BMAL_590_Quantitative_Research_Techniques_and_Statistics.doc

QUIZ Section 3 Bayes’ Law is used to compute . Posterior Probabilities The classical approach describes a probability . In terms of the proportion of times that an event can be theoretically expected to occur If a set of events includes all possible outcomes of an experiment these events are considered to be . Exhaustive Which statement is not correct? If event A does not occur, then its compliment A i will also not occur QUIZ Section 4- The concept 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 . The central limit theorem Each of the following are characteristics of the sampling distribution of the mean except . If the original population is not normally distributed, the sampling distribution of the mean will also be approximately normal for large sample sizes -Suppose you are given 3 numbers that relate to the number of people in a university sample. The three numbers are 10,20,30. If the standard deviation is 10, the standard error equals . 5.77 You are tasked with finding the standard deviation. You are given 4 numbers. Numbers are 5, 10, 15, and 20. The standard deviation equals. 6.455 Two methods exist to create a sampling distribution. Once involves using parallel samples from a population and the other is to use the . Rules of probability QUIZ Section 5 The hypothesis of most interest to the researcher is the . Alternative Hypothesis A Type I error occurs when . Reject a true null hypothesis Statisticians can translate p values into several descriptive terms. Suppose you typically reject H 0 at a level of .05. Which of the following statements is incorrect ? If the p value <.01, there is overwhelming evidence to infer that the alternative hypothesis is false. In a criminal trial where the null hypothesis states that the defendant is innocent a type I error is made when . An innocent person is found guilty To take advantage of the information of a test result using the rejection region method and make a better decision on the basis of the amount of statistical evidence we can analyze the. P Value Quiz Section 6 An unbiased estimator is . A sample Statistic, which has an expected value equal to the value of the population parameter

Descriptive Statistics • Descriptive Statistics - is one of two branches of statistics, which focuses on methods of organizing, summarizing, and presenting data in a convenient and informative way. o One form of descriptive statistics uses graphical techniques, which allow statistics practitioners to present data in ways that make it easy for the reader to extract useful information. ▪ Histogram (bar graph) can show if the data is evenly distributed across the range of values, if it falls symmetrically from a center peak (normal distribution), if there is a peak but more of the data falls to one side (skewed distribution), or if there are two or more peaks in the data (bi-or multi-modal) ▪ Numerical Techniques- rather than providing raw data the professor may only share summary data with the student. One such method used frequently calculates the average or mean • Measure of central location- the mean (average) is one such measure, it is the sum of all data values divided by the number of values • Range - the simplest measure of variability, is calculated by subtracting the smallest number from the largest. • Median - midpoint of the distribution where 50% of the data values are high and 50% are lower. (not that the mean and median will not necessarily be an observed test score). • Mode - the most frequently occurring data value • Variance- the average squared deviation to the mean. To compute the difference between each data value and the mean is calculated and squared. If differences are not squared sum will always be 0. • Standard deviation • - simply the square root of the variance and gets the variability measure back to the same units as the data • Negatively skewed if mean is to the left (point is to the right), positively skewed if the mean is to the right (point is to the left) Inferential Statistics • Inferential statistics is a body of methods used to draw conclusions or inferences about characteristics of population based on sample data o Example of inferential statistics is exit polling during elections o Practitioners can control the fraction of the size of the sample with between 90-99% Key Statistical Concepts • Statistical inference problems involve three concepts: o population - the group of all items of interest to a statistics practitioner. Frequently very large and may in fact be infinitely large. Does not necessarily refer to a group of people ▪ parameter - descriptive measure of a population, represents the information we need o sample – set of data drawn from the population o statistical inference- we use statistics to make inferences about parameters. Statistical inference is the process of making an estimate, prediction, or decision about a population based on sample data. ▪ Build in measure of reliability • Confidence level- proportion of times that an estimating procedure will be correct ▪ Significance level- measures how frequently the conclusion will be wrong in the long run • Statistic- a descriptive measure of a sample • Populations have parameters while samples have statistics

▪ Advantage is ability to make inferences within each stratum to compare strata (ex looking at lowest income group favors tax increase or compare highest and lowest income groups to determine whether they differ in support of tax increase) ▪ Stratifications must be done where the strata are mutually exclusive, meaning that each member of the population must be assigned exactly one stratum ▪ After population has been stratified, we use simple random sampling to generate complete sample o Cluster sampling ▪ Simple random sample of groups or clusters of elements versus a simple random sample of individual objects ▪ Useful when it is difficult or costly to develop a complete list of the population members, also useful when population elements are dispersed geographically ▪ Ex- randomly select block within a city to gather data from (rather than getting lists of households to use) ▪ Cluster sampling reduces costs ▪ Increased sampling error, as may have many similarities in those you sample Larger sample size usually means more accurate sample estimates Sampling Error • Two major types of error when sample is taken from a population: sampling error and non- sampling error • Sampling error - refers to the differences between the sample and the population that exists only because of the observation that happened to be selected for the sample. o Error that we expect to occur when we make a statement about a population that is based only on the observation contained in a sample taken from a population o Difference between the true (unknown) value of the population mean and its estimate, the sample mean, is the sampling error. Size of the deviation may be large simply due to bad luck that a particularly unrepresentative samples happened to be selected Non-Sampling Error • Non-sampling error- more serious than sampling error because taking a larger sample wont diminish the size or the possibility of occurrence of this error o Result from mistakes that are made in the acquisition of the data an from the sample observations being selected improperly o Three types of non-sampling errors: o 1-Data Acquisition errors- arise from the recording of incorrect responses. May be result of incorrect measurement taken because of faulty equipment, mistakes made during transcription from primary sources, inaccurate recording of data due to misinterpretation of terms or inaccurate responses to questions concerning sensitive issues o 2- Non-Response Error- refers to error or bias introduced when responses are not obtained from some members of the sample. When this happens sample observations may not be representative of the target population resulting in biased results. ▪ Response rate- the proportion of all people selected who complete the survey, key survey parameter and helps in understanding the validity of the survey and sources of non-response error o 3- Selection bias - occurs when the sampling plan is such that some members of the target population cannot possibly be selected for inclusion in the sample. Together with non- response error selection bias • When responses are not received from a sampled person bias is introduced QUIZ Section 2

• One of the objectives in calculating a conditional probability is to determine if the two events are related. In particular we would like to know if thy are independent events. • Two events are said to be independent if: P(A|B)=P(A) or P(B|A)=P(B) • Independent - two events are independent if the probability of one event is not affected by the occurrence of another event. • Ignore mutually exclusive combinations • In each combination in the example the two events are independent, in this type of problem when one combination is dependent all 4 will be dependent or visa versa. This rule does not play to any other situation. Union • Union is another combination of events, the Union of events A and B is the event that occurs when either A or B or both occur denoted as A ○ B • Ex. To determine that a randomly selected fund outperforms the market or the manager graduated from a top MBA program, we will need to compute the union of the two events. Union occurs when: o Fund out performs the market and the manager graduated from a top mba program o Fund outperforms the market and the manager did not graduate from a top mba program o Fund does not out perform the market and the manager graduated from a top mba program Complement Rule • Complement of event A is the event that occurs when event A does not occur. Complement of event A is denoted as A c . Event consisting of all sample points that are “not in A” o The compliment of the rule defined here derives from the fact that the probability of an event and the probability of the events complement must sum to 1 • Compliment rule is P(A c )=1-P(A) for any event A o Ex. Roll of die, probability the number “1” is rolled is 1/6, the probability that some other number than “1” will be rolled is 1-1/6=5/6 Multiplication Rule • Multiplication rule is used to calculate the join probability of two events. It is based on the formula for conditional probability defined earlier. o P(A|B)=P(A 𝗇 B)/P(B) o We derive the multiplication symbol my multiplying both sides by P(B) o Joint probability of ay two events is P(A 𝗇 B)= P(A)xP(B|A) o Ex. Course has 7 male and 3 female students, professor wants to select 2 students at random. o Probability that the 1 st student is female- P(A)= 3/10 = .30 o Probability after that event that the second student is female 2/9=.22 o To determine P(A and B) ▪ =(3/10)(2/9) which is =6/90, then = .067 i.e. there is 6.7% chance the professor will choose two female students from the graduate class 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 joint probability of two independent events is simply the product of the probability two events • Multiplication rule for independent events- P(A 𝗇 B)=P(A)xP(B) o Ex. Probability of choosing females to answer questions in classes (if different teachers) is (3/10)=9/10=.09 Addition Rule • Example two newspapers, Sun and Post 22% subscribe to Sun and 3% to post, 6% to both. • P(A 𝗇 B)= P(A)+P(B)- P(A 𝗇 B) o =.22+.35-.06

