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Instructions for analyzing customer purchase data from a large population using a random sample of 30 customers. The goal is to estimate the population mean for anticipated purchases in 2006 (muy) and calculate the 95% confidence interval. Two methods are used: sample mean (ybar) and regression estimator. The document also discusses the importance of random sampling and potential issues with sample concentration.

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Download Estimating Customer Purchases: Statistical Analysis & Confidence Intervals and more Assignments Probability and Statistics in PDF only on Docsity! BONUS Recitation Assignment KEY 2-23-06 This assignment is outside the usual recitation and exam scoring system and will count for up to 4 bonus points towards your course total. Work for clarity, good write-up, accuracy and detail. It is not required that you type your solutions, but you can if you wish. They must be stapled and every page must have your name on it. 1. Consult the file POP_1.pdf file listing data pairs (x, y) with x = purchases 2005, y = purchases 2006 for a large population of customers. In the file an entry of {644, {924.353, 1253.12}} means customer #644 has purchases x = 924.353 in 2005 and is projected (by an audit) to purchase y = 1253.12 in 2006. a. Select a random with-replacement equal-probability sample of n = 30 customers from this population. Take care to explain exactly how you are using Table 14 of random digits to ensure that each customer listed has an equal chance of being drawn into your sample. Do NOT use systematic sampling (e.g. skipping through the file in big steps) or any other method than bona-fide random sampling with the clearest connection to Table 14. There are particular dangers to just grabbing a chunk of 30 from the file since the data of the file is not itself randomly arranged. Here is MY sample of 30: {{848, {781.121, 1089.74}}, {393, {413.987, 802.836}}, {768, {1067.81, 1339.48}}, {1859, {1350.8, 1418.45}}, {4325, {2769.81, 2031.7}}, {954, {1233.25, 1422.75}}, {4523, {1311.99, 1529.73}}, {3189, {2134.45, 1782.54}}, {3770, {1017.99, 1367.98}}, {3469, {1852.4, 1673.79}}, {3118, {2283.99, 1844.66}}, {391, {652.538, 1083.83}}, {2731, {923.814, 1247.25}}, {3853, {2180.92, 1810.25}}, {495, {896.985, 1293.57}}, {2572, {2101.85, 1775.26}}, {2378, {922.76, 1220.3}}, {1128, {1035.48, 1254.37}}, {876, {1007.79, 1265.42}}, {632, {732.591, 1079.93}}, {1016, {1015.57, 1250.96}}, {1607, {1474.41, 1496.74}}, {3861, {878.538, 1323.}}, {4068, {1064.8, 1408.74}}, {152, {507.352, 1154.28}}, {3221, {795.379, 1233.98}}, {620, {1015.29, 1346.65}}, {1752, {1437.01, 1468.11}}, {4583, {2049.46, 1787.2}}, {2699, {1485.71, 1496.3}}} Important. Choose your particular starting point for your use of Table 14 in the following very specific way (so we can check your work if need’s be). Examine the last two digits of your student number. Choose a column of Table 14 by stepping off the columns in blocks of 4 digits according as the last digit of your student number is 1, 2, through 9. Then choose a row of Table 14 by stepping off blocks of 5 rows according as the second-last digit of your student number is 1 through 9. So a person whose student number ends with “62” will begin with the 30th row (6 times 5) and the 8th column (2 times 4). They would start with this position in the table then continue either by running through ensuing rows (moving right then b. Estimate the population mean MUy (anticipated purchases for 2006) using the sample mean yBAR and give a 95% CI for muy based upon this estimator. This CI yBAR +/- 1.96 sy / root(30) evaluates to 1409.99 +/- 1.96 275.452 / root(30) = {1311.42, 1508.56} and does indeed cover MUy = 1391.29, the POP1 mean. Around 5% of students would, due to luck of the draw, miss MUy with their interval. c. Re-estimate the population mean for score y using the regression estimator of y on x and give a 95% CI for muy based upon this regression estimator. To do this you need to know that for this population the mean purchase in 2005 is mux = 1297.03. Show a plot of the (x, y) pairs for your sample of 30 customers and how you use it to re-estimate muy with the aid of this plot and your knowledge that mux = 1297.03 (i.e. how you get the regression estimator). regr est = yBAR + (MUx – xBAR) rhoHAT sy / sx = 1409.99 + (1297.03 – 1279.86) .970545 .. .. 