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# Business Statistics STAT 2160

WMU

GPA 3.54

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This 205 page Class Notes was uploaded by Alia Gerhold on Wednesday September 30, 2015. The Class Notes belongs to STAT 2160 at Western Michigan University taught by Jung Wang in Fall. Since its upload, it has received 22 views. For similar materials see /class/216952/stat-2160-western-michigan-university in Statistics at Western Michigan University.

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Date Created: 09/30/15

Chapter 4 Part 1 Normal Distribution JC Wang Department of Statistics Western Michigan University cpartmcnt of Statist39 n l zzzz xw aur writ b 39 Goal and Objective of Chapter 4 gt To learn how to compute normal and binomial probabilities Departmentof Sta 39 Outline Computing Normal Probabilities Using StemandLeaf Display Using TlCalculator Computing Normal Percentiles Using StemandLeaf Display Using TlCalculator Review of Empirical Rule Normal Review Questions Questions Answers TVMore Example Question 1 gt Assumption demand quotfollowsquot normal bellshaped curve with mean 36 and SD 8 TV sets per month evidenced by past observations and assumed or believed remains the same present and future gt Q If the store orders 40 sets what is the chance that it runs out of stock Department of Statistics noquot TVMore Example con nued Probability under Normal Curve actual probability a approximate probability 16 20 24 28 32 36 40 44 48 52 56 of TV Sets Dc artmcnt of Statistics TVMore Example using siemandIeaf display Normal Curve StemiandiLeaf Plot 1 mmmbbbwwwNNNNiI 6 gt Find 89 012 34455 66777888 9990001111 2222333344444 0bsgtgt40 5555566666677777 PX gt 40j f 21 8888899990000 1111222333 44455566 77889 012 34 6 g 033 0bsgt40 PX240 n 100 gt Find 29 gt The desired probability PX 2 40 is 033 Z 029 031 Department of Statistics Using TI8384 to calculate area under normal curve gt PX 2 40 lt2ndgtltDISTRgtLnormalCDF409999368ltENTERgt 03085 031 gt How many standard units are we away from the mean 2 XM4073605 PX 2 40 PZ 2 5 normalcdf59999 03085 x 031 Area Under Normal Curve M36c8 u 002 0 03 0 04 0 05 r r r 4 001 r 000 0 Department of Statistics nun TVMore Example Question 2 gt Given the cost of running out of stock and the storage cost of keeping too many TVs the manager decides to order enough TV sets to satisfy customer demands 90 of the time gt Q How many TV sets should they order Inverse Normal Curve using stemandIeaf display Normal Curve StemiandiLeaf Plot 1 6 1 89 2 012 2 34455 gt Probability not satisfying customer 2 66777888 demand is PXgt a 010 2 9990001111 a Lets count backwards 3 2222333344444 3 5555566666677777 gt ngtlt010100gtlt01010 3 8888899990000 gt From 56 count 10 numbers backwards 4 1111222333 gt end at47 PXgt47010 4 44455566 a47 4 77889 5 012 5 34 5 6 Department of Statistics rm 39 Using TI8384 to calculate normal percentiles PXgta01 PXa09 Note PX gt a 01 ltgt PX g a 09 lt2ndgtltDlSTRgtlinvnorm9 36 8 and get a 4625 or z 46 or 47 Normal Percentiles Exercise TV sales p 36 a 8 gt a 427 when PX lt a 08 gt a 293 when PX lt a 02 gt a 36 when PX lt a 05 D partmcnrof Statistics mm mm Standard Normal Curve or Z curve x is how many sd above or below the mean Xiu 039 2 Properties of the Zcurve gt Zcurve has mean 0 and sd 1 gt P71ltZlt168 gt Pi2 lt Z lt 2 9545 gt P73 lt Z lt 3 997 partmcnt of Statistics mmun m m mmmn rm Empirical Rule If data histogram z bellshaped you should expect the following gt 68 of the observations will fall within 1 SD of the mean gt 95 of the observations will fall within 2 SD of the mean gt 997 of the observations will fall within 3 SD of the mean D partmentof Statistics rm 39 Empirical Rule