o Thus a randomly selected value of X (mean of the umber of spots observed in say 5 throws of dice), is likely to be closer to the mean value of 3.5 than is a randomly selected value X (number of spots observed in one throw) o AS the number of throws increases, probability of the sample mean will also increase o Thus we observe the sampling distribution of X becomes narrower or more concentrated about the mean, as sample size n increases o As n gets larger the sampling distribution of X becomes increasingly bell shaped. • Sampling distribution of the mean of random sample drawn from any population is approximately normal for a sufficiently large sample size. o The larger the sample size the more closely the sampling distribution of X will resemble a normal distribution. • Accuracy of the approximation alluded to in the central limit theorem depends on the probability distribution of the population and on the sample size. Sampling Distribution of the Sample Mean • Statisticians have shown that the mean of the sampling distribution is always equal to the mean of the population and that the standard error is equal to  / √ n for infinitely large populations. If the population is finite, the standard error is • Where N is the population size and √ N-n/N-1 is called the finite population correction factor. An analysis revealed that if the population size is large relative to the sample size, the finite population correction factor is close to 1 and can be ignored • As a rule of thumb, we treat any population that is at least 20x larger than the sample size as large. • In practice most applications involve populations that qualify as large because if the population is small it may be possible to investigate each member of the population and in doing so calculate the parameters precisely o As a consequence the finite population correlation factor is usually omitted • If x is normal, X (with line) is normal. If X is non-normal, X (with line) is approximately normal for sufficiently large sample sizes. The definition of sufficiently large depends on the extent of non- normality of X. Creating the Sampling Distribution Empirically • To create the sampling distribution empirically, we can actually toss the dice repeatedly, calculating the mean for each sample, counting the number of times each value of X occurs and computing the relative frequencies to estimate the theoretical probabilities. • Disadvantages are excessive amount of time Contents of a 32-oz bottle • Ex. Foreman at a bottling plant observed that the amount of soda in a 32oz bottle is actually normatively distributed random variable, with a mean of 32.2oz and a standard deviation of . 3oz o We want to find P(X>32) where X is normally distributed and  =32.2 and  =.3 o P(Z>.67)=1-.2514=.748 o There is about a 75% chance that a bottle of soda contains more than 32 oz Salaries of business school graduates • We want to fin the probability that the sample mean is less than $750 (earned per week for grad school grads)- P( X <750) • The distribution of X , the weekly income, is likely to be positively skeed but not sufficiently so to make the distribution of X non normal. As a result we may assume that X is normal with the mean  x =  800 and standard deviation is  x =20 • Thus =P(Z<-2.5) or =.5-.4938 which is =.0062 • The probability of observing a sample as low as $750 when the population mean is $800 is extremely small. Because the event is quite unlikely Using the Sampling Distribution for Inference

are almost always unknown because they represent descriptive measurements about extremely large populations. o Statistical inference addresses this problem, it does so by reversing the direction of the flow of knowledge. o We will assume that most population parameters are unknown. The statistics practitioner will sample from the population and compute the required statistic. The sampling distribution of the statistic will enable us to draw inferences about the parameter ▪ Sampling Distribution in inference: statistic Sampling distribution> parameter ▪ The sampling distribution of a statistic is created by repeated sampling form one population. We introduced the sampling distribution of the mean, the proportion, and the difference between the two means. We described how these distributions are created theoretically and empirically. QUIZ Section 4- The concept 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 . The central limit theorem Each of the following are characteristics of the sampling distribution of the mean except . If the original population is not normally distributed, the sampling distribution of the mean will also be approximately normal for large sample sizes -Suppose you are given 3 numbers that relate to the number of people in a university sample. The three numbers are 10,20,30. If the standard deviation is 10, the standard error equals . 5.77 You are tasked with finding the standard deviation. You are given 4 numbers. Numbers are 5, 10, 15, and 20. The standard deviation equals. 6.455 Two methods exist to create a sampling distribution. Once involves using parallel samples from a population and the other is to use the . Rules of probability Section 5 Introduction to hypothesis testing Concepts of Hypothesis testing • There are a variety of non statistical applications of hypothesis testing for example criminal trial. • IN a trial, the jury conducts a test of hypothesis, there are two hypothesis tested o Null hypothesis- represented by H 0 (H-nought- british for zero), it is H 0 if the defendant is innocent o Alternative or research hypothesis and is denoted by H 1 , in a criminal trial it is H 1 if the defendant is guilty.

o 1) rejection region method which can be used in conjunction with the computer, but is mandatory for those computing statistics manually o 2) p-value approach, which generally can be employed only in conjunction with a computer and statistical software • Rejection Region method o To make decision about close sample mean o Range of values such that if the test statistic falls into the range, we decide to reject the null hypothesis in favor of the alternative. o Produces a yes or no response to the question: “is there sufficient statistical evidence to infer that the alternative hypothesis is true?” ▪ Implication is that the result of the test will be converted automatically into one of two possible courses of action: • 1- action as result of rejecting the null hypothesis in favor of the alternative, • 2- result of not rejecting the null hypothesis in favor of the alternative. Rejection of the null hypothesis • seems to imply the new billing system will be installed. P Value of a test • Several drawbacks to the rejection region method, foremost the type of information provided by the result of the test. • P value provides what is needed to take full advantage of the information available from the test result and make much better decision to the amount of statistical evidence supporting the alternative hypothesis so that it can be weighed in relation to the other factors, especially financial ones. Interpreting the P value • To properly interpret results of an inferential procedure, must remember that the technique is based on the sampling distribution • Sampling distribution allows us to make probability statements about a sample statistic assuming knowledge of the population parameter o For the example above, 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 an unlikely event, so we seriously doubt the null hypothesis is true. Consequently we have a reason to reject the null hypothesis and support the alternative. o We cannot make a probability statement about a parameter as it is not a random variable/ • P value of a test provides valuable information because it measures the amount of statistical evidence that supports the alternative hypothesis. • The smaller the p value the more statistical evidence supports the alternative hypothesis Describing the P Value • How small does the p value have to be to infer that the alternative hypothesis is true? o Answer depends on the number of factors including costs of making type I and type II errors. o Type I error would occur if the manager adopts the new billing system when it is not cost effective. If cost error is high we would attempt to minimize its probability ▪ Rejection region method, we would do this by setting the significance level quite low say 1% ▪ P value method we would insist the p value be quite small, providing sufficient evidence to infer that the mean monthly account is greater than $170 before proceeding with 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.

o 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 o If the p value is between .05-.10 there is weak evidence to indicate that the alternative o hypothesis is true. Over .1 no evidence. The P Value and rejection region methods • We can choose to use the p value to make the same type of decisions we make with the rejection region method • Rejection region method requires the decision maker to select a significance level from which the rejection region is constructed • We then decide to reject or not reject the null hypothesis • Another way of making the decision is to compare the –value with the selected value of significance level • If the p value is less than 𝛼 we judge the p value to be small enough to reject the null hypothesis. If the p value is greater then 𝛼 we do not reject the null hypothesis. o Since P value=.0069< 𝛼 =.05 we reject H 0 in favor or H 1 • Use p value when computer is available, use rejection region when computing manually Interpreting the results of a test • In our example we rejected the null hypothesis. This does not prove that the alternative hypothesis is true because our conclusion is based on sample data, not the entire population. We can never prove anything by using statistical inference. • We summarize the test by saying “there is enough statistical evidence to infer that the null hypothesis is false and that the alternative hypothesis is true” • If the value does not fall into the rejection region ( p value is too large) rather we say we accept the null hypothesis, we say we do not reject the null hypothesis and we conclude that not enough evidence exists to show that the alternative hypothesis is true. • The conclusion is based on the alternative hypothesis. o If we reject the null hypothesis we conclude there is enough statistical evidence to infer that the alternative hypothesis is true o If we do not reject the null hypothesis we conclude there is not enough statistical evidence to infer that the alternative hypothesis is true • Alternative hypothesis is focus in conclusion, it represents what we are investigating o Why it is called a research hypothesis o You have 3 choices for alternative hypothesis the parameter is greater than, less than or not equal to the value specified in the null hypothesis. SSA Envelope Plan • Fed ex wants to improve their payment withing30 days. Current mean is 24 and standard deviation is 6 days. CFO thinks included stamped envelope will improve cash flow by a 2- day increase in payment period. • Objective is to draw a conclusion about the mean payment period. Parameter tested is the population mean. We want to know if there is enough statistical evidence to show the population mean I less than 22 days. o Alternative hypothesis is H 1 :  < 22 o Null Hypothesis is H 0 ;  = 22 • To solve manually we need to define the rejection region which requires us to specify a significance level. 10% is significance level 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. o As a result we locate the rejection region in the left tail of the sampling distribution o Remember we are trying to determine if there is enough statistical evidence to infer the mean is less than 22 ( alternative hypothesis)

o If we observe a large sample mean ( hence a large value of z) do we want to reject the null hypothesis in favor of the alternative? NO. ▪ It is illogical to think that if the sample mean is say 30 that there is enough evidence to conclude that the mean payment for all customers would be less than 22. ▪ We want to reject the null hypothesis only if the sample mean ( and value of test statistic z) is small. Which is determined by the significance level and rejecton region. ▪ Direction of inequality in the rejection region matched the direction of inequality in the alternative hypothesis. o Value of test statistic is -.91 and p value is .181 which does not allow us to reject the null hypothesis o Because we were not able to reject the null hypothesis we say there is not enough evidence to infer that the mean payment period is less than 22 days One tail test • The rejection region is located in only one tail of the sampling distribution • Right tail test – right side positive Two tail test • Two tail testing is used when we want to test a research hypothesis that a parameter is not equal to some value AT&T Example • Scenario: several companies have formed that offer competition against AT&T for long distance calls. Al advertise their rates are lower than AT&T, resulting in lower bills • AT&T responded arguing customers will see no difference in billing o Statistics practitioner for AT&T determined that mean is $17.09 and standard deviation is $3.87 o He takes 100 customers and recalculates their last month bills using rates quoted by leading competitors. o Assuming the same standard deviation of this population can we conclude at the 5% significance level that there is no difference between the average AT&T bill and the one of the leading competitor? o In this problem we want to know whether the mean monthly long distance bill is different from $17.09. o Consequently we set up the alternative hypothesis to express this condition H 1 :  ≠ 17.09 o So the null hypothesis is H 0 :  = 17.09 o Rejection region (left small tail or right small tail) is set up so we can reject the null hypothesis when the test statistic is large or when it is small ▪ Total area in the rejection region must sum to a 𝛼 , so we divide this probability by 2 ▪ At a 5% significance level (i.e. 𝛼 =.05), we have a 𝛼 /2 =.025 thus z .025 =1.96 ▪ Z<-1.96 or z>1.96 o Since z=1.19 is not great than 1.96 or less than -1.96 we cannot reject the null hypothesis in favor of H1. That is there is insufficient evidence to infer that there is a difference between the bills • We can also compute the p value of the test. Because it is a 2-ailed test, we determine the p value by finding the area in both tails o The p value= P(Z<-1.96 + P(>1.19)= .1170+.1170=.2340 o Or more simply multiple the probability in one tail by 2. In general the p value in a 2 tail test is determined by p value=2P(Z>|z|) where z is the actual value of the statistic and |z| is its absolute value o There is not enough evidence to infer that the mean long distance bill is different from AT&Ts mean of $17.09

