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## STAT METH FOR RSRCH

by: Giovani Ullrich PhD

5

0

11

# STAT METH FOR RSRCH STAT 401

Giovani Ullrich PhD
ISU
GPA 3.5

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
11
WORDS
KARMA
25 ?

## Popular in Statistics

This 11 page Class Notes was uploaded by Giovani Ullrich PhD on Saturday September 26, 2015. The Class Notes belongs to STAT 401 at Iowa State University taught by Staff in Fall. Since its upload, it has received 5 views. For similar materials see /class/214407/stat-401-iowa-state-university in Statistics at Iowa State University.

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Date Created: 09/26/15
48 A Useful Discrete Random Variable The Binomial 430 432 439 440 Use the binomial formula Py 5v53v for y 0 1 2 3 with the probability of a head being 05 x 0375 P2 0375 P3 0125 a Py 3 13quot2387 0201 b Py 2 4762 03456 c Py 12 7 34 0204 a a Py s 4 P0 P1 P2 P3 P4 0168 00896 02090 02787 02322 08263 b Pygt4 1 Pys4 1 08263017 7 c Py g 7 1 P8 1 4868E 1 00007 09993 d Py gt 6 2 P0 2 7 Py s 4768 7 456H 00085 at Bar graph of Py be P0122 Py0 Py 1Py2 05 ci Py 2 7 Py 7 Py 8 0 9 Plty 10 013 d P1 gyg 5Py1Py2Py3Pg4Py5071 a Py 2 3 1 1 Pg lt 3 1 006 014 016 064 b P030 Py 2Py3Py4Py5 0026 016014012010008 060 c Py gt s Py 9 Py 10 004 003 007 No No people may not answer the question 0 in quotum an t iimmteiietiu u 1 n which would lead to a biased atimate of the overall percentage Binomial experiment with 1110 and 77 060 e Py 0 00001 b Py 6 2508 c Py 2 6 Py lt 6 1 P0 P1 P2 P3 P4 P5 1 03669 06331 d Py 10 00060 Binomial experiment with 1110 and 7r 030 a Py 0 3 71 00282 P0 6 16quot3574 00368 P lt 6 Py 2 6 H0 P1 P2 P3 P4 P5 1 00527 00473 Py 10 g3w70 0000006 b With 71 1000 and 7r 3 Py s 100 22 1 9 3quot71 7i This would be at length calculation In Section 413 we will provide an approx imation which will greatly reduce the amount of calculations However We can also consider that mr 10003 2 3000 2 1mr1 1r 2 1100037 1449 1 i so 300 l 3 1449 25653 34347 Thus the chance of observing the event y g 100 is very small 441 Py 0 321quot2A 00016 7 1 fs12a 00016 y g 1 Py 0 P 1 00272 3 1 P0 2 2 Py 2 P0 3 9211 0310 0972 443 Demonstration exercise 444 Py 0 Shims P0 1 2 WW P0 2gtZgt2gt1 P0 3 36quot 445 1 Mr 10005 2 500 a s 100055 1581 1 i 30 45257 54743 446 P0 100 232 6gtlt41 7 447 a P0 2 f170a 01937 b Py 2 2 1 Py 0 Py 1 1 13 1 91 1199 02639 c 7r Peither outstanding or good 2 085 Py S 10858152 02759 d 7r Punsatisfwctory 005 Py 0 10 005 05 05987 448 No The trials are not identical 449 Binomial with It 151r 12 9 Py 0 0512 8815 01470 b Py 21 1 Py 0 08530 0 Ply 2 2 1 Py 0 Py 1 1 01470 03006 05524 450 Yes since the experiment ful lls the 5 conditions for a binomial experiment 451 Binomial with n 507r 17 a Py 3 210 2 007i0935 i 05327 using a computer program b The posting of price changes ere independent with the some probability 007 of being posted incorrectly new lm an aoI 49 Probability Distributions for Continuous Random Variables 452 a 09032 05000 04032 05000 00287 04713 7580 05000 02580 05000 01151 03849 09015 05000 04015 05000 02206 02794 09115 v 04168 04947 08849 06443 02406 09778 09236 00542 02266 00409 01857 457 00401 458 08729 459 Zn 0 460 2 196 461 z 237 462 za 2 1645 5 w 1 tenancy o Cquot y 100 P zgt 1m1 Pzgt0 05 e P100 lt y lt 108 P 08413 05000 03413 465 a P500 lt y lt 090 