Scanned by CamScanner Sampling and Sampling Distribution (Ch. 5 & 9) BA1605-07 (Sampling) 1 Polls BA1605-07 (Sampling) 2 Statistical Inference Involves Estimation Population? Hypothesis testing Purpose Make decisions about population parameters. Proper sampling methods provides \"good\" estimates. 3 BA1605-07 (Sampling) Examples The owner of the Seadogs' Arena, a recently opened gourmet restaurant, hopes to estimate the average demand for French Fries after a home hockey game. Using the data from the last two years, he found the average of 242. A manufacturer claims that the defective rate of his process is at most 2%. To find out whether the statement is true, we randomly drew a sample of 30 items and found 5 defective items. BA1605-07 (Sampling) 4 A manufacturer claims that the defective rate of his process is at most 2%. To find out whether the statement is true, we randomly drew a sample of 30 items and found 5 defective items. If the manufacturer's claim is correct, the sampling distribution of the number of defective items in the 30 sample follows binomial distribution with n = 30 and p = 0.02. Binomial Dist with n=30, p=0.02 0.6 0.4 P(X5)=0.0003 0.2 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 # Defectives Do you think that manufacturer's claim is true? BA1605-07 (Sampling) 5 5.3 Sampling A sample survey is designed to ask questions of a small group of people in the hope of learning something about the entire population. 1. (Population ) consists of all items of interest. 4. make an estimate of the population (parameter). \"the key numbers in mathematical models to represent reality\" BA1605-07 (Sampling) 2. Select a small group to survey (sample ). 3. Collect data from the sample (statistic). \"Any summary found from the data \" 6 Defining the Population The population is determined by the ( ) of the study. Unfortunately, the sample is just those we can reach to obtain responses - the ( ) of the study. 1. 2. 3. (Statistical Study) You should think about the population of interest. You must specify the ( ). Define your target samplethe individuals for whom you intend to measure responses. 4. Finally, there is your samplethe ( ) respondents. (Example) Mail survey for the IT usage in dairy farms. 7 BA1605-07 (Sampling) Population vs Samples Population: The group you are interested in Students in UNB BA1605 class University Students in Canada Sample Group(s) from a population that were ( ) Becomes the basis of ( ) about the wider population BA1605-07 (Sampling) 8 Sampling Methods Simple Random Sampling Stratified Random Sampling Cluster Sampling Systematic Sampling Random Sampling Convenience Sampling Judgment Sampling 9 BA1605-07 (Sampling) Simple Random Sample Ensures that everyone in the population has an __________________________________. Procedure Draw from a hat (equal size, thickness) o With or without replacement Table of Random numbers Computer-generated random number BA1605-07 (Sampling) 10 Simple Random Sample Procedure to use \"Random Number\" 1. A sampling frame is a list of individuals from which the sample will be drawn. 2A. We assign an ID to each individual in the sampling frame and draw random numbers (related to ID). 2B. An alternative method: to assign a random number to each member, sort the random numbers, then pick a random sample of any size off the top of the sorted list. Sample-to-sample differences in the values for the variables we measure are called sampling variability. 11 BA1605-07 (Sampling) Use Excel to Generate Random Samples No NAME 1 2 3 4 5 6 7 8 9 10 11 12 13 STUDENT ID Random Al Khatam, Maryam Hussain Al-Mansour, Ali Jufaysh Al-Talaq, Ali Alassaf, Saif Mohammed Aldobas, Mohammed Salom Alhaidary, Abdullah Abdulaz Alhawas, Meshal Nasser Aljanabi, Abdullah Hasan Almansour, Bader Basem a Almegayel, Mohammed Khal Almoqabqab, Rowan Mohamm Alosimi, Abdullah Mohamme Alsadiq, Hadeel BA1605-07 (Sampling) 4562 1703 5169 0669 2882 9564 9934 2025 2846 4536 3852 0866 4195 0.0020194 0.0864722 0.813735 0.8113618 0.943355 0.5168319 0.9044879 0.7142984 0.8230147 0.2358325 0.7692187 0.4717374 0.6584824 Random2 0.3525472 0.250671 0.0503776 0.4004631 0.5569 0.8477231 0.6018644 0.0250872 0.6599076 0.9797069 0.9574089 0.308012 0.8198479 12 Stratified Random Sampling Assures that the sample reflects the population w.r.t. a particular _______________ (called gender, age, ... ). Need a list of the total population PLUS the characteristic(s) with known statistics. Select samples from each group proportionally Reduced sampling variability is the most important benefit of stratifying. 