When do we conduct two tail tests • A two tail test is conducted whenever the alternative hypothesis specifies that the mean is not equal to the value stated in the null hypothesis, that is when the hypothesis assume the following form o H 0 :  =  0 o H 1:  ≠  0 • There are two one-tail tests. We conduct a one tail test focused on the right tail of the sampling distribution whenever we want to know whether there is enough evidence to infer that the mean is greater than the quantity specified in the null hypothesis, that is when our hypothesis are: o H 0 :  =  0 o H 1:  >  0 • The second one tail test involves the left tail of the sampling distribution. It is used when the statistics practitioner wants to determine whether there is enough evidence to infer that the mean is less than the value of the mean stated in the null hypothesis o H 0 :  =  0 o H 1:  <  0 Testing Hypothesis and Confidence Interval Estimators • The test statistic and confidence internal estimator are both derived from the sampling distribution • We can use the confidence interval estimator to test hypothesis • LCL- lower confidence limit • UCL- Upper confidence limit • LCL- 16.79 and UCL is 18.31 • Because $17.09 lies between we cannot conclude that there is sufficient evidence to infer that the population mean differs from 17.09 • For the department store billing example, the 95% confidence level interval estimate is LCL=171.63 and UCL+ 184.37 o Since the interval estimate includes 170 allowing us to conclude that the population mean account is not equal to $170 • The confidence interval estimator can be used to conduct tests of hypothesis • This process is equivalent to the rejection region approach, however instead of finding the critical values of the rejection region and determining whether the test statistic falls into the rejection region, we compute the interval estimate and determine whether the hypothesizes value of the mean falls into the interval Advantage • Interval estimator to test hypothesis has the advantage of simplicity. Apparently we don’t need a formula for the test statistic we need only the interval estimator Disadvantage • When conducting a one tail test or conclusion may not answer the original question o Example in the department store billing example, we wanted to know whether there was enough evidence to infer that the mean is greater than 170. The estimate concludes the mean differ from 170. In attempting to draw the conclusion that the entire interval is greater than 170 there is enough statistical evidence to infer the population mean is greater than 170, however we run into the problem of determining the procedures significance level. Is it 5% or 2.5%? ▪ We may be able to overcome this problem with the use of one sided confidence level estimators. However if the purpose of using confidence interval estimators instead of test statistic is simplicity, one sided estimators are contradicted.

• Confidence interval estimator does not yield a p value which we have argued is better way t draw inferences about a parameter. Using the confidence interval estimator to test hypothesis forces decision maker into making a reject/don’t reject decision rather than providing information about how much statistical evidence exists to be judged with other factors in the decision process o Furthermore we only postpone the point in time when a test hypothesis must be used • As in the case with the confidence level estimator, the test of hypothesis is based on the sampling distribution of the sample statistic. The result of a test of hypothesis is probably a statement about the sample statistic. o We assume that the population mean is specified by the null hypothesis. o We then compute the test statistic and determine how likely it is to observe this large (or small) a value when the null hypothesis is true o If the probability is small we conclude that the assumption that the null hypothesis is true is unfounded and we reject it • When we (or the computer) calculate the value of the test statistic, were also measuring the difference between the sample statistic (x hat) and the hypothesized value of the parameter in terms of the standard error ( 𝛼 / √n ) o The unit of measurement of the difference is the standard error o In the AT&T example, we found the value of the test statistic was z=1.19. this means that the sample mean was 1.19 standard errors above the hypothesized value of  o The standard normal probability table told us that this value is not considered unlikely. As a result we did not reject the null hypothesis. Probability of a Type II Error • To properly interpret the results of a test hypothesis requires that you be able to specify an appropriate significance level or to judge the p value of a test • It also requires that you have an understanding of the relationship between Type I and Type II errors, that is how the probability of a Type II error is calculated and its interpretation o Recall in the dept store billing example, where we conducted a test using the sample mean as the test statistic and we computed the rejection region (with 𝛼 =.05) as x(hat)>175.34 • A type II error occurs when a false null hypothesis is not rejected. o In our example, if h(hat) is less than 175.34 we will not reject the null hypothesis. I we do not reject the null hypothesis, we will not install the new billing system o Thus the consequence of a type II error in this example is that we will not install the new system when it would be cost effective. The probability of this occurring is the probability of a type II error. It is defined as 𝗉 = P(X <175.34, given that the null hypothesis is false) • The condition that the null hypothesis is false tells us only that the mean is not equal to 170. o If we want to compute 𝗉 we need to specify the value for  ▪ Suppose that when the mean account is at least $180 the new billing system savings becomes so attractive the manager would have to make te mistake not installing it. As a result she would lik o determine the probability of not installing the system when it would produce large cost savings. Because calculating probability from an approximately normal sampling distribution requires a value of  (as well as 𝛼 and n) we will calculate the probability of not installing the new system when  is equal to 180: 𝗉 = P(X <175.34 given that  =180) • We know that x(hat) is approximately normally distributed with mean  and standard deviation ?/ √? , to proceed we standardize x(hat) and use the standard normal table • This tells us that when the mean account is actually $180 the probability of incorrectly not rejecting the null hypothesis is .0764

o Notice that to calculate the probability of a type II error we had to express the rejection region in terms of the unstandardized test statistic x(hat) and we had to specify a value for  other than the one shown in the null hypothesis. In this illustration the value of  was based on a financial analysis indicating that when  is at least $180 the cost savings would be very attractive. o Suppose that in the previous illustration we had used a significance level of 1% instead of 5%. The rejection region expressed in terms of the standardized test statistic would be • The probability of a type II error when  = 180 is P(Z<-.75)=.2266 o By decreasing th significance level from 5% to 1% we have shifted the critical value of the rejection region to the right and thus enlarged the area where the null hypothesis is not rejected. The probability of a tpe II error increases from .0764 to .2266 o this calculation illustrates the inverse relationship between the probabilities of Type I and Type II errors ▪ its important to understand this relationship form a practical point of view it tells us that if you wnt to decrease the probability of a Typ I error (by specifying a small value of 𝛼 ) you increase the probability of a type II error ▪ in applications where the cost of a type II error is relatively large, a significance level of 5% or more is appropriate. ▪ Unfortunately there is no simple formula to determine what the significance level should be. It is necessary for the manger to consider the costs of both mistakes in deciding what to do. Judgment and knowledge of the factors in the decision are crucial. Effecting of Changing the Type I 𝛼 and Type II errors 𝗉 Decreasing the significance level , increases the value of 𝗉 and vice versa • Shifting the critical value line to the right to decrease a will mean a larger area under the lower curve for B and vice versa Judging the Test • A statistical test of hypothesis is effectively defined by the significance level and the sample size, both of which are selected by the statistics practitioner • We can judge how well the test functions by calculating the probability of a type II error at some value of the parameter • To illustrate, in the dept store billing example, the manager chose a sample size of 400 and a significance level of 5% on which to base her decision. With those selections, er found B to be . 0764 when the actual mean is 180 o If we believe the type II error is high and this that the probability is too large, we have two ways to reduce the probability ▪ We can increase the value of a, however this would result in an increase chance of making a type I error, which is very costly ▪ Alternatively we can increase the sample size • For this example if sample size was taken up tp 1000, we reduce the probability of not installing the system wen the actual mean account is $180 to virtually 0. IE when we increase the sample size we reduce the probability of a type II error Developing an understanding of statistical concepts • The sampling distribution of the mean is narrower because the standard error of the mean 𝛼 / √n becomes smaller as n increases • Narrower distributions represent more information o Increased information is reflected in a smaller probability of a type II error • The calculation of the probability of a Type II error for n=400 and for n=100 illustrates a concept whose importance cannot be overstated.

o By increasing the sample size, we reduce the probability of a Type II error , by reducing the probability we make tis type of error less frequently. Hence longer sample sizes allow us to make better long term decisions. • This finding lies at the heart of applied statistical analysis and reinforces “statistics is a way to get information from data” o The more information the better decision o Without such information decisions must be based on guesswork, instinct and luck. o “Without data you’re just another person with an opinion”- Deming • Another way of expressing how well a test performs is to report its power: the probability of its leading us to reject the null hypothesis when its false. Thus the power f a test 1-B. • When more than one test can be performed in a given situation, we would naturally prefer to use the test that is correct more frequently. • If (given the same alternative hypothesis, sample size and significance level) one test has a higher power than a second test, the first test is said to be more powerful and the preferred test. Determining the Alternative Hypothesis to define Type I and Type II errors • Alternative hypothesis represents the condition were investigating. We wanted to know whether there was sufficient evidence to infer that the new billing system would be cost effective, that is, whether the mean monthly account is greater than $170. • In real life however the manger will be asking and answering the question. In general the question can be posed in two ways o We asked whether there was evidence to conclude that the new system would be cost effective o Another way of investigating the issue is to determine whether there is sufficient evidence to infer that the new system would not be cost effective. ▪ Remember in the criminal trial, the burden of proof falls on the prosecution to prove the defendant is guilty. In other countries with less emphasis on indiv rights, the defendant is required to prove his or her innocence • In a statistical test where we are responsible for asking the question and answering it, we must ask the question so that we directly control the error that is more costly. • We control the probability of a type I error by specifying its value (significance level). o There are two possible errors: conclude the billing system is cost effective when it isn’t and conclude that the system is not cost effective when it is Determining the Alternative Hypothesis to define Type I and Type II errors • Suppose we believe the cost of installing the new system that is not cost effective is higher than the potential loss of not installing an effective system • The error we wish to avoid is the erroneous conclusion that the system is cost effective. We define this as a type I error • As a result the burden of proof is placed on the system to deliver sufficient statistical evidence that the mean account is greater than $170 • However if we believe the potential loss of not installing the new system when it could be cost effective is a larger cost, we would place the burden of proof on the manager to infer that the mean monthly account is less than $170 • This discussion emphasizes the need in practice to examine the costs of making both types of error before setting up the hypothesis Derivations • The process that produces the formulas