P 5053quot lt z lt M P0 lt z lt 196 09750 05000 0475 b Py gt 696 Pz gt 695330quot Pz gt 196 z 1 1 09750 0025 c P304 lt y lt 696 NW lt z lt 235quot P 196 lt z lt 190 09750 7 00250 x 095 d P500 kltylt500kP Wltzlt t g P 01k lt z lt 0110 050 Ram Table 1 we nd P 845 lt z lt 345 060 Thus 01 845 4 k s45 5 n F 5 F 90939 E y gt 325 gt 4 gt Py lt 130 Pz lt 2 09772 F E P PP PP PP PP PP PP Py lt12o Pz lt1 130 100 15 2 z lt 2 825 100 15 gt 2 gt 117 ylt130gt13lt 1 Pz g 117 03790 Py gt 32 5 Pz gt 117 Py gt 106 Pz gt a Pz gt 04 1 Pz g 04 03446 Py lt 94 Pz lt 5 Pz lt 04 03446 H94 lt y lt 106 P 04 lt z lt 04 1 03446 03445 03103 30 00 Pz gt 2 1 Pz s 2 00223 Py lt 70 Pz lt 70 Pz lt 2 00223 P70 lt y lt 130 P 2 lt z lt 2 1 00223 00223 2 09544 09332 05000 04332 09641 05000 04641 09750 00250 095 09901 7 00099 09302 Pz gt 196 1 09750 0025 Pz gt 221 1 09364 00136 Pz gt 233 1 09979 00021 Pz gt 073 1 07673 02327 Pz lt 712 01151 Pz lt 262 00044 Pz lt 134 9671 Pz lt 217 09350 0 39a 6 Py gt 50 Pz gt 39 Pz gt 133 1 09664 00336 Since 55 is 55739 267 std dev above 2 39 thus Py gt 55 Pz gt 267 3 x 00038 We would then mnclude that the voucher has been lost 500 a39 100 Py gt 600 Pz gt 313300 Pz gt 1 01537 Py gt 700 Pz gt 7001750 Pz gt 2 00223 Py lt 450 Pz lt 453330 03035 P450 lt y lt 300 P6525300 P 05 lt z lt 1 05323 50 Py gt 200 Pz gt Py gt 220 2 Pz gt 20035150 Pz gt 143 00764 2 50 Pz gt 2 00223 T5quot Pz lt 7086 01949 P100 lt y lt 200 130003751 lt z lt 2003515quot P 143 lt z lt 143 03473 4 500 a 100 Find k such 11135130 gt k 010 Pz gt 1235 010 5 k 500 1235100 62850 or greater These commands yield a histogram with a normal curve superimposed and the following 39 t summary sums cs Variable N Mean Median TrHean StDev SE Mean C20 500 59982 60033 59980 1215 0054 Variable Minimum Maximum Q 1 Q3 C20 56340 64805 59 125 60818 490 2 930 a 130 a P800 lt y lt 1100 H803 lt z lt 11033393quot P 1 lt z lt 131 09049 01537 7462 b Py lt 800 Pz lt 501333 01587 c Py gt 1200 Pz gt 1203093quot Pz gt 203 1 09811 00189 491 a Pz lt 128 0190 i y9 930 l28130 10964 b Pz lt 06745 Pz gt 06745 025 g 1125 9307016745130 84231 1175 93006745130 101769 IQR 101769 4 84231 17538 492 a 125 i 32 should contain approximately 68 of the Weeks 125 i 64 should contain approximately 95 of the weeks 125 l 96 should contain apprmdmately 997 of the weeks b Py gt 160 Pz gt 160321 Pz gt 109 1 08621 01379 493 Pz gt 1645 005 k 125 164532 17764 a Facility size should be at least 178 Pz gt 2326 001 gt 1 125 232632 1994 gt Facility size should be at least 200 494 p 5 a 113 a Py gt 7 Pz gt g Pz gt 154 00613 b Pg gt 55 Pz gt Pz gt 86 m 0 The results of the survey am not consistent 405 1 21 a 03 a Py gt 27 Pz gt 23 Pz gt 2 00228 b Pz gt 06745 025 gt M 21 0674503 230 0 Let pm be the new value of the mean We need Py gt 27 g 005 From Table 10105 Pz gt 1645 and 005 Py g 27 H gt 21 5 Law 1645 gt p 27 031645 22065 496 Mean for eet is up 15021 315 standard deviation for the eet is g 15003 367 497 Individual baggage Weight has l 95 a 35 Total weight has mean up 20095 19000 and standard deviation W 20035 49497 Therefore Py gt 20 000 Pz gt 3 0233300 Pz gt 202 00217 498 5 P39y s 150 Pz g Pa 5 05 03035 160 160 7 7 b Py 5150 Pz g m e Pz g 112 7 01314 c Pz g 2320 001 a 1373 2326 gt 11 2164 At least 22 measure ments would be needed Supplementary Exercises 499 No The last date may not be representative of all days in