects it is considerably less than 50%. Next Page Descriptive Statistics A quick analysis reveals that if its current market share were 25%, then, with an exclusivity agreement, Pepsi would sell 88,000 (22,000 is 25% of 88,000) cans per week or 3,520,000 cans per year (over the 40 weeks of university operation). The profit or loss can be calculated. The only problem is that we do not know how many soft drinks are sold weekly at the university. Pepsi assigned a recent university graduate to survey the university's students to supply the missing information. Accordingly, she organizes a survey that asks 500 students to keep track of the number of soft drinks they purchase over the next 7 days. The information we would like to acquire is an estimate of annual profits from the exclusivity agreement. The data are the numbers of cans of soft drinks consumed in 7 days by the 500 students in the sample. We can use descriptive techniques to learn more about the data. In this case, however, we are not so much interested in what the 500 students are reporting as we are in knowing the mean number of soft drinks consumed by all 50,000 students on campus. To accomplish this goal, we need the second branch of statistics called inferential statistics. Next Page Inferential Statistics Inferential statistics is a body of methods used to draw conclusions or inferences about characteristics of populations based on sample data. The population in question in this case is the soft drink consumption of the university's 50,000 students. The cost of interviewing each student would be prohibitive and extremely time consuming. Statistical techniques make such endeavors unnecessary. Instead, we can sample a much smaller number of students (the sample size is 500) and infer from the data the number of soft drinks consumed by all 50,000 students. We can then estimate annual profits for Pepsi. When an election for political office takes place, the television networks cancel regular programming and instead provide election coverage. When the ballots are counted, the results are reported. However, for important offices such as president or senator in large states, the networks actively compete to see which will be the first to predict a winner. Winner predictions are made by using exit polls, wherein a random sample of voters who exit the polling booth is asked for whom they voted. From the data the sample proportion of voters supporting the candidates is computed. A statistical technique is then applied to determine whether there is enough evidence to infer that the leading candidate will garner enough votes to win. The exit poll results from the state of Florida during the 2000 year elections were recorded (only the votes of the Republican candidate George W. Bush and the Democrat Albert Gore). The network analysts would like to know whether they can conclude that George W. Bush will win the state of Florida. Next Page Inferential Statistics Exit polls are a very common application of statistical inference. The population the television networks wanted to make inferences about is the approximately 5 million Floridians who voted for Bush or Gore for president. The sample consisted of the 765 people randomly selected by the polling company who voted for either of the two main candidates. The characteristic of the population that we would like to know is the proportion of the total electorate that voted for Bush. Specifically, we would like to know whether more than 50% of the electorate voted for Bush (counting only those who voted for either the Republican or Democratic candidate). Because we will not ask every one of the 5 million actual voters for whom they voted, we cannot predict the outcome with 100% certainty. A sample that is only a small fraction of the size of the population can lead to correct inferences only a certain percentage of the time. You will find that statistics practitioners can control that fraction and usually set it between 90% and 99%. Incidentally, on the night of the United States election in November 2000, the networks goofed badly. Using exit polls as well as the results of previous elections, all four networks concluded at 8:00 p.m. that Al Gore would win the state of Florida. Shortly after 10:00 p.m., the networks reversed course and declared that George W. Bush would win the state of Florida. By 2:00 a.m., another verdict was declared: The result was too close to call. Next Page Key Statistical Concepts Statistical inference problems involve three key concepts: the population, the sample, and the statistical inference. A population is the group of all items of interest to a statistics practitioner. It is frequently very large and may, in fact, be infinitely large. In the language of statistics, population does not necessarily refer to a group of people. It may, for example, refer to the population of diameters of ball bearings produced at a large plant. A descriptive measure of a population is called a parameter. In most applications of inferential statistics, the parameter represents the information we need. A sample is a set of data drawn from the population. A descriptive measure of a sample is called a statistic. We use statistics to make inferences about parameters. Confidence Level+ Significance Level=1Confidence Level+ Significance Level=1 Confidence Level plus Significance Level equals one. Consider a statement from polling data you may hear about in the news: "This poll is considered accurate within 3.4 percentage points, 19 times out of 20.“ In this case, our confidence level is 95% (19/20 = 0.95), while our significance level is 5%. A 5% significance level means, that in the long run, this type of conclusion will be wrong 5% of the time. Next Page Methods of Collecting Data Statistics is a tool for converting data into information: But where then does data come from? How is it gathered? How do we ensure it is accurate? Is the data reliable? Is it representative of the population from which it was drawn? Methods of Collecting Data The question arises: Where does data come from? The answer is that there are a large number of methods that produce data. Data are the observed values of a variable. That is, we define a variable or variables that are of interest to us and then proceed to collect observations of those variables. There are many methods used to collect or obtain data for statistical analysis. Three of the most popular methods are direct observation (number of customers entering a bank per hour), experiments (new ways to produce things to minimize costs), and surveys. The simplest method of obtaining data is by direct observation. When data are gathered in this way, they are said to be observational. For example, suppose that a researcher for a pharmaceutical company wants to determine whether aspirin does reduce the incidence of heart attacks. Observational data may be gathered by selecting a sample of men and women and asking each whether he or she has taken aspirin regularly over the past 2 years. Each person would be asked whether he or she had suffered a heart attack over the same period. The proportions reporting heart attacks would be compared and a statistical technique would be used to determine whether aspirin is effective in reducing the likelihood of heart attacks. There are many drawbacks to direct observation. One of the most critical limitations of this data collection method is that it is difficult to produce useful information in a meaningful way. Next Page Methods of Collecting Data For example, if the statistics practitioner concludes that people who take aspirin suffer fewer heart attacks, can we conclude that aspirin is effective? It may be that people who take aspirin tend to be more health conscious, and health conscious people tend to have fewer heart attacks. The one advantage to direct observation is that it is relatively inexpensive. A more expensive but better way to produce data is through experiments. Data produced in this manner are called experimental. In the aspirin illustration, a statistics practitioner can randomly select men and women. The sample would be divided into two groups. One group would take aspirin regularly and the other group would not. After 2 years, the statistics practitioner would determine the proportion of people in each group who had suffered a heart attack and again statistical methods would be used to determine whether aspirin works. If we find that the aspirin group suffered fewer heart attacks, we may more confidently conclude that taking aspirin regularly is a healthy decision. One of the most familiar methods of collecting data is the survey, which solicits information from people concerning such things as their income, family size, and opinions on various issues. The majority of surveys are conducted for private use. Next Page Methods of Collecting Data An important aspect of surveys is the response rate. The response rate is the proportion of all people who were selected who complete the survey. A low response rate can destroy the validity of any conclusion resulting from the statistical analysis. Statistics practitioners need to ensure that data are reliable. Many researchers feel that the best way to survey people is by means of a personal interview, which involves an interviewer soliciting information from a respondent by asking prepared questions. A personal interview has the advantage of having a higher expected response rate than other methods of data collection. In addition, there will probably be fewer incorrect responses resulting from respondents misunderstanding some questions, because the interviewer can clarify misunderstandings when asked. But, the interviewer must also be careful not to say too much, for fear of biasing the response. The main disadvantage of personal interviews is that they are expensive, especially when travel is involved. A telephone interview is usually less expensive, but it is also less personal and has a lower expected response rate. Unless the issue is of interest, many people will refuse to respond to telephone surveys. This problem is exacerbated by telemarketers trying to sell services or products. Next Page Methods of Collecting Data Another popular method of data collection is the self-administered questionnaire, which is usually mailed to a sample of people. This is an inexpensive method of conducting a survey and is, therefore, attractive when the number of people to be surveyed is large. But self-administered questionnaires usually have a low response rate and may have a relatively high number of incorrect responses due to respondents misunderstanding some questions. Next Page Questionnaire Design Whether a questionnaire is self-administered or completed by an interviewer, it must be well designed. Proper questionnaire design takes knowledge, experience, time, and money. Over the years, a lot of thought has been put into the science of the design of survey questions. Key design principles include: 1. Keep the questionnaire as short as possible. Most people are unwilling to spend much time filling out a questionnaire. 