o 1) factors determine which parameter were interested in (e.g.  ) o 2) each parameter has a “best” estimator (statistic) (e.g. xhat) o 3) the statistic has a sampling distribution (eg. Z=xhat-  / 𝛼 / √n) o 4) the formula representing the sampling distribution is often the formula for the test statistic o 5) with a little bit of algebra, the confidence interval estimator can be derived from the sampling distribution ▪ xhat +- z 𝛼 /2 𝛼 / √n QUIZ Section 5 The hypothesis of most interest to the researcher is the . Alternative Hypothesis A Type I error occurs when . Reject a true null hypothesis Statisticians can translate p values into several descriptive terms. Suppose you typically reject H 0 at a level of .05. Which of the following statements is incorrect ? If the p value <.01, there is overwhelming evidence to infer that the alternative hypothesis is false. In a criminal trial where the null hypothesis states that the defendant is innocent a type I error is made when . An innocent person is found guilty To take advantage of the information of a test result using the rejection region method and make a better decision on the basis of the amount of statistical evidence we can analyze the. P Value Section 6 Inference about a population Concepts of Hypothesis testing • Population mean  • Population proportion P Inference about a population mean when the standard deviation is unknown • If the population mean is known we use the confidence interval estimator and test statistic derived from the sampling distribution of the same mean with  known • If the population mean is unknown so is the population standard deviation. Consequently the previous sampling distribution cannot be used. o Instead we substitute the sample standard deviation s in place of the unknown population standard deviation  . The result is called a t statistic because that is what Gosset called it. T distributed when the sampled population is normal (Gosset published this under ne “student” hence the Student t distribution • When  is unknown we use its point estimator • The z statistic is replaced by the t statistic where the number of degrees of freedm is v is n-1 • Tables and calculations or the t-distributions are available online and in computer statistical programs • When the population standard deviation is unknown and the population is normal the test statistic for testing hypothesis about u is : o t= xhat-u/s/√n o which is stuent t distribution with v=n-1 degrees of freedom. The confidence interval estimator of u is given by ▪ xhat +- t a/2 8/√n Newspaper Recycling Plant Example • problem objective is to describe the population of the amount of newspaper discarded by each household in the population

• the data are quantitative, indicating that the parameter to be tested is the population mean. The analyst needs to determine whether the mean is greater than 2.0 lbs the alternative hypothesis is: o H 1 :  > 2.0 • Null hypothesis states the mean is equal to the value listed in the alternative hypothesis o H 0 :  = 2.0 • The manager believes the cost of the type I error (concluding that the mean is greater than 2 when it isn’t) is high. Consequently he sets the significance level at 1% Tax Collected from Audited Returns • IRS examined that .77% of 1,008,000 returns to determine if they were correctly done. To determine auditors performance, 209 of them were drawn and additional tax reported. • Problem objective is to describe the population of additional income tax, data are interval hence the population man u. question asks us to estimate the parameter o LCL- $5610, UCL- $6392 • When we introduced the student t statistic, we pointed out that the t statistic is student t distributed if the population from which we’ve sampled is normal. o Statisticians have shown the mathematical process derived from the student t distribution is robust which means if the population is non normal the results of the t test and confidence interval estimate are still valid provided the population us not extremely non- normal. o To check this requirement, we draw the histogram of the data and determine whether it is far from bell shaped. o If a histogram is extremely skewed ( say in exponential distribution), that could be considered extremely non-oral and hence the t-statistic would not be valid in this case. Estimating the tools of finite populations • Most populations are finite (infinite populations are usually the result of some endlessly repeatable process such as flipping a coin or selecting items with replacement) • When the population is small, we must adjust the test statistic and interval estimator using the finite population correction factor o However in large populations relative to sample size we can ignore the correction factor. Large populations are at least 20x the sample size • Finite populations allow us to use the confidence interval estimator of a mean to produce confidence interval estimator of the population total. o To estimate this total we multiply thr lower and upper confidence limits of of the estimate of the mean by the population size. • Example- 500 households in a city of 1 MM reveals a 95% confidence interval estimate that the household mean spent on Halloween candy lies between $20-30 o We can estimate a total amount spent in the city by multiplying the lower and upper confidence limits by the total population ▪ =1,000,000[20…30] ▪ thus we estimate the total amount spent on Halloween candy in the city lies between $20-30MM Developing an understanding of statistical concepts • Degrees of Freedom • Student t distribution is based on using the sample variance to estimate the unknown population variance • Sample variance (s 2 )- the difference between each data value and x-bar is squared, the squared difference s are summed and the total is divided by n-1 • To compute the sample variance, we must determine x-bar(hat). Recall that sampling distributions are derived by repeated sampling from the same population. To repeatedly take samples to

compute sample variance, we can choose any numbers for the firs n-1 observations in the sample. However we have no choice on the nth value, because the mean sample must be calculated first. o Ex. Suppose n=3 and x-bar=10. We can have x 1 and x 2 assume any values without restriction. However x 3 must be such that x-bar=10. o Ex- if x 1 = 6 and x 2 =8 then x 3 must equal 16 Therefore there are only 2 degrees of freedom (freedom to choose numbers) in our selection since we have to calculate x- bar o Notice the denominator in the calculation of sample variance is equal to the number of degrees of freedom and this is not a coincidence • The t-statistic like the z-statistic measures the difference between the sample mean x- bar and the hypothesized value of u in terms of the number of standard errors. o However when the population standard deviation is unknown we estimate the standard error by s/√n o When we introduce student t distribution, we pointed out that it is more widely spread out than the standard normal. The circumstance is logical the only variable the z statistic is the sample mean x-bar which will vary from sample to sample • The t statistic has two variables: the sample mean xbar and the sample standard deviation s, both of which vary from sample to sample. Because of the greater uncertainty, the t statistic will display greater variability • Factors that identify the t-test and estimator of u are: o 1- problem objective: describe a population o 2- data type: interval o 3- type of descriptive measurement: central location Inference about a population Variance • We presented the inferential methods about population mean where we were interested in acquiring information about the central location of the population. • As a result we tested and estimated the population mean • If we are interested in drawing inferences about a populations variability, the parameter we need to investigate is the population variance  2 o inference about the variance can be used to make decisions in a variety of problem because variance is a measure of the risk • example- operations management uses variance to ensure size, weight or volume of product is consistent • best estimator- has a sampling distribution from which we produce the test statistic and the interval estimator Statistic and sampling distribution • If we are interested in drawing inferences about a populations variability, the parameter we need to investigate is the population variance or  2 • The estimator of  2 is the sample variance, the sample variance (s 2 ) is unbiased, consistent and efficient point estimator of  2 • Statisticians have shows that the sum of squared deviations from the mean E(x i -xbar) 2 equal to (n- 1)s 2 divided by the population variance is chi-squared distribution with v=n-1 degrees of freedom provided that the sampled population is normal • Test statistic used to test hypothesis about  2 is o X 2 = (n-1) s2 /  2 or n-1 times the sample variance divided by the population variance o Which is called the chi-squared statistic (x 2 -statistic) distributed with v=n-1 degrees of freedom when the population random variable is normally distributed with variance equal to  2 . X 2 calculators and tables are available online and in stats calculators Consistency of a container-filling machine Part 1

• Example- container filling machines are used to package liquids. Amount of liquid should vary only slightly. • Company has developed a new type of machine that will fill 1 liter (1000 cubic centimeter) containers so consistently that the variance of the fills will be less than 1 cubic centimeter • To examine veracity of claim, a random sample of 25 1 liter fills was taken and results recorded. Do these data allow them to make this claim at the 5% significance level? • Problem objective is to describe the population of 1-liter fills from this machine. The data interval we are interested in the variability of the fills. It follows that the parameter of interest is the population variance  2 • Because we want to determine whether there is enough evidence to support the claim, o the alternative hypothesis is H 1 :  2 <1 o the null hypothesis is H 0 :  2 =1 • Value of the test statistic is 15.20 and the rejection region is 13.85. • Since 15.20 is not less than 13.85, we cannot reject the null hypothesis in favor of the alternative. There is not enough evidence to infer tat the claim is true. • As discussed, the result does not say the variance is equal to 1, merely states that we are unable to show that the variance is less than 1. • Consistency of a container-filling machine Part 2 • If we look at this example with a 99% confidence the variance of fills in the previous example we find that LCL is .3336 and UCL is 1.537. Part of this interval is above 1 which tells us that the variance may be larger than 1 confirming the conclusion we reached in the previous example. o We may be able to use the estimate to predict the percentage of overfilled and under filled bottles. This may allow us to choose between competing machines • Like the t-test and estimator of u, the chi-squared test and estimator of  2 theoretically require that the sample population be normal • In practice the technique is valid as long as the population is not extremely non-normal. We can gauge the extent of non-normality by drawing the histogram. We conclude that the normality requirement is not seriously violated. • Factors that identify the chi-squared test and estimator  2 of are: o 1- problem objective: describe a population o 2- data type: interval o 3- type of descriptive measurement: central location Inference about a Population Proportion • Look at problems describing a population, of populations of nominal data o Nominal data means that the population consists of nominal or categorical values ▪ Ex- brand preference survey, stats practitioner asks consumers of a particular product which brand they purchase. Values of the random variable are the brands. ▪ If there are 5 brands, values could be represented by their names, by letters, or by a number. When numbers are used should be understood that the numbers only represent the name of the brand and are arbitrarily designed and cant be treated as real numbers to calculate means and variances. o When data are nominal, all that we are permitted to do is describe the population or sample is count the number of occurrences of each value. ▪ For counts, we calculate proportions ▪ Thus the parameter of interest in describing a population of nominal data is the population proportion p ▪ Parameter was used to calculate probabilities based on the binomial experiment ▪ Binomial experiment – one characteristic is there are only two possible outcomes per trial