the month 4100 a 1 mr 100000001 10 b a W 316 Py lt 5 Py 5 4 m Pz s 4330 Pz lt 174 00409 c Py lt 2 P0 g 1 Pz g 15150 Pz lt 269 00036 4101 a Py lt 5 5 lt452 G 5 5gt 9 5 5 23quot535 ZEN345 0059 200505 2236 32gt 24 00125 b For the binomial distribution Py lt 5 Py S 4 Thus we have Py lt 5 Py s 4 a Pz 5 45231quot Pz lt 246 00069 c P8ltylt14P9y13 29 15 5 lt335 lt51 i5gt 095 5gt82251007 06906 Using normal approximation with correction P8 lt y lt 14 P9 g y g 13 Py g 13 Py g 8 3 Pz 5 131336 Pz g 823quot Pz lt 157 Pz lt 067 06904 4102 n 101 5 0131430 6 Py4Plty 5Py6 3 5 5 5555 000605 065625 b 1 1005 5u 100505 158 y 6 Pz lt T Hz lt L53 z lt 63 Pz lt 63 04714 It did not work well 4103 P4 g y g 6 Pz lt Pz lt P0 lt 95 Pz lt 95 06579 With the correction the approximation is very accurate 4104 m 11 500025 1250a 150002575 3062 ys1ooopzg whng s m b Py 2 3000 Pz 2 W Pz 2 5714 m 0 4105 a Py gt 2265 Pz gt M PZ gt 147 00707 1 Approximately normal with mean 2250 and standard deviation z 263 4105 ag 9 253 4107 Pg 2 2268 Pz 2 W9 Pz 2 684 z o 4108 No them is strong evidence that the new fabric has a greater mean breaking strength 4109 a The normal probability plot and box plots are g39van here Nolmnl Pmbnbility Pm Ming 91mm 9 W Mun smut m 11H1 Note that the plotted points deviate from the straightline b Since the population distribution is skewed to the right for this sample and with the sample size only 2 the sampling distribution for y will be skewed to the right 4110 my 5 1 7 35 7 3 1 00308 The probability that one or fewer of the ii is is bo t 3 c ances in 39 able occurrence is a rare event Having actually observed this event we could draw one of two can lusions Either we have observed an unlikely event or the board is currently admitting with probably less than 07 We would tend to accept the latter conclusion 4111 Could use random sampling by having a computer randomly generate social security num to match income tax returns 4112 a535 0012 a Py gt log250 Py gt 552 Pz gt iffy 5 00078 078 P009050 lt y lt 109950 P501 lt y lt 552 P 5 5 quot1535 lt z lt 5 5535 09194 9194 c Py gt lag300 Py gt 57 Pz gt 3335 00018 018 o 4113 a The random variable y is the number of people out of 1000 who responded that they were planning to buy the reformulated drink a Y is a binomial variable and so has mean mr and variance mr1 1r We do not know the value of the parameter 1r so We cannot compute the true mean and variance We can only estimate them by using the sample proportion as an estimate a Using the sample estimate of 1r we can compute P y S 250 by the normal approx mation if 7m and n1 7r are both greater than 5 4114 a The random variable Y representing the number of sales in 5 calls is binomial With 1r 001 The event that the rst sale occurs in the rst ve calls is the complement of the event that no sales occur in the rst ve calls Thus the probability of interest is 1 Py 0 1 g01 995 0049 If the rst sale occurs after 10 calls the first ten calls must have resulted in no sales Let y be the number of sales in the 10 calls Py o 3901 991 09044 6 4115 n400 7r02 a p 2 Mr 2 40020 8017 2 40028 8 a r 2 g 6875 e 0 The ad is not successful With 1r 20 we expect 80 positive responses out of 400 but We observed only 25 The probability of getting so few positive responses is virtually 0 if 1r 20 We therefore conclude that 1r is much less than 020 o l U 4116 n 200001r 