2. Ask short, simple, and clearly worded questions to enable respondents to answer quickly, correctly, and without ambiguity. 3. Start with demographic questions to help respondents get started and become comfortable quickly. 4. Use dichotomous (yes/no) and multiple choice questions because of their simplicity. 5. Use open-ended questions cautiously because they are time consuming and more difficult to tabulate and analyze. 6. Avoid using leading questions that tend to lead the respondent to a particular answer. 7. Trial a questionnaire on a small number of people to uncover potential problems, such as ambiguous wording. 8. Think about the way you intend to use the collected data when preparing the questionnaire. First determine whether you are soliciting values for an quantitative variable or a categorical variable. Then consider which type of statistical techniques, descriptive or inferential, you intend to apply to the data to be collected, and note the requirements of the specific techniques to be used. Next Page Sampling Recall that statistical inference permits us to draw conclusions about a population based on a sample. The chief motives for examining a sample rather than a population are cost and practicality. Statistical inference permits us to draw conclusions about a population parameter based on a sample that is quite small in comparison to the size of the population. It’s less expensive to sample 1,000 television viewers than 100 million TV viewers. For example, suppose a public opinion survey is conducted to determine how many people favor a tax increase. A stratified random sample could be obtained by selecting a random sample of people from each of the four income groups we just described. We usually stratify in a way that enables us to obtain particular kinds of information. In this example, we would like to know whether people in the different income categories differ in their opinions about the proposed tax increase because the tax increase will affect the strata differently. We avoid stratifying when there is no connection between the survey and the strata. For instance, little purpose is served in trying to determine whether people within religious strata have divergent opinions about the tax increase. Next Page Figure 2: Stratified Random Sampling A stratified random sample is obtained by separating the population into mutually exclusive sets or strata , and then drawing simple random samples from each stratum. We can acquire information about the total population, make inferences within a stratum (gender), or make comparisons across strata (gender and age). Figure 2 outlines gender within a stratum. It highlights gender and age as an example comparison that can occur across data. Next Page Stratified Random Sampling Besides acquiring information about the entire population, another advantage of stratification is that inferences can be made within each stratum to compare strata. For instance, we can estimate what proportion of the lowest income group favors the tax increase or we can compare the highest and lowest income groups to determine whether they differ in their support of the tax increase. Any stratification must be done in such a way that the strata are mutually exclusive, which means that each member of the population must be assigned to exactly one stratum. After the population has been stratified in this way, we can use simple random sampling to generate the complete sample. For example, we can draw random samples from each of the four income groups according to their proportions in the population. Thus, if in the population, the relative frequencies of the four groups are as listed in the chart in Figure 3, our sample will be stratified in the same proportions. If a total sample of 1,000 is to be drawn, we will randomly select 250 from stratum 1, 400 from stratum 2, 300 from stratum 3, and 50 from stratum 4. However, the problem with this approach is that if we want to make inferences about the last stratum, a sample of 50 may be too small to produce useful information. In such cases, we usually increase the sample size of the smallest stratum to ensure that the sample data provide enough information for our purposes. An adjustment must then be made before we attempt to draw inferences about the entire population. Next Page Figure 3: Stratified Random Sampling After the population has been stratified, we can use simple random samplingto generate the complete sample: Figure 3 shows 4 columns, which are income category, population proportion, sample size n = 400, and sample size n = 1000. Income under $25,000 has a population proportion (PP) of 25%, $25,000-$39,999 has a PP of $40%, $40,000-$60,000 has a PP of 30%, and income over $60,000 has a PP of 5%. If we only have sufficient resources to sample 400 people total, we would draw 100 of them from the low income group (under $25,000), which has a population proportion of 25%. …if we are sampling 1000 people, we’d draw 50 of them from the high income group (over $60,000), which has a population proportion of 5% Next Page Cluster Sampling A cluster sample is a simple random sample of groups or clusters of elements versus a simple random sample of individual objects. Cluster sampling is particularly useful when it is difficult or costly to develop a complete list of the population members (making it difficult and costly to generate a simple random sample). It is also useful whenever the population elements are widely dispersed geographically. For example, suppose we want to estimate the average annual household income in a large city. To use simple random sampling, we would need a complete list of households in the city from which to sample. To use stratified random sampling, we would need the list of households and we would also need to have each household categorized by some other variable (such as age of household head) in order to develop the strata. A less expensive alternative would be to let each block within the city represent a cluster. A sample of clusters could then be randomly selected, and every household within these clusters could be questioned to determine income. By reducing the distances the surveyor must cover to gather data, cluster sampling reduces the cost. Next Page Cluster Sampling But, cluster sampling also increases sampling error because households belonging to the same cluster are likely to be similar in many aspects, including household income. This can be partially offset by using some of the cost savings to choose a larger sample than would be used for a simple random sample. Whichever type of sampling plan you select, you still have to decide what sample size to use. We can rely on our intuition which tells us that the larger the sample size is the more accurate we can expect the sample estimates to be. Next Page Sampling Error Two major types of error can arise when a sample of observations is taken from a population: sampling error and non-sampling error. Anyone reviewing the results of sample surveys and studies, as well as statistics practitioners conducting surveys and applying statistical techniques, should understand the sources of these errors. Sampling error refers to differences between the sample and the population that exists only because of the observations that happened to be selected for the sample. Sampling error is an error that we expect to occur when we make a statement about a population that is based only on the observations contained in a sample taken from the population. Illustration 4. Experiment: Outcome: Illustration 5. Experiment: Outcomes: Illustration 6. Experiment: Outcomes: Record student evaluations of a course Poor, fair, good, very good, and excellent Measure the time to assemble a computer Number whose smallest possible value is 0 seconds with no predefined upper limit Record the party that a voter will vote for in an upcoming election Party A, Party B, … Requirements of Probabilities The first step in assigning probabilities is to produce a list of the outcomes. The listed outcomes must be exhaustive, which means that all possible outcomes must be included. Additionally, the outcomes must be mutually exclusive, which means that no two outcomes can occur at the same time. A list of exhaustive and mutually exclusive outcomes is called a sample spaceand is denoted by S. The outcomes are denoted by O1,O2,....,OkO1,O2,. ,Ok. A sample space of a random experiment is a list of all possible outcomes of the experiment. The experimental outcomes must also be exhaustive and mutually exclusive. Using set notation we represent the sample space and its outcomes as: S={O1,O2,....,Ok}S={O1,O2,.. .,Ok} Once a sample space has been prepared, we begin the task of assigning probabilities to the outcomes. There are three approaches to assign probability to outcomes. Each of these approaches must follow the two rules governing probabilities: 1. The probability of any outcome must lie between 0 and 1. That is: 0≤P(Oi)≤10≤P(Oi)≤1 2. The sum of the probabilities of all the outcomes in a sample space must be 1. That is: Σki=1P(Oi)=1Σi=1kP(Oi)=1 Next Page Three Approaches to Assigning Probabilities The classical approach is used by mathematicians to help determine probability associated with games of chance. If an experiment has n possible outcomes, this method would assign a probability of 1/n to each outcome. For example, the classical approach specifics that the probabilities of heads and tails in the flip of a balanced coin are equal to each other. Because the sum of the probabilities must be 1, the probability of heads and the probability of tails are both ½ or 50%. Similarly, the six possible outcomes of the toss of a balanced die have the same probability; each is assigned a probability of 1/6. In some experiments, it is necessary to develop mathematical ways to count the number of outcomes. When rolling two dice, there are 36 different outcomes but only 11 possible totals: The total 2 comes from a 1 on the first die and a 1 on the second die or (1,1); 3 from (1,2), (2,1); 4 from (1,3), (2,2), (3,1); 5 from (1,4), (2,3), (3,2), (4,1); 6 from (1,5), (2,4), (3,3), (4,2), (5,1); 7 from (1,6), (2,5), (3,4), (4,3), (5,2), (6,1); 8 from (2,6), (3,5), (4,4), (5,3), (6,2); 9 from (3,6), (4,5), (5,4), (6,3); 10 from (4,6), (5,5), (6,4); 11 from (5,6), (6,5); and 12 from (6,6). Note that the roll (1,2) and (2,1) are different outcomes. Figure 1 shows the classical approach when rolling two dice. Next Page Figure 1: Classical Approach Experiment: Rolling two dice and observing the total Outcomes: {2, 3, …, 12} Examples: To find P(2), only 1 cell in the matrix shows a total of 2 so P(2) = 1/36. For P(6), there are 5 cells with a total of 6 so P(6) = 5/36. For P(10), there are 3 cells with a total of 10 so P(10) = 3/36. The outer row and column of the matrix show the six possible results (1-6) of rolling one die. The inner cells of the matrix list the totals resulting from each of the 36 possible combinations when rolling two dice and range from 2 to 12. Next Page Three Approaches to Assigning Probabilities The relative frequency approach defines probability as the long-run relative frequency with which an outcome occurs. For example, suppose that we know that of the last 1,000 students who took the statistics course you’re now taking, 200 received a grade of A. The relative frequency of A’s is then 200/1000 or 20%. This figure represents an estimate of the probability of obtaining a grade of A in the course. It is only an estimate because the relative frequency approach defines probability as the “long-run” relative frequency. One thousand students do not constitute the long run. The larger the number of students whose grades we have observed, the better the estimate becomes. Next Page Figure 2: Relative Frequency Approach Bits & Bytes Computer Shop tracks the number of desktop computer systems it sells over a month (30 days): The table shows the number of days and the number of desktops sold on those days. For example there was 1 day that zero laptops were sold; 2 days where 1 laptop was sold; 10 days where 2 laptops were sold; 12 days where 3 laptops were sold and 5 days where 4 laptops were sold. From this we can construct the probabilities of an event (the # of desktop computer systems sold on a given day) Next Page Figure 3: Relative Frequency Approach A∪B (read as A union B) Next Page Figure 4: Intersection Of Two Events The intersection of events A and B is the set of all sample points that are in both A and B. Figure 5: Union Of Two Events The union of two events A and B, is the event containing all sample points that are in A or B or both: Union of A and B is denoted: A∪BA∪B Figure 5 shows a Venn diagram with the two circles of A and B intersecting. Next Page Figure 6: Joint Probability Why are some mutual fund managers more successful than others? One possible factor is where the manager earned his or her MBA. The following table compares mutual fund performance against the ranking of the school where the fund manager earned his or her MBA: The table shows 2 categories: the top 20 MBA programs and not top 20 MBA programs in relation to the mutual fund outperforming the market or not outperforming the market. In the example the top 20 MBA programs and the mutual fund outperforming the market shows 0.11 for this joint probability. This is the probability that a mutual fund outperforms the market and the manager was in a top-20 MBA program. The joint probability of top 20 MBA program manager with mutual funds that don’t outperform the market is 0.29. The probability of a manager not from a top 20 MBA program and the mutual fund outperforming the market is 0.06. The probability of a manager not from a top 20 MBA program and the mutual fun does not outperform the market is 0.54. Next Page Joint Probability The table in Figure 6 tells us the joint probability that a mutual fund outperforms the market and that its manager graduated from a top-20 MBA program is 0.11. That is, 11% of all mutual funds outperform the market and their managers graduated from a top-20 MBA program. The other three joint probabilities are defined similarly. That is, the probability that a mutual fund outperforms the market and its manager did not graduate from a top-20 MBA program is 0.06. The probability that a mutual fund does not outperform the market and its manager graduated from a top-20 MBA program is 0.29. The probability that a mutual fund does not outperform the market and its manager did not graduate from a top-20 MBA program is 0.54. To help make our task easier, we'll use notation to represent the events. Let: A1A1 = Fund manager graduated from a top-20 MBA program A2A2 = Fund manager did not graduate from a top-20 MBA program B1B1 = Fund outperforms the market B2B2 = Fund does not outperform the market Thus, P(A1∩B1)=0.11 P(A2∩B1)=0.06 P(A1∩B2)=0.29 P(A2∩B2)=0.54P Next Page Marginal Probability The joint probabilities allow us to compute various probabilities. Marginal probability is a measure of the likelihood that a particular event will occur, regardless of whether another event occurs. Marginal probabilities, computed by adding across rows or down columns, are so named because they are calculated in the margins of the table as seen in Figure 7. Adding across the first row produces P(A1∩B1)+P(A1∩B2)=0.11+0.29=0.40P(A1∩B1)+P(A1∩B2)=0.11+0.29=0.40. Notice that both intersections state that the manager graduated from a top-20 MBA program (represented by A1). Thus, when randomly selecting mutual funds, the probability that its manager graduated from a top-20 MBA program is 0.40. Expressed as relative frequency, 40% of all mutual fund managers graduated from a top-20 MBA program. Adding across the second row produces P(A2∩B1)+P(A2∩B2)=0.06+0.54=0.60P(A2∩B1)+P(A2∩B2)=0.06+0.54=0.60. This probability tells us that 60% of all mutual fund managers did not graduate from a top-20 MBA program (represented by A2A2). Notice that the probability that a mutual fund manager graduated from a top-20 MBA program and the probability that the manager did not graduate from a top-20 MBA program add to 1. Next Page Marginal Probability Adding down the columns produces the following marginal probabilities: Column 1: P(A1∩B1)+P(A2∩B1)=0.11+0.06=0.17P(A1∩B1)+P(A2∩B1)=0.11+0.06=0.17 Column 2: P(A1∩B1)+P(A2∩B2)=0.29+0.54=0.83P(A1∩B1)+P(A2∩B2)=0.29+0.54=0.83 These marginal probabilities tell us that 17% of all mutual funds outperform the market and that 83% of mutual funds do not outperform the market. Next Page Figure 7: Marginal Probability A2A2 = Fund manager did not graduate from a top-20 MBA program B1B1 = Fund outperforms the market B2B2 = Fund does not outperform the market Thus, there is a 27.5% chance that a fund will outperform the market given that the manager graduated from a top-20 MBA program. Next Page Conditional Probability Suppose that in the previous example, we select one mutual fund at random and discover that it did not outperform the market. What is the probability that a graduate of a top-20 MBA program manages it? We wish to find a conditional probability. The condition is that the fund did not outperform the market (event B2B2), and the event whose probability we seek is that the fund is managed by a graduate of a top-20 MBA program (event A1A1). Thus, we want to compute the following probability: P(A1|B2) (read as “probability of A sub 1 given B sub 2)P(A1|B2) (read as “probability of A sub 1 given B sub 2) Using the conditional probability formula, we find: P(A1|B2)=P(A1∩B2)/P(B2)=0.29/0.83=0.3494P(A1|B2)=P(A1∩B2)/P(B2)=0.29/0.83=0.3494 Thus, 34.94% of all mutual funds that do not outperform the market are managed by top-20 MBA program graduates. The calculation of conditional probabilities raises the question of whether the two events, the fund outperformed the market and the manager graduated from a top-20 MBA program, are related, a subject we tackle next. Next Page Independence One of the objectives of calculating conditional probability is to determine whether two events are related. In particular, we would like to know whether they are independent events. Two events A and B are said to be independent if: P(A|B)=P(A)P(A|B)=P(A) or P(B|A)=P(B)P(B|A)=P(B) Put another way, two events are independent if the probability of one event is not affected by the occurrence of the other event. Let’s determine whether the event that the manager graduated from a top-20 MBA program and the event the fund outperforms the market are independent events. We saw that P(B1| A1)=0.275P(B1|A1)=0.275. The marginal probability for B1B1 is: P(B1)=0.17P(B1)=0.17 Since P(B1|A1)≠P(B1)P(B1|A1)≠P(B1), B1B1 and A1A1 a re not independent events. Stated another way, they are dependent. That is, the probability of one event (B1B1) is affected by the occurrence of the other event (A1A1). Note that there are three other combinations of events in this problem. They are (A1∩B2) (A1∩B2), (A2∩B1)(A2∩B1), (A2∩B2)(A2∩B2) [ignoring mutually exclusive combinations (A1∩A2)(A1∩A2) and (B1∩B2)(B1∩B2)]. In each combination the two events are dependent. In this type of problem where there are only four combinations, if one combination is dependent, all four will be dependent. Similarly, if one combination is independent, all four will be independent. This rule does not apply to any other situation. Next Page Union Another event that is the combination of other events is the union depicted in Figure 10. The union of events A and B is the event that occurs when either A or B or both occur. It is denoted as: A∪BA∪B Determine the probability that a randomly selected fund outperforms the market or the manager graduated from a top-20 MBA program. We want to compute the probability of the union of two events P(A1∪B1)P(A1∪B1). The union A1A1 or B1B1 consists of three events. That is, the union occurs whenever any of the following joint events occurs: 1. Fund outperforms the market and the manager graduated from a top-20 MBA program. 2. Fund outperforms the market and the manager did not graduate from a top-20 MBA program. 