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Intro. to Statistics I

by: Miss Adolphus Adams

Intro. to Statistics I STAT 2231

Miss Adolphus Adams
GPA 3.7

Anna Oganyan

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Anna Oganyan
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This 105 page Class Notes was uploaded by Miss Adolphus Adams on Monday October 12, 2015. The Class Notes belongs to STAT 2231 at Georgia Southern University taught by Anna Oganyan in Fall. Since its upload, it has received 31 views. For similar materials see /class/222000/stat-2231-georgia-southern-university in Statistics at Georgia Southern University.


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Date Created: 10/12/15
Homework 2 Submitted by lindsmu1109aolcom on 932011 124405 PM Points Awarded 32 Points Missed 0 Percentage 100 Make a stemplot of the average travel times to work in the table below Use whole minutes as your stems Because the stemplot preserves the actual value of the observations it is easy to find the midpoint 26th of the 51 observations in order and the spread What are they STATE A lnhnmn 3quot I n 5121 Arizona a lrkzmsas California Cc ih ll altli Cunnucriuut Delaware F L Jr him in irg i a lawAil ILlilhu lll innis ml i and lawn Ku mus Kent Utk39g TIME 236 I71 250 313 268 4 II 5 3 39 Um b ut 39 ru cu ru 11 rd Eu 4 Ema 1 Match your answers below 1 144 6 2 155 7 3 218 8 4 221 9 5 225 10 234 228 290 304 309 A Center point B Largest value STATE lenuisizinn Malini Maryland Mnssuchuwns Michigan Minnesota h39li sietsippi Missouri Iklumnnu Nuhrzisku Pacvudzi Ncw Hampshire Nunquot Jcrwv New Mexico New York North Carolina North Dakota TIME 251 323 WI Hiin 134 120 140 119 176 STAT E Dim Oklahoma Oregon PUI IHSHTFIJHZH Rhoda Island South Crimli n 21 South lilzilmm Tennessee Tums Unih i urnmnr Virginia W ashinmun Wear Virginia W39iscni m in Ruining Flirt of Cnlumhi C Smallest value From the stem plot it is easy to count that the midpoint 26th entry is at 227 The spread is from 152 to W Points Earned 33 Your Response A26 B10 C2 Burning fuels in power plants or motor vehicles emits carbon dioxide C02 which contributes to global warming The table below displays COZ emissions per person from ountries with of at least 20 million Table 16Carbon dioxide emissions metric tons per person Country COZ Country COZ Country COZ Algeria 26 Iran 60 Poland 7 8 Argentina 36 Iraq 29 Romania 42 Australia 184 Italy 7 8 Russia 10 8 Bangladesh 03 Japan 95 Saudi Arabia 13 8 Brazil 18 Kenya 03 South Africa 70 Canada 170 Korea North 33 Spain 03 China 39 Korea South 93 Sudan 01 Colombia 13 Malaysia 55 Tanzania 01 Congo 02 Mexico 37 Thailand 33 Egypt 20 Morocco 14 Turkey 30 Ethiopia 0 1 Myanmar 0 2 Ukraine 6 3 France 6 2 Nepal 01 United Kingdom 8 8 Germany 99 Nigeria 04 United States 196 Ghana 0 3 Pakistan 08 Uzbekistan 4 2 India 11 Peru 10 Venezuela 54 Indonesia 16 Philippines 09 Vietnam 1 0 Data set 2 Why do you think we choose to measure emissions per person rather than total COZ emissions for each country A Measuring emissions per person allows us to compare between countries with different populations B Measuring emissions per person allows us to deduce how efficiently each country is converting fossil fuels to useable energy C Presenting the data as emissions per person allows us to understand various factors such as the fuel efficiency of cars in each country D Using emissions measured per person gives us information on the rate of deforestation in each particular country The correct explanation is provided in answer A B This information cannot be deduced from the data C This information cannot be deduced based only on the data D This explanation is totally irrelevant Points Earned 11 Your Response A 3 Make a stemplot to display the data Use Whole tons per person for the stems and tenths for the leaves Compare your result to the stemplots below and choose the correct one Stemplot I Stemplot II 0 111223333489 0 1235789 1 0013468 1 2478 2 069 2 358 3 033679 3 6789 4 22 4 68 5 45 5 1 6 023 6 18 7 0889 7 36 8 8 8 18 9 359 9 17 108 102 11 11 12 12 138 13 14 14 15 15 16 16 17 17 184 18 196 199 Stemplot III Stemplot IV 0 001122223357899 0 001122223357899 1 02478 1 02478 2 3558 2 3558 3 67899 3 67899 4 68 4 68 5 1 5 1 6 18 6 18 7 36 7 36 8 018 8 018 9 017 9 017 1002 1002 110 110 12 120 13 130 14 140 15 150 160 160 170 170 18 180 199 199 A Stemplot I B Stemplotll C Stemplot HT D Stemplot TV g Points Earned 11 Your Response A 4 How would you best describe the shape of the distribution A Smooth and symmetrical B Slightly uneven and skewed to the right C Smooth and skewed to the left D Totally random The best description is given in B y Points Earned 11 Your Response B 5 By counting your entries in the stemplot what is the value of the center There are 48 countries in the table so take the center to be the 24th entry in your stemplot A 20 B 23 C 33 D 38 If we count 24 entries from the beginning of the stemplot we reach 28 v Points Earned 11 Your Response C 6 Ignoring the outliers what is the spread of the distribution A 01 to 102 B 00 to 102 C 00 to 199 D 01 to 108 By examining the correct stemplot HI we see that the spread without outliers is from 00 to 110 Points Earned 11 Your Response D 7 Which countries are outliers Classify the countries in the list below 1 Is an outlier 2 Not an outlier A United States B Congo C Australia D Canada E Mexico F Russia G Thailand H Germany The three countries that are explicit outliers are Canada Australia and the US Points Earned 88 Your Response A 1 Here are data on the average tuition and fees charged by public fouryear colleges and universities for the 1976 to 2007 academic years Because almost any variable measured in dollars increases over time due to in ation the falling buying power of a dollar the values are given in constant dollars adjusted to have the same buying power that a dollar had in 2007 Tuition T uition 2197 2426 4221 1977 2225 2532 2001 4411 1978 1986 2656 2002 4715 1979 1986 2699 2003 5231 1980 1939 2721 2004 5624 1981 2018 2792 2005 5814 2194 2977 5918 2358 3187 6185 Data Set 8 True or false The time plot below correctly displays the average tuition and fees 500m 50004 4000 3000 2000 KM 1000 1quot1 39l I Iquotr 1975 1980 1985 1990 1995 2000 2005 Year Average tuiion and fees J Points Earned 11 Your Response True 9 What overall pattern does your plot show A Strong periodic variations B An upward trend C A downward trend D None of the above The dominant feature of the time plot is the upward trend in the average tuition v Points Earned 11 Your Response B 10 Some possible deviations from the overall pattern are outliers periods of decreasing charges in 2007 dollars and periods of particularly rapid increase Which are present in your plot and during which years Match the correct description for each time period below 1 Increase only 2 Decrease only 3 Fluctuates A 1976 to 1980 B 1981 to 1985 n mm m mu 1va w man 1 mm w mm mm B m nan am mm mm m w mm wdhs m mm mm mama mu w m m hi m m yz mmsmm maskmlf yudsmz hada m mu m Omega3 my acids as a hadron Momegws any in God oris on on n11 named Wan Cam Wm germ vbm Mustard Gray see Smdm Mmhmsn alum smug Mavnnnms 3mmquot Cad um 1m bmn Shortemng household Bum Shortemng mdusmnl Sun ower 1m e m om Sesame 51m m Cmmuem Sundmver 012K Palm m1me 1mm Cozm mm mm mmmmmanmamme a mm mm mm ma m an a WW Whydummxmmvhm xnmmm mm mdlm H mm manw vfdum warmlqu ma Indus ram WWW mwyw mm Mark mmmmzmmm mywa mmdmusuz n75 hh mm mw m a than Emma m mummmmmm lugummmmann 39 umgmmbahwlwwsumd 15krwwlmhuzmmsfmms Mm quuellcy J 1n vfmmyltemmzanszmdhy mwmmm HARn um w mzsnwmnnxnzss mm m 1n vfmmyltammwnszmdhy mMmmm HARn um w mmwmmm mm m eHomework 12 Points Awarded W 6 Points Missed I Percentage 83 Published reports of research work are terse They often report just a test statistic and P value For example the conclusion of Example 149 describing executive39s blood pressures might be stated as quot2 109 P 02758quot Find the values of the onesample ZStatistic needed to complete these conclusions 1 Example 147 describes a study of sweetness loss in colas The study tests the hypotheses H0 u 0 Ha ugt 0 For the first cola the 10 tasters found mean sweetness loss X 03 Suppose we know that for any cola the sweetness loss scores vary from taster to taster according to a Normal distribution with standard deviation 7 1 For the first cola in Example 147 Z 7 P 01711 Give the value of the test statistic rounded to 4 decimal places Fill in the blank X 03 i 0 1 09487 own 1710 g Points 11 Earned Your Response 0948 7 2 For the second cola in Example 147 the data gave X 102 Z 7 P 00006 Give the value of the test statistic rounded to 4 decimal places Fill in the blank X 102 i 0 Z 0 32255 1an 1710 g Points Earned 11 Your Response 32255 3 Example 148 describes a study of job satisfaction of assembly workers The study tests the hypotheses Suppose we know that differences in job satisfaction scores in the population of all workers follow a Normal distribution with standard deviation 7 60 Data from 18 workers give