200 40 60 300 BA1605-07 (Sampling) Male Male Female Female 13 Cluster (Multistage) Sampling When the total population cannot be easily listed. Ex) all graduating high school students in province We select \"________\" randomly, and perform a census. Randomly select a sample of high schools (clusters) Survey all the graduating students at those schools Save time & money, but can result in a biased sample High schools in NB SIMONDS HIGH Ex) low-income district Sampling schemes that combine several methods are called ______________samples. BA1605-07 (Sampling) 14 Systematic Random Sample Select individuals at _____________________. Need to know the total # of the population and the # you want to sample. Example: 250/75 = 3 But, Be careful with any systematic pattern in the list of names (e.g. by department, ethnic groups). To make sure our sample is random, we must start this method with a randomly selected individual. BA1605-07 (Sampling) 15 Sampling Error (or Variability) Sampling Error refers to _____________________. We cannot remove it. The best we can do is to state the probability that the sampling error is less than a certain amount (by random sampling) Non-sampling errors result from mistakes in sampling. Errors in data acquisition ____________________(especially in survey) _____________________(by nonrandom sampling) such as convenience Sampling [\"Person-on-the-street\" interviews by reporters and journalists] Non-sampling errors will weaken the basis for generalizing from sample to population. BA1605-07 (Sampling) 16 Role of Sampling Distributions Inference Sampling distribution Sample BA1605-07 (Sampling) 17 Sampling Distribution of the Mean Population Distribution Summary Measures = 0.3 SD = = 0.2 0.1 0 1 2 3 4 X 1 2 3 4 P(X) 0.25 0.25 0.25 0.25 (X-)2 (X-)2*P(X) 2 = Var(X) = [(X - E(X))2P(X)] BA1605-07 (Sampling) 18 Sampling Example with n = 2 16 Sample Means 1st Obs. x1 x 2 x 2 2nd Observation 1 2 3 Sampling Distribution 0.3 4 0.2 1 2 0.1 3 0 1 4 x 1.5 2 2.5 3 3.5 4 1 . 0 1 . 5 1 . 5 2 . 0 ... 4 . 0 16 (1 . 0 2 . 5 ) 2 (1 . 5 2 . 5 ) 2 ... ( 4 . 0 2 . 5 ) 2 16 x BA1605-07 (Sampling) 19 Sampling Example: Comparison Population Sampling Distribution about the sample MEAN 0.3 0.3 0.2 0.2 0.1 0.1 0 0 1 2 2.5 1.12 BA1605-07 (Sampling) 3 1 4 = 1.5 2 2.5 3 3.5 4 x 2.5 x 0.79 20 Standard Deviation of Sample Mean Measures variability in sample mean Referred to as the \"__________________\" of the sample mean Less than population standard deviation Formula x n Do you remember this standard error was in the Descriptive Statistics In EXCEL? 21 BA1605-07 (Sampling) Sampling without Replacement 1st Obs 2nd Obs Sample Mean 1 1 1 2 2 3 2 3 4 3 4 4 1.5 2 2.5 2.5 3 3.5 x 2.5 x 0.65 You don't have to remember this formula. But, remember that without replacement is less than with replacement Finite Population Correction Factor x BA1605-07 (Sampling) N n 1.12 4 2 1.12 n N 1 2 4 1 3 22 _ Sampling Distribution of Sample Mean (X) Population Distribution Sampling Distribution Normal Normal Approximately Normal (when n > 30) Non-Normal ? (when n < 30) BA1605-07 (Sampling) 23 Central Limit Theorem (CLT) In selecting simple random samples of size n from a population, the sampling distribution of the sample mean can be approximated by a normal distribution as the sample size becomes large ( ). We know the behavior of the sample mean without knowing the shape of the population. Furthermore, it permits us to use sample statistics to make inferences about the population parameter for practically any populations as long as the sample size is large. BA1605-07 (Sampling) 24 An Illustration of the CLT 25 BA1605-07 (Sampling) Example The average length of a hospital stay in general or community hospitals in the United States is 5.9 days. Assuming a population