3. Fund does not outperform the market and the manager graduated from a top-20 MBA program. Their probabilities are: P(A1∩B1)=0.11P(A1∩B1)=0.11 P(A2∩B1)=0.06P(A2∩B1)=0.06 P(A1∩B2)=0.29P(A1∩B2)=0.29 Next Page Figure 10: Union Of Two Events The union of two events A and B, is the event containing all sample points that are in A or B or both: Union of A and B is denoted: A∪BA∪B Figure 10 shows a Venn diagram with the two circles of A and B intersecting. Next Page Union The probability of the union, the fund outperforms the market or the manager graduated from a top-20 MBA program, is the sum of the three probabilities. That is, P(A1∪B1)=P(A1∩B1)+P(A2∩B1)+P(A1∩B2)=0.11+0.06+0.29=0.46P(A1∪B1)=P(A1∩B1) +P(A2∩B1)+P(A1∩B2)=0.11+0.06+0.29=0.46 as seen Figure 11. Notice in Figure 12 that there is another way to produce this probability. Of the four probabilities in the table, the only one representing an event that is not part of the union is the probability of the event the fund does not outperform the market and the manager did not graduate from a top- 20 MBA program. The complement of event A is defined to be the event consisting of all sample points that are “not in A”. Complement of A is denoted by AcAc (read A complement) The Venn diagram with A and AcAc below illustrates the concept of a complement. P(A)+P(Ac)=1P(A)+P(Ac)=1 Next Page Multiplication Rule The multiplication rule is used to calculate the joint probability of two events. It is based on the formula for conditional probability defined earlier. That is, from the following formula: P(A|B)=P(A∩B)/P(B)P(A|B)=P(A∩B)/P(B) we derive the multiplication rule simply by multiplying both sides by P(B). The joint probability of any two events A and B is P(A∩B)=P(B)×P(A|B)P(A∩B)=P(B)×P(A|B) or altering the notation, P(A∩B)=P(A)×P(B|A)P(A∩B)=P(A)×P(B|A). Selecting Two Students Without Replacement: A graduate statistics course has seven male and three female students. The professor wants to select two students at random to help her conduct a research project. What is the probability that the two students chosen are female? Next Page Multiplication Rule Let A represent the event that the first student chosen is female and B represent the event that the second student chosen is also female. We want the joint probability P(A and B). Consequently, we apply the multiplication rule: P(A∩B)=P(A)×P(B|A)P(A∩B)=P(A)×P(B|A) Because there are three female students in a class of ten, the probability that the first student chosen is female is: P(A)=3/10=0.30P(A)=3/10=0.30 Let B represent the event that the second student is female. After the first student is chosen, there are only nine students left. Given that the first student chosen was female, there are only two female students left. It follows that: P(B|A)=2/9=0.22P(B|A)=2/9=0.22 Thus, we want to answer the question: What is P(A and B)? P(A∩B)=P(A)×P(B|A)=(3/10)(2/9)=6/90=0.067P(A∩B)=P(A)×P(B|A)=(3/10) (2/9)=6/90=0.067 “There is a 6.7% chance that the professor will choose two female students from her graduate class of 10.” Next Page Multiplication Rule For Independent Events If A and B are independent events, P(A|B) = P(A) and P(B|A) = P(B). It follows that the joint probability of two independent events is simply the product of the probabilities of the two events. We can express this as a special form of the multiplication rule. The joint probability of any two independent events A and B is P(A ∩ B) = P(A) • P(B). Selecting Two Students With Replacement: The professor who teaches the graduate statistics course is suffering from the flu and will be unavailable for two classes. The professor's replacement will teach the next two classes. The replacement’s style is to select one student at random and pick on him or her to answer questions during that class. What is the probability that the two students chosen are female? We wish to compute the probability of choosing two female students. However, the experiment is slightly different. It is now possible to choose the same student in each of the two classes the replacement teaches. Thus A and B are independent events, and we apply the multiplication rule for independent events: P(A ∩ B) = P(A) • P(B). The probability of choosing a female student in each of the two classes is the same. That is, P(A) = 3/10 and P(B) = 3/10. That is, the probability of choosing a female student given that the first student chosen is unchanged since the student selected in the first class can be chosen in the second class. P(A ∩ B) = P(A) • P(B) = (3/10)(3/10) = 9 / 100 = 0.090. Next Page Addition Rule In a large city, two newspapers are published, the Sun and the Post. The circulation departments report that 22% of the city's households have a subscription to the Sun and 35% subscribe to the Post. A survey reveals that 6% of all households subscribe to both newspapers. What proportion of the city's households subscribe to either newspaper. We can express this question as, “What is the probability of selecting a household at random that subscribes to the Sun, the Post, or both?” What is P(Sun or Post)? Another way of asking the question is, “What is the probability that a randomly selected household subscribes to at least one of the newspapers?” It is now clear that we seek the probability of the union, and we must apply the addition rule illustrated in Figure 14. Let A = the household subscribes to the Sun and B = the household subscribes to the Post. We perform the following calculation: P(A B)=P(A)+P(B)−P(A∩B)=0.22+0.35−0.06=0.51∪ P(A∪B)=P(A) +P(B) −P(A∩B)=0.22+0.35−0.06=0.51 The probability that a randomly selected household subscribes to either newspaper is 0.51. Expressed as relative frequency, 51% of the city's households subscribe to either newspaper. If the first selection, P(F) is 3/10 and the second selection, P(F|F) is 2/9, the joint probability is P(FF) calculated by multiplying 3/10 times 2/9. Next Page Figure 17: Probability Tree For Selecting Two Students With Replacement Suppose we have our graduate class of 10 students again, but make the student sampling independent, that is “with replacement” – a student could be picked first and picked again in the second round. Our tree and joint probabilities now look like: P(F) is 3/10 and P(M) is 7/10. Following the probability of picking a female first, the second pick would be P(F|F) equals 3/10 and the P(M|F) is 7/10. The probability tree shows P(FF) is the product of 3/10 times 3/10. The P(FM) is the product of 3/10 times 7/10. Following the probability of picking a male first, the second pick is female is P(F|M) or 3/10 and the P(M|M) is 7/10. The probability tree shows P(MF) equals 7/10 times 3/10. The P(MM) is 7/10 times 7/10. Next Page Probability Trees The advantage of a probability tree on this type of problem is that it restrains its users from making the wrong calculation. Once the tree is drawn and the probabilities of the branches inserted, virtually the only allowable calculation is the multiplication of the probabilities of linked branches. An easy check on those calculations is available as shown in Figure 18. The joint probabilities at the ends, of the branches must sum to 1, because all possible events are listed. In Figure 16 and Figure 17, notice that the joint probabilities do indeed sum to 1. The special form of the addition rule for mutually exclusive events can be applied to the joint probabilities. In both probability trees, we can compute the probability that one student chosen is female and one is male simply by adding the joint probabilities. For the tree in Figure 16, we have P(F∩M)+P(M∩F)=21/90+21/90=42/90P(F∩M) +P(M∩F)=21/90+21/90=42/90. In the probability tree in Figure 17, we find P(F∩M)+P(M∩F)=21/100+21/100=42/100P(F∩M) +P(M∩F)=21/100+21/100=42/100. Next Page Figure 18: Probability Trees The first selection on the probability tree shows P(F) is 3/10 and P(M) is 7/10. 3/10 plus 7/10 is 10/10 or 1. Each branch on the tree follows the same. In the second selection P(F|F) is 2/9 and P(M|F) is 7/9. 2/9 plus 7/9 is 9/9 or 1. The same is true for the node or branch of the P(M) branch. Next Page Figure 19: Probability Trees There is no requirement that the branch splits be binary, nor that the tree only goes two levels deep, or that there be the same number of splits at each sub node. Next Page Bayes’ Law Conditional probability is often used to gauge the relationship between two events. In many of the examples and exercises you've already encountered, conditional probability measures the probability that an event occurs given that a possible cause of the event has occurred. We calculated the probability that a mutual fund outperforms the market (the effect) given that the fund manager graduated from a top-20 MBA program (the possible cause). There are situations, however, where we witness a particular event and we need to compute the probability of one of its possible causes. Bayes’ Law is the technique we use as explained in Figure 22. The Graduate Management Admission Test (GMAT) is a requirement for all applicants of MBA programs. There are a variety of preparatory courses designed to help improve GMAT scores, which range from 200 to 800. Suppose that a survey of MBA students reveals that among GMAT scorers above 650, 52% took a preparatory course, whereas among GMAT scorers of less than 650 only 23% took a preparatory course. An applicant to an MBA program has determined that he needs a score of more than 650 to get into a certain MBA program, but he feels that his probability of getting that high a score is quite low - 10%. He is considering taking a preparatory course that costs $500. He is willing to do so only if his probability of achieving 650 or more doubles. What should he do? Next Page Figure 22: Bayes’ Law Bayes’ Law is named for Thomas Bayes, an eighteenth century mathematician. Next Page Bayes’ Law Let A=GMATA=GMAT score of 650 or more, hence AC=GMATAC=GMAT score less than 650. Our student has determined the probability of getting greater than 650 (without any prep course) as 10%, that is: P(A)=.10P(A)=.10. It follows that P(AC)=1–0.10=0.90P(AC)=1– 0.10=0.90. Let B represent the event “take the prep course” and thus, BCBC is “do not take the prep course”. From our survey information, we’re told that among GMAT scorers above 650, 52% took a preparatory course, that is: P(B|A)=0.52P(B|A)=0.52. (Probability of finding a student who took the prep course given that he or she scored above 650). But our student wants to know P(A|B)P(A|B), that is, what is the probability of getting more than 650 given that a prep course is taken? If this probability is > 20%, he will spend $500 on the prep course. Among GMAT scorers of less than 650 only 23% took a preparatory course. That is: P(B| AC)=.23P(B|AC)=.23. (Probability of finding a student who took the prep course given that he or she scored less than 650). The conditional probabilities are P(B|A)=.52P(B|A)=.52 and P(B|AC)=.23P(B|AC)=.23. Again using the complement rule, we find the following conditional probabilities: P(BC| A)=1−0.52=0.48P(BC|A)=1−0.52=0.48 and P(BC|AC)=1−0.23=0.77P(BC|AC)=1−0.23=0.77. We are trying to determine P(A|B)P(A|B). The definition of conditional probability from earlier will assist us: P(A|B)=P(A∩B)/P(B)P(A|B)=P(A∩B)/P(B) We don’t know P(A∩B)P(A∩B) and we don’t know P(B)P(B). We can construct a probability tree as shown in Figure 23 and 24. Next Page Figure 23: Bayes’ Law In order to go fromP(B|A)=0.52P(B|A)=0.52 to P(A|B)=??P(A|B)=??we need to apply Bayes' Law. Graphically: Creating a probability tree with the test score is greater than or equal to 650 as the first branches, the first branch shows P(A) equals .10 and P(AC) is .90. The second branch builds off the A branch, P(B|A) is .52 and P(BC|A) is .48. The joint probability of A ∩ B is .10 times .52 or .052. The joint probability of A ∩ BC is .48. Following the AC branch we have B|AC is .23 and BC| AC is .77. Therefore the P(AC∩B) is .90 times .23 or .207. The P(AC ∩ BC) is .90 times .77 or . 