275.452 / 587.153 = 1417.81 Notice that sample mean xBAR = 1279.86 < MUx = 1297.03. Thinking I have a low-spending sample I’ve INCREASED my estimate of MUy from yBAR = 1409.99 to regre est = 1417.81. Comparing: yBAR = 1409.99 regr est = 1417.81 MUy = 1391.29 The relevant 95%CI for the regression estimator is regr est +/- 1.96 (sy / root(30)) root(1 – rhoHAT2) = 1410 +/- 1.96 (275.452 / root(30)) root(1 - .9705452) = {1394.93, 1440.69} fails to cover MUy = 1391.29. The regression estimator for this data fails utterly. Not only is it actually further from the true value than the ordinary estimate yBAR, but its CI also fails to cover MUy whereas that for yBAR did cover MUy. Other samples will yield different results. If everything is working properly the regression CIs for different samples will be around 24% as wide as the yBAR CIs and each method will cover MUy around 95% of the time. d. By what factor is CI (c) narrower than CI (b)? This involves calculating the sample correlation between x and y scores. Show how you do that. The CI based upon the regression estimator is narrowed by the factor root(1 – rhoHAT2) = 0.241 (for my sample). *********************************************** REPEAT THE ABOVE FOR A 2nd SAMPLE OF 30. {{4443, {2847.41, 2059.94}}, {175, {387.541, 892.948}}, {3958, {2244.92, 1836.44}}, {575, {946.318, 1301.44}}, {2776, {2233.22, 1828.73}}, {2036, {1142.87, 1310.99}}, {4260, {1168.22, 1460.54}}, {3089, {2155.55, 1791.21}}, {4274, {2741.74, 2021.32}}, {4608, {2529.46, 1952.68}}, {2382, {572.257, 1053.57}}, {4120, {2731.44, 2016.63}}, {1289, {677.042, 1012.99}}, {2295, {1024.48, 1263.83}}, {1003, {1257.26, 1429.94}}, {2969, {2187.48, 1805.44}}, {1957, {1205.17, 1341.36}}, {1401, {1422.59, 1482.76}}, {253, {114.58, 422.021}}, {3535, {1605.03, 1579.09}}, {2501, {1253.49, 1383.06}}, {2440, {1445.16, 1470.72}}, {2263, {1222.82, 1358.93}}, {553, {452.69, 823.938}}, {3278, {2074.15, 1758.56}}, {171, {533.73, 1158.42}}, {2343, {1970.32, 1721.11}}, {914, {655.11, 987.652}}, {706, {829.922, 1151.51}}, {2976, {1867.75, 1669.07}}} x, y sample means = {1449.99, 1444.89} x, y sample SDs = {776.857, 399.912} sample correlation rhoHAT = 0.97258 Compare these with the POP1 values x, y means = {1297.03, 1391.29}. population SDs for x, y = {672.047, 348.926}. population correlation rho = 0.967864. For this data regression is closest to MUy regr est = 1368.31 closest yBAR = 1444.89 furthest MUy = 1391.29 truth Here are the CI using yBAR {1301.79, 1588.} wide, covers using regr est {1336.05, 1400.58} narrow, covers For this new data the regr est is doing everything right. For my data this is {1172.66, 1387.06}. See (b) for the similar CI for MUy. It does contain the population mean MUx = 1297.03. h. Refer to (g). Has your sample of 30 included any customers more than once? Detail. What is the FPC you are entitled to use for an equal probability without- replacement sample of 30? Would it make any difference to use it? There were no repeats in my sample of 30 so I am entitled to view it as a sample without replacement. But what will I gain by doing so? The FPC is root((4612-30)/(4612-1)) = 0.99685 ~ 1 (negl. effect) i. Use the table of random digits (from your starting point) to produce a with-replacement random sample of 10 listings from the white pages of a Lansing telephone directory for the score x = first three digits of phone number. Describe exactly what you are doing. Don’t phone them. Describe any practical difficulties or “fudges” in the method that you feel you must make in order to carry this off with reasonable effort yet still do a good job of it. Remember, the goal is to avoid a sample that may fall into unfortunate synchrony with the variable x of interest. Suppose there were 893 pages, each with three columns. To produce one sample listing: Sample a page at random (use digit blocks 001 to 893) then one column on that page (use digits blocks 1, 2, 3) then take the next two digits (e.g. 42) as “go 42% down the column.” This method is imperfect because not all pages have the same number of listings (nor do columns) and we’re “roughing it” when we go (say) 42% of the way down a column. Nonetheless, it is a completely adequate procedure in the practical sense because it is difficult to imagine circumstances in which there could be much danger in using it. j. Do you think that just walking up to 30 students on campus at 12 noon and asking them for score x = 1 (on way to lunch) or 0 (not) would yield comparable results to what would be learned from randomly sampling 30 students beforehand and shadowing them and “pouncing on them” at noon? Why? People on their way to lunch are out “on their way.” We may catch some walkers on our “walk about” but probably not ones comprising a random sample in which each target person has the same chance of inclusion (e.g. lunch could be over before we get far along so we’ve missed everyone but those on our little trajectory). In particular, anyone driving will be missed altogether, etc. The “identify and pounce” method is much more likely to produce an accurate result but of course it is impractical. Perhaps we could interview those identified as pounce-sample persons the next day? k. Refer to (j). What if the score were y = last digit of student number? Would the two sampling methods produce similar results? Why? With last digit of student number it is hard to see how we could get a less than random sample no matter what we do. It is almost as if each person we encounter were assigned a digit at random 0 through 9, independently. True, the digits are really not random, being assigned serially (one would suppose) at some point in the admitting process. Conspiracy theorists can imagine that some of us are assigned last digit”2” (say) because we frequently walk to our lunch. If so, our walk about sample method is biased.