TVMore Example revisited 002 mm mm nos l l l l 002 mm mm nmnn am i l nnn nm Questions exam completion time example Time required to complete a final examination in a particular college course is normally distributed with a mean and standard deviation of 80 and 10 minutes respectively L l O 4 What is the probability of completing the exam in one hour or less ie PX g 60 What is the probability a student will complete the exam in a time between 60 and 75 minutes ie P60 g X g 75 What is the probability of completing the exam in 88 minutes or more ie PX 2 88 What is the interquartile range IQR for completion times ie Po1 g X g as Department of Statistics m Answers using Tl calculator wN K PX g 60 normalCDFi9999 60 80 10 0228 P60 g X g 75 normalCDF60 75 80 10 2858 PX 2 88 normalCDF88 9999 80 10 2119 Find first quartile 01 invnorm25 80 1073255 Find third quartile 03 invnorm75 80 1086745 IQFr 86745 7 73255 1349 Department of Statistics mmtm agm nmmn Chapter 7 Confidence Intervals JC Wang Department of Statistics Western Michigan University cn f StaKiSthS Goal and Objectives Goal to learn confidence intervals Objectives 9 To understand that each interval has two endpoints lower and upper bound and Interpret the confidence interval Depanmcnr of Starisucs mqnmvp ummrlnm he Goal and Objectives Goal to learn confidence intervals Objectives 9 To understand that each interval has two endpoints lower and upper bound and Interpret the confidence interval 0 To compute the confidence interval a point estimate i the margin of error Dcpanmcnr of Stansucs Goal and Objectives Goal to learn confidence intervals Objectives 9 To understand that each interval has two endpoints lower and upper bound and Interpret the confidence interval 0 To compute the confidence interval a point estimate i the margin of error 0 To determine the sample size Dcpanmcnr of Stansucs Outline 9 Introduction 0 Applications Definations and Notation 9 Confidence Intervals 0 Computation 0 zConfidence Intervals 9 tConfidence Intervals 0 Sample Size Determination 0 An Example 9 Comparing Two Populations 0 Comparing Means of Two Independent Populations 0 Comparing Means of Two Dependent Populations 9 Confidence Interval for Proportion 0 Confidence Interval for Population Proportion 0 Sample Size Determination Department of Starisucs Outline a Introduction 0 Applications Definations and Notation Department of Su sm Applications of Estimation in Business examples 9 Store inventory value De anmcm of Statistics Applications of Estimation in Business examples 9 Store inventory value 9 Manufacture process Department of Sta stics Applications of Estimation in Business examples 0 Store inventory value 0 Manufacture process 0 Distribution process Department of Su sm Applications of Estimation in Business examples 0 Store inventory value 0 Manufacture process 0 Distribution process a Drug delivery Department of Su sm Applications of Estimation in Business examples 0 Store inventory value 0 Manufacture process 0 Distribution process a Drug delivery 0 Auditor Department of Su sm Definitions 0 Sample statistic a value computed from the sample ie from data Department of Statistics m npmm Definitions 0 Sample statistic a value computed from the sample ie from data 0 Point estimate ptest a single sample statistic that estimates the population parameter such as the mean or proportion Dcpanmcm of Statistics mummy mum39s Definitions 0 Sample statistic a value computed from the sample ie from data 0 Point estimate ptest a single sample statistic that estimates the population parameter such as the mean or proportion o Interval estimate of the true population parameter takes into account the sampling distribution of the point estimate where we have an upper bound