Pr (X = 0) = Pr (X = 1) =… = Pr (X = 9) = 0.1POINT ESTIMATIONESTIMATIONInferences about the properties of an underlying distributionEstimation: Values of the specific parametersHypothesis Testing: Do values of the parameter equal some specific value or are parameters from two or more populations the same?Point Estimate: Specific valueInterval Estimate: Narrow range in which parameter value is likely to fallRandom Sample: Selection in which each member is independently chosen with a known probability of being selectedSimple Random Sample: Each group member has the same probability of being chosenCluster sampling and stratified sampling are other possible sampling methods. Population: Group we want to studyRandom Number: A random variable, X, that takes on values 0, 1, 2, …, 9 with equal probabilityPr (X = 0) = Pr (X = 1) =… = Pr (X = 9) = 0.1Random Number Table—Table 4 in Appendix. This table consists of 1000 computergenerated random digits. Example: 1. Let’s take the first 20 people in this room. Assign them numbers 01, 02, …, 20. We want to select 4 individuals. 2. Go to Table 4 and start anywhere. Let’s start at top of columns 6 and 7.22, 90, 43, 11, 70, 58, 56, 47, 35, 01, 04, 94, 87, 36, 79, 79, 17This is a random selection process.Random Assignment: 1. Suppose we have 10 subjects and want to assign 5 to Treatment A and 5 to Treatment B. 2. Number subjects 0, 1, 2, … ,9. 3. Go to Table 4. Let’s start at column 10 at the top.4, 4, 5, 4, 9, 6, 7Randomized Clinical Trial: This is a special type of research design to compare two or more treatments. Random assignment is used to assign treatments to subjects. If the sample sizes are large, then groups tend to be similar on many characteristics. Blinding: A clinical trial is a double blind trial if neither the physician nor the patient knows which treatment he or she is getting. A trial is single blind if only the patient is blinded to treatment assignment. A trial is unblinded if both are aware of the treatment assignment. POINT ESTIMATIONLet x1, x2, …, xn be a random sample to estimate , 2. nXXnii1Note: We could have drawn several samples and obtained a sample mean for each. Let’s call these,...x,x21We have created the sampling distribution of X. This is the distribution of values of the sample means for all possible samples of size n from the reference population. The average of sample means will be approximately  if we take a large number of samples. It will be exactly equal to  if we take all possible samples. Therefore,μ)XE( Xis an unbiased estimator of .Θ)ΘE(ΘΘΘˆˆˆifunbiasedisparameterofestimateparameterLetLarge samples are better. Means for repeated samples of size 20 are less variable than means for samples of size 5.   nσ)XVar()Xsd(nσnσnσ...σσn)Var(Xn)XVar(n)XVar(ii2222222221111This is called the standard error of the mean. We estimate this by ns/ since s estimates . Example: The data on page 155 represent the birthweights (oz) from 1000 consecutive deliveries at Boston City Hospital. Suppose we take a sample of 10 individuals and get the following data:101 114 79 120 115117 106 86 110 119468416231314101314131470106....sem.s.XThis is a measure of the variability of the distribution of sample means. Obtained from repeated random samples of size n. As n increases, SEM decreases. This is proportional to 1/n.As  increases, SEM increases.Larger sample sizes provide more precise estimates of .Example of a Sampling Distribution: We can construct the entire sampling distribution fora small population.1. Draw all possible samples of size n.2. Compute the statistic of interest.3. Make a Frequency Table of values of the statistic.As N increases, this is a big job and impossible if the population is infinite.If N = 600 and n = 4,samplespossible85016434654600,,,Let N = 6, n = 2.x1 = 2, x2 = 4, x3 = 6, x4 = 8, x5 = 10, x6 = 12 = mean = 72 = (X - )2/ N = 70/6 = 11.6667s2 = (X - )2/ (N – 1) = 70/5 = 14Draw all possible samples of size 2 with replacement1st draw 2nd draw2 4 6 8 10 122 2 3 4 5 6 74 3 4 5 6 7 86 4 5 6 7 8 98 5 6 7 8 9 1010 6 7 8 9 10 1112 7 8 9 10 11 12Data within the table are the means for each possible sample of size n = 2.Frequency TableMean f Relative Frequency2 1 1/363 2 2/364 3 3/365 4 4/366 5 5/367 6 6/368 5 5/369 4 4/3610 3 3/3611 2 2/3612 1 1/36Note that the sum of the relative frequencies = 1.Xμ mean of sampling distribution= Xi / n = (2 + 3 + … + 12) / 36 = 252 / 36 = 7Var of X= 3622/)μΣ(XσXiX = [(2 – 7)2 + (3 – 7)2 + … + (12 – 7)2] = 210 / 36 = 5.83332Xσ = 2 / n = 11.6667 / 2 = 5.8333nσ/σX This is called the Standard error of the mean or sometimes the standard error.CENTRAL LIMIT THEOREM)nσ~N(μX)σX~N(μIf,,22If X is not normally distributed, we still want to make a statement about sampling distribution of sample means.Let X1, X2, …, Xn be a random sample from a population with mean  and variance 2. For large n, X ~ approximately N(, 2/n) even if the underlying distribution of individual observations in the population is not normally distributed. Example:Suppose diastolic blood pressure in children is N(56, 64). What is the probability that the mean DBP from a sample of 20 children will be between 53 mm Hg and 59 mm Hg?907006771677178891565378891565959537889185620.).Φ().Φ(.Φ.Φ)X(.nσ,,σμ,nPrWhat is the probability that a single individual would have DBP between 53 and 59?296103750375085653856595953.).Φ().Φ(ΦΦ)X(PrExampleHospital wages for a type of hospital employee are ~ N(4.50, 0.25)What is the probability that a single individual will earn more than $4.25 per hour?Pr(X > 4.25)= 1 – Pr(X  4.25)= 1 – Pr(z  (4.25 – 4.50) / (0.50)= 1 – Pr(z  -0.50)= 1 – 0.3085= 0.6915What is the probability that the mean for a random sample of 16 employees will be greater than