693. Now we just need P(B) Next Page Figure 24: Bayes’ Law Bayes’ Law Formula Conditional probabilities are P(B|A1)=0.52P(B|A1)=0.52 and P(B|A2)=0.23P(B|A2)=0.23 Substituting the prior and likelihood probabilities into the Bayes' Law formula yields the following: P(A1|B)=P(A1)P(B|A1)P(A1)P(B|A1)+P(A2)P(B|A2)=(.10)(.52)(.10)(.52)+(.90) (.23)=0.52.052+.207=.052.259=0.201P(A1|B)=P(A1)P(B|A1)P(A1)P(B|A1)+P(A2)P(B| A2)=(.10) (.52)(.10)(.52)+(.90)(.23)=0.52.052+.207=.052.259=0.201 As you can see, the calculation of the Bayes' Law formula produces the same results as the probability tree. Next Page Identifying the Correct Method Although it is difficult to offer strict rules on which probability method to use, nevertheless we can provide some general guidelines. The key issue is whether joint probabilities are provided or are required. Where the joint probabilities were given, we can compute marginal probabilities by adding across rows and down columns. We can use the joint and marginal probabilities to compute conditional probabilities, for which a formula is available. This allows us to determine whether the events described by the table are independent or dependent. We can also apply the addition rule to compute the probability that either of two events occurs. We need to apply some or all of the three probability rules in circumstances where one or more joint probabilities are required. We apply the multiplication rule (either by formula or through a probability tree) to calculate the probability of intersections. In some problems we're interested in adding these joint probabilities. We're actually applying the addition rule for mutually exclusive events here. We also frequently use the complement rule. In addition, we can also calculate new conditional probabilities using Bayes' Law. The first step in assigning probability is to create an exhaustive and mutually exclusive list of outcomes. The second step is to use the classical, relative frequency, or subjective approach and assign probability to the outcomes. There are a variety of methods available to compute the probability of other events. These methods include probability rules and trees. An important application of these rules is Bayes' Law, which allows us to compute conditional probabilities from other forms of probability. Next Page Section 4 Sampling Distributions ampling Distribution of the Mean Sampling distributions describe the distributions of sample statistics. A sampling distribution is created by, as the name suggests, sampling. There are two ways to create a sampling distribution. The first is to actually draw samples of the same size from a population, calculate the statistic of interest, and then use descriptive techniques to learn more about the sampling distribution. The second method relies on the rules of probablility and the laws of expected value and variance to derive the sampling distribution. We'll demonstrate the latter approach by developing the sampling distribution of the mean of two dice. The population is created by throwing a fair die infinitely many times, with the random variable X indicating the number of spots showing on any one throw. The probability distribution of X is: x 1 2 3 4 5 6 P(x) 1/6 1/6 1/6 1/6 1/6 1/6 The table shows x can be 1, 2, 3, 4, 5, or 6 The P(x) is one sixth for each of these options. and the mean, variance, and standard deviation are calculated as well: μ=∑xP(x)=1(16)+2(16)+...+6(16)=3.5μ=∑xP(x)=1(16)+2(16)+...+6(16)=3.5 σ2=∑(x−μ)2P(x)=(1−3.5)2(16)+...+(6−3.5)2(16)=2.92σ2=∑(x−μ)2P(x)=(1−3.5)2(16)+... +(6−3.5)2(16)=2.92 σ=σ2−−√=2.92−−− −√=1.71σ=σ2=2.92=1.71 Figure 1 shows all the possible samples and their corresponding values of x¯x¯. Figure 2 shows the resulting sample distribution x¯x¯. Next Page Figure 1: Sampling Distribution of Two Dice A sampling distribution is created by looking at all samples of size n=2 (i.e., two dice) and their means. Sample x¯x¯ Sample x¯x¯ Sample x¯x¯ 1,1 1.0 3,1 2.0 5,1 3.0 1,2 1.5 3,2 2.5 5,2 3.5 1,3 2.0 3,3 3.0 5,3 4.0 1,4 2.5 3,4 3.5 5,4 4.5 1,5 3.0 3,5 4.0 5,5 5.0 1,6 3.5 3,6 4.5 5,6 5.5 2,1 1.5 4,1 2.5 6,1 3.5 2,2 2.0 4,2 3.0 6,2 4.0 2,3 2.5 4,3 3.5 6,3 4.5 2,4 3.0 4,4 4.0 6,4 5.0 2,5 3.5 4,5 4.5 6,5 5.5 2,6 4.0 4,6 5.0 6,6 6.0 The table shows for a sample of 2 the mean is as follows: for a 1 and 1 the mean is (1+1)/2 = 1.0. For 1 and 2 the mean is (1+2)/2) = 1.5. For 1 and 3 the mean is (1+3)/2 = 2.0. The progression continues through all 36 possible combinations of rolling two dice. For example, a roll of 4 and 3 has a mean of 3.5 and a roll of a 6 and 2 has a mean of 4. While there are 36 possible samples of size 2, there are only 11 values for x¯x¯, and some (e.g., x¯=3.5x¯=3.5) occur more frequently than others (e.g., x¯=1x¯=1). Next Page Figure 2: Sampling Distribution of Two Dice The sampling distribution of x¯x¯ is shown below: with the sampling distribution of x¯x¯. The figure shows two bar graphs. The first is the population distribution of X, the result of rolling a single die. Each of the possible values from 1 to 6 had the same probability so the graph is a series of six bars of the same height. The second is the sampling distribution of the sample means, x-bar. This graph has values from 1 to 6 in increments of one-half. The values on either of the graph are small and increase symmetrically toward the center as discussed earlier. Comparing the computations done earlier, note that the mean of x-bar is the same as the mean of the population while the variance of x-bar is half the variance of the population. As well, note that: μx¯=μμx¯=μ σ2x¯=σ2/2σx¯2=σ2/2 Next Page Standard Error We can generalize the mean and variance of the sampling of two dice: μx¯=μμx¯=μ σ2x¯=σ2/2σx¯2=σ2/2 …to n-dice: σx¯=σn−−√σx¯=σn μx¯=μμx¯=μ σ2x¯=nσx¯2=n The standard deviation of the sampling distribution is called the standard error: σx¯=σn√σx¯=σn Next Page Central Limit Theorem As you can see in Figure 3, the variance of the sampling distribution of X¯X¯ is less than the variance of the population we’re sampling from for all sample sizes. Thus, a randomly selected value of X¯X¯ (the mean of the number of spots observed in, say, five throws of the dice) is likely to be closer to the mean value of 3.5 than is a randomly selected value X (the number of spots observed in one throw). Indeed, this is what you would expect, because in five throws of the die you are likely to get some 5s and 6s and some 1s and 2s, which will tend to offset one another in the averaging process and produce a sample mean reasonably close to 3.5. As the number of throws of the die increases, the probability that the sample mean will be close to 3.5 also increases. Thus, we observe that the sampling distribution of X¯X¯becomes narrower (or more concentrated about the mean) as the sample size, n, increases. Another thing that happens as n gets larger is that the sampling distribution of X¯X¯ becomes increasingly bell shaped. This phenomenon is summarized in the central limit theorem. The sampling distribution of the mean of a random sample drawn from any population is approximately normal for a sufficiently large sample size. The larger the sample size, the more closely the sampling distribution of X¯X¯ will resemble a normal distribution. The accuracy of the approximation alluded to in the central limit theorem depends on the probability distribution of the population and on the sample size. If the population is normal, then X¯X¯ is normally distributed for all values of n. If the population is non-normal, then X¯X¯ is approximately normal only for larger values of n. In many practical situations, a sample size of 30 may be If a customer buys one bottle, what is the probability that the bottle will contain more than 32 ounces? We want to find P(X>32)P(X>32), where X is normally distributed and μ=32.2μ=32.2 and σ=0.3σ=0.3. P(X>32)=P(X−μσ>32−32.20.3)=P(Z>0.67)=1−0.2514=0.7486P(X>32)=P(X−μσ>32−32.20.3)= P(Z>0.67)=1−0.2514=0.7486 “There is about a 75% chance that a single bottle of soda contains more than 32 oz.” (The Z probability comes from a probability table for positive Z-scores. A table which includes negative Z values would give the probability directly.) Next Page Contents of a 32-Ounce Bottle The foreman of a bottling plant has observed that the amount of soda in each “32-ounce” bottle is actually a normally distributed random variable, with a mean of 32.2 ounces and a standard deviation of .3 ounce. If a customer buys a carton of four bottles, what is the probability that the mean amount of the four bottles will be greater than 32 ounces? We want to find P(X¯>32)P(X¯>32), where X is normally distributed with μ=32.2μ=32.2 and σ=0.3σ=0.3. Things we know: 1. X is normally distributed, therefore so will X¯X¯. 2. μx¯=μ=32.2 oz.μx¯=μ=32.2 oz. 3. σx¯=σ/n−−√=0.3/4–√=0.15σx¯=σ/n=0.3/4=0.15 P(X¯>32)=P(X¯−μX¯σX¯>32−320.15)=P(Z>1.33)=0.9082P(X¯>32)=P(X¯−μX¯σX¯>32−320. 