• Most practical inferences about p involve more than 2 outcomes. However in most cases were interested in only one outcome, which we label a success and other outcomes are labeled as failures. • Ex- in brand preference surveys, we are interested in our company’s brand • Ex- in political surveys we wish to estimate or test the proportion of voters who will vote for a particular candidate, likely the one who had paid for te survey. Statistic and Sampling Distribution • Logical statistic used to estimate and test the population proportion is the sample proportion defined as p hat= x/n where x is the number of successes in a sample and n is the sample size • The sampling distribution of p hat is approximately normal with mean p and standard deviation √p(1-p)/n provided that np and n(1-p) are greater than 5. We express this sampling distribution as: z= p hat –p/√p(1-p)/n • The formula summarizes the sampling distribution also represents the test statistic • The test statistic for p is z= p hat –p/√p(1-p)/n which is approximately normal for p and n(1- p) grater than 5. • We attempt to derive the confidence interval estimator of p from the sampling distribution • The result is: p hat +- z 𝛼 /2 √p(1-p)/n • This formula although technically correct is useless. To understand why, examine the standard error of the sampling distribution √p(1-p)/n o To produce the interval estimate we must compute the standard error, which requires us to know the value of p (the parameter we wish to estimate) o This is the first of several statistical techniques where we face the same problem, how to determine the value of the standard error. o In this application the problem is easily and logically solved ▪ Simply estimate the value of p with p hat thus e estimate the standard error with √p hat(1-p hat)/n • The confidence level estimator of p is p hat +- z 𝛼 /2 √p hat(1-p hat)/n • Which is valid provided that np hat and n(1-p hat) are greater than 5 Election day exit poll • Tv stations predict winners in election through exit polls, where a random sample of voters who exit the polling booth are asked for whom they voted. o From the data, the sample proportion of voters supporting the candidates is computed. A statistical technique is applied to determine where there is enough evidence to infer that e leading candidate will garner enough votes to win. • EX- suppose from state election pollsters recorded only the votes of the 2 candidates who had any chance of winning, Dem candidate (code 1) Rep Candidate (code 2). Polls close at 8pm • Can the network conclude from the data that the republican candidate will win? Should the network announce at 8:01 the candidate will win? • The problem objective is to describe the population of votes in the state • The data are nominal because the values are democrat or republican • Thus the parameter to be tested is the proportion of votes in the entire state that are for the republican candidate • Because we want to determine whether the network can declare the republican the winner at 8:01, the alternative hypothesis is H 1 : p>.5 which makes the null hypothesis H 0 : p=.5 • Appears that this is a standard problem which requires a 5% significance level. Thus the rejection region is z> z 𝛼 =z .05 = 1.645

• From the file we count the number of successes (number of votes cast for republican) and find x=47 sample size is 765 hence the proportion is o P hat= x/n or 407/765=.532 o The value of the test statistic is z= phat-p/ √p(1-p)/n = .532-.5/√.5(1-.5)/765 = 1.77 o Since the test statistic is approximately normally distributed, we can determine the p value. ▪ Pvalue= P(z>1.77)=1-p(Z<1.77)=1-.9616=.384 ▪ There is enough evidence at the 5% significance level that the republican candidate has won. The value of the test statistic z=1.77 and the ne teail p value=.0382. ▪ Using a 5% significance level we reject the null hypothesis and conclude there is enough evidence to infer that the candidate won the election. Is this the right decision? Exit Poll • One of the key issues to consider is the cost of Type I and Type II errors. • Type I errors occur if we conclude that the republican will when when in fact he has lost. Such an error would mean the network would announce at 8:01 the wrong winner and have to later admit the mistake. If only one error made this mistake it would cast doubt on their integrity and possibly affect their number of viewers o This happened in 2000 when all networks declared al gore had won Florida. A couple hours later they had to admit the mistake that Bush had won. o Considering the cost of Type I and II errors it would have been better to use at 1% significance level Missing Data • In real statistical applications we occasionally fund that the data set is incomplete • In some instances, the statistics practitioner may have failed to properly record some observations or some data may have been lost • In other cases respondents may refuse to answer o EX- political surveys asking who someone intends to vote for, some may refuse to answer or say they haven’t decided. o If nonresponses are high the results of our analysis may be invalid because the sample is no longer truly random. o Ex. – suppose people in the top quarter of household incomes regularly refuse to answer questions about their incomes. The resulting estimate of the population household income mean will be lower than the actual value. • There are several ways to compensate for non-responses. o The simplest method is to eliminate them o Surveyors record results for 1-candidate A, 2- Candidate B, 3- don’t know, 4-refuse to say o If we wish to infer something about the undecided voters we can simply omit codes 3- 4 when analyzing the data o Missing data are the nonresponses we wish to eliminate collectively Estimating Totals for Large populations • As was the case with the inference about a mean, the techniques in this section assume infinitely large populations o When the populations are small it is necessary to include the finite population correction factor • A population is small when it is less than 20x the sample size

o When a population is large and finite we can estimate the total number of successes in the population by taking the product of the size of the population (N) and the confidence interval estimator o The produce the confidence interval estimator of the total, we multiply the lower and upper confidence limits of the interval estimator of the proportion of successes by the population size. The confidence interval estimator of the total number of successes in a large finite population is N[p hat +- z 𝛼 /2 √p hat(1-p hat)/n] o The Nieslen ratings (used to measure tv audiences) use this technique. Results from a small sample audience (5k viewers) is extrapolated to the total number of tv households 110M Nielsen Ratings • Statistical techniques play a vital role in helping advertisers determine how many viewers watch the shows that they sponsor o Nielsen ratings are based on random sample of approximately 5000 of the 110M households in the US watching tv o A meter attached to the televisions in selected households keeps track f when the tvs are turned on and what channels they watch o Problem objective is t describe the population of tv shows watched by viewers in the country, data are nominal o Combination of problem objective and data type make the parameter to be estimated the proportion of the entire population that watched deal or no deal (code 3) Selecting the Sample Size • When we introduce the sample size selection method to estimate a mean, we pointed out that the sample size depends on the confidence level and the bound on the error estimation that the statistics practitioner is willing to tolerate • When the parameter to be estimated is a proportion, the bound on the error of estimation is o B= z 𝛼 /2 √p hat(1-p hat)/n] • Solving for n we produce the required sample size to estimate p and where B is the bound on the error of estimation o N=( z 𝛼 /2 √p hat(1-p hat)/B) 2 • Unfortunately we do not know the value of p hat. To illustrate the use of this formula, suppose that in a brand preference survey, we want to estimate the proportion of consumers who prefer our company’s brand to within .03 with 95% confidence. This means that bound on the error of estimation is B=.03 Since: o 1- 𝛼 =.95, 𝛼 =.05, 𝛼 /2=.025, and z 𝛼 /2 = z.025= 1.96 o to solve for n we need to know p hat, unfortunately this value is unknown because the sample has not yet been taken ▪ at this point we can use either of the two methods to solve for n • Method 1: o If we have no knowledge of event thru appropriate value of p hat, we let p hat=.5. we choose p hat=.5 because the product of phat(1-phat) equals its maximum value at phat=.5 ▪ This in turn results in a conservative value of n, and as a result the confidence interval will be no wider than the interval p hat +-.03. If when the sample is drawn, p

hat does not equal .5 the confidence interval estimate will be better (that is narrower) than planned • N=(1.96 √(.5)(.5)/.03) 2 = (32.67)2=1,068 ▪ It turns out that p hate =.05 the interval estimate is p hat +-.03 if not the interval estimate will be narrower. For instance it turns out that p hat=.2, the estimate is phat +-.024 which is better than we had planned o Often used to determine the sample size use in public opinion surveys reported by newspapers, magazines tv etc. These polls usually estimate proportions to within 3% with 95% confidence (media often say with confidence level of 19 times out of 20) o Polls use proportion within 3% because of the size required to estimate to within 1% ▪ Increased sample size to get to 1% would be very costly, increased confidence does not usually overcome increased costs ▪ Confidence interval estimates with 5% or 10% bounds are generally considered too wide to be useful. Thus 3% provides a reasonable compromise • Method 2: o If we have some idea about the value of p hat we can use the quantity to determine n ▪ For ex- if we believe that p hat will turn out to be approximately .2 we can solve for n as follows • N=(1.96 √((.2)(.8)/.03) 2 = (26.13) 2 = 683 • Notice that this produces a smaller value of n (thus reducing the sampling costs) tan does method 1. If p hat actually lies between .2 and .8 howeve the estimate will not be as good as we wanted bc the interval will be wider than desired Wilson Estimators • When applying confidence interval estimator of a proportion when success is a relatively rare vent, it is possible to find no successes. Especially if the sample size is small • This implies that if we find no successes in the sample then there is no chance of finding a success in the population. Drawing such a conclusion from virtually any sample size is unacceptable o Remedy may be a suggestion made by Edwin Wilson. o Wilson estimate denoted that p~ (p tilde which is above p) is computed by adding 2 to te number of successes in the sample and 4 to the sample size ▪ P~= x+2/n+4 ▪ Standard error or p~ is 𝛼 p~ √p~(1-p~)/n+4 o The confidence interval estimator of p using the Wilson estimate is ▪ P~+- z 𝛼 /2 √p~(1-p~)/n+4 o Factors that identify the z-test and interval estimator of p are: ▪ 1) problem objective: describe a population ▪ 2) data type: nominal o Required conditions for z-test and interval estimator of p are ▪ Np> (or equal to) 5 and n(1-p)>(or equal to) 5 for test ▪ Np hat> (or equal to) 5 and n(1-p hat)>(or equal to) 5 for estimate Quiz Section 6 An unbiased estimator is . A sample Statistic, which has an expected value equal to the value of the population parameter