and a lower bound Department of Stansucs mum mumm Notations to be discussed and used later 0 CI confidence interval Depanmcm of s 39 Notations to be discussed and used later 0 Cl confidence interval 0 CVal critical value cn f Starisucs Notations to be discussed and used later 0 Cl confidence interval 0 CVal critical value 0 ME margin of error Depanmcm of s 39 Notations to be discussed and used later 0 Cl confidence interval 0 CVal critical value 0 ME margin of error 0 SE standard error Depanmcm of s 39 Notations to be discussed and used later 0 Cl confidence interval 0 CVal critical value 0 ME margin of error 0 SE standard error 0 SD standard deviation Department of Starisucs m emu mm Notations to be discussed and used later 0 Cl confidence interval 0 CVal critical value 0 ME margin of error 0 SE standard error 0 SD standard deviation 0 ptest point estimate Depanmcm of Starisucs vm mmw Notations to be discussed and used later 0 Cl confidence interval 0 CVal critical value 0 ME margin of error 0 SE standard error 0 SD standard deviation 0 ptest point estimate 0 Zaz normal distribution critical value use invnorm Deparrmcnr of Starisucs mum mvmmw Notations to be discussed and used later 0 Cl confidence interval 0 CVal critical value 0 ME margin of error 0 SE standard error 0 SD standard deviation 0 ptest point estimate 0 Zaz normal distribution critical value use invnorm 0 171 students tdistribution critical value with n1 degrees of freedom use math solver or the ian Dcpanmcm of Stansucs Outline 9 Confidence Intervals 0 Computation 0 zConfidence Intervals 0 tConfidence Intervals 0 Sample Size Determination 0 An Example Department of Starisucs Computation of confidence intervals 0 ptest i ME 0 Where the point estimate estimates population mean p by X or population proportion p by p O marginOtError criticalValue x standardError Depanmcnr of Starisucs mqhmip umlmlnm he Computation of confidence intervals 0 ptest i ME 0 Where the point estimate estimates population mean p by Y or population proportion p by 3 c marginOtError criticalValue x standardError 0 In other words ME CValSE Deparrmcnr of Starisucs mummymm w Standard Error 0 Most of the time we will not have the SD of population mean but we can compute sample SE of the mean 3 SE7 7 x W Depanmcm of Statistics mm mw Standard Error 0 Most of the time we will not have the SD of population mean but we can compute sample SE of the mean 3 SE7 7 X W 9 Also we will not have the SD of population proportion but we can compute sample proportion SE I7 Dcpanmcm of Statistics u um Critical Value 0 z for normal distribution Department of Statistics Critical Value 0 z for normal distribution 0 tfor students tdistribution Department of Staxisucs Critical Value 0 z for normal distribution 0 ttor students t distribution o The students t distribution has n7 1 degrees of freedom df n7 1 Depanmcm of StaKiSUCS mm mm zCritical Value O Notation 2042 upper 100 x a2th standard normal percentile Department of Su sm zCritical Value 0 Notation 2042 upper 100 x a2th standard normal percentile 0 That is PZ gt 2042 042 E PZ g zaz17 042 SO 2042 invNorm1 7 Oz 2 Depanmcm of Statistics mm mm w zCritical Value 0 Notation 2042 upper 100 x 042th standard normal percentile 0 That is PZ gt 2042 042 E PZ g zaz17 042 SO 2042 invNorm1 7 042 0 Example 95 confidence interval will give 25 in each tail of the bellshaped curve therefore the zCVal ZCV 2025 invNorm17025 invNorm975 196 Department of Statistics zCritical Value continued areatothele ofzgwa 17 Department sf 5 tCritical Value using Tl calculators 0 math solver tcdfL7 U7 D 7 AT where o L ta to be solved 0 U 9999 o D df n71 0 A a error rate a T number of tails 2 forci 9 or use ian17 042 df Department of Starisucs Cereal Box