15)=P(Z>1.33)=0.9082 P(X>32)=P(X−μσ>32−32.20.3)=P(Z>0.67)=1−0.2514=0.7486P(X>32)=P(X−μσ>32−32.20.3)= P(Z>0.67)=1−0.2514=0.7486 “There is about a 91% chance the mean of the four bottles will exceed 32 oz.” Figure 4 illustrates the distributions used in the 32-ounce bottle example. Next Page Figure 4: Contents of a 32-Ounce Bottle What is the probability that one bottle will contain more than 32 ounces? What is the probability that the mean of four bottles will exceed 32 oz? The figure compares the graphs of the distribution of X, a single 32-ounce bottle with the distribution of X-bar, the mean of a sample of four 32-ounce bottles on the same graph. The distribution of both variables are normal curves with the same mean of 32.2, however the distribution of X is more spread out with a lower peak than the distribution of X-bar. Consequently, the area under the curves to the right of 32 are not the same; the probability of a single bottle having more than 32 ounces is almost 75% while the probability of four bottles having an average greater than 32 ounces is almost 91%. Next Page Salaries of a Business School’s Graduates Deans and other faculty members in professional schools often monitor how well the graduates of their programs fare in the job market. Information about the types of jobs and their salaries may provide useful information about the success of the program. In the advertisements for a large university, the dean of the School of Business claims that the average salary of the school's graduates one year after graduation is $800 per week with a standard deviation of $100. A second-year student in the business school who has just completed his statistics course would like to check whether the claim about the mean is correct. He does a survey of 25 people who graduated one year ago and determines their weekly salary. He discovers the sample mean to be $750. To interpret his finding, he needs to calculate the probability that a sample of 25 graduates would have a mean of $750 or less when the population mean is $800 and the standard deviation is $100. After calculating the probability, he needs to draw some conclusions. We want to find the probability that the sample mean is less than $750. Thus, we seek: looking for the proportion of successes. (Success is getting the outcome in which we’re interested even if it’s broken items.) Since there are only two outcomes (success and failure), these situations are called binomial experiments and have a binomial distribution. Because the number of successes can only take on whole number value, the binomial distribution is a discrete distribution. The binomial distribution parameter is p, the probability of success in any trial. In order to compute binomial probabilities, we have to assume that p was known. However, in the real world p is unknown, requiring the statistics practitioner to estimate its value from a sample. The estimator of a population proportion of successes is the sample proportion. That is, we count the number of successes in a sample and compute p^=Xnp^=Xn (P^P^ is read as p-hat) where X is the number of successes and n is the sample size. When we take a sample of size n, we’re actually conducting a binomial experiment and as a result, X is binomially distributed. Thus, the probability of any values of P^P^ can be calculated from its value of X. Sampling Distribution of a Proportion For example, suppose that we have a binomial experiment with n=10n=10 and p=.4p=.4. To find the probability that the sample proportion P^P^ is less than or equal to .50, we find the probability that X is less than or equal to 5 (because 5/10 = .50). P(P^≤.50)=P(X≤5)=0.8338P(P^≤.50)=P(X≤5)=0.8338 We can calculate the probability associated with other values of P^P^ similarly using a binomial distribution calculator or table or a computer. Discrete distributions such as the binomial do not lend themselves easily to the kinds of calculation needed for inference. And inference is the reason we need sampling distributions. Fortunately, we can approximate the binomial distribution by a normal distribution. Next Page Normal Approximation to the Binomial Distribution The normal distribution can be used to approximate a binomial distribution. We developed the density function by converting a histogram so that the total area in the rectangle equaled 1. We can do the same for a binominal distribution. To illustrate, let X be a binomial random variable with n=20n=20 and p=.5p=.5. We can easily determine the probability of each value of X, where X = 0,1,2…,19, 20. A rectangle representing a value of x is drawn so that its area equals the probability. We accomplish this by letting the height of the rectangle equal the probability and the base of the rectangle equal 1. Thus the base of each rectangle for x is the interval x – .5 to x + .5. Figure 5 depicts this graph. As you can see, the rectangle representing x = 10 is the rectangle whose base is the interval 9.5 to 10.5 and whose height is P(X=10)=.1762P(X=10)=.1762. If we now smooth the ends of the rectangles, we produce a bell-shaped curve as seen in Figure 6. Thus, to use the normal approximation, all we need do is find the area under the normal curve between 9.5 and 10.5. To find normal probabilities requires us to first standardize x by subtracting the mean and dividing by the standard deviation. The values for μμ and σσ are derived from the binomial distribution being approximated. In this case μ=10μ=10 and σ=2.24σ=2.24. Next Page Figure 5: Normal Approximation to the Binomial Distribution Binomial distribution with n=20n=20 and p=.5p=.5 with a normal approximation superimposed (μ=10μ=10 and σ=2.24σ=2.24). Next Page Figure 6: Normal Approximation to the Binomial Distribution Binomial distribution with n=20n=20 and p=.5p=.5 with a normal approximation superimposed (μ=10μ=10 and σ=2.24σ=2.24). μ=np,σ2=np(1−p),σ=np(1−p)−−−−−−−−√μ=np,σ2=np(1−p),σ=np(1−p) P(X=10)≈P(9.5<Y<10.5)P(X=10)≈P(9.5<Y<10.5) Where Y is a normal random variable approximating the binomial random variable X. Next Page Sampling Distribution of a Sample Proportion Using the laws of expected value and variance, we can determine the mean, variance, and standard deviation of P^P^ (The standard deviation of P^P^ is called the standard error of the proportion. P^P^ is approximately normally distributed provided that npnp and n(1−p)n(1−p) are greater than or equal to 5. E(P^)=pV(P^)=σ2P^=p(1−p)nσP^=p(1−p)/n−−−−−−−− −√E(P^)=pV(P^)=σP^2=p(1−p)nσP^=p(1−p)/n Sample proportions can be standardized to a standard normal distribution using this formulation: Z=P^−pp(1−p)/n−−−−−−−−−√Z=P^−pp(1−p)/n Next Page Political Survey In the last election, a state representative received 52% of the votes cast. One year after the election, the representative organized a survey that asked a random sample of 300 people whether they would vote for him in the next election. If we assume that his popularity has not changed, what is the probability that more than half of the sample would vote for him? The number of respondents who would vote for the representative is a binomial random variable with n=300n=300 and p=0.52p=0.52. We want to determine the probability that the sample proportion is greater than 50%. That is, we want to find P(P^>50)P(P^>50). We now know that the sample proportion P^P^ is approximately normally distributed with mean np=0.52np=0.52 and standard deviation = p(1−p)/n−−−−−−−−−√=(.52)(.48)/300−−−−−−− −−−−√=0.0288p(1−p)/n=(.52)(.48)/300=0.0288. Thus, we calculate P(P^>.50)=P(p^−pp(1−p)/n−−−−−−−− −√>.50−.52.0288)=P(Z>−0.69)=−P(Z<−0.69)=1−0.2451=0.7549P(P^>.50)=P(p^−pp(1−p)/n>.5 0−.52.0288)=P(Z>−0.69)=−P(Z<−0.69)=1−0.2451=0.7549 If we assume that the level of support remains at 52%, the probability that more than half the sample of 300 people would vote for the representative is 75.49%. Next Page Sampling Distribution of the Difference Between Two Means Another sampling distribution that you will encounter is that of the difference between two sample means. The sampling plan calls for independent random samples drawn from each of two normal populations. The samples are said to be independent if the selection of the members of one sample is independent of the selection of the members of the second sample. We are interested in the sampling distribution of the difference between the two sample means. The central limit theorem states that in repeated sampling from a normal population whose mean is μμ and whose standard deviation is σσ, the sampling distribution of the sample mean is normal with mean μμ and standard deviation σ/n−−√σ/n. Statisticians have shown that the difference between two independent normal random variables is also normally distributed. Thus, the difference between two sample means X¯1−X¯2X¯1−X¯2 is normally distributed if both populations are normal. If the two populations are not both normally distributed, but the sample sizes are “large” (more than 30), the distribution of X¯1−X¯2X¯1−X¯2 is approximately normal. Next Page Sampling Distribution of the Difference Between Two Means Through the use of the laws of expected value and variance, we derive the expected value and variance of the sampling distribution of X¯1−X¯2X¯1−X¯2. Thus, it follows that in repeated independent sampling from two populations with means μ1μ1 and μ2μ2 and standard deviations σ1σ1 and σ2σ2, respectively, the sampling distribution of X¯1−X¯2X¯1−X¯2 is normal with mean μ1−μ2μ1−μ2. μX¯1−X¯2=μ1−μ2μX¯1−X¯2=μ1−μ2 and standard deviation (which is the standard error of the difference between two means). σX¯1−X¯2=σ21n1+σ22n2−−−−−−−−√σX¯1−X¯2=σ12n1+σ22n2 If the populations are non-normal, then the sampling distribution is only approximately normal for large sample sizes. The required sample sizes depend on the extent of non-normality. However, for most populations, sample sizes of 30 or more are sufficient. Next Page Starting Salaries of MBAs Suppose that the starting salaries of MBAs at Wilfred Laurier University (WLU) are normally distributed with a mean of $62,000 and a standard deviation of $14,500. The starting salaries of MBAs at the University of Western Ontario (UWO) are normally distributed with a mean of $60,000 and a standard deviation of $18,300. If a random sample of 50 WLU MBAs and a random sample of 60 UWO MBAs are selected, what is the probability that the sample mean starting salary of WLU graduates will exceed that of the UWO graduates? University 1 University 2 Mean 62,000 $/yr 60,000 $/yr Sampling Distribution in Inference: Statistic −→−−−−−−−−−− −Sampling distribution→Sampling distribution Parameter The sampling distribution of a statistic is created by repeated sampling from one population. We introduced the sampling distribution of the mean, the proportion, and the difference between two means. We described how these distributions are created theoretically and empirically. Next Page Section 5 Introduction to Hypothesis Testing Next Page Concepts of Hypothesis Testing The term hypothesis testing is likely new to most readers, but the concepts underlying hypothesis testing are quite familiar. There are a variety of non-statistical applications of hypothesis testing, the best known of which is a criminal trial. When a person is accused of a crime, he or she faces a trial. The prosecution presents its case and a jury must make a decision on the basis of the evidence presented. In fact, the jury conducts