Thirty-six months were randomly sampled and the discount rate on new issues of 91-day Treasury Bills was collected. The sample mean is 4.76% and the standard deviation is 171.21. What is the unbiased estimate for the mean of the population? 4.76% a 98% confidence interval estimate for a population mean is determined to be 75.38 to 86.52. If the confidence level is reduced to 90%, the confidence interval for the population mean Becomes Narrower Suppose the population of blue whales is 8,000. Researchers are able to garnish a sample of oceanic movements from 100 blue whales from within this population. Thus Researchers can ignore the finite population correction factor In the sample proportion, represented by p=x/n the variable x refers to: The number of succeses in the sample Section 7- Analysis of Variance Analysis of Variance • The technique presented in this section allows statistics practitioners to compare two or more populations interval of data • Analysis of variance- determines whether differences exist between population means o Ironically the procedure works by analyzing the sample variance o Extremely powerful and widely used procedure • First used in 1920 to determine whether treatments of fertilizer produced different crop yields. Terminology form this first experiment is still used. No matter what experiment, the procedure is designed to determine whether there are significant difference between the treatment means • The analysis of variance is a procedure that tests to determine whether differences exist between two or more population means • The name of the technique derives from the way the calculations are performed. That is the technique analyzes the variance of the data to determine whether we can infer that the population means differ o Experimental design is a determinant in identifying proper method of to use • One way analysis of variance is used to describe the procedure to apply when the samples are independently drawn One-way Analysis of Variance • Independent samples are drawn from K populations • These populations are referred to as treatments. It is not a requirement that n 1 =n 2 …=n k • The mean and variance of population j(j=1,2,….k) are labeled  j and  2j o Both parameters are unknown o For each population we draw independent random samples for each sample we an compute the mean: x hat j and the variance s 2j

o X- response variable and its values are responses o X ij – refers to the i th observation in the j th sample size (ex x 35 is the third observation of the 5 th sample) o The grand mean x double bar is the mean of all the observations where the sum of all observations is divided by the total of all observations from all samples o Population classification criterion is called a factor ▪ Each population is a factor level Portion of total assets invested in stocks • Percentage of total assets invested in the stock market is the response variable • Parameters are the 4 population means • Analysis of variance determines whether there is enough statistical evidence to show that the null hypothesis is false. Test statistic • Test statistic measures the proximity of the sample mean to each other would also be of interest • Test statistic is computed in accordance with the following rationale o If the null hypothesis is true, the population means would all be equal o We would then expect the sample means would be close to one another o If the alternative hypothesis is true however, there would be large differences between some of the sample means o Statistic that measures the proximity of the sample means to each other is called the between treatments variation, denoted SST which stands for Sum of Squares Treatment o Or the difference between each sample mean and the grand mean is calculated, squared and multiplied by the sample size. The value for all samples are then summed o Large SST indicates large variation between sample means which supports the alternative hypothesis. • SST gave us the between treatments variation • Within Treatments Variation- measured by SSE the Sum of Squares for Error o SSE can be calculated by squaring the difference between each observation and its corresponding sample mean and summing the total. • Alternatively each sample variance is multiplied by the degrees of freedom (n j -1) and the results from each sample are summed • It is easier to see that SSE provides a measure of the amount of variation we can expect from the random variable we’ve observed • As you can deduce the formula, the sample means are close to each other all of the sample means would be close to the grand mean and as a result SST would be small o SST achieves its smallest value (zero) when all the sample means are equal. • It follows that a small value of SST supports the null hypothesis, H 0 :  1 =  2 =  3 =  4 . o A large value of SST supports the alternative hypothesis • If large differences between the samples means, at least some sample means differ considerably from the grand mean, producing a large value of SST • it is then reasonable to reject the null hypothesis in favor of the alternative hypothesis • key question is: “how large does the statistic have to be for us to justify rejecting the null hypothesis?” • To answer this in our example, we need to know how much the variation exists in the percentage of assets, which is measured y the within treatments variation which is denoted by SSE (sum of squares error) • The within treatments variation provides a measure of the amount of variation in the response variable that is not caused by treatments • All of the variables (along with others we are unable to identify) are sources of variation which we would group together and call it an error.

o The source of variation is measured by the sum of squares for error Mean Squares • Next step is to compute quantities called mean squares o Mean square for treatments (MST) is computed by dividing SST by the number of treatments minus 1. o the mean square for error (MSE) is determined by diving SSE by the total sample size (labeled n) minus the number of treatments o finally the test statistic, F, is defined as the ratio of the two mean squares MST/MSE • The test Statistic is F-distributed with k-1 and n-k degrees of freedom provided that the response variable is normally distributed. o The ratio F=MST/MSE is the ratio of the two samples variances o The degrees of freedom for this application are the denominators in the mean squares that is v 1 = k-1 and v 2 = n-k ▪ The degrees of freedom are: • v 1 = k-1=4-1=3 • v 2 = n-k = 366-4=362 o Purpose of calculating the f-statistic is to determine whether the values of SST is large enough to reject the null hypothesis ▪ If sst is large, f will be large hence we reject the null hypothesis only if : • F>F 𝛼 ,k-1,n-k • Tables and calculators for the F distributions are available online and in computer statistical programs • We found value of F to be 2.79. Thus there is enough evidence to infer that the mean percentage of total assets invested in the stock market differs between the 4 age groups. • The p value of the test is P(F>2.79) a computer is required to calculate the value which is .0405 o The result of the analysis of variance is usually reported in an analysis of variance (ANOVA) table • Terminology in the ANOVA table is based on the partitioning of the sum of squares o Such partitioning is derived from the following equation, whose validity can be demonstrated using the rules of summation • Total variation of all the data is denoted SS (total) if we divide SS (Total) by the total sample size minus 1 (that is, by n-1) we could obtain the sample variance (assuming the null hypothesis is true) • First term on the right of the equal sign is SST the second term is SSE o Total variation SS (total) is partitioned into two sources of variation ▪ The sum of squares for treatments (SST) is the variation attributed to the differences between the treatment means ▪ Whereas the sum of squares for error SSE measures the variation within the samples o SS(Total)= SST+SSE • The test is then based on the comparison of the mean squares of SST and SSE • In the analysis of variation the sum of squares for treatments explains the variation attributed to the treatments (age categories) • The sum of squares for error measures the amount of variation that is unexplained by the different treatments • If SST explains a significant portion of the total variation, we conclude that the population means differ • The value of the test statistic is F=2.79 and its p value is .0405 which means there is evidence to infer that the percentage of total assets invested in stocks are different in at least two of the age categories

• We reject the null hypothesis in favor of the alternative hypothesis o Not that in this example the data are observational o We cannot conduct a controlled experiment • When data are obtained through a controlled experiment in the one way analysis of variance, we call the experiment design completely randomized design of the analysis of variance. Checking the required conditions • F test of the analysis of variance requires that the random variable be normally distributed with equal variances o The normality requirement is easily checked graphically by producing histograms for each sample • The equality of variances is examined by printing the sample standard deviations or variances o The similarity of the sample variances allows us to assume that the population variances are equal • If the data are not normally distributed, we can replace the one-way analysis of variance with its nonparametric counterpart, which is the Kruskal Wallis Test • If the population variances are unequal, we can use several methods to correct the problem ANOVA and t-Tests of 2 means • T test measures whether there is evidence of a difference between two population means o We don’t want to use multiple t tests because many additional calculations would be needed and we increase the probability of making a type I error. ▪ We could decrease the significance level to account for this, however then we increase the probability of a type II error. ▪ Performing multiple t-tests increases the likelihood of making mistakes • When we want to compare more than two populations of interval data, e use the analysis of variance • Factors that identify the one-way analysis of variance o 1) problem objective: compare two or more populations o 2) Data type: interval o 3) Experimental design: independent samples o required conditions: populations are normal with equal variances, if they are non normal apply Kruskal-Wallis test Multiple Comparisons • when we conclude from the one-way analysis of variance that at least two treatment means differ, we often need to know which treatment means are responsible for these differences • Multiple comparisons – the technique used to determine if the sample means are different from the population means • When we conclude fro the one-way analysis of variance that at least two treatment means differ (i.e. we reject the null hypothesis) we often need to know which treatment means are responsible for these differences • Three statistical inference procedures that allow us to determine which population means differ: o 1) Fishers least significant difference (LSD) o 2) Bonferroni adjustment o 3) Tukey’s multiple comparison method • all three apply to one-way experiment only • Two means are considered different if the absolute difference between the corresponding sample means is larger than the critical number • The larger sample mean is then believed to be associated with a larger population mean