Packaging Example Consider a cereal packaging plant in Battle Creek that is concerned with putting 368 gram of cereal into a box 0 What are the costs associated with putting too much cereal in a box Depanmcm of Statistics mm mm w Cereal Box Packaging Example Consider a cereal packaging plant in Battle Creek that is concerned with putting 368 gram of cereal into a box 0 What are the costs associated with putting too much cereal in a box o What are the costs associated with putting too little cereal in a box Depanmcm of Statistics mummymm w Cereal Box Packaging Example continued 0 Suppose sample size n 25 0 Suppose sample average Y 365 grams 0 Suppose SD is a process SD therefore a 15 grams 0 Suppose we want a 95 confidence interval 0 Therefore the critical value is 20 196 Depanmcm of Statistics mm mm w Cereal Box Packaging Example continued margin of error 0 Recall ME CVal x SE 0 The critical value CVal for 95 Cl means that the area under the curve of one tail is 5 2 or 0025 therefore ch invNorm1 7 025 invNorm975 196 S 15 SE773 t25 o ME196X3588 Department of Statistics ruqhnmtb tmrumtt he Cereal Box Packaging Example continued confidence interval 0 Since the confidence interval is the ptest i ME 0 CI 365 i 588 35912 37088 0 Therefore we are 95 confident that the population mean is between 359 and 371 0 Since 368 the value that is printed on the box indicates the manufacturing process is working properly is within the interval there is no reason to conclude that anything is wrong with the process Department of Statistics zConfidence Interval Using Tl Calculators example Let39s use Tl calculator 0 Do this STAT a TESTS a Zinterval a STATS la 15 liz365 l n25 l CiLevel 95 l CALCULATE Department of Su sm example zConfidence Interval Using TI Calculators Let39s use Tl calculator 0 Do this STAT a TESTS a Zinterval a STATS la 15 liz365 l n25 l CiLevel 95 l CALCULATE o READOUT Zinterval 35912 37088 365 Department of Statistics Y n i 25 Since 368 the target of the package is within the interval production should continue Note on zConfidence Intervals o The value of 2 selected for constructing such a confidence interval is called the critical value for the distribution Depanmcm of Starisucs Note on zConfidence Intervals o The value of 2 selected for constructing such a confidence interval is called the critical value for the distribution 0 There are different critical values for each level of confidence or confidence level CL 1 7 a where a significance level SL or error rate 0 Frequently Used 20V SL CL 2tailed CVal 10 90 1 645 5 95 1 96 1 99 258 Dcpanmcnr of Stansucs t um mt u u um Note on zConfidence Intervals o The value of 2 selected for constructing such a confidence interval is called the critical value for the distribution 0 There are different critical values for each level of confidence or confidence level CL 1 7 a where a significance level SL or error rate 0 Frequently Used 20V SL CL 2tailed CVal 10 90 1 645 5 95 1 96 1 99 258 Note There is a trade off between the width of the confidence interval and the level of confidence Deparrmcnmsmms u m Problem When SD is Unknown We have been dealing with Nu a where a population or process SD is known What happens when standard deviation 0 is not from a population or process SD Is this requirement rigid Can we compute standard deviation from the sample Let us review some history first Department of Statistics History of the Student t Distribution William Gosset an employee of Guinness Breweries in Ireland had a preoccupation with making statistical inferences about the mean when SD was unknown Since the employees of the company were not allowed to publish their scientific work under their own name He chose the pseudonym Student Therefore his contribution