a test of hypothesis. There are actually two hypotheses that are tested. The first is called the null hypothesis and is represented by H0H0 (pronounced H-nought: nought is a British term for zero). It is H0H0: The defendant is innocent. The second is called the alternative or research hypothesis and is denoted H1H1. In a criminal trial, it is H1H1: The defendant is guilty. The hypothesis of most interest to the researcher is the alternative hypothesis. Of course, the jury does not know which hypothesis is correct. They must make a decision on the basis of the evidence presented by both the prosecution and the defense. There are only two possible decisions. Convict or acquit the defendant. In statistical parlance, convicting the defendant is equivalent to rejecting the null hypothesis in favor of the alternative. That is, the jury is saying that there was enough evidence to conclude that the defendant was guilty. Acquitting a defendant is phrased as not rejecting the null hypothesis in favor of the alternative, which means that the jury decided that there was not enough evidence to conclude that the defendant was guilty. Notice that we do not say that we accept the null hypothesis. In a criminal trial, that would be interpreted as finding the defendant innocent. Our justice system does not allow this decision. Next Page Type I Error and Type II Error There are two possible errors. A Type I error occurs when we reject a true null hypothesis. A Type II error is defined as not rejecting a false null hypothesis. In the criminal trial, a Type I error is made when an innocent person is wrongly convicted. A Type II error occurs when a guilty defendant is acquitted. The probability of a Type I error is denoted by αα, which is also called the significance level. The probability of a Type II error is denoted by ββ (Greek letter beta). The error probabilities αα and ββ are inversely related, meaning that any attempt to reduce one will increase the other. In our justice system, Type I errors are regarded as more serious. As a consequence, the system is set up so that the probability of a Type I error is small. This is arranged by placing the burden of proof on the prosecution (the prosecution must prove guilt - the defense need not prove anything) and by having judges instruct the jury to find the defendant guilty only if there is "evidence beyond a reasonable doubt." In the absence of enough evidence, the jury must acquit even though there may be some evidence of guilt. The consequence of this arrangement is that the probability of acquitting guilty people is relatively large. Next Page Critical Concepts in Hypothesis Testing The five critical concepts in hypothesis testing are as follows: 1. There are two hypotheses. One is called the null hypothesis and the other the alternative or research hypothesis. 2. The testing procedure begins with the assumption that the null hypothesis is true. 3. The goal of the process is to determine whether there is enough evidence to infer that the alternative hypothesis is true. 4. There are two possible decisions: Conclude that there is enough evidence to support the alternative hypothesis. Conclude that there is not enough evidence to support the alternative hypothesis. 5. Two possible errors can be made in any test. A Type I error occurs when we reject a true null hypothesis and a Type II error occurs when we don't reject a false null hypothesis. The probabilities of Type I and Type II errors are P(Type I error)=αP(Type I error)=α P(Type II error)=βP(Type II error)=β Next Page Critical Concepts in Hypothesis Testing: Concept 1 There are two hypotheses. One is called the null hypothesis and the other thealternative or research hypothesis. The usual notation is: H0H0: — the ‘null’ hypothesis H1H1: — the ‘alternative’ or ‘research’ hypothesis The null hypothesis H0H0 will always state that the parameter equals the valuespecified in the alternative hypothesis H1H1. Next Page Critical Concepts in Hypothesis Testing: Concept 1 The Doll Computer Company makes its own computers and delivers them directly to customers who order them via the Internet. To achieve its objective of speed, Doll makes each of its five most popular computers and transports them to warehouses from which it generally takes 1 day to deliver a computer to the customer. This strategy requires high levels of inventory that add considerably to the cost. To lower these costs, the operations manager wants to use an inventory model. He notes demand during lead time is normally distributed and he needs to know the mean to compute the optimum inventory level. He observes 25 lead time periods and records the demand during each period: 235 374 309 499 253 P(Type I error)=αP(Type I error)=α P(Type II error)=βP(Type II error)=β αα is called the significance level. Next Page Types of Errors A Type I error occurs when we reject a true null hypothesis (Reject H0H0 when it is TRUE). The graphic shows a grid with two rows and two columns. The columns of the grid list the two possible values of the null hypothesis (True or False) and the rows list the possible statistical conclusions (Reject the null hypothesis and Do Not Reject the null hypothesis). The four cells in the grid show that Type I and Type II errors occur when the conclusion and the actual value conflict. When the conclusion and the actual value agree there is no error. A Type II error occurs when we don’t reject a false null hypothesis (Do NOT reject H0H0 when it is FALSE). Next Page Testing the Population Mean When the Population Standard Deviation Is Known The manager of a department store is thinking about establishing a new billing system for the store's credit customers. After a thorough financial analysis, she determines that the new system will be cost-effective only if the mean monthly account is more than $170. A random sample of 400 monthly counts is drawn, for which the sample mean is $178. The manager knows that the accounts are approximately normally distributed with a standard deviation of $65. Can the manager conclude from this that the new system will be cost-effective? This example deals with the population of the credit accounts at the store. To conclude that the system will be cost-effective requires the manager to show that the mean account for all customers is greater than $170. Consequently, we set up the alternative hypothesis to express this circumstance: H1:μ>170H1:μ>170(install new system). If the mean is less than or equal to 170, the system will not be cost-effective. The null hypothesis can be expressed as H0:μ≤170H0:μ≤170 (do not install new system). We will actually test μ=170μ=170, which is how we specify the null hypothesis: H0:μ=170H0:μ=170. Is a sample mean of 178 sufficiently greater than 170 to allow us to confidently infer that the population mean is greater than 170? Next Page Rejection Region Method There are two approaches to answering this question. The first is called the rejection region method. It can be used in conjunction with the computer, but it is mandatory for those computing statistics manually. The second is the p-value approach, which in general can be employed only in conjunction with a computer and statistical software. It seems reasonable to reject the null hypothesis in favor of the alternative if the value of the sample mean is large relative to 170. If we had calculated the sample mean to be say, 500, it would be quite apparent that the null hypothesis is false and we would reject it. On the other hand, values of x¯x¯ close to 170, such as 171, do not allow us to reject the null hypothesis because it is entirely possible to observe a sample mean of 171 from a population whose mean is 170. Unfortunately, the decision is not always so obvious. In this example, the sample mean was calculated to be 178, a value apparently neither very far away from nor very close to 170. To make a decision about this sample mean, we set up the rejection region. The rejection region is a range of values such that if the test statistic falls into that range, we decide to reject the null hypothesis in favor of the alternative hypothesis. Suppose we define the value of the sample mean that is just large enough to reject the null hypothesis as x¯Lx¯L. The rejection region is x¯>x¯Lx¯>x¯L. Next Page Rejection Region Method It seems reasonable to reject the null hypothesis in favor of the alternative if the value of the sample mean is large relative to 170, that is if x¯>x¯Lx¯>x¯L. α=P(Type I error)α=P(Type I error) α=P(reject H0 given that H0 is true)α=P(reject H0 given that H0 is true) α=P(x¯>x¯L)α=P(x¯>x¯L) The graph shows a normal distribution with mu equals 170. The value of x-bar L is marked on the right tail of the distribution such that the area under the curve to the right of x-bar L is equal to sigma. A value of x-bar which falls into that right tail is in the rejection region. Next Page Rejection Region Method All that's left to do is calculate x¯Lx¯L and compare it to 170. P(x¯−μσ/n−−√>x¯L−μσ/n−−√)=P(Z>x¯L−μσ/n− −√)=αP(x¯−μσ/n>x¯L−μσ/n)=P(Z>x¯L−μσ/n)=α Since P(Z>zα)P(Z>zα) is equal to αα, we see that Figure 1: The Big Picture Again Because 2.46 is greater than 1.645, reject the null hypothesis and conclude that there is enough evidence to infer that the mean monthly account is greater than $170. The graphic is the normal Z distribution with mu = 0. The Z 0.05 value of 1.645 is indicated on the right tail with the rejection area falling to the right. Since the calculated standardized test statistic of z = 2.45 falls into the rejection area, the null hypothesis is rejected in favor of the alternate hypothesis. Next Page p-Value of a Test There are several drawbacks to the rejection region method. Foremost among them is the type of information provided by the result of the test. The rejection region method produces a yes or no response to the question, “Is there sufficient statistical evidence to infer that the alternative hypothesis is true?” The implication is that the result of the test of hypothesis will be converted automatically into one of two possible courses of action: one action as a result of rejecting the null hypothesis in favor of the alternative and another as a result of not rejecting the null hypothesis in favor of the alternative. The rejection of the null hypothesis seems to imply that the new billing system will be installed. In fact, this is not the way in which the result of a statistical analysis is utilized. The statistical procedure is only one of several factors considered by a manager when making a decision. The manager