standard deviation of 3.5 days and a simple random sample of 50 patients, what is the probability that the average length of stay for this group of patients will be no more than 6.7 days? P( X 6.7) If the sample size had been 20 patients, can we accurately assess the probability? BA1605-07 (Sampling) 26 Class Practice (Ex 9.1) The bottling manager has observed that the amount of soda in 32-oz. bottle is actually a normally distributed random variable, with a mean of 32.2 oz. and a standard deviation of .3 oz. a. If a customer buys one bottle, what is the probability that the Z-Table bottle will contain more than 32 oz? b. If a customer buys a carton of four bottles, what is the probability that the mean amount of the four bottles will be greater than 32 oz? 27 BA1605-07 (Sampling) Sample Proportions Categorical variable (e.g., gender) Sample proportion Proportion of population having the property/attribute If two outcomes, the sample proportions follow a binomial distribution. (ex) Possess or don't possess property _ number of success p = = sample size BA1605-07 (Sampling) 28 Sampling Distribution of Proportion Required Condition ( ) condition ( ) condition: np 5 and nq 5 (# Success) (# Failure) where p = the population proportion, and q = 1 - p Expected Value: E ( p ) p Standard Deviation (Standard error of sample proportion) Finite Population Infinite Population p BA1605-07 (Sampling) pq n N n N 1 Bernoulli Process has 2 = pq. Thus, /(n) = (pq/n) 29 Example 9.41 A psychologist believes that 80% of male drivers when lost continue to drive hoping to find the location they seek rather than ask directions. To examine this belief, he took a random sample of 350 male drivers and asked each what they did when lost. If the belief is true, determine the probability that less than 75% said they continue driving. (Solution) Given parameters success rate p = ,n= Check the conditions Problem solving: Sampling Dist. p ~ P(p 0.75) BA1605-07 (Sampling) 30 Class Practice A bank analyst investigates 90 mortgages in some region. The analyst found that 11 mortgages went into default. The finance department assumed that no more than 15% of the mortgages would go into default. Any amount above that will result in losses for the bank. From the package of above 90 mortgages, what's the probability that there will be more than 15% foreclosures in the region? (Define the success rate) p = (Check the conditions) (Problem solving) Sample distribution p ~ N (Conclusion) 31 BA1605-07 (Sampling) Summary Statistical Inference: Sample vs. Population Sampling Methods Simple or Stratified Random Sampling Cluster Sampling Systematic Sampling Nonrandom Sampling Sampling Error (Variability) vs. Non-sampling Errors Central Limit Theorem (n 30) Sampling Distribution: Sample mean N , n BA1605-07 (Sampling) or Proportion pq N p, n 32 Answers #6: Population, Sample, Statistic, Parameter #7: WHY, WHO, sampling frame actual #8: actually studied, generalizations #10: equal chance of being selected #12: 1. =rand() 2. Paste by value 4. Choose the first n students. 3. Sort by this pasted value #13: characteristic , strata #14: Cluster multistage #15: every fixed interval #16: differences among samples Nonresponse bias Selection bias #18: = 2.5, SD = = 1.12 #19: x 2 . 5 , x 0 . 79 #20: x 0 . 79 1 . 12 / 2 33 BA1605-07 (Sampling) Answers (2) #21: standard error #24: n 30 6.7 5.9 #26: P Z 3.5 / 50 = P(Z1.62) =0.9474; we don't know the population distribution, due to the small sample size #27: a) P(X > 32) =0.7486 b) P ( X 32) = P(Z > -1.33)=0.9082 #28: = n/x pq #29: Independence, Success/Failure p n #30: n = 350, p = 0.80; Independence, Success/Failure np=280, nq=705 p ~ N 0.8, 0.021 P ( p 0.75) P ( Z 2.38) .0087 #31: p = 11/90 = 0.122; Check the conditions; p ~ N p 0.122, pq / n 0.0345 P( p 0.15) P(Z > 0.81) = 20.9% Thus, There exists 20.9% chance of exceeding 15% default rate. BA1605-07 (Sampling) 34 Standard Normal Table z z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 .0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 .1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753 .2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 .3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 .4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 .5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224 .6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549 .7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852 .8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133 .9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389 1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621 1.1 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810 .8830 1.