is still known as Student39s tDistribution Dcpanmcm of Statistics u um Comparing Standard Normal Curve with 1 curves Comparison of Standard Normal with t Curv s densry o N tConfidence Interval for the Mean summer quiz example Construct a 95 CI for the mean score for Summer Quiz Data of 14 students Given 95 CL Y 25 s 10777 17 14 SE 28803 m CVal Ta2n1 02513 iHVT17025713 21604 ME 21604 x 28803 62225 ptest 25 950 ptest i ME 1877831222 Deparrmcm of Staxisucs mummymm w tConfidence Interval for the Mean using Tl calculators 0 Do this STAT a TESTS l TInterval a STATS liz25lsx10777 lnzl4lciLevelz95l CALCULATE Department of Statistics tConfidence Interval for the Mean using Tl calculators a Do this STAT a TESTS l TInterval a STATS liz25lsx10777 lnzl4lciLevelz95l CALCULATE o READOUT Tinterval 18778 31222 25 14 Dxln Department of Staxisucs tConfidence Interval for the Mean using Tl calculators 9 Do this STAT a TESTS l TInterval a STATS liz25lsx10777 lnzl4lciLevelz95l CALCULATE o READOUT Tinterval 18778 31222 Y 25 n 14 a We are 95 confident that the true mean quiz score is between 188 and 312 Department of Su sm Sample Size Determination based on confidence intervals 0 What sample size should we use for the average quiz score determination if we want 95 confidence ME 5 and a 10777 178z 18 2202 1962 x 107772 n 7 7 7 2 7ME2 7 5 Department of Statistics Slow Wave Sleep Example page 100 problem 1 21202279142391025151711 0 aY 156154 and s61310 Depanmcm of Sta Slow Wave Sleep Example page 100 problem 1 21202279142391025151711 0 aY 156154 and s61310 o b population average and SD not possible Department of Staxisucs Slow Wave Sleep Example page 100 problem 1 21202279142391025151711 Y156154ands61310 o a o b population average and SD not possible o 0 he the sample average will miss the population average by t i Department of Statistics ruqhnmtb tmnvmtt he Slow Wave Sleep Example page 100 problem 1 21202279142391025151711 o Y156154ands61310 8 b population average and SD not possible o c the sample average will miss the population average by the SE 0 d 8 s mamm 17 Depanmcm of Statistics mm mm w Slow Wave Sleep Example page 100 problem 1 21202279142391025151711 Y156154ands61310 8 b population average and SD not possible o c the sample average will miss the population average by the SE 0 d 8 s mamm 17 0 0 G 6 ME gtlt Tvg75131gtlt17 21788 gtlt17 3704 Department of Statistics mqhmip umlmlnm he Slow Wave Sleep Example page 100 problem 1 21202279142391025151711 Y156154ands61310 8 b population average and SD not possible o c the sample average will miss the population average by the SE 0 d 8 s 61310N 17 0 0 G 6 ME gtlt Tvg75131gtlt17 21788 gtlt17 3704 0 1 95 Cl is 156154 i 3704 11911932 Depanmcm of Statistics m1mvmw w Slow Wave Sleep Example continued 0 1 continued can also do this assuming data have been entered into list 1 L1 STAT a TESTS L tInterval a DATA LList L1 L CALCULATE Department of Staxisucs Slow Wave Sleep Example continued 0 1 continued can also do this assuming data have been entered into list 1 L1 STAT a TESTS l tInterval a DATA lL lst L1 l CALCULATE o g If the confidence level is reduced to 90 the new interval will be shorter Depanmcm of Starisucs mm mm w Slow Wave Sleep Example continued 0 1 continued can also do this assuming data have been entered into list 1 L1 STAT a TESTS l tInterval a DATA lL lst L1 l CALCULATE o g If the confidence level is reduced to 90 the new interval will be shorter o h 90 CI a 12585 18646 Depanmcnr of Starisucs mqhmip umlmlnm he Slow Wave Sleep Example continued 0 1 continued can also do this assuming data have been entered into list 1 L1 STAT a TESTS l tInterval a DATA lL lst L1 l CALCULATE o g If the confidence level is reduced to 90 the new interval will be