X 17 New um 7 mm swam Mmmummam mmmg hm m mwmmmmw Huvm mmmm szzosxs mst mu m Wxnwpasadmluw mum Dammmmg quot mamm mw murkyxsmm m mwmm mskstmmywmdmmuhnnguuwe wwmmmmm 5 mm nummmmmmmmm m mwmwmmmmwy 0 sun mm mmnmdmd m Ismmammmnmmummnmuimwrmm v mm 5 mm 1mmka mum p11 w H H 311 km a m 2 w v a 35wemltmamdmmndmmmfm5namdmmmsammau wasde in nu uIwn A 1 sunI mummmsmMmMWhm Ednumduusmm mmmmxwzdum mm mmwa 5a mmmmmmmmmms yxxzmumm um l min mm m mo mm s 1 21m in mm mm mm m 19 l ammo mm mm w m mm mm m mmMuf n mum s sm xm yd mm m mm m mph WMMM Wham WE nah u 1nquot m mMnmmmmgmmmmmmmmmmmmm WWW AnsammmsmuWmmwmnwammamlmhkmmahmx aquot s h Inmaukswnhmnsnmmu C H0 ult 18 The null hypothesis states that utakes on the quotdefaultquot value 18 seconds J Points Earned 11 Your Response A 11 Experiments on learning in animals sometimes measure how long it takes mice to find their way through a maze The mean time is 18 seconds for one particular maze A researcher thinks that a loud noise will cause the mice to complete the maze faster She measures how long each of 10 mice takes with a noise as stimulus The sample mean is X 165 seconds In Exercise 1431 you stated the null hypothesis for the significance test The alternative hypothesis for the significance test is A H u 18 B H ult 18 C H u 165 The researcher believes that loud noises will decrease the completion time so the alternative is onesided Points Earned 11 Your Response B 12 You use software to carry out a test of significance The program tells you that the P value is P 0031 This means that A the probability that the null hypothesis is true is 0031 B the value of the test statistic is 0031 C a test statistic as extreme as these data give would happen with probability 0031 if the null hypothesis were true A P value gives the likelihood of observing results at least as extreme as those in our data when we assume H0 is true g Points Earned 11 Your Response C 13 A test ofHo u 1 against H ugt 1 has test statistic Z 1776 Is this test significant at the 5 level a 005 Is it significant at the 1 level a 001 1 Significant 2 Not significant A At least 5 level B At least 1 level For a onesided test Z 1776 is significant at 5 but not at 1 To see this either compute the P value for Z 1776 which is P 00379 or note that 1776 falls between the 5 and 1 critical values from Table C 1645 and 2326 respectively V Points Earned 22 Your Response A1B2 14 Atest ofHo 471 against HR 41 1 has test statistic Z 1776 Is this test significant at the 5 level a 005 Is it significant at the 1 level 1 Significant 2 Not significant A At the 5 level B At the 1 level For a twosided test Z 1776 is not significant at either level To see this either compute the Pvalue for Z 1776 which is P 00757 or note that 1776 is smaller than the critical value 1960 from Table C which corresponds to the twosided Pvalue 005 Thus Pgt 005 393 Points Earned 22 Your Response A22 B2 15 When asked to explain the meaning of quotthe Pvalue was P 003 a student says quotThis means there is only probability 003 that the null hypothesis is truequot Explain what P 003 really means in a way that makes it clear that the student39s explanation is wrong A P 003 means that the outcome is at least 3 higher among one group than among the other group P 003 means that there is a probability of 003 that the null hypothesis is false P 003 means that the probability assuming the null hypothesis is true that the test statistic will take a value at least as extreme as that actually observed is 003 P 003 means that the probability assuming the null hypothesis is true that the test statistic will take a value at least as extreme as that actually observed is less than 003 053 U quotP 003 docsmean that H0 is unlikely but only in the sense that the evidence from the sample would not occur very often if Ho were true P is a probability associated with the sample not with the null hypothesis either H0 is true or it isn39t Points Earned 01 Your Response D Athletes performing in bright sunlight often smear black eye grease under their eyes to reduce glare Does eye grease work In one study 16 student subjects took a test of sensitivity to contrast after 3 hours facing into bright sun both with and without eye grease This is a matched pairs desi n Here are the differences in sensitivity with eye grease minus without eye grease 007 064 012 7005 7018 014 016 003 005 002 043 024 011 028 005 029 We want to know whether eye grease increases sensitivity on the average 16 What are the null and alternative hypotheses n n c m MIAn m n Museum mm mmsuf wmgyaykwnhmmmmmltmm mmmstauwNmamwmnmmwm zmmnmm mm an Fm ammummm s39mn Marl1k Inkva w mm guumummmumm mm mm 9mm 9mmquot smm Impnanny mum w 1 hi 1 y m 1 mm w mm mm mud nym a sunI mm mm u huulgam39r mmmm mum mum hm n lnlstawunmmx mu m an suLvE mmd mxm Pm MAW Evmymlxmswumzdumlybus nummm Wampmndsmquumanmm mph M 2mm mpmmn lnn mm mm mm 5va WW pmmm w v u u 2 2 z y a x2 x W me m mxqusuumpa mu m n ma mwAwn m awusmu 5n swam nu um m3 mm mums uan a 95 MW 5mm am mump mung me nu mm aswmmmwwnpwmamuvmq 7m unmanme aw mum mm mm mph m mm mm mm a n u m 1 uu f Wm mum asu umpqm m 4 mm my mme 5M6 idmls nu mm mm Wwsssm ms mmmm mmvan M v mum m mm Janmm u m n nu an an X mud mu wrap wmmxmmnzwndnwmxumasn m 1n suwmmm mxm mm MAW mmmmmmm numb Useymlxnkm ummImmvdmufunm mnum x may ml 35 summon mmmmmn A m dnadnmumm gnaammmmmmsmhmmmams menhmZSAQl 1 mph msmmwmmmmmmmmwm mum st nmmu msmrgmmmmmmmmmshnmmgvmumz m n msupswxmynwpumvhmmmsmsn wiyndmm mtadms mm Emlmamm mm m BMWMHW mmmm WsmmmshnMBgamm 25M m m mn a as my mammfmmfa mlgzmd m n mmlvmfmm 1anme mmde 9W 1n n mmmmmmmmwmmmw 11 mm Lzm mm mm m m mmu Wuhwwdum mam nm m39hbk mm winn D sml nu mu ymmmsks snzsmnkuhumm 51mph n mxm mam rmmmmmm nmmwvm meuhwwvlu m mde m 1 mm gm mm m Yn39 a mmmm amuwm mmhIIgmbb uyufn um me Mum mum utknsswummnuummhnmmndmTafmmumuwmmnkmmbu m I mbmvlu or may n15 mm mm gm napalm run ms dammdlpammhnwhwpumxamsamdwm Huv sma ummmunmsmu m Tan2nwhjuumnvhayhwaemgwnssnmmmkup1m mmmmmmm5mmmsbwmmmymmmmm s amm zhnkasmnmmmdaun bzhnxesbawemImdmmnmmm mh mfwdwyumsy hqhkmmy a uehyexmd mmes lase mquotKKP jmmmmus mm WWW damaka 3753mm mum mmmm zmmmmz mm 1n525n77 MUMva kMeWW mjmdwmn mummquot vamammmmmmmn m mmm 51615 me wmm w mmnmmmmum mam n7sxx am 57 no 7x v 1 hi 1 n n A n Mun mm mum mmmamm msmmmdm m 39 Emu m quotmama mmme mmyaxammuk m mmmmmm mm m mam mm xmuxnvsr mmsz mun vhlyvuhwwVht mmm m durum u 1 mm mm mug lm am m 5125th yum1m mom 1 v 1 u wmmm ifuwfnkmm mm a n aquot mx mam mm mmbhady sm a mm yummammm mm eHomework 4 Points Awarded 24 Points Missed 2 I Percentage 92 1 After the physical training required during World War II the distribution of mile run times for male students at the University of Illinois was approximately Normal with mean 7 ll minutes and standard deviation 074 minutes What proportion of these students could run a mile in 5 minutes or less A 00022 B 09978 C 07032 D Less than 00022 The proportion of students who could run a mile in 5 minutes or less is about 00022 Xlt 5 corresponds to 5 r 711 Zlt 285 074 for which Table A gives 00022 g Points Earned ll Your Response A 2 Emissions of sulfur dioxide by industry set off chemical changes in the atmosphere that result in quotaci r in The acidity of liquids is measured by pH on a scale of 0 to 14 Distilled water has pH 70 and lower pH values indicate acidity Normal rain is somewhat acidic so acid rain is sometimes defined as rainfall with a pH below 50 The pH of rain at one location varies among rainy days according to a Normal distribution with mean 54 and standard deviation 054 What proportion of rainy days have rainfall with pH below 50 A Less than 07704 B 07704 C Less than 022 D About 02296 The proportion of rainy days with rainfall pH below 50 is about 02296 Xlt 50 corresponds to 50 i 54 Zlt 7074 054 for which Table A gives 02296 Points Earned 11 Your Response D aasaayaxamm 9mm vfmhmmswlkmuwaMfm mms mxhanlmzsmba yummm m mmmemmaMam aammmm mmmmstmm smmsmmmmwmzanmm Wymmamummsluwhamxm am 13m Aaaaaay I my ma Mam mum amsma mmam azaaam Ma mammmmw Ba 0 Z xa aawaaamaams Ham nmxx sxv hunt m u a an In 0 Whumuma lu mamas am a ubww B 39r nu I39ll m n l c 5 a wpmmm a mmmammmmfknwihdmmmm a ins a uxml 39 u mam Indus am Wy mlhngmzm mam Wannava Nnmsn a aammm 39 u aauaama dlnwslmwldzhsbemm mnmnn mam whwrvpm a m mm spumnmmsquot Ahmnn am Ma a auaaaammma x72nltxltn X7X zmwpmm nxnnrnxun lt lt nmsnrnxun nnnu nun a 7 5nltslt25n a aawaaamugm am aaaaz am 1 hi 1 Ayhzrmzamdmnfunrufm mmmmmm mmum m m um wmMWWW Minna mun awymdwmuhhsmsbmwmm mm Dan1mm m was mm am typing mu mquot m m 122 kg Wymm mm x mm m mum Mama mum mmmmm mm ll7ngultxlt122zmgpmw um uer m m 7 5 lt1lt25 n2 n2 39 mmhmugm um um mmemPmmwmx nxnnmmmsmmmaammnmxm Whululumdmm mmmmmm Wrmmm mxhwquot Gmymlxmmmm hmnmw mums nummm mum mmmmxsdusmuvhmm mum y nun t m up um mmmm Maw Us mummmmmmmm quotmm W mywa n 7mm mtmwmxmvmwmzswuiwammmmmmua umndsmuuhmn mg m m um mmm qu Us I AIII QWymezwwu39 mummsmb mmmg nummm XSZZm mqmuwmmyam mimmmnmanmwmm rwwroaruro rmrnm r m mu 1 m Yh nsszz Us mummmm mmumwummwmm anamme Useduwhavfsm39hhh Annuamsdnsuwsmm mm H Mnmhmmmmmmmmmwwwmmum