discovered that there was enough statistical evidence to conclude that the mean monthly account is greater than $170. However, before taking any action, the manager would like to consider a number of factors including the cost and feasibility of restructuring the billing system and the possibility of making an error, in this case a Type I error. What is needed to take full advantage of the information available from the test result and make a better decision is a measure of the amount of statistical evidence supporting the alternative hypothesis so that it can be weighed in relation to the other factors, especially the financial ones. The p-value of a test provides this measure. Next Page p-Value of a Test The p-value of a test is the probability of observing a test statistic at least as extreme as the one computed given that the null hypothesis is true. In the case of our department store example, the p-value is the probability of observing a sample mean at least as large as 178 when the population mean is 170. Thus, p-value=P(X¯>178)=P(x¯−μσ/n−−√>178−17065/400−− −√)=P(Z>2.46)=1−P(Z<2.46)=1−0.9931=0.0069p- value=P(X¯>178)=P(x¯−μσ/n>178−17065/400)=P(Z>2.46)=1−P(Z<2.46)=1−0.9931=0.0069 Next Page Interpreting the p-Value To properly interpret the results of an inferential procedure, you must remember that the technique is based on the sampling distribution. The sampling distribution allows us to make probability statements about a sample statistic assuming knowledge of the population parameter. Thus, the probability of observing a sample mean at least as large as 178 from a population whose mean is 170 is .0069, which is very small. In other words, we have just observed an unlikely event, an event so unlikely that we seriously doubt the assumption that began the process, that the null hypothesis is true. Consequently, we have reason to reject the null hypothesis and support the alternative. Students may be tempted to simplify the interpretation by stating that the p-value is the probability that the null hypothesis is true. Don't! As was the case with interpreting the confidence interval estimator, you cannot make a probability statement about a parameter. It is not a random variable. The p-value of a test provides valuable information because it is a measure of the amount of statistical evidence that supports the alternative hypothesis. Values of x¯x¯ far above 170 tend to indicate that the alternative hypothesis is true. Thus, the smaller the p-value, the more the statistical evidence supports the alternative hypothesis. Next Page Describing the p-Value This raises the question, “How small does the p-value have to be to infer that the alternative hypothesis is true?” In general, the answer depends on a number of factors, including the costs of making Type I and Type II errors. A Type I error would occur if the manager adopts the new billing system when it is not cost-effective. If the cost of this error is high, we attempt to minimize its probability. In the rejection region method, we do so by setting the significance level quite low, say 1%. Using the p-value method, we would insist that the p-value be quite small, providing sufficient evidence to infer that the mean monthly account is greater than $170 before proceeding with the new billing system. Statistics practitioners can translate p-values using the following descriptive terms: If the p-value is less than .01, there is overwhelming evidence to infer that the alternative hypothesis is true. The test is highly significant. If the p-value lies between .01 and .05, there is strong evidence to infer that the alternative hypothesis is true. The result is deemed to be significant. If the p-value Next Page Interpreting the Results of a Test Consequently, if the value of the test statistic does not fall into the rejection region (or the p- value is large), rather than say we accept the null hypothesis (which implies that we're stating that the null hypothesis is true), we state that we do not reject the null hypothesis, and we conclude that not enough evidence exists to show that the alternative hypothesis is true. Although it may appear to be the case, we are not being overly technical. Your ability to set up tests of hypotheses properly and to interpret their results correctly very much depends on your understanding of this point. The point is that the conclusion is based on the alternative hypothesis. In the final analysis, there are only two possible conclusions of a test of hypothesis. If we reject the null hypothesis, we conclude that there is enough statistical evidence to infer that the alternative hypothesis is true. If we do not reject the null hypothesis, we conclude that there is not enough statistical evidence to infer that the alternative hypothesis is true. Observe that the alternative hypothesis is the focus of the conclusion. It represents what we are investigating. That is why it is also called the research hypothesis. Whatever you're trying to show statistically must be represented by the alternative hypothesis (bearing in mind that you have only three choices for the alternative hypothesis - the parameter is greater than, less than, or not equal to the value specified in the null hypothesis). Next Page SSA Envelope Plan Federal Express (FedEx) sends invoices to customers requesting payment within 30 days. The bill lists an address and customers are expected to use their own envelopes to return their payments. Currently the mean and standard deviation of the amount of time taken to pay bills are 24 days and 6 days, respectively. The chief financial officer (CFO) believes that including a stamped self-addressed (SSA) envelope would decrease the amount of time. She calculates that the improved cash flow from a 2-day decrease in the payment period would pay for the costs of the envelopes and stamps. Any further decrease in the payment period would generate a profit. To test her belief, she randomly selects 220 customers and includes a stamped self-addressed envelope with their invoices. The numbers of days until payment is received were recorded. Can the CFO conclude that the plan will be profitable? The objective of the study is to draw a conclusion about the mean payment period. Thus, the parameter to be tested is the population mean ?. We want to know whether there is enough statistical evidence to show that the population mean is less than 22 days. Thus, the alternative hypothesis is H1:μ<22H1:μ<22. The null hypothesis isH0:μ=22H0:μ=22. The test statistic is the only one we've presented thus far. It is: z=x¯−μσ/n−−√z=x¯−μσ/n Next Page SSA Envelope Plan To solve this problem manually, we need to define the rejection region, which requires us to specify a significance level. A 10% significance level is deemed to be appropriate. We wish to reject the null hypothesis in favor of the alternative only if the sample mean and hence the value of the test statistic is small enough. As a result, we locate the rejection region in the left tail of the sampling distribution. To understand why, remember that we're trying to decide whether there is enough statistical evidence to infer that the mean is less than 22 (which is the alternative hypothesis). If we observe a large sample mean (and hence a large value of z), do we want to reject the null hypothesis in favor of the alternative? The answer is an emphatic "no." It is illogical to think that if the sample mean is, say, 30, there is enough evidence to conclude that the mean payment period for all customers would be less than 22. Consequently, we want to reject the null hypothesis only if the sample mean (and hence the value of the test statistic z) is small. How small is small enough? The answer is determined by the significance level and the rejection region. Thus, we set up the rejection region as z<−zα=−z0.10=−1.28z<−zα=−z0.10=−1.28. Note that the direction of the inequality in the rejection region (z<−zα)(z<−zα)matches the direction of the inequality in the alternative hypothesis (μ<22)(μ<22). Also note that we use the negative sign, because the rejection region is in the left tail (containing values of z less than 0) of the sampling distribution. Next Page SSA Envelope Plan From the data, we compute the sum and the sample mean. They are: Σxi=4,759(the sum of all the payment periods for the SSA plan)x¯=Σxi220=4,759220=21.63Σxi=4,759(the sum of all the payment periods for the SSA plan)x¯=Σxi220=4,759220=21.63 We will assume that the standard deviation of the payment periods for the SSA plan is unchanged from its current value of σ=6σ=6. The sample size is n=220n=220, and the value of μμ is hypothesized to be 22. We compute the value of the test statistic as: z=x¯−μσ/n−−√=21.63−226/220−−−√=−0.91z=x¯−μσ/n=21.63−226/220=−0.91 Because the value of the test statistic, z = -0.91, is not less than -1.28, we do not reject the null hypothesis and we do not conclude that the alternative hypothesis is true. There is insufficient evidence to infer that the mean is less than 22 days. We can determine the p-value of the test as follows: p-value=P(Z<−.91)=0.5−0.3186=0.1814p-value=P(Z<−.91)=0.5−0.3186=0.1814 In this type of one-tail (left-tail) test of hypothesis, we calculate the p-value as P(Z<z)P(Z<z) where z is the actual value of the test statistic. Next Page SSA Envelope Plan The value of the test statistic is -0.91 and its p-value is 0.1814, a figure that does not allow us to reject the null hypothesis. Because we were not able to reject the null hypothesis, we say that there is not enough evidence to infer that the mean payment period is less than 22 days. Note that there was some evidence to indicate that the mean of the entire population of payment periods is less than 22 days. We did calculate the sample mean to be 21.63. However, to reject the null hypothesis, we need enough statistical evidence; and in this case, we simply did not have enough reason to reject the null hypothesis in favor of the alternative. In the absence of evidence to show that the mean payment period for all customers sent a stamped self-addressed envelope would be less than 22 days, we cannot infer that the plan would be profitable. A Type I error occurs when we conclude that the plan works when it actually does not. The cost of this mistake is not high. A Type II error occurs when we don't adopt the SSA envelope plan when it would reduce costs. The cost of this mistake can be high. As a consequence, we would like to minimize the probability of a Type II error. Thus, we chose a large value for the probability of a Type I error; we set α=0.10α=0.10.