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8997 .9015 1.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177 1.4 .9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319 1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441 35 BA1605-07 (Sampling) Standard Normal Table 2 z z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441 1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545 1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633 1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706 1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767 2.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812 .9817 2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857 2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887 .9890 2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .9916 2.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .9936 2.5 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .9949 .9951 .9952 2.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .9964 2.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .9972 .9973 .9974 2.8 .9974 .9975 .9976 .9977 .9977 .9978 .9979 .9979 .9980 .9981 2.9 .9981 .9982 .9982 .9983 .9984 .9984 .9985 .9985 .9986 .9986 3.0 .9987 .9987 .9987 .9988 .9988 .9989 .9989 .9989 .9990 .9990 BA1605-07 (Sampling) 36 Interval Estimation for 10 & 12-1. Means 12-3. Proportions Estimation Point Estimator: What is the best value to represent? Single value of point Interval Estimator: Provides range of values Based on observations from 1 sample Gives information about closeness to unknown population parameter Stated in terms of probability BA1605-08 (Estimation) 2 Confidence Interval Knowledge of the sampling distribution enables us to make probability statements about the sampling error even though the population proportion \"p\" is not known. Sampling Distribution Precision Statement: Probability that the maximum sampling error is a certain value. Sampling Error (Margin of Error) p p p Example: Gallup says with 95% confidence that Obama has 48% of presidential vote, with the maximum margin of error 2 % points in Election Survey in Nov. 2012. (It says the results based on the total sample of 2,854 voters.) (http://www.gallup.com/poll/158519/romney-obama-gallup-final-election-survey.aspx) BA1605-08 (Estimation) 3 Confidence Interval (Example) In 2010, a Gallup Poll found that 1012 out of 2976 respondents thought economic conditions were getting better - a sample proportion of p = 1012/2976 = 34.0%. We'd like use this sample proportion to say something about what proportion, p, of the entire population thinks the economic conditions are getting better. We know from that the shape of the sampling distribution is approximately Normal and we can use p to find the standard error (SE). SE ( p ) p BA1605-08 (Estimation) 4 Confidence Interval Z-Table Z.025 = Value= p +Zp = Sampling Distribution P ~ N(0.34, 0.0087) 95% of values 2.5% 2.5% p=.34 ^ p BA1605-08 (Estimation) 5 Confidence Interval Sampling Distribution /2 1-of all p values /2 p z / 2 p pz pq pq p pz n n BA1605-08 (Estimation) 6 Margin of Error Confidence Interval (CI): p z / 2 p The extent of that interval on either side of p . The general confidence interval can now be expressed in terms of the ME. (Example) In the Gallup example, p = 1012/2976 = 34.0 %. What is the Margin of Error in 95% Confidence Level? ME = BA1605-08 (Estimation) 7 Margin of Error You have more confident for wide range. The more confident we want, the larger the ME is. In Gallup example, we are 100% confident that any proportion is [0%, 100%], but we are not very confident that the proportion is [33.8%,34.2%]. Sampling