mmwu mkmmam ms I ms mm mm lam m mum Awafm mw mmmMmmmmbmnmghuman B 57 MIAn 1 mm 2 n mmmumuhumunv maummxmmummmm gmum m I m mm m39hbk a mum mayan vfx nwevu39mdmnsawe fmgwwhu xmwmyny mm mm mum wwwme mm n25wmhmmmuin5vx7 mums1mmmmyynymmmmwmm m m nu n a u malphmm maximum mmmznnxmauwhuksw Ehvmmmml nwzm gmygmdmmuszwmks mm unmmememmy muslst H mm mm qu mmddymhdnwysumge m m uzanmmmuwmmmmm7msmghmam deWOKEPS mam AhmIUZWVy mm KIX763demxlt25zmaauzdsm 25 r m p m oz mm mm A gm my my n h m yum n A 2 maximum mmmznnxmauwhuhsw Wham ngsdmmuszwmks n x mmmmmmmwm gusme H mm WWW mmddymhdnwyasmgeatm m hsxwmmuiy mwihmm xx 7msyuglllmMQm Hnwhdnmmznnxwhuhk gunkgun nmduwh n uwp my mmw mdmnm mugqume mmmw mpg mm mm mm Afmuuyrvpmumzhsumn my 1 2m memm mum mm my mwms mmmhmmge mm m mm mm IhwthAmm xx mmxmm mmm m ht m Iu n D Wmmm lemmwmmm mummyhymn 1mm WWW mm wk sum dun wills B mum mm m sarmm m m m h u mznnhm m inan mnh Sanade mm Imam Wymm vi mnswnd 7m WW m glimmmmnkms l szmeMadamdmw smus mm Yunmnrmym 18mmde Zdumyhzas m 1 warm wwwmmmmmmm m m mamdmmmmwmw 1875aweluw H mm mmth I h m r Ian 11 mum smmm mu141 de now un mm Wymm vfwmn and 7m mbmg39r mawtmmyuumgemwmdqmmn sm MW WW mum 7 smyummnrmyun zmmam Zdumyhzas Inme 7 2m wwwmmmmmmm vxxl ammmmmmwvxzzxsmm H vxx7n unzth m ht m n u A W a mum Pammufm mm m s mum msYhl mm mmm Wm mm m 3 mm mamast m mbe zw mzmmmy ma mmsm F vfessmn wudm39s mm m afsm mymm mag sub n Maummsyam szmsn Iyn my ma yammsm m ksuhm numd c mmmkmmm mummy Wmmmmm mm mmn H mm mm I425n mngl7v7 c 2mm 7 n mmmxzdndxkdhymmgdu mwmms ummwmmwzxn Pwmum mum WWW msamfm39 mmw m 7 Wm a m a um um nmmmsutmmm xmwmmm o a 7 054 054 051 554 554 054 554 055 055 055 555 555 501 o z w mw m W I Ade xemd ym mm m vi 15mm mumwme mun wihm mums mm m m mm mg mbdmv Histograml WE TE 3 12 8 4 BUD D 131 I D Em DEE D7 D75 DE DES D9 D95 ma Lengmmm Histogram III E E El D5 DEE I7 I75 DE DES DB EI39BE DES Lenglllhnm W I ng mn C Histogram 111 D Histogram IV The correct answer is histogram 11 Histogram ll 3915 5 12 2 ET 8 LI 1 I 15 DEE ELquot DEE DB ELSE DB I195 Length mm quot PointsEarned 11 Your Response B 19 Find the mean median standard deviation and quartiles for these a Match your results below 1 00600 6 08004 11 08000 16 07600 2 00400 7 00312 12 07512 17 08500 3 04749 8 09400 13 08600 18 00050 4 00782 9 00590 14 07764 19 07605 5 08453 10 05306 15 00681 20 07800 A Mean B Median C Standard deviation Dr 01 E 03 eHomework 11 Points Awarded W 7 Points Missed I Percentage 83 The National Assessment of Educational Progress NAEP gave a test of basic arithmetic and the ability to apply it in everyday life to a sample of 840 men 21 to 25 years of age Scores range from 0 to 500 for example someone with a score of 325 can determine the price of a meal from a menu The mean score for these 840 young men was X 272 We want to estimate the mean score u in the population of all youn men Consider the NAEP sample as an SRS from a Normal population with standard deviation 7 60 1 If we take many samples the sample mean xvaries from sample to sample according to a Normal distribution with mean equal to the unknown mean score u in the population What is the standard deviation of this sampling distribution Give your answer to 4 decimal places Fill in the blank The standard deviation is a 60 2 0702 V11 V840 39 39 v Points 11 Earned Your Response 20702 2 What is the 95 confidence interval for the population mean score ubased on this one sample A 268 to 276 B 270 to 274 C 266 to 278 D 152 to 392 The 95 confidence interval is Xi 20 272 120 268 to xn i V840 276 y Points Earned 11 Your Response A The critical value 2 for confidence level C 975 is not in Table C 3 Use software or Table A of standard Normal probabilities to find 2 A Z 196 132k 0838 Ir m nr222 m mam rummumumus a rszsrar c Ln39sm mgdmw mmmm mm A Jammy cm wwwqu mmm rm mam mm mmwnhnrpmwm leammmmzmdmm m m mammxnmmmmm mums auhmlusmxuvf msamwmmm mW mmm Wm szn rm WNW HZ 717 m x m m mm mm Awe mum B mum r 2 2 NW mum mumth m wash Wm m munm a w m ht m Iu n Wham mammgmmmm mum mm mumMlmuzh mm amineWsm mm m m A Sketch 2 The critical value is 2 224 and the area of each tail is 00125 such that the total area under the curve is 2 X 00125 0975 1 v Points Earned 11 Your Response D The National Institute of Standards and Technology NTST supplies quot standard materialsquot whose physical properties are supposed to be known For example you can buy from NTST a liquid whose electrical conductivity is supposed to be 5 The units for conductivity are microsiem ens per centimeter Distilled water has conductivity 05 Of course no measurement is exactly correct NTST knows the variability of its measurements very well so it is quite realistic to assume that the population of all measurements of the same liquid has the Normal distribution with mean u equal to the true conductivity and standard deviation 7 02 Here are 6 measurements on the same standard liquid which is supposed to have conductivity 5 532 488 510 473 515 475 NTST wants to give the buyer of this liquid a 90 confidence interval for its true conductivity What is this interval Follow the fourstep process as illustrated in Example 143 5 STATE What is the practical question that requires estimating the parameter A What level of confidence will be reasonable for estimating the conductivity of this liquid mums m mummmmmmn m mmmmmamwmm mm vi wmasv hswquot mm m m ht 1 run 9 1 sunI mm WWW llama mam A Yes 0 mu mm mm mm mm x 5mm weluwmsksmm uyvm mmuf mmm vi m mmw mxummmhmsmm mm m n I A I sunI Kma Emu m mam maulme a wrrhuYIIIXY Ham 5 m I mm 3173 m mm 1226 n 5 m3 mmmu h my s m mummmm mun A m mam 1 mmm 1m mm mm m B r m mmnuwumtm n V6 v m Your Response C Here are the IQ test scores of 31 seventhgrade girls in a 1 dwest school district 114100104 89102 91114114103 1 108130120132111128118119 86 72 111103 74112107103 98 9611211293 Data set 9 These 31 girls are an SRS of all seventhgrade girls in the school district Suppose that the standard deviation of IQ scores in this population is known to be a 15 We expect the distribution of IQ scores to be close to Normal Below is the distribution39s stemplot 7 24 8 69 91368 10 023334578 11 1122244489 12 08 13 02 True or False There are no major departures from Normality You have now checked the quotsimple conditions to the extent possible g Points Earned 11 Your Response True 10 Estimate the meale score for all seventhgrade girls in the school district using a 90 confidence interval A 7200 to 13200 B 10141 to 11027 C 10056to 11112 D 10584 to 13200 A 90 confidence interval for uis Xi 1645an Here it is 10584 i 1645 X 1531 10141 to 11027 Points Earned 01 Your Response C 11 To give a 96 confidence interval for a population mean L1 you would use the critical va ue A 2 1960 B 2 2054 C 2 2326 Table C or software shows that 2 2054 for 96 confidence Points Earned 11 Your Response B mm nummmmmmm vi n nlg m yum mm um nu mde delydm bmdmnhmmew m mmmw mmumn mgmnm m mam mmme umwwzggw Rommun A mwa mmy m quotmama mmmm 257mm 31m mum m ml v n h m a n Ammummvpam mnkhmmmthtymadm m 25 mm 1 hi 1 n u a mum m m mm mm mmbadyms mxmmm vfymmg m Mm mm mansyi v v mummmm Embawunz mm 5 m MW mu mmsz mm I m mnunv wabuh iymnmwmmnmmuz xmn mvsv mamwm m as ammumvswu mmzmdmm mamyum m s mm quotmm m PVmIb wnmm 1 MB m u Amm mwm mam Mfmwmmbadymsmx vfymmg mxmmxz xi mwmmmmw mm wwwmmxmmmmmmmm m mums w mbavsmmmmnmm smphsufymmgwmnwm mmm mbmmzmmz humawhmmwmq bwkmym x A mm I m WW3 mzmnu m mm smkswmhm mammm vi m N mm c m WW3 mzmnu m mm smkswmhm mm mwhxhx mm a mu mum mmmn mum lwhmmnduyxwlmuamzm m vimn smmwmma mum mmmmwihm1n aim mg 1mm mmmmzs x mm mmwmmmmm zhsammnm mswh mmnhymmyubsamwm n Mewkmmumwmu m mmmmmmmsmmmmm mm m ht 1 mm a Mn N Inaz mmmdmswmmmmm mxmnmunso Wmn m m mm m m imp ms wanna a s m Sksfm ummmammmm Wadumum 5 17 Give three confidence intervals for the mean BMI uin this population using 90 95 and 99 confidence Match your answers below 1 2620 to 2740 6 2650 to 2710 2 2604 to 2756 7 2705 to 3355 3 2623 to 2737 8 2632 to 2728 4 2678to2682 9 2235to3l25 5 2672to 2688 1 2510to 2850 0 A 90 confidence interval B 95 confidence interval C 99 confidence interval The three confidence intervals are given in the table below in Xi 268 i 029332 Vn Cnnf bowl lrnl39cwnl 2631 m 3738 3613 m 1737 4 m 1756 x Points Earned 13 Your Response A23 B28 Oz 18 What are the margins of error for 90 95 and 99 confidence Match your answers below 1 07555 5 00125 2 06132 6 08915 3 03108 7 05748 4 04824 A For 90 confidence B For 95 confidence C For 99 confidence in The margin of errors are given in the table V n C 1 ran a Points Earned 33 Your Response A24 B7 C1 19 How does increasing the confidence level change the margin of error of a confidence interval when the sample size and population standard deviation remain the same True or False quot The larger the confidence level the larger the margin of error of a confidence interval The margins of error increase as confidence level increases 9 Points Earned 00 Your Response True Example 141 page 360 described NHANES survey data on the body mass index BMI of 654 young women The mean BMI in the sample was X 268 We treated these data