shorter o h 90 CI a 12585 18646 0 i Interpret the 95 Cl We are 95 percent confident that the true population average is between 12 and 19 Dcpanmcnr of Statistics Slow Wave Sleep Example continued e j Does the 95 CI suggest that elderly men over 60 spend 20 of their sleep in REM No since 20 is not in the 95 CI Department of Staxisucs Slow Wave Sleep Example continued 0 j Does the 95 Cl suggest that elderly men over 60 spend 20 of their sleep in REM No since 20 is not in the 95 Cl 0 k What sample size should we use if we change the ME to 25 CVa2 x 802 1962 x 6132 use niwiTi lo 24 Department of Statistics mummymm w Outline 9 Comparing Two Populations 0 Comparing Means of Two Independent Populations 0 Comparing Means of Two Dependent Populations Department of Starisucs Comparing Means of two independent populations o We are not limited to comparing an average to a constant Suppose we want to compare the means of two independent populations Department of Statistics m npmm Comparing Means of two independent populations c We are not limited to comparing an average to a constant Suppose we want to compare the means of two independent populations 0 Parameter of interest 6 m 7 pg 0 Recall Cl is ptest i ME Department of Statistics mwm mmw Comparing Means of two independent populations c We are not limited to comparing an average to a constant Suppose we want to compare the means of two independent populations 0 Parameter of interest 6 m 7 pg 0 Recall Cl is ptest i ME ptest H Y1 7 Y2 ME CVal x SE where CVal 071an SE iSE12 85 Q Department of Starisucs Example battery example A statistics student designed an experiment to see if there was any real difference in battery life between brandname AA batteries and generic AA batteries He used six pairs of AA alkaline batteries from two major battery manufactures a well known brand name and a generic brand He measured the length of battery life while playing a CD player continuously He recorded the time in minutes when the sound stopped Department of Statistics mm manm Battery Example continued Generic Brand Name Y 206 1874 8 103 146 n 6 6 Want 95 CI 0 a What is the standard error 0 b What is the 95 CVaI o c What is the ME 0 d What is the 95 CI Depanmcm of Statistics mummymm w Battery Example continued Generic Brand Name Y 206 1874 8 103 146 n 6 6 Want 95 CI 0 a What is the standard error 0 b What is the 95 CVal o c What is the ME 0 d What is the 95 Cl 0 e Does this confidence interval suggest that generic AA batteries will last longer than brandname AA batteries Department of Statistics u um Battery Example continued Generic Brand Name Y 206 1874 8 103 146 n 6 6 Want 95 CI 0 a What is the standard error 0 b What is the 95 CVal c What is the ME d What is the 95 Cl e Does this confidence interval suggest that generic AA batteries will last longer than brandname AA batteries 0 t Interpret the 95 CI Departmcnmsmms u um Battery Example continued answers 3 71 2 7 1 2 p00ed 30 W 1267 I71 I72 7 2 1262 1262 SE 6 6 727 o b CVal 1 an ian97510 22281 0 0 ME CVal x SE 22281 x 727 162527 0 d 95 CI a 235 3485 Depanmcnr of Staxisucs mqhmip umlmlnm he Battery Example continued answers 0 e Does this confidence interval suggest that generic AA batteries will last longer than brandname AA batteries Yes because zero is not Within the interval 0 t Interpret the 95 Cl We are 95 confident that the true mean difference is between 2 and 35 Department of Stansucs Battery Example continued using Tl calculator 0 Do this STAT a TESTS l 2SampTInt a STATS l Y1z206OlSXllO3ln1z6lY2zl874lSX2zl46l n2 6 l CiLevel 95 l Pooled Yes l CALCULATE Department of Staxisucs Battery Example continued using Tl calculator 0 Do this STAT a TESTS l 2SampTInt a STATS l Y1z206OlSXllO3ln1z6lY2zl874lSX2zl46l n2 6 