• There are several statistical inference procedures that deal with this problem Fisher’s Least Significant Difference (LSD) • The critical number is the confidence interval estimator of u 1 -u 2 • If the interval excludes 0 we can conclude that the population means differ. Conduct a two-tail test to determine • However, a better estimator of pooled variances is MSE, Mean for Squared Error, this allows us to compare the difference between means to the Least Significant Difference LSD • LSD will be the same for all pairs of means if all k sample sizes are equal. If some sample sizes differ, LSD must be calculated for each combination Cost of repairing Car Bumpers • Manufacturer is considering several new bumpers • To test how well they react to low speed collisions, 10 bumpers of each of 4 different types were installed on mid-size cars which were driven into a wall at 5MPH o Cost of repairing damage in each case was assessed/ • Problem objective is to compare 4 populations, the data are interval and the samples are independent • The correct statistical method is the one way analysis of variance • A simple way of determine whether differences exist between each pair of population means is to compare the absolute value of the difference between their two sample means and LSD • We calculate the absolute value of the differences between means and compare them to LSD Fisher’s Least Significant Difference • LSD will be the same for all pairs of means if all k sample sizes are equal, if some sample sizes differ LSD must be calculated for each combination. • Fisher’s method is flawed because it will increase the probability of committing • a type I error. That is, it is more likely than the analysis of variance to conclude that a difference exists in some of the population when in fact none differ o We calculated in our example that is k=6 and all population means are equal, the probability of erroneously inferring at 5% significance level that at least two means differ was about 54%. The 5% figure is now referred to as the comparison-wise Type I error rte ▪ The true probability of making at least one Type I error is called the experiment wise Type I error rate denoted 𝛼 E • Experiment-wise Type I error rate can be calculated as 𝛼 E = 1-(1- 𝛼 ) C • Here C is the number of pairwise comparisons which can be calculated by C=k(k-1)/2 o 𝛼 E ≤ C 𝛼 • Which means that if we want the probability of aking at least one Type I error to be no more than 𝛼 E we simply specify 𝛼 = 𝛼 E /C • The resulting procedure is called the Bonferroni Adjustment Bonferroni adjustment to LSD Method • The adjustments made by dividing the specified experiment-wise Type I error rate by the number of combinations of pairs of population means • If we perform the LSD procedure with the Bonferroni adjustment, the number of pairwise comparisons is 6 • Now no pair of means differ since all the absolute values of the differences between sample means are less than 139.19 • Drawback to LSD procedure: o We increase the probability of at least one Type I error o Bonferroni adjustment corrects this problem, however the probabilities of Type I and Type II errors are inversely related. Tukey’s Multiple Comparison Method

• A more powerful test is Tukey’s multiple comparison method o Tis technique determines a critical number similar to LSD for Fishers test denoted by 𝜔 (green letter omega), such that if any pair of sample means has a difference greater than 𝜔 , we conclude that the pair’s two corresponding population means are different. o The test is based on the studentized range , which is defined as a variable. • Theoretically this procedure requires that all sample sizes be equal. However if the sample sizes are different, we can still use this technique provided that the sample sizes are at least similar • The value of n g used is the harmonic mean of the sample sizes Which Multiple Comparison Method to Use? • There are two absolute values larger than 133.45 hence we conclude that u 2 , u 4 , u 3 and u 4 differ. The other four pairs do not differ • Using Bonferronis adjustment of Fisher’s LSD method we discover that none of the bumpers differ. • Tukey’s method tells us that bumper 4 differs from both bumpers 2 and 3 o Based on the sample bumper 4 appears to be the lowest cost of repair o Because there was not enough evidence to conclude that bumpers 1 and 4 differ, we would consider using bumper 1 if there are advantages that it has over bumper 4 • No one procedure works best with all types of problems. • Most statisticians agree to the following guidelines: o If you have identified two or three pairwise comparisons that you wish to make before conducting the analysis of variance, use the Bonferroni with C=2 o If you plan to compare all possible combinations use Tukey o If the purpose of Analysis is to point to areas that should be investigated further, Fisher’s LSD is indicated o To employ Fisher’s LSD or the Bonferroni adjustment you must first perform the analysis of variance o Tukey’s method can be employed instead of the analysis of variance Analysis of Variance Experimental Designs • The experimental Design- has been one of the factors that determines which technique to use • Statistic practitioners often design experiments to help extract the information they need t assist them in making decisions • The one-way analysis of variance is only one of many different experimental designs of the analysis of variance • For each type of experiment, we can describe the behavior of the response variable using a mathematical expression or model • The criterion by which we identify populations is called a factor. • The experiment described in the proportion of total assets invested stock example is a single factor analysis of variance because it addresses the problem of comparing two or more populations defined on the basis of only one factor o The proportion of total assets invested in stock example, is a single factor design because we had one treatment, age of the had of household. That is the factor is the age and the 4 age categories were the levels of this factor. o For another study if we added gender of the head of household, we would then develop a two-factor analysis of variance where the first factor age has 4 levels and the second factor gender has 2 levels • Multifactor experiment- is one where there are two or more factors that define the treatments Independent Samples and Blocks • When the problem objective is to compute more than two populations, the experimental design that is the counterpart of the matched pairs experiment is called the randomized block design • Randomized block design experiment reduces variation within the samples, making it easier to detect differences between populations

o The term block refers to a matched group of observations from each population • We can also perform a blocked experiment by using the same subject (person, plant or store) for each treatment. o EX- we can determine whether sleeping pills are effective by giving 3 brands of pills to the same group of people to measure the effects o Such experiments are called repeated measures designs • Technically this is a different design than the randomized block, however the data are analyzed in the same way for both designs o Hence we will treat repeated measures as randomized block designs • The randomized block experiment is also called the two-way analysis of variance, not to be confused with the two-factor analysis of variance Fixed and Random Effects • If our analysis includes all possible levels of a factor, the technique is called a fixed- effects analysis of variance • If the levels included in the study represent a random sample of all the levels that exist, the technique is called a random-effects analysis of variance o In the cost of car bumpers example, there were only 4 possible bumpers. Consequently the study is a fixed effects experiment ▪ However if there were other bumpers besides the 4 described, and we wanted to know whether there were different repair costs between all bumpers, the application would be a random effects experiment. • Ex- looking at if there is a difference in the number of units produced by a machine at a large factory, 4 machines out of 50 in the plant are randomly selected for a study. o The number of units each produces per day for 10 days will be recorded o The experiment is a random-effects experiment because we selected a random sample of 4 machines and therefore the statistical results will allow us to determine whether there are differences between the 50 machines • In some experimental designs, there are no differences in calculations of the test statistic between fixed and random effects, however in others the calculations are different Randomized Block Analysis of Variance • The primary interest in designing a randomized block experiment is to reduce the within treatments variation to more easily detect differences among the treatment means • In the one-way analysis of variance, we partitioned the total variation into the between treatments and the within treatments variation • In the randomized block design of the analysis of variance , we partition the total variation into 3 sources of variation: o SS(Total)= SST+ SSB+SSE o SSB the sum of square blocks measures the variation between the blocks o When the variation associated with the blocks is removed, SSE is reduced, making it easier to determine whether differences exist between the treatment means • In addition to k treatments, we introduce notation for b blocks into our experimental design Sum of Squares: Randomized Block • The definition of SS(total) and SST in the randomized block design are identical to those in independent samples design. • SSE in the independent samples design is equal to the sum og SSB and SSE in the randomized block design. Squaring the distance fro the grand mean, leads to the following set or formulae • This test is conducted by determining the mean squares which are computed by dividing the ums of squares by their respective degrees of freedom

• Can summarize information in an ANOVA table for randomized bloc analysis of variance as we can with one way experiment Comparing Cholesterol-Lowering Drugs • Pharm company has introduced 4 drugs to lower cholesterol. To determine whether any differences exits in their benefits an experiment was organized. o Company selected 25 groups of 4 men who had cholesterol over 280. In each group men were matched according to age and weight. Drugs were administered over 2 moth period, and cholesterol reduction was recorded o Problem objective is to compare 4 populations and the data ate interval ▪ Because researchers recorded the cholesterol reduction for each drug for each member of similar groups of men we identify the experimental design as a randomized block ▪ Response variable is the cholesterol reduction, the treatments are the drugs and the blocks are the similar groups of men ▪ Null hypothesis H 0 : u 1 =u 2 =u 3 =u 4 ▪ alternative hypothesis H 1 : at least two means differ o Each of the 4 drugs can be considered a treatment ▪ Each group can be blocked, because they are matched by age and weight ▪ By setting up the experiment this way, we eliminate variability in cholesterol reduction related to different combinations of age and weight ▪ This helps detect differences I the mean cholesterol reduction attributed to the different drugs ▪ A type I error occurs when you conclude that differences exist when, in fact, they do not. ▪ A type II error is committed when the test reveals no difference when at least two means actually differ ▪ Both errors are equally costly, accordingly we just the p value against the standard of 5% ▪ Because the p value is .0094, there is sufficient evidence to infer that t least two of the drugs differ. An examination reveals that cholesterol reduction is greatest using drugs 2 & 4 ▪ Checking the required conditions, the F-test of randomized block design of the analysis of variance has the same requirements as the independent samples design. • That is the random variable must be normally distributed and the population variances must be equal • Histogram appear to support the validity of results, the reductions appear to be normal • Equality of variances also appear to be met ▪ Violation of the required conditions: when the response is not normally distributed, we can replace the randomized block analysis of variance with the Friedman test Criteria for Blocking • The purpose of blocking is to reduce the variation caused between the experimental units. By grouping experimental units into homogenous bocks with respect to the response variable. o The statistics practitioner increases chances of detecting actual differences between the treatment means o Hence we need to find criteria for blocking that significantly affect the response variable. ▪ Ex- suppose that a stats professor wnts to determine which of the 4 methods of teaching stats is best