as an SRS from a Normally distributed population with standard deviation 7 75 20 Suppose that we had an SRS of just 100 young women What would be the margin of error for 95 confidence A 075 B 147 C 154 D 123 Wichk 196 and a 75 the margin of error is X 212 Va 7100 1394739 v Points Earned 11 Your Response B 21 Find the margins of error for 95 confidence based on SRSs of 400 young women and 1600 young women Match your answers below 1 0617 4 03675 2 0385 5 0098 3 0735 6 0308 1 ya x 7 5 man 735 m z n m 2 WWW mmm Hnwdmsmusngdu imp s mmm mm ummma mm th mm meymhmmnmnmm my mudUse W an mu m bxgu mgnvfm39mvfum mmu nunl Mngxlufmdunuas 0mm makuynmmmhmnmm s mum mm mm mm w 1nquot m 2 Mn N mg zanmmammmm mex mhnnmxm mnmmmmm mm 75 mummummmmmmmmmmsmmwmmw mmuwnhvswzmmm Emuymlxmswu sw m A v 3quot m m 1mm rmpm u Ma lmszms mmmzsymmm ymufduummdmsesmm wimummulhvgw AnnuifalhwNmm mmdmmmpsn 1mm mmmmmwmnmmm muggmsxsammmmww Emu mswensuwhnhmlmhu39 mm Mark 1mm m Akbmmszdarhwmwmmdmmmvf n nlg mmnyamd mm um mmde mummydmmAdMnhmmw m mmmwmha mum Mmmsmwnmd mumnmmmwgu mmvfmmm mmnnnswnhxnv Yummy x nusum nu a Inc vi mm m Tarxmhq 39xs39n mm Waw Wm m dumxlnualu smhugdmhmmufm Nm wmmmSWmmmmnddAvmn mmnnstxm n m umth mammth mm anNurmkmwahnmkmm mm m mmmmmzm durum methum umam1 mmsa uzmadm 255 337 418 5 582 61 478 484 492 5 508 51 risme z n Nm wnhmmS my magnum um v n h m mmmma vxuhssmn 25mmbewhmmdmmm u E mugvalust ymmmmamumummamnmm memmmammms spumdlymwglmmxuu I llu m hn a mhamlsn swplus Wawwm nmumlmuuimmhmnr ghmkmmviyr wpasedwbuxxdM khnxusmwmmnm mwzmmmvf Wummhtm m TmmsduzmmIzYInly um lemdmmn39ma smvhmgVly whyms39rnkassmmms 1mmmmmmmmsammAPWmemyassmummm mymnmh mm m Mammy mmm sfmmmdyufahzmdmmnmdasmmdm u mumn 5 m mm 167ubww as mm mm mdmm 5W H pm a m W x mmmumxms mmanwmwm m mammm mint 1 n c m mm mung mm m sn dkl g d gsamwmmxm 3 sn MWessmmswlmumssmBPamuummmuk mum mm mmxmmxmmmzmsumk m 1 smm MWMmbmmmmnmmmuf mm mushsaduWiessmzmpueswxmmxmwnh summiwmhsasmma mm H saw a 4lt5n summmsmmwmmmmmmmmmwumsm m l hk 1 run D u mwnghw lxwuammummmbzms ymwwmwmmmmmhwuimwmtm gmwmmm Mszhalud mmmmmmmwmge ym Whipssz mm mm mumvhuhsww Bmmmrmmmmmysamme m mamaquot mm 1 1 1n n A Mn N s new Mm mm mm m n abs Hummus mi mm hsmmmmmmm mm s an mm m mszmdw whmu mm mm m m mm quotLgmum Mumhdym ewumm a m mmmkmm wtglmm m mum mam DWIht m vi mm we knwdumr my m vi m W mmsshs mu swammum H n H pa 79wsmammswam lassfm39mwhs mm9mmms39szms mm mm mm mm mm mm abs VT 17 m 1 3a nwmqnmen a m m on am no 1 me m and ms Wm x z w 5m g mmmma mm m umxm m mumm mn m Punuth Manama m mmmmym mbemuhutmkasuv m lugging 5 n2 n mxm mmdmhmmufmn m ammypemm gm u a Wmm Wmmm imm 1mm mnwmmmm I um dlvumm il unmaldmbmmwnhmmmrdmddamnnzmz Dmmdmhmmwnhmmmrdmddgmnnl mmmmhypmgxwzn mu1m 31 sawh n4 vhadm bnwnvfxx mum A Yu n39u n m ummmuywmm mm PmeunhA m an m p n z r n Mm 124 n 1 m 1 n n In mm mm my 5mm mems 9a m m swam vi summing ammmm s mm mame mymsm mbmhsamgmnmmmdu m s mmm mist m 1mm vi mam mam mm mm mm ywm n wnuf mmxy m my jabsmuummmk u memsummu H w n H w n n m WWW mxmamm may Swan we hwwvhtd fuwes mm swam ms unnamth mud m ywmIJAwnvf mkusfallnw Nmammmmwnhmud mm P an amxmgm K n 11 mmm mum sJPadmvmmmmdu mg 5mm mm mm mm Pm m mbb uyvf mmxuhm 5 mm n mmmum Wm 17 mm mmmm amp m m 1 him a 1 arm mm m hm u 7mm ypmmmnwmm ximmmwe mm Pita u 0 18 Update H0u 0 PH 11 0 IT R tl f Hn u G ESE PVatue shaded area 02302 4093 10 00 CI 00 2000 40 00 I have data and the observed is Y 17 39 lhe P value is the sum of the two shaded areas under the Normal curve It is P 02302 Values as far from 0 as X 17 in either direction would happen 23 of the time when the true population mean is 0 An outcome that would occur so often when Ho is true is not good evidence against H0 The P value in Example 148 above is the probability taking the null hypothesis u 0 to be true thatkaes a value at least as far from 0 as 17 34 What is the sampling distribution of xwhen u 0 This distribution appears in Figure 147 A Normal with mean 0 and standard deviation 60 B Normal with mean 0 and standard deviation 33333 C Normal with mean 0 and standard deviation 141421 D Normal with mean 17 and standard deviation 01852 With a 60 and 11 18 the standard deviation is aV18 141421 so when u 0 the distribution of Xis M0141421 y Points Earned 11 Your Response C 35 Do a Normal probability calculation to find the P va ue A 01151 B 02302 C 03849 Points Awarded 35 Points Missed 4 Percentage 90 You are preparing to study the televisionviewing habits of college students Describe two categorical variables and three quantitative variables that you might measure for each student 1 Select the appropriate categorial variables from the list below A Age B Humanities major C Number of roomates D Sex E Engineering major F Time spent watching TV G Favorite type of TV show Age A quantative variable Humanities major Could possibly be a specific value for a categorial variable such as quotField of majorquot Number of roomates This is a quantative variable that may in uence TV habits Sex The students39 sex is a classical categorial variable Engineering major see quotHumanities major above Time spent watching TV This is a quantative variable that could be of interest in this study Favorite type of TV show This is a categorial varaible that may interest you a Points Earned 22 Your Response D G 2 Select the appropriate quantative variables from the list below A Age B Humanities major C Number of roomates D Sex E Engineering major F Time spent watching TV G Favorite type of TV show Age A quantative variable Humanities major Could possibly be a specific value for a categorial variable such as quotField of majorquot Number of roomates This is a quantative variable that may in uence TV habits Sex The students39 sex is a classical categorial variable Engineering major see quotHumanities major above Time spent watching TV This is a quantative variable that could be of interest in this study Favorite type of TV show This is a categorial varaible that may interest you J Points Earned 33 Your Response A How can we help wood surfaces resist weathering especially when restoring historic wooden buildings In a study of this question researchers prepared wooden panels and then exposed them to the weather 3 Here are some of the variables recorded Which of these variables are categorical and which are quantitative 1 Categorical variable 2 Quantative variable A Type of wood ellow poplar pine cedar E Type of water repellent solventbased waterbased C Paint thickness millimeters D Paint color white gray light blue E Weathering time months Variables that relate the individual paint to one of several groups are categorical variables In this case these are the type of wood type of water repellent and paint color Variables that take numerical values that may also have a unit of measure are quantative variables Here these are the paint thickness and weathering time Points Earned 55 Your Response A21 B21 C2 Dl E2 Here is a small part of a data set that describes Major League Baseball players as of opening day of the 2005 season Player Team Position AgeHeightWeight Salary Ortiz David Red Sox Outfielder 2964 230 5250000 Nix Laynce Rangers Outfielder 2460 200 316000 Perez AntonioDodgers lnfielder 25511 175 320500 Piazza Mike Mets Catcher 3663 21516071429 Rolen Scott Cardinalslnfielder 3064 24010715509 Data Set 4 What individuals does this data set describe A Baseball players B Baseball teams C Baseball positions D The Major League The baseball players are the individuals described by the variables in the table 9 Points Earned 11 Your Response A 5 In addition to the player s name how many variables does the data set contain There are 6 variables describing the individual baseball players39 attributes They are given by the second to seventh column headings J Points Earned 11 Your Response 6 6 Which of these variables are categorical and which are quantitative 1 Categorical variable 2 Quantative variable 3 Not a variable A Player B Team C Position D Age E Height F Weight G Salary Player is not a variable it is just the heading for the individuals column The categorical variables are the team and position The quantative variables are all the rest of the column headings Points Earned 77 Your Response A23 B21 C2 D2 E2F2 G2 Births are not as you might think evenly distributed across the days of the week Here are the average numbers of babies born on each day of the week in 2003 Day Births Sunday 7563 Monday 1 1733 Tuesday 13001 Wednesday12598 Thursday 12514 Friday 12396 Saturday 8605 Data Set 7 Select the bar graph that correctly presents the data 9mm Ear Graph 9mm Bar Graph i w 6 quot A A b s 9 e9 8 9 E 99 an 3 2 f uv cannon Ear Graph Ill sk w A A b s 99 6 gt 9 x 3 g9 46 cf uv Bar Graph IV 9mm x A A f9 w bo 9 4 a my mm tsmsmgummv nrdzy wk mum m mmaammmn whm Hmmawnunlsnkhnmwwgm w s 41 my mme mint 1 n x mm mm mm m um mm m dmmxybaymemd slyu mm nmmsmmmak m mummyka bmmuhazmmmmxmmummbmymmwk nusum m 9 Suggest some possible reasons why there are fewer births on weekends Select the most plausible and logically consistent explanation below A Caesarian sections are usually performed as elective