00001 There are 2 possible outcomes and each birth is an independent event We cannot use the normal approximation because Mr 2000000001 2 lt 5 We can use the binomial formula Py 2 1 1 Py 0 1 2030 0001 99992 m0 8647 4117 The sampling distribution of the sample mean consists of the following values for 37 and t eii39 frequency of occurrence Ps7 37 H 2 P07 17 P01 37 Pl 1 70 1700 1 70 2150 270 2600 170 3575 170 725 1050 170 1725 170 2175 170 2625 0 1125 170 170 2225 270 2650 1150 170 270 2250 270 2675 1200 170 170 2325 270 2750 1225 170 1875 270 2350 170 2325 1300 270 1900 170 2375 170 2900 1400 270 170 2400 170 3000 1475 170 170 2425 270 3075 1550 270 270 2475 170 3100 1625 170 270 2500 270 3150 1650 170 2075 270 2550 170 1675 170 2125 170 2575 170 3250 170 thD b IOND AND b lb h Eh hh hhhh oooooccooo 4118 a The normal probability plot is given here Normal Probabilily Plot for the Sample Mean saw man r2 4 o 1 Qunlles m sumn1 Nmmal Note that the plotted points deviate only slightly from the straightLine Thus even though the sampl size is very small the ormal approximation ts quite well The reason for the closeness of the approximation is that the population has a somewhat symmetric distribution b Both the population mean and the mean of the 70 3 5 are equal to 215 4119 The sampling distribution of the sample median consists of the following values for the median M and their frequency of occurrence M PM 75 5 7o 9 o Amwwwmuewwmwm hhhhhh hhhh occoocooocooo 340 570 4120 a The normal probability plot is given here Normal Probability Plot lor the Sample Median 51m Mun 29 I v y l 72 4 o Quanllles m Slenoam Novrnal Note that the plotted points deviate considerably from the straightline Thus the sampling distribution is not approximated very well by a normal distribution If the samp e size was much larger than 4 the approximation would be greatly improved b The population median equals 1 2315 185 whereas the mean of the 70 values of the sample is 194536 The values differ by a signi cant amount due to the fact that the sample size Was on y 4 41121 alb The mean and standard deviation of the sampling distribution of 37 are given when the population distribution has values a 100 o Sample Size Mean Standard Deviation e As the sample size increases the sampling distribution of g concentrates about the true value of 4 or n 5 and 20 the values of 37 could be a considerable distance from 100 4122 a o The probabilites are given here Sample Size P0 2 105 l Pi7 s 95 P95 g g g 105 5 02 02280 05439 20 00680 00680 08640 80 00014 00014 09971 m i L o i i i very rapidly e sample size increases Of course the rate of increase also depends on the population standard deviation or 4123 n 36i 40 12 a The sampling distribution of y is approximately normal with a mean of 40 and a standard deviation of 715 2 b Py gt 36 Pz gt Pz gt 2 09772 c Py lt 30 Pz lt 3 A0 Pz lt 5 287z10397 d Pz gt 1645 005 2gt k 2 40 16452 43129 n 4124 a The sampling distribution of 2 y is approximately normal with a mean of 36401440 and a standard deviation of muE 72 b P2iy gt 1440 Pz gt W Pz gt 0 05 c HEM gt 1540 Pz gt Egg 14quot Pz gt 139 00823 d P 196 lt z lt196 095 a h 1440 19672 129888 gt 70 1440 19672 158112 4125 a Mean 10 Standard Deviation 7135 2 b Mean 10 Standard Deviation 711 1 4 c Mean 10 Standard Deviation 739 p

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