l CiLevel 95 l Pooled Yes l CALCULATE o READOUT ZisampTInt 23471 34853 dflO Sxp 12 6342788 Department of Staxisucs glg Battery Example continued using Tl calculator 0 Do this STAT a TESTS l 2SampTInt a STATS l Y1z206OlSXllO3ln1z6lY2zl874lSX2zl46l n2 6 l CiLevel 95 l Pooled Yes i CALCULATE o READOUT ZisampTInt 2 3471 34853 dflO Sxp 12 6342788 that there Department of Staxisucs quotmy mww a Zero is not within this interval we can conclu is a difference between the two means Comparing Means of two related groups 0 We are not limited to comparing two averages of independent populations Suppose we want to compare the means of two related populations Depanmcm of Statistics vm mmw Comparing Means of two related groups 0 We are not limited to comparing two averages of independent populations Suppose we want to compare the means of two related populations 0 Recall Cl is ptest i ME Department of Statistics mwm mmw Comparing Means of two related groups 0 We are not limited to comparing two averages of independent populations Suppose we want to compare the means of two related populations 0 Recall Cl is ptest i ME ptest Y17Y2 ME CVal x SE where CVal ta2n1 1an1 7771 SE 3 Department of Statistics Example computer stock prices We want to compare January 2002 prices vs January 2003 prices of computer companies see page 92 Computer Stock Prices Jan 02 Jan 03 Diff Y 2591 1796 7946 s 634 565 61426 n 5 5 5 Department of Stansucs 01mm Computer Stock Prices Example continued 0 What is Standard Error 0 What is 95 Critical Value 0 What is 95 Margin of Error 9 What is a 95 Confidence Interval Depanmcm of Starisucs Computer Stock Prices Example continued 0 What is Standard Error 0 What is 95 Critical Value 0 What is 95 Margin of Error 0 What is a 95 Confidence Interval 0 Does this confidence interval suggest a difference in stock prices between Jan 2002 and Jan 2003 Depanmcnr of Starisucs mqhmip umlmlnm he Computer Stock Prices Example continued 0 What is Standard Error 0 What is 95 Critical Value 0 What is 95 Margin of Error 0 What is a 95 Confidence Interval 0 Does this confidence interval suggest a difference in stock prices between Jan 2002 and Jan 2003 o Interpret the 95 Cl Deparrmcnr of Starisucs mummymm w Computer Stock Prices Example answers 0 6 1426 sdiff SE 7 7 27471 W V5 0 025 1 iHVT1 7 02574 27764 0 ME 27764 X 27471 76271 0 95CI A 03189 15573 Department of Staxisucs Computer Stock Prices Example answers 0 6 1426 SW SE 7 7 27471 W V5 0 025 1 iHVT1 7 02574 27764 0 ME 27764 x 27471 76271 95CI 03189 15573 o Does this confidence interval suggest a difference in stock prices between Jan 2002 and Jan 2003 Yes because zero is NOT Within Cl Depanmcm of Staxisucs mummymm w Computer Stock Prices Example answers 0 6 1426 sdtt SE 7 7 27471 5 0 025 1 iHVT1 7 02574 27764 0 ME 27764 x 27471 76271 c 95CI 03189 15573 o Does this confidence interval suggest a difference in stock prices between Jan 2002 and Jan 2003 Yes because zero is NOT Within Cl 0 Interpret the 95 CI We are 95 confident that the true difference is between 3 and 156 Dcpanmcntof Scansucs Computer Stock Prices Example answers using TI calculators 0 Do this STAT a EDIT and Enter data into L1 and L2 then place cursor on L3 do 211012 7 2nd1ieL2 7 L1 a STAT a TESTS t tTntervaT a DATAL List Lat CiLevel 95 t CALCULATE Department of Su sm Computer Stock Prices Example answers using TI calculators 0 Do this STAT a EDIT and Enter data into L1 and L2 then place cursor on L3 do 211012 7 2nd1ieL2 7 L1 a STAT a TESTS t tInterval a DATAL List Lat CiLevel 95 t CALCULATE READOUT TInterval 03189 15573 Y 7946 SX 61426 n 5 Department of Su sm Computer Stock Prices Example answers using Tl calculators 0 Do this STAT a EDIT and Enter data into L1 and L2 then place cursor