operations on weekdays and thus increase the birth count on weekdays B There are more weekdays than the two days of the weekend so of course there will be more weekday births C The data doesn39t take into account the fact that due to the partial staff present on weekends the neonatal death rate is higher A This is the only logically consistent answer and it is also plausible B This choice doesn39t explain why there are more births per day during weekdays C The explanation isn39t relevant to the number of births v Points Earned 11 Your Response A Portable MP3 music players such as the Apple iPod are popularibut not equally popular with people of all ages Here are the percents of people in various age groups who own a portable MP3 player Age groupPercent owning an years MP3 player 1217 27 1824 18 2534 20 3544 16 4554 10 5564 6 65 2 Data Set To help you plan advertising for a Web site for downloading MP3 music files you want to know what percent of owners of portable MP3 players are 18 to 24 years old The data in the above table do not tell you what you want to know Why not 10 Select the correct answer below A The data only gives the percentage of MP3 owners per age group If we would also know the total number of MP3 owners we would be able to calculate the percentage of MP3 owners in the 18 to 24 age bracket CD Indeed the data in the table does not give us the information we want directly However with some simple arithmetic we can extract the required percentage The data only gives the percentage of MP3 owners per age group If we would also know the total number of people surveyed we would be able to calculate the percentage of MP3 owners in the 18 to 24 age bracket 0 U The data only gives the percentage of MP3 owners per age group If we would also know the number of people in each age group we would be able to calculate the percentage of MP3 owners in the 18 to 24 age bracket Answer D is the only one that is mathematically correct The calculation is as follows 1 Knowing the size of each age group we can calculate the actual number of owners in each group based on the percentages given in the table 2 Now we add up all of the number of owners in each group to find the total number of MP3 users 3 The required percentage is calculated from the ratio between the number of MP3 owners in the 18 to 24 age group and the total number of owners Note that had we known the information stated as missing in choices A and C we still would not have been able to carry through with the above calculation v Points Earned 11 Your Response D Here are data on the most popular colors for luxury cars made in Europe Color Per cent Black 30 Silver 24 Gray 1 9 Blue 1 4 Green 3 White pearl 3 Data Set 11 What percent of European luxury cars have other colors The data in the table adds up to 93 Thus 7 are missing a Points Earned l 1 Your Response 7 1 Make a graph of these data Select the graph below that correctly presents the data 2 Percent 35 30 25 20 15 10 5 Bar Graph D Black Silver Grey Ellue White Green Color Pie Graphl Silver 211 E l 3 I L A Bar graph 1 B Bar graph H C Pie graphl D Pie graph H Bar graphl is the only one that is correct Note that the quotOtherquot category is not required in a bar graph Bar graph H the category labels are wrong Pie graph I is missing the quotOthersquot category Pie graph H is missing the quotBlackquot category Points Earned 01 Your Response B 13 Here is the distribution of the most popular colors for 2005 model luxury cars made inNorth America Color Per cent Silver 20 White pearl 18 Black 16 Blue 13 Light brown 10 Red 7 Yellow gold 6 Data Set What are the most important differences between color preferences in Europe and North America Select the correct answers below A There are no appreciable differences B The Americans have a marked preference for warm yellow tan gold colors that the Europeans lack C The Europeans don39t like white cars D The Americans don39t like gray cars All of the specified differences exist Points Earned 3 3 Your Response B C Below is a pie chart prepared by the Census Bureau to show the origin of the 353 million Hispanics in the United States according to the 2000 census E I Cu Ml All HI ICE 14 About what percent of Hispanics are Mexican Choose the best estimate below A 10 B 25 C 30 It is clear that the quotMexicanquot pi e slice is much less than a quarter thus the best estimate is 10 However 39it is hard to tell by sight if the correct answer is closer to 10 or 15 PointsEamed 11 Your Response A 15 About what percent of Hispanics are Puerto Rican Choose the best estimate elow A 40 B 50 C 60 It is clear that the quotPuerto Rican pie slice is slightly more than a half thus the best estimate is 60 J Points Earned 11 Your Response C You see thatit is hard to determine numbers from a pie chart Bar graphs aremucheasier to use 16 In drawing a histogram which of the following suggestions should be followed A No bar should be taller than it is wide B Generally bars should be square so both the height and width equal the class count C The scale of the vertical axis should be that of the variable whose distribution you are displaying D None of the above Correct Points Earned 1 1 Your Response D How long must you travel each day to get to work or school The table below gives the average travel time to work for workers in each state who are at least 16 years old and don t work at home Make a histogram of the travel times using classes of width 2 minutes starting at 15 minutes Make this histogram by hand even if you have software to be sure you understand the process You may then want to compare your histogram with your software s choice verage travel time to work minutes for adults employed outside the home Time State Time State 236 Louisiana 251 Ohio 177 Maine 223 Oklahoma 250 Maryland 306 Oregon 20 7 Massachusetts 26 6 Pennsylvania a 268 Michigan 227 Rhode Island 239 Minnesota 234 South Carolina cut 241 Mississippi 240 South Dakota 236 Missouri 229 Tennessee 259 Montana 176 Texas 273 Nebraska 177 Utah 255 Nevada 242 Vermont 201 New Hampshire 246 Virginia 279 New Jersey 291 Washington 223 New Mexico 209 West Virginia 182 New York 309 Wisconsin 185 North Carolina 234 Wyoming r 224 North Dakota 155 District of Columbia Data set 1 Compare your histogram to those below Select the correct one 7 Number of States 25 Histogram 1E 1 3 22 25 25 31 Average travel time to work minutes Number of States 50 Histogram 3913 15 IT 1 9 2123252 29 31 M x mammmwm Yin Emumwmsammm Wu uvmgmlumsmww xmw m guns m ggginb m 9 m Wubnmg 5 mm a Mum a mum My m 15minka mm mm aummwmbm l 1m 1m mymm mamaswag Emu mmImymmdwnhszwmmxmSWskmm m at Nmulmnx mm human mmmnammwammm mm mcmm ymmmmn nun m m c u Ahmluvhm m mmmmmm dxh kluluvlu Wamm m mksl mmgmmammamummm 1 13 6 25 2 15 7 27 3 17 8 31 4 19 9 33 5 23 10 35 A Center point B Largest value C Smallest value The distribution is rougly centered around 23 minutes The spread is described by the range of 15 to 31 minutes x Points Earned 13 Your Response A25 B10 C1 Figure 19 shows the distribution of the state percents of women aged 15 and over who have never been married r a n o utlier is a n o bservaticu r1 that Falls 1 ea Fly outside the werequot patter r1 Number nfstates f O I I l I I I I I 10 14 23 32 36 40 44 13 51 Percent ofwo men one r age 15 who never mar ried 20 The main body of the distribution is slightly skewed to the right There is one clear outlier the District of Columbia eHomework 6 Points Awarded 21 Points Missed O I Percentage 100 A political scientist wants to know how college students feel about the Social Security system She obtains a list of the 3456 undergraduates at her college and mails a questionnaire to 250 students selected at random Only 104 questionnaires are returned 1 What is the population in this study Be careful what group does she want informa from about A The students who responded B The students who were surveyed C The entire student body D All college students The population in a statistical study is the entire group of individuals about which we want information In this case the scientist studies the opinion of all college students hence this is the population of the study a Points Earned 11 Your Response D 2 What is the sample Be careful from what group does she actuallyobtam informa tron 7 A The students who responded B The students who were surveyed C The entire student body D All university students A sample is a part of the population from which we actually collect information W Points Earned 11 Your Response A You see a woman student standing in front of the student center now and then stopping other students to ask them questions She says that she is collecting student opinions for a class assignment 3 Why is this sampling method is almost certainly biased A Not all students may travel by the student center with the same frequency and thus the entire student population is not represented uniformly B The questions she is asking may influence the responses C The students may be in uenced to respond positively since they are in a public setting A