Distribution N p, pq / n N .34,.0087 /2 1-of all p values /2 p BA1605-08 (Estimation) ME 8 Confidence Interval Sampling Distribution 1-of all ^ values p /2 /2 p p Intervals extend from p z / 2 p to p z / 2 p BA1605-08 (Estimation) 9 Critical Values Z-Table Level of Confidence (1 - )% A probability that the population parameter falls somewhere within the interval Denoted by (1 - )% Typical values are 99%, 95%, and 90%. Critical Value (z/2) = P(the interval fails to include the true value) This failure happens in both sides (too big or too small) It follows the Normal Dist'n, we denote it z and call it Critical Value. Confidence Limits Upper and Lower Confidence Limits (UCL & LCL) p z / 2 p p z / 2 SE ( p) BA1605-08 (Estimation) 1% 5% 10% CL 99% 95% 90% z/2 10 Interval for Population Mean Sampling Distribution x ~ N , n 1- of all x values /2 /2 z / 2 x P z x z 1 n n BA1605-08 (Estimation) 11 Factors Affecting Intervals Intervals extend from x z / 2 x to x z / 2 x where x s n Data dispersion Measured by s Sample size x / n (Consistency) As n , sampling error More efficient! Level of confidence (1 ) Affects z value BA1605-08 (Estimation) 12 Interval Estimate of a Population Mean: Large-Sample Case (n 30) With Known x z / 2 x z / 2 n n With Unknown x z / 2 s s x z / 2 n n where: s is the sample standard deviation BA1605-08 (Estimation) 13 Example: National Superstore National Superstore has 260 retail outlets throughout Canada. National evaluates each potential location for a new retail outlet in part on the mean annual income of the individuals in the marketing area of the new location. A sample of 100 individuals in the location resulted in a sample mean of $41,100 and a sample standard deviation of $12,000. Set up a 95% confidence interval estimate of the population mean. BA1605-08 (Estimation) 14 Example: National Superstore (2) Interval Estimate of the Population Mean x z / 2 s s x z / 2 n n Z-Table Since sample size > 30, we apply the central limit theorem. _ x = 41,100, s = 12,000, n = 100 = 5% z/2 = 0.025 z/2 Margin of Error: E = Lower limit (LCL) = Upper limit (UCL) = BA1605-08 (Estimation) 15 Sample Size for an Interval Estimate of Population Mean Let E = the maximum sampling error (or margin of error) We have E z / 2 z n /2 n E Solving for n we have BA1605-08 (Estimation) 16 Example: National Superstore (3) Suppose that National's management team wants an estimate of the population mean such that there is a .95 probability that the sampling error is $1,500 or less. Assume the standard deviation has been quite stable at 12,000. How large a sample size is needed to meet the required precision? z n /2 E 2 BA1605-08 (Estimation) 17 Class Practice #1 (Ex 10.1) The Doll Computer makes its own computers and delivers them directly to customers who order them via the Internet. The computers are made at the factory and transported to the warehouses, from which it generally takes 1 day to deliver a computer to the customer. The operations manager finds that the demand during lead time (that is, time elapsed from order to delivery) is normally distributed with the standard deviation is 75. He observes 20 lead time periods and records the demand during each period. These data are listed here. 235 394 374 439 309 348 499 344 253 421 330 261 361 374 514 302 462 466 369 530 The manager would like a 95% confidence interval estimate of the mean demand during lead time. (Hint: X = 7350.) (Sol) X~N(, =75) E z / 2 CI : x z / 2 BA1605-08 (Estimation) n n 18 Class Practice 2 (Ex 12.1) Although most products from recycled material are more expensive than those from material found in the earth, newspapers are an exception. One research showed that the paper company would make a profit if the mean weekly newspaper collection from each household exceeded 2.0 pounds. To determine the feasibility of a recycling plant, a random sample of 125 households was drawn from a large community: The data have a mean of 2.18 and sample standard deviation of 0.981. What is a 90% confidence interval estimate of the mean weekly collection? (Sol) E z / 2 s n CI : x z / 2 s n BA1605-08 (Estimation) 19 t Distribution Similar to the standard normal distribution