• In a one-way experiment, he might take 4 samples of 10 students, teac each sample by a different method, frade the students at the end of the course and perform an F-test to determine whether differences exist. o However it is likely that there are very large differences between the students at the end of the course may hide the differences between classes o To reduce this variation, the stats professor must identify variables that are linked to a students grade in stats ▪ For example, overall ability of students, completion of math courses, exposure to other stats courses ▪ The experiment could be performed in the following way: ▪ Stats prof. selects 4 students at random whose average grade before stats is 95-100, he randomly assigns the students to 1 of 4 classes, he repeats the process with students whose average is 90-95, 85-90,….,50-5, final grades would be used to test for difference between the classes ▪ Any characteristics that are related to the experimental units are potential blocking criteria ▪ If experimental units are people, we may block by age, gender, income, work experience, intelligence, residence, etc. ▪ If experimental unit is a factory we measuring number of units produced hourly blocking criteria would include workforce experience, age of ant and quality of suppliers Developing and Understanding of Statistical Concepts • As we explained previously, randomized block experiment is an extension of the matched pairs experiment • In the matched pairs experiment, we simply remove the effect of the variation caused by differences between the experimental units. o Effects of this removal is seen in the decrease in the value of the standard error (compared to the standard error in the test statistic produced from independent samples) and increase the value of the t-statistic • In randomized block experiment of the analysis of variance, we actually measure the variation between the block by computing SSB (sum of squares block measures the variance between blocks) • The sum of squares error (SSE) is reduced by SSB making it easier to detect differences between the treatments • Additionally we can test to determine whether the blocks differ- a procedure we were unable to perform in the matched pairs experiment. • Factors that identify the randomized block of analysis of variance: o 1) problem objective: compare two or more populations o 2) data type: interval o 3) experimental design: blocked samples o Required conditions: populations are normal, equal variances, if non-normal apply Friedman test o Test statistic- F= MST/MSE Two factor analysis of variance • Experiment features two factors, the general term for data gathering procedures with 2 factors is factorial experiment o We can determine the effect on the response variable of two or more factors

o We can use the analysis of variance to determine whether the levels of each factor are different from one another o We will look at fixed effects only Comparing Lifetime number of jobs and education level • Looking at gender and education levels as two factors • Begin by treating the ex as one-way analysis of variance. There are 8 treatments but they are defined by 2 factors: gender (which has 2 levels) and educational attainment (which has 4 levels) • A complete factorial experiment is an experiment in which the data for all possible combinations of levels of the factors are gathered- also known as two-way classification . o This means we measured the number of jobs for all 8 combinations. This experiment is called a complete 2x4 factorial experiment • One of the factors is referred to as factor A, number of factors of this level is denoted by 𝛼 • The other factor is called factor B and its levels are denoted by b • To test differences between the levels of Factor A: o H 0 : the means of the 𝛼 levels of a factor A are equal o H 1 : at least two means differ • Are there differences in the mean number of jobs between men and women? o H 0 : u men =u women o H 1 :at least two means differ In an ANOVA table- sample refers to factor B and columns refers to factor A Conducting the Analysis of Variance for the Complete Factorial Experiment • We conduct a one-way analysis of variance to determine whether differences exist between the 8 treatment means o This was done primarily for pedagogical reasons to enable you to see that when the treatment means differ we need to analyze the reasons for differences o However in practice we generally do not conduct this test in the complete factorial experiment (some practitioners prefer this to the “two stage” strategy we recommend using the two-factor analysis of variance o We conducted a the test of each factor and then test for interaction ▪ However if there is evidence of interaction, the tests of the factors are irrelevant ▪ There may or may not be differences between the level of factors • Accordingly we would change the order of conducting the F-tests • Order of testing in the two-factor analysis of variance o 1) test for interaction o 2)if there is enough evidence to infer that there is interaction, do not conduct the other tests o 3) if there is not enough evidence to conclude there is interaction, proceed to conduct the F- tests for factors A and B Developing an Understanding of Statistical Concepts • Two-Factor Experiments and Randomized Block Experiment o Calculations are identical for the two • In general the difference between the two experimental designs is that the randomized block experiment, blocking is performed specifically to define that blocks are always characteristics of the experimental units • Consequently, factors that are characteristics of the experimental units will be treated not as factors in a multifactor study but as blocks in a randomized block experiment • Factors that identify the independent samples in two-factor analysis of variation o 1) prob objective: compare two or more populations (population defined as the combination of the levels of two factors)

o 2) data type: interval o 3) experimental design: independent samples o required conditions: populations are normal, equal variances, if non normal apply Kruskall wallis test QUIZ Section 7 Distribution of the test statistic for the analysis of variance is the . F-distribution In Fisher’s least significant difference (LSD) multiple comparison method, the LSD value will be the same for al pairs of means if . All Samples are the same One-way ANOVA is applied to 3 independent samples having means 10, 13, and 18 respectively. I each observation in the 3 rd sample were increased by 30, the value of the F statistic would . Increase Assume a null hypothesis is found to be true. By dividing the sum of squares of all observations or SS (Total) by (n-1) we can retrieve the . Sample Variance Which of the following is true about a one-way analysis of variance? N1=n2…=nk it not required Section 8- Decision Analysis Decision Analysis • Hypothesis testing concludes with either rejecting or not rejecting some hypothesis concerning a dimension of a population • In decision analysis we deal with the problem of selecting one alternative from a list of several possible decisions • In hypothesis testing the decision is based on the statistical evidence • In decision analysis there may be no statistical data or if there are data, the decision may depend only party on them • Finally, costs (and profits) are only indirectly considered (in the selection of a significance level or in interpreting the p-value) in the formulation of a hypothesis test • Decision analysis directly involves profits and losses o Because these are major differences, the topics covered previousl that are required for an understanding of decision analysis are probability (including bayes’ Law) and expected value The investment Decision • Man wants to invest $1M for 1 year o After analyzing he has narrowed his choice to 1 of 3 alternatives o These alternatives are referred to as acts o Acts are controllable, they reflect our choices. Acts are denoted as a i ▪ Invest in guaranteed income cert. paying 10% ▪ Invest in bond with coupon value of 8% ▪ Invest in well-diversified portfolio of stocks • What comes to pass in the future, an outcome, is referred to as a state of nature o States of Nature are uncontrollable

o When enumerating states of nature we should define all possible outcomes ▪ This list should be mutually exclusive and collectively exhaustive. State of nature denoted as s i • Interest rates increase • Interest rates stay the same • Interest rates decrease o Payoff table used to show combination of an act and state of nature • Opportunity loss - is the difference between what the decision makers profit for an act is and what the profit could have been had the best decision been made o Calculated row wise, taking the combination of act and state of nature with the highest value and then subtracting this maximum value from the all payoffs in the row ▪ If done correctly we will be left with a zero (where max payoff was located) and a positive number for all other acts/state of nature combinations Decision Trees • Most problems involving simple choice of alternatives can readily be resolved by using the payoff table (or opportunity loss table) • In other situations, the decision maker must choose between sequences of acts • In these cases, a payoff table will not suffice to determine the best alternative, instead we require a decision tree • The probability tree is useful for computing probabilities , with this all the branches represent stages of events • With the decision tree, branches represent both acts and events (states of nature) o A square node represents a point where a decision is to be made ▪ A point where in the statue of nature occurs is represented by a round node Expected Monetary Value Decision (EMV) • It is possible to assign probabilities to the states of nature o Ex if we must decide whether to replace a piece of equipment that has frequently broken down in the past, we can assign probabilities to the basis of the relative frequency breakdowns o In many other instances, formal rules and techniques of probability cannot be applied • Because probabilities are subjective, we would expect another decision maker to produce a completely different set of probabilities o Ex- stock market would rarely have buyers and sellers if this were not true, everyone would be a seller. • After determining the probabilities of the states of nature, we can address the expected monetary value decision o Now we calculate what we expect will happen for each decision o We generally measure consequences of each decision in monetary erm, we compute the expected monetary value EMV for each act ▪ We calculate the EMV by multiplying the values of the random variables by their respective probabilities and summing the products • In general, the expected monetary values do not represent possible payoffs • If an investment is made a large number of times, with exactly the same payoffs and probabilities, the expected monetary value is the average payoff per investment • EX- How many investments will be made? If one, the payoff and probabilities of the states of nature will undoubtedly change from year to year • Expected value decision is the only method that allows us to combine the two most popular factors in the decision process: payoffs and probabilities o It seems inconceivable that when both factors are known the investor would want to ignore either one

• Typical decision makers make a large number of decisions over their lifetime. By using the expected value decision, the decision maker should perform at least as well as anyone else. Despite the problem interruption, we advocate the expected monetary value decision • Calculated EMV is placed on round node in tree Expected Opportunity Loss Decision (EOL) • We can calculate the expected opportunity loss (EOL) of each act by using the opportunity los table and our probabilities for states of nature • Because w want to minimize losses we choose the act that produces the smallest expected opportunity loss which is a 1 • EMV is the same as the EOL decision o Not a coincidence because the opportunity loss table was produced directly from the payoff table • Rollback technique is the process of determining the EMV decision o At each square node in the decision tree, we make a decision by choosing thr branch with the largest EMV Decision Making with Additional Information • Conditional probabilities or likelihood probabilities • I i are referred to as experimental outcomes- the process by which we gather additional information is called the experiment. Prior, Likelihood and Posterior Probabilities • We can assume what forecasts will actually take place and choose accordingly o Drawbacks to this: ▪ Puts the investor in the position of ignoring whatever knowledge (in the form of subjective probabilities) he had concerning the issue ▪ Instead the decision make should use this information to modify the initial assessment of the probabilities of the states of nature ▪ Use Bayes’ Law to incorporate the investors subjective probabilities with the consultants forecast • Prior probabilities- original probabilities- because they were determined prior to the acquisition of additional information o In this example based on investors experience • Likelihood probabilities- from the company providing the information. • Posterior or Revised probabilities are the set of probabilities we want to compute • We can use revised or posterior probabilities in the same way as used the prior probabilities once calculated Preposterior Analysis • By performing the computations described, the investor can determine whether he should hire the IMC, that s he can determine whether the value of IMC’s forecast exceeds the cost of its information this determination is called Preposterior analysis o Objective is to determine whether the value of the prediction is greater or less than the cost of the information o Posterior refers to the revision of probabilities and the “pre” indicated that this calculation is performed before paying the fee Expected Monetary Value • Notice that these probabilities sum up to 1. • EVSI expected value of sample information – is the difference between the expecyed monetary value with the additional information and the expected monetary value without the additional information

QUIZ section 8 A tabular presentation that shows the outcome for each decision alternative under the various states of nature is called a . Payoff table Which of the following statements is false regarding the expected monetary value (EMV)? In general, the expected monetary values represent possible payoffs In the context of an investment decision, is the difference between what the profit for an act is and the potential profit given an optimal decision. An opportunity Loss The branches in a decision tree are equivalent to . Events and Acts Which of the following is not necessary to compute posterior probabilities? EMV A payoff table lists the monetary values for each possible combination of the . event (state of nature) and act (alternative)