simple random sample SRS of size 11 consists of 11 individuals from the population chosen in such a way that every set of 11 individuals has an equal chance to be the sample actually selected It is biased if not every set of 11 individuals has an equal chance to be selected Answers B and C are wrong because they deal with response bias instead of sample bias y nun t m nun m A 0 Ymueyhmgnymmwmmmmuamgemledumamsekzuhne Wmmumkxzs mm Wmmmmm u mu 5mgx In m Wumummu um mfa mgPmmzmphws mdmmmxmm m mdthIyypmnmubh wp mmmmmm wxd39li mva Wmva WM mmyvul mm Scullram Bum Pmuiymms an mme mm mm In mm m manmmmu 5mm gwsufdla rm wwwaumx w my mm mm mkmmmmm 1 hi 1 n u n m lazulgmadngnlsamv m m swam ww nhumvb mull 55w mymmumsuumw humbmmxm mxms n m cmsm mm mm m 515 wt m luka w am am mywmm mg g x mm mnnnnnmmu mgan 6911mm mmmmnnnnmsw mama m m was 1 Emmi eunuch mums m5 mm Mmekmammmdmnnnmmsmo Us mummum Izmmdwasuhfmfmnzmkamvh 5R5 mmmmuaqu A zmsmzxzmsum my zmmzwz165x5nm265mo 35576zmlmzs 35 35ozm n nussswzawzmsxsnmum mm a m 1 m m m rm 35mzmlnozs 3msm um m h m n n c 1 A m mums m smame m an yml mm ms mymlx mm WI Msmmmgsmr ammny39m mywm mmmmu s mmmmuu mu m m7 mmm mmd Inkt m n n a x A lug xmmm u was Wuemzudb kwmmmmmmm mmmmmmmmmmmmummy m mm zmplluiy swam mums 1mg b ml pas mm mun mm erhhmleurvlmlmum y mynm m m Lmemdmmuinmhpesmdm smphxsmalnnnmhpasm admng gmm mywm mm nnnmbpasm edmmgvmhmnmmswhxmm Messeth mm umbnml pessekmdmmsmkuw mnnmmma mugmum mymm mun mbpesMedmmgvmhmn mm mmmwpwma v Mann 1 mm 5 lewmw sk 5 pr mug mm mumm nwmdndnymmmmm mmmmnymimmm m1 Msmnmmmmmw ym rm1m mmwmmP mgmm m mm mmmmw m 15 mm x Amusembmg Glam u mm mm Demsmmh Em pm Hm I a 5mm Immhma Vehsw m nmmn x 5 mm mm sum Mugn mm mm Mm Emu aws m saw m mm u mu amth 117 Math magswamde mkmdmmmmmmw mmm x mxmuuwuuw l mauuwuw Euzukndx1glu anmgwdlnuElandmgmHIBnEmu glmn 1 Wm mmmwmmmh ammxnwzm mwnmm mm x bummM meymb khakdtmnlmonwxgmdkw wtamm mum vfzhnasmgm gamma imam 1 III 1 n Inn An Upmw um mummany mumsmm m mam sks w M wnhmmmmbu ufvhhmnhn m mmwu39 sks wamnymvusluw ymlwmhadmmvu mme 115112 W7 1 WPVmIJmndnyvummw m mm A lbw gnu l Ymmngpk m an be me mmme m mm zdh39m mmnsmnsmmus m w M Mmm mm m n c u m n mm mm Wmlu nu mum vi mqmq znnnmmusmdudmussmpmmznnn1365335 admmpmmmnnmn m I hi 1 12 human mm m B maywa mum my my mymmmlwwmlymmsthtyhxwmmdumm 1y I m mummmmrwmm mm sme39hymmm mmyw W s n w memndmsmxdhudu and Euwmsmamywmdmmuamywphwhnmmma mmmmmmymsmxmmmmammmmm msahmluyn mm ltmnmmvunth c m n D A u Bump x mm m cwmltmmmmmamm nun mum m mmmw mun ms ymammammamm MM 51 m m mm 1mm anng 5 twhk 5447mm w um mm m gum mm mm m mmmmmmm A sum 41 quotmummmm m ummm m M my I sum myam mny mmw nnmm mugwpmms mm Myanmmmls whs DEu awhmmm mph mxmuuubk mmmwm WWW mglml Akmmesmm haw m w gwmuvluhhym vfwhlu Wm I hi 1 a mum my mysm memwmmmww A mm mmgmwmmm mdnmmlukpknm mg mm um Suns nio ysmdhmmsammus5nszmmde u mink vfmmspumafmmuswh m ym unwisth mm Nmmmmawm nbabm mudkmnm l n m w m gvmvhunzv Pwa memd mm ummw 327 mm mm dummy mm m w W m Martians m B nu WVPmumufywph m mmwmmm mymsysmv 1 hi 1 non A Six y mam 1 ma 1 sum Wm Dams m mm ammmm mum wen mam mwmm mmawm vth My summymniy mm wen mm saubahwhAnYhqw kd m m m yxlmg m m mmymmumsmry bmmmm in mm 5m x mm mm m absu39wd nkaianesmdu my a kyvmlh mrmmf mmngmmmmw 1m smamymmeymynpmmmmmmmmmm mmmmammmmu n m dnvu39 mamsehzudumm m mm m mam m dnvu39s m mum mm mm w mm m y Illn a u Daymlewazlbnsmma m mmnmm hmmm A Yes I 0 m Mm mm Inkpmme mm dun mummm m ywph bdlm um mmmsaubakuwmmsm Whaman mwmammmmmmmdmumi mm mm m h m mymm mammmxmmmmxmx mm m mm m2 was wa manningsz muvuw ammsmmmx 1 mm gwnavu39ymmnmuvhymvh m 4mm mm mgwuma manram Mn aXnmmmWtrdwasehz13ufmamxmnmiymuuh Mmmswhsumwf lmbekhzkdhwsazmgwMenznm m wlmhwedamsZWhnurhmmmmmsgwhasmwfnlwbuekmdumn mgmnwyewnmuuuwymdmlus m mmma nun m m Whyxymlxsmphmmsxsn Yul1 mm sumph mmmm m ukm mm mame nmhsmew zmewbemswk mm m m 1 Ian mammalianwatvahywm n mufwdmmkmm manyrhnsmmgllam mmmwnmalmmwm mmm m my munan mmumwmm mmm nvnnnszhmm sum ynnnnnnum m mmummw Mummmhmvmy osmm mum mmummm 1mm Pam msuynsm23 xllstxgrdsamnpw I Us 4 mm mm mm mm gy7x mm Wmmmmu viiammmmm znn mm um um m n ma mm m mum m blanks uwmwwm mu uan mm ammnmmmmymmm 1mm 59mm mm mm mm 3575115155mm rhnsm 39mduuhh y mm mg 3 hi 1 n m m skaWynn mu gm mm m 4mm 1mm mmwmm mm Ammuwpmmm 5m humaer mm m Mamba chasm an my hanksMmsmphmane mxymwmmmdu m mmme b 4mm Mum mm x mama 9w vi mm 5m m mm vi mmmmmmmmm mm m alum eHomework 7 Points Awarded 48 Points Missed O I Percentage 100 The postlunch dip is the drop in mental alertness after a midday meal Does an extract of the leaves of the ginkgo tree reduce the postlunch dip Assign healthy people aged 18 to 40 to take either ginkgo extract or a placebo pill After lunch ask them to read seven pages of random letters and place an X over every 6 Count the number ofmisses 1 What are the subjects A Ginkgo extracts and placebo pills B The number of 6 s identified C The healthy people aged 18 to 40 D The pill given to the subjects The subjects are the healthy people aged 18 to 40 a Points Earned 11 Your Response C 2 What are the factors A The pill given to the subject B The age of each subject C The number of 6 s identified by each subject D Postlunch dip The factor is the pill given to the subject v Points Earned 11 Your Response A 3 What are the treatments A Midday meals B The pill given to the subject C Ginkgo D Ginkgo or placebo E The number of 6 s identified by each subject The treatments are ginkgo or placebo 39939 Points Earned 11 Your Response D 1 mm vxnbh n ma axfmmn vi a damnivy nah subrzl quotshun nun9 Ammmrnmmdmmmmmhmmsu emmwmnmymed Wmdnwnmmummsmmbmuhmm supmsemsHamsmuimshdnmnmmmbmmsam1621 mumpsqu mmmypeduauhzmmm hwm mud Umaugmmmvmmnymmmmmm smmmmm Lhausedhanu l m gammmm mm D mmmvfhlwmrpud mmuwm quotshun wmmuutmmgv Lam vf lph 39limn Immde c m mu mhm39u u Mmmwmammm mummmmwmmmmmmmm mamm my a A man Mmmwwmwmmmw Eskwmgnp WWW W mm mm Hxlcpxmm n 5 hm up n x mmwmmmm m m mm m Estrth m m euh lhkummlnd mm WWW megenhnllnu human 1 s Wmushbmmsrpudmummwm mh dywpwlknmdlpmmuf awmsamb msdmmwu xv vfyewphwhntmdzkmtmmmphymnm wwwmmmumk Asmewmm u q muss nnwywpk meWKMummmmumm Myexemwhndnsamnues m77 aws m 31mm um mmmmmmmwdm mm l mmmmmwmmU mpwa m wwwmmmmmnm m mu mmmmmmmmm quot1mm mummm smmm vfywphwhnmdzhmmmmphymmmmmmx m yaw mm x mng WWW yummy er mm 1 hi 1 n a mm m awnAmman kmmmxmmwm m m mmmwummmmm Ammm vfduahunumpmy ablazummzhl mgummlng m m ma m Whammerqu mm mm Anyama Inn 1 my a Axum mwxmwmmmmmmmm Am yv hnmmslrmbammxsm nmmmm hm mh gzwu mmaum kwkmmmmkxmlzx hgsm wwwmamm mm rimyhmgumudmhs was km W n mg1m Wmmuwm mm mmammuhlyvfhmgzwumbmhgmmmmm mmamw mm mums u myme s mabsmummdyxmmmawamm m sums mmmmm mm mm 1mmth Mmhxwm39 u13m mmm wen uswd n mwd Kmum mmmmmmsumdmmgvmmevu xymm hmmw yzmmmwmmnwxgmwmd Kmxm mwyummm nun MWMW Ammmmmsmm mmumkgmumm I Am ummVQ I nuse farm musedlmgnmummuu mymwummbmmmwmm n m mummuywnmmba awamdm mm mnkmgwmxwu be 3 vi 1 hi 1 n n c As mmhwmm wk 9mm 1mm quot2mm mhmba yms mm mm mmme m mm m WWW mam mum A Six y mm mum mm 237mm aged n m xn m hm memmm lka animumnnam WykmmurPbuba m mm m Jazmina mrmwmmmmm maxmm Wu amm hmmmamm mumu thmauhvfmsummsmmwlwhwwsmms I 39thhbm mmsmnm mammal mykzebahsswdw mmm um 911mm m swam anquot vi mm quotDmlhh Dmemm numum m Wham sswdm mm WWW w my mum Wm M massng mmss withingmm u mmwmwmm Nb 37 gm anquot m mm wen mmwnm mm m hwyme m mm ammm 91mm swamWhjulwlswhwmm x7th mama mm m swam mmms vfduunmm 1 hi 1 n n c u 3mm mmmgvfdu m x rdnmd lwumm m um ma um huh mm m kmdeng Pva IYwnvfm 39ld nmxnwmhwmhwnmumnam 1m mmmmnm39h iymwen xmmwamuw mm tsunami mmquot mm x 3mm mmmgvfdu m 1me quotmm an A mm mm mm m gmnybuba mm m sum um nutmme w mamd mms quotmm mm amen mm mm m m egmnybzabas scyasswimmdxdmmswm mm WWW m n An mmmm mam Whlmdmmnmus wwwmm m mm mmde alums mam kwlafmmlly 11mm whjamwm my m mwmm mmamumumnhmam mum mngmmhuam mmqu m manmwmwhxdhss may 1 m I mwmmmmmwmm mmme m QWWWMhmn Wmea Ma me whjansmnmxmn ymwdwm ms in mm mmmm mm dmihwwwlmhmmmYhq x mm y 1 hi 1 n n c 1 mmmmmbmss zmddbusYhuvpmtdnndu A Mumsle mama anyth PM whwlvf m bln xmmmmmgmnm m 7de mmsmmxwmsbns a m mm hmwhuhlxummYhqwue mm mmmm mama mm mm mth PM bwkwluf mm mm beMLhnmdmmnnvhus mum may wmanm 1 hi 1 n u a vhx lmguwxdlymrgummm MxmrdlihmdyewHYmummmb mm m amnyymg minivanssz m mm mm n w m x m smuemg u Hummummm m mestbww x ThuMum mu m m m mm n mm wavxmmamwmm mpuh mnahvfmwmmm 5mg lnxd mdywph 5 mm yamer mmmmm mm slt n m mu 39 I mum a Samanthak manningdig mmamm mm vi 1m amp c 1mm magnum 1 mmmmmmm mm mm x Ind w W 1 mm sax mm x mpml AmmdPxs man mm m mum mm Pu m mum magma gm m Mam me MAD mm m ht m n n m Wymmswms mm mmmd man mm mpn Wmmvk Mm mbks mm mm m Wm mmsm 1m ummmmmm alt mmuwmm man In mm mm uh Xhznns mm mm mm mm m m hznms mum hznns xmxmnm sum 1mm mm m mum LxIhI h u 1 Hanan s hilly 2 mum mm 2 Human 7 hilly deium x mum deamummdmmmmbmmw y Hum cd srundurd lt 5 Chris39enszn Subjects lectures Had zld X 39 Eu mulwmedm 1 m randy standard Comm HVH UVI ml mdin mm My quot5quot 7 standard Comm a 39 mummedia 5 quot Sumects leclurzs H df M smndard Comm 5 mulvimedm 39 Mquot smndard 5mm Av uquot T u 7 W mul mzdm Graupl Treuvmml HlH an Standard 1 Group 2 Treatment 2 4 cnnmnsen Mummdm Random