Symmetric, Bell-shaped, and Mean = 0 Has fatter tail than the standard normal distribution Depends on the ________________________. Standard Normal Distribution t (d.f. = 10) t (d.f. = 5) BA1605-08 (Estimation) 0 20 Degrees of Freedom (d.f.) Number of observations that are free to vary after sample statistic has been calculated Example Sum of three numbers is 6. x1 = 1 (or any number) x2 = 2 (or any number) x3 = 3 (cannot vary) Sum = 6 Degrees of freedom of t distribution BA1605-08 (Estimation) 21 Interval Estimation of a Population Mean: Small-Sample Case (n < 30) with Unknown The interval estimate is given by: x t / 2, n 1 s s x t / 2, n 1 n n t/2 = TINV(, n 1) 1 /2 BA1605-08 (Estimation) /2 22 Example: Apartment Rents t-Table A sample of 10 one-bedroom units within a half-mile of campus resulted in a sample mean of $450 per month and a sample standard deviation of $30. Let us provide a 95% confidence interval estimate of the mean rent per month for the population of one-bedroom units within a half-mile of campus. We'll assume this population to be normally distributed. (Sol) Small-Sample Case (n < 30) with Unknown x t / 2,n 1 s n BA1605-08 (Estimation) Class Practice 2-2 (Ex 12.1) 23 t-Table Although most products from recycled material are more expensive than those from material found in the earth, newspapers are an exception. One research showed that the paper company would make a profit if the mean weekly newspaper collection from each household exceeded 2.0 pounds. To determine the feasibility of a recycling plant, a random sample of 25 households was drawn from a large community: The data have a mean of 2.18 and sample standard deviation of 0.981. What is a 90% confidence interval estimate of the mean weekly collection? (We assume the weekly collection follows the normal distribution.) (Sol) Small sample with unknown Use t-dist'n with d.f = n-1. E t / 2,n 1 x t / 2 s n s n BA1605-08 (Estimation) 24 Interval Estimation for Population Mean The factors for determining sampling distributions of the sample mean 1. Sample size 2. Normal or non-normal population distribution 3. Known or unknown population standard deviation N x, E z / 2 Yes n n known? Large (n 30) No Yes Yes known? No Sample Size Is the population Small normal? No s s t n 1 x , E t / 2,n 1 n n Sampling dist is not available. 1% 5% 10% BA1605-08 (Estimation) s s N x, E z / 2 n n z 2.326 1.645 1.282 z/2 2.576 1.960 1.645 25 Interval of Population Proportion Check Success/Failure Condition: The interval estimate is given by: Compare it with x z / 2 BA1605-08 (Estimation) x z / 2 n n 26 Example: Political Science, Inc. Political Science, Inc. (PSI) specializes in voter polls and surveys designed to keep political office seekers informed of their position in a race. In a recent election campaign, PSI found that 220 registered voters, out of 500 contacted, favored a particular candidate. Develop a 95% confidence interval estimate for the proportion of the population of registered voters that favors the candidate. BA1605-08 (Estimation) 27 Example: Political Science, Inc. 1) Given Information: p x n , z/2 = 2) Check Conditions: , nq = np = 3) Interval Estimate of a Population Proportion E z / 2 pq n CI : p z / 2 BA1605-08 (Estimation) pq n 28 Sample Size for an Interval Estimate of a Population Proportion Let E = the maximum sampling error We have E pq pq E z / 2 z / 2 n n 2 E pq z n /2 Solving for n we have Note: If no information is available about p, then 0.5 is often assumed because it provides the highest possible sample size. BA1605-08 (Estimation) 29 Example: Political Science, Inc. Suppose that PSI would like a 95% probability that the sample proportion is within .03 of the population proportion. At 95% confidence, z.025 = 1.96 . p 0.1 2 z n / 2 pq E BA1605-08 (Estimation) 385 0.2 How large a sample size is needed to meet the required precision? n 683 0.3 897 0.4 1025 0.5 1068 0.6 1025 0.7 897 0.8 683 0.9 385 30 Class Practice - Nielsen 1 (Nielsen TV Rating) Nielsen is monitoring the television shows watched by viewers across the country, by a special device preinstalled to the panel member's TV. In a specific time, they observe 275 watching \"Big Bang Theory\