aSSI mnan 5 sump 39 Trenmem Had eld Standard Eruup4 Trwmmz Tnl zy 4 Mulhmzdlu Eroup 1 4 Trzmmem 1 4Jec7ur25 Smndard Randnm nsswgnmem Emupz mummy 41am Mummedia A awn in mm dam Imam5mm vi 2mm mum 3 mm srpmuywum nah bmmmmm mm m 1 hi 1 a an WWW amass m waRYwsmmskdn kmplmiymmnmd mmmnwm m Emma n my m mpkuiymmnmdewumn mm mm x mmummm mm mmmmm mm in m u n n had 45 humambmkdmgm meduzmhlynmmed Mn mmmmmwnm gym Mamesbawumh mlmsmxmwd nun t m YlI I m wuluwnmdz mmummmmmmmmmm y Mummmmmmmms lewmw mmpuuhu eds Wmms mam Mm mem 1 Whv h fallmlgmnlmsasudAslgmfzmhkiynmmedewemm arm 1 100 mm Trzatmzn Trzmmem 2 Randnm 7 GroupZ drinking A drmkmg nsswgnmzm 00 mm just usr wine spirits Gruup a 100 duHS Grou Ier v harm d justwm Eran nu mm karma asmgnmzm 39 Erm 5b Drm g Randum henrtd J2 Juscbezr assignmem am no hear u Grm Drinking a Random 39 m Justsplms nssxgnmzm sruu no hurt Trzahnem 1 drink Groupi hmquot disease f Trzarmzmz drunku Tremmem 3 drinkh Randam assignmem Treatmem l drinki M hfgjfglgm 4 mumm 2 arm Tremmzm 3 drinki Group 1 Treatment 1 100 adults drinking just win Rundam Group 2 I Treatment 2 assngnmant 100 adults drinking just be Group 3 Treatment 3 100 adults drinking just Spit A Design 1 B Design 2 C Design 3 D Design 4 See figure 94 for an example of a completely randomized experiment No such experiment has been done because subjects aren39t willing to have their drinking regulated for years g Points Earned ll Your Response D You have recruited 300 adults aged 45 to 65 who are willing to follow your orders about alcohol consumption over the next five years You want to compare the effects on heart disease of moderate drinking of just wine just beer or just spirits Women as a group develop heart disease much later than men We can improve the completely randomized design of this experiment by using women and men as blocks Your 300 subjects include 120 women and 180 men 2 Here is a partial outline of a block design for comparing the effect of drinking wine beer and spirits on 3 heart disease among men and women Match the missing terms to the appropriate items A 120 B 6039 C just beer D 180 E just spirits F heart disease G 40 H wOmen The blocks are the 180 men andthe 120 women each one of them is divided into 3 groups randomly and the effect on heartdisease is compar d quot PointsEarned 8 8 Your Response A22 B27 C24 D21 E3F5 Gz8 H16 Choose a student at random from a large statistics class Describe a sam ple space Sforeach of the following In some cases you may haversome freedom in specifying S mulan mwmmsm Mmammmhymoms IMWszmu am In mu m mmmm mmumm mam Lndmusulummuvf mmmm mamulna mmmy mun m mu m mmmm mmumm mam Lndmusawemnk n m Namymysmym m mm mummyth hm w Mam mum hmmwmmmm Dimu Yesmanyquot Yes nunm NamzrdYex Yes Dimu NamzrdYex Yes bumu NamzrdYex Nu m Mmmmmmmmmmmmmmm mm WugmmA 1 smnmm WNWme W m m muWM m mums m m mam m mmxmm vi mm m x will Mswhn m zmx ymph i mm an WWW mwwm x mu m m m summxmamm Wbukgmm gmxshlhs m2 mm mummy m m m msmvhyrabuh iyu nyemmsbmmnmnLurmyexmagenmumbemm m m mnv m mu guns mm mwmmm mu m m 1 91 mm Medan m usm mbmmsgnv m m 5v mnv mm m In u 1m 1 mm W x mm Wm mum mm mumyum mmmmbuh uyufdu Jaman mam m mbwm s wnhyrabm iy m mamme m m mum vi mm mummy m Myums m m 5v 1 hi 1 n u 5 m 1 mume mm m mum mmmm media mm mum myqu Manama mm waspuwuymmmsmampammmmwnhmm v H Jlsnymyhmlmgugewhnmmm m39E B A N am mmdmvh Pasmzhnsm s mem abesa m mmdmvh Pusuzm mmwm am N am mmdmvh Pasmzhnsm s mum mg mm myesmzhnsmrmabase 91mm Imquot 1 hi 1 n u A mm mm m mummyan N Wimmzmnses mm W n sums mme Nnnm smmnzemdz mm E s2n c mnv nwm n aws W m mmm n Wm mumm mmsw gm m Mm m 1mm 4mm mmm m m mm an wnh zb u mambk mmmmmam mu 1 x n n 1 a n mmmmumx n WrmmmaIHJQKJquot um masmwAmE nun u 1 my mammawa 2 HXE3PJ zoPJ onuon2 w u Wham imammm mm m mmmgmynmxm vdmsu hmmmvmabk x mm M IX EX1 3 nm yumL a 1 m mmmmmm mm mmmgmwmum Em HXlt2PJ noPJ 1 mm mm v Innhm m yum Ab sknm ylyuwmxmem mugm m mph gamma m mmM 1 hi 1 n n a r aquot x mwmmmmm madly mmmmwmm n ma m l w u A I AI mwutumn mmmmmmwm Amxmh wa Abhadmba meubmmnml I am Points Awarded 48 Points Missed 3 Percentage 94 1 What percent of the observations in a distribution lie between the first quartile and the third quartile By definition the first quartile has 25 of the observations below it While the third quartile has 75 of the observations below it This leaves us with 50 of the observations between the first and the third quartile Points Earned 11 Your Response B 2 To make a boxplot of a distribution you must know A all of the individual observations B the mean and the standard deviation C the fivenumber summary A boxplot is a graphical representation of the fivenumber summary The boxplot makes it easier to compare between the fivenumber summaries of different distributions v Points Earned 11 Your Response C 3 Here are the amounts of money cents in coins carried by 10 students in a statistics c ass 50350977600872365 The fivenumber summary of these data is A 0 0 425 76 97 B 0 29 575 815 97 C 0 29 425 75 97 The fivenumber summary is Min 0 01 0 M 425 03 76 Max 97 w Points Earned 1 Your Response A Table 11 gives the percent of residents in each state who were born outside the United States TABLE LI Percent of state population born outside the United STATE Alabama Alaska Ar izona Arkanms California Colorado Connectic th Delawm e Florida Georgia Hawaii Idaho Illinois Indiana Iowa Kansas Kentu c qr PERCENT Lu U1 Pql 39A 05 C3 1 39 J nJ 1 C bu 129 STATE Louisiana Maine Maryland Massach usetts Michigan Minnesota M ississippi Missouri Montana Nebraska N evada New Hampshire N ew Jerseyr New Mex in 0 New York N orrh Carol ina North Dakota P C NT Dbipbjbnb STATE Ohio Oklahoma Oregon Pennsylvania Rhode Island South Carolii South Dakota Tennessee Texas Utah Vermont Virginia W ashi gton W39est Virginia Wrist 0115 in W39mming Dist ochlur 1n Tab1e 11 we see that California 272 stands slightly above the rest of the distribution 1s California a suspected outlier by the 15 X I QR ru1e Start from the stemp1ot inFigure 110 page 20 which arranges the observations in increasing order u 1690ummgtumA o A m Am n mka m mm 0 En m m myaxmm m aumw mm Mark mfmq mukrz v y mm m n 38 5 mumde a mmmmmwmm Mm m mywa mandqu my v Points Earned 11 Your Response 126 6 1s California a suspected outlier by the 15 X QR A Yes B No Yes The OR is Q i 01 126 i 38 88 so we would consider any numbers greater than Q 15 X IQR 126 132 258 to be outliers J Points Earned 11 Your Response A We asked the students in a large firstyear college class how many minutes they studied on a typical weeknight Here are the responses of random samples of 30 women and 30 men from the class Women Men 180120180360240 90120 30 90200 120180120240170 90 45 30120 75 150120180180150150120 60240300 200150180150180240 60120 60 30 120 60120180180 30230120 95150 90240180115120 0200120120180 Data set The most common methods for formal comparison of two groups use xand sto summarize the data 51 What kinds of distributions are best summarized by xand s 7 A Skewed distributions without outliers B Distributions that are fairly symmetric and free of outliers C Symmetric distributions outliers make no difference D Distributions of economic variables since they are usually skewed to the right Both the mean and the standard deviation are not resistant measures meaning that they are highly influenced by outliers and skewedness Therefore only symmetric distributions without any outliers are good candidates for using the mean and standard deviation Answer B V Points Earned 11 Your Response B 8 One overzealous student in each group claimed to study at least 300 minutes five hours per night Let39s check their in uence on xand 5 By how much does removing these observations change Xfor the men39s group Note that negative results indicate a decrease in xwhen the over zealous student was removed A 1286 B 736 C 630 D 736 The mean for all of the men is 11717 while removing the overzealous student gives 11086 for an overall change of 11086 11716 630 39 u mxmxxxmxxxmxu xmxxxxmmx m mmmmx xxxxxxx x 39 u mmfm39 x xxx My xxx xmxxxx nmvmgdu m zubm mmgw xxx an m mmxm xx 15x55 716517 672 nun m m xn x xx Wmmzhdmsxmlgm m ndmu mmdmge xxxuxxmx gww 39 u mxmxxxmxxwxmxxx xxxxxxxxxxmxxgxxxx m zubm xxxxxxgm xxs furmwvu39dkhmg x xxx 517 xxxx nun t m n x Hm mxx mexmamxnx mungftme mjuudwihmumws xmmmxmxxxxmm mexxxx m m xxxmexxxxxxxxxxxxxmmxxxxxmmmxym skrwedwmu li oz 55 xx xx xx xx xx xx xx 7x 7 7y xx xx xx xx xx xx xx x xx xx xx xx xx xx v7 xx xx xxx xx xxx xx xxx xxx xxx 1m 1m xxx xxx xxx m xxx xxx xxx xxx xxx 137 138 139 144 145 147 156 162 174 178 179 184 191 198 211 214 243 249 329 380 403 511 522 598 Dataset 1 Make a histogram of the distribution using classes 50 days Wide for example the second class has 2 values 50 lt days S 100 Which of the histograms below correctly describes the distribution Histogram i 35 35 r in 3D in 30 9 2 L a 25 f 25 5 15 5 15 a in a 10 1 D E 5 E 5 Jmmmm z a D 100 EDD 300 400 500 600 Survival Time days Histogram 39 Number of Guinea Pigs Number at Guinea Pigs 1100 200 300 400 500 500 Survival Time days A Histogram l B Histogram H C Histogram llT D Histogram IV The correct choice is Histogram HI Make sure you chose the classes exactly as specified Note that the second class 50 lt days S 100 has 30 guinea pigs and Histogram III is the only one that re ects this Points Earned 11 Your Response C 13 Describe the distribution39s main features Mark the appropriate features below A Right skewed B Symmetrical C Left skewed D Single peaked E Double peaked F None of the above The distribution is best described as right skewed with a single main peak v Points Earned 22 Your Response A D 14 Which numerical summary would you choose for these data A Mean and standard deviation B Fivenumber summary C Neither of the above Since the distribution is single peaked a numerical summary is applicable The skewedness of the distribution means that the fivenumber summary is better suited than the mean and standard deviation both of which are not resistant to skewed tails and outliers Points Earned 11 Your Response B 15 Calculate your chosen summary Mark numerical measures that are not relevant to your numerical summary as so Note that the fivenumber summary may vary slightly depending on the definitions used by different calculatorsoftware applications Therefore if applicable calculate it manually exactly as described by the procedures in the text As for the standard deviation if it39s relevant make sure that you calculate it as defined in the text dividing by n i 1 and not by n as done by some calculatorssoftware applications 1 42 5 825 9 1515 2 43 6 1025 10 153 3 435 7 103 11 598 4 815 8 1035 12 NotRelevant B Standard deviation C Minimum D First Quartile E Median F Third Quartile G Maximum The correct numerical measure is the fivenumber summary Refer to examples 23 and 25 for explanations on how to calculate the median and quartiles e The table below gives the mean number of births in the United States on each day of the week during an entire year Day Sunday Monday Tuesday Wednesday Thursday Friday Saturday Data set Points Earned 77 Your Response A21 2 Births 7374 11704 13169 13038 13013 12664 8459 Bl 2 C2 D5 120 100 Number of births EU Monday Tuesday Wednesday Thursday Friday Day ufweek 16 Based on these boxplots give a more detailed description of how births depend on the day of the week Mark the correct answers below A There is a marked drop in weekend birthrates with at least 75 percent of the weekday observations not overlapping with at least 75 of the weekend observations B There is a marked drop in weekend birthrates with no overlap between the weekend and weekday observations C There is a marked drop in weekend birthrates with an overlap of more than 75 between the weekend and weekday observations D All of the days have highly skewed distributions E The weekend days have similar distributions F Most weekdays have similar distributions The correct answers are A E andF A Note that there is no overlap between weekend observations below the third quartile and weekday mm mm m m mmmnmugvhn mum uh mm WWW am umlwpmabsewmmbammm mkm 1mm an m Iwuzgsmmidxyhmhxiymmdammm mm seahhymamdnnhllmg mwmmmm m m wwssbk mme mushyyhdnm dwa MW 2 Amum mgemdmwukdzydmfbmms mkybemmthmsemsmsdnm mmmhm mm m h A sum wwwsmmm mmmmtamm 51m to WannhmPak mmm mmznmxuf mhmxm mum new 1 Densityr Curve I 0420 CI 20 015 I115 010 010 005 I105 000 r 0 r30 0 5 1D 15 20 25 30 35 4 0 Density Curve Ill 020 U 20 015 0115 110 U 1039 005 U 05 000 D 5 1t 15 20 25 30 35 4 1 Density Curve 1 2 Density Curve H 3 Density Curve 1H 4 Density Curve 1V A Symmetric with two peaks B Single peaked and skewed to the left The symmetric density curve with two peaks is Density Curve H1 Note that although Density Curve 11 has two peaks it is not symmetric Density Curve 1V is single peaked and leftskewed as evident by the long lefthand tail v Points Earned 22 Your Response A3 134 Examining the location of accidents on a level 3mile bike path shows that they occur uniformly along the length of the path Figure 34 displays the density curve that describes the distribution of accidents height 153 I O 1 2 nu Distance along bike path Inilesll 18 Explain why this curve satisfies the two requirements for a density curve A Because it is a smooth curve on or above the horizontal axis B Because it is on the nonnegative part of the Xaxis and the area beneath the curve is 1 C Because it is on or above the horizontal axis and the area beneath the curve is 1 D Because it is symmetrical with respect to the vertical line at X 15 and the area beneath the curve does not extend 1 It is on or above the horizontal axis everywhere and because it forms a 13 X 3 rectangle the area beneath the curve is 1 g Points Earned 11 Your Response C 19 The proportion of accidents that occur in the first mile of the path is the area under the density curve between 0 miles and 1 mile What is this area A 16 E 13 C 23 D 1 10 Onethird of the accidents occur in the first mile this is a 13 X 1 rectangle so the proportion is 13 Points Earned 11 Your Response B m mmmmmmmmmmmsmmmm mu m Muwrwpmufuzmhaypmn mvf au39syrvpm n I n mm Dn27 m mmummumw Su sywpuly mm m x nzmwsmmmgnx mm 1 1 YIq A mfmbahwdwphyiw mum mmmmm mam 13 7 1 stance along bike pm mih mmmmm Mmmmmw m mum m nusmmmmm ukmhmm mmwm HWBWMWWWWMWW nun I m ms Dn75 1 mamamum m uw m swm 5mm um Wm imam Wham hi 1 m1meme v Your Response C The figures below display three density curves each with three points marked on them At which of these points on each curve do the mean and the median fall Match your answers below for each curve 23 Density Curve a ABC 1 PointA 2 PointB 3 Point C A Mean B Median Point A cannot be the median since it is obvious that the area to its right is larger than the area to its left We also know that for a skewed distribution the mean gets pulled in the direction of the tail In the present case this implies that the mean is to the right of the median so that the median cannot be at Point C Thus we are left with the median at Point B and the mean to its right at Point C Points Earned 22 Your Response A23 B2 24 Density Curve b 1 Point A 2 PointB 3 Point C A Mean B Median This is a symmetric density curve Thus both the mean and the median fall exactly in the middle Point B y Points Earned 22 Your Response A2 i32 25 Density Curve c ABC 1 PointA 2 PointB 3 Point C A Mean B Median Point C cannot be the median since it is obvious that the area to its right is smaller than the area to its left We also know that for a skewed distribution the mean gets pulled in the direction of the tail In the present case this implies that the mean is to the left of the median so that the median cannot be at Point A Thus we are left with the median at Point B and the mean to its left at Point A g Points Earned 22 Your Response A21 B2 The summer monsoon brings 80 of India39s rainfall and is essential for the country39s agriculture Records going back more than a century show that the amount of monsoon rainfall varies from year to year according to a distribution that is approximately Normal with mean 852 millimeters mm and standard deviation 82 mm Use the 687957997 rule to answer the following questions 26 Between what values do the monsoon rains fall in 95 of all years A Between 688 and 1016 mm B Between 770 and 934 mm C Between 606 and 1098 mm D Between 164 and 1540 mm In 95 of all years monsoon rain levels are between 688 and 1016 mm two standard deviations above and below the mean 852 i 282 688 to 1016 mm sun nun v mm m mung I A n Manuemmamnm mum m M ww LLassxhn m Lassvhn xx m anle mamnm km hsgmskamumamz Wamnmnw Mm Mm 628 252 v unn an a mmxms x mmmsym ms mammufsa39lmxms mnnmmmmmmummmm human m Assesmmm qmm s 51mm 27 garmezan my mum mnnmmdmnmn n wxmm stzm7kmmymlxmswum2dumlybus mmdsmwnmdmmmim seru xn7515m 5mmdmzdumykas mumbmnphznmmdamk v Inn 1 mm m 17 um 51an Wanmwmmmmm 2 mm mm dswxa B ndeme mu my mammm mumbmnphznmmdamk Ingmar Amwam we amm dszmuiunwmmmmammmmwhngmusn y bill11 n s dasvfwnmngldz wz unmmy mlmmmmmsm dem ms Mmmsmgehxwhugnsmnn kammub Waxmmmmwmmmmm m n mxm nsszmm feuw39rkmmymnmswuw mum mm um x x a x 12 72 W m m Wammmm 727mm ymmamzaumms farm dunk mun um 11 mm pswnfuruwmm fau 7mmymlxmswuw Hum 111 mmmnmu a x 12 72 W m m Wammmm normarmran yszmdmzmmlymas mum MhZHm39m dunk mm m n Iq m n SIym slap nguAg mmmm mm m unk umullwg an m A m mm dmlgve 5 my mmmmwemudthwmnm vwhugmwmnumhmla MRS c m mm mm mm mm mprumba Wadumum mam 3mmdea umrmwedndummuwm d ammltse mmmmmmmmwmmm 9mmMB m whjez szmpnmdu wmjazlwnhmdux gumme mmmmmm mmrmwadnm mdhnmms smummmzwmmmgm MAME a kmsmemdm mmmmBmmdmm mdm 5 awmgsmuha mummqu dlvumm mmamm 1 hi 1 n u n n w vhlywphwlw x mmlyabese m 1 um dun lunxywph m mth mg mg mg mksmalymmu mm mm mm Madalynbeseywphmmsvfuumumduzmdmgw u mummm mm vi mm 1 erpr mm m m m mmmmam ummsimvsv u hywph meuhw r Us m mm 7 m 5quot mm m m m M m munyam gmwmv Tnke ummymmmm wasan mm m Msmm mm m yumu 5 mmmmam ummsimvsv u hywph mm hmgwp hm m m mmmm 91mm wasan m m 1n73mu mm 5mm wuwmmmm w 0mg mummy m Mm7nmnmmmmmmwmxwwmmwm mum m uddzvuumstmdummMmmsmmmm mmmsvm ymdmm mawmbamsmemumlmmrmuu m wwwmasm mmsz


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