on L3 do 211012 7 2nd1ieL2 7 L1 a STAT a TESTS l tInterval a DATAl List Lgl CeLevel 95 l CALCULATE o READOUT TInterval 03189 15573 Y 7946 SX 61426 n 5 that there Department of Statistics quotmy mww a Zero is not within this interval we can conclu is a difference between the two means West Michigan Telecom Example problem 13 on page 104 Some stock market analysts have speculated that parts of West Michigan Telecom might be worth more that the whole For example the company39s communication systems in Ann Arbor and Detroit can be sold to other communications companies Suppose that a stock market analyst chose nine 9 acquisition experts and asked each to predict the return in percent on investment ROI in the company held to the year 2003 if i it does business as usual or ii if it breaks up its communication system and sells all its parts Their predictions follow Dcpanmcm of Statistics um mt u m West Michigan Telecom Example continued Expert123456789 NotBreak 12 21 8 2O 16 5 18 21 1O BreakUp 15 25 12 17 17 10 21 28 15 9 SE sciquotNB 28626 f 09542 0 CVal MN 10253 ian1 7 025 8 23060 0 ME 2306 x 9542 22004 0 95CI 10218 54226 Deparrmcm of Staxisucs mm mm 18 West Michigan Telecom Example continued Expert123456789 NotBreak 12 21 8 2O 16 5 18 21 1O Break Up 15 25 12 17 17 10 21 28 15 9 SE sdm 28626 f 09542 0 CVal 04mm 0253 ian1 7 025 8 23060 0 ME 2306 x 9542 22004 0 95CI 10218 54226 0 Does this confidence interval suggest a difference between breaking up the company or not Yes because zero is NOT Within Cl Depanmm of Starisucs West Michigan Telecom Example continued Expert123456789 NotBreak 12 21 8 2O 16 5 18 21 1O Break Up 15 25 12 17 17 10 21 28 15 9 SE sciquotNB 28626 f 09542 0 CVal 04mm 0253 ian1 7 025 8 23060 0 ME 2306 x 9542 22004 0 95CI 10218 54226 0 Does this confidence interval suggest a difference between breaking up the company or not Yes because zero is NOT Within Cl 0 Interpret the 95 CI We are 95 confident that the true difference among the experts is between 10 and 54 Department of Statistics mu m1 Wm West Michigan Telecom Example continued using Tl calculators o STAT a EDIT and Enter data into L1 and L2 and place cursor on L3 do 211012 7 2nd1ieL2 7 L1 then do STAT gt TESTS l tTnterval gt Data l List Lgl CiLevel 95 l CALCULATE Department of Staxisucs West Michigan Telecom Example continued using Tl calculators o STAT a EDIT and Enter data into L1 and L2 and place cursor on L3 do 211012 7 211011 ie L2 7 L1 then do STAT gt TESTS l tTnterval gt Data l List Lgl CiLevel95LCALCULATE o READOUT TTnterval 102 542 Y 322 SX 28626 n 9 Department of Staxisucs West Michigan Telecom Example continued using Tl calculators o STAT a EDIT and Enter data into L1 and L2 and place cursor on L3 do 211012 7 211011 ie L2 7 L1 then do STAT gt TESTS l tTnterval gt Data l List Lgl CiLevel95LCALCULATE o READOUT TTnterval 102 542 Y 322 SX 28626 n 9 that there Department 01 Statistics 0 Zero is not within this interval we can conclu is a difference between the two means Outline 9 Confidence Interval for Proportion 0 Confidence Interval for Population Proportion 0 Sample Size Determination Department of Starisucs Confidence Interval for population proportion Suppose we want to estimate the population proportion using intervals 0 Recall CI is ptest i ME Department of Staxisucs Confidence Interval for population proportion Suppose we want to estimate the population proportion using intervals 0 Recall CI is ptest i ME 0 Therefore use success X Lest 7 p sampleSIze n Depanmcm of Statistics mm mmw Confidence Interval for population proportion Suppose we want to estimate the population proportion using intervals 0 Recall CI is ptest i ME 0 Therefore use 7 success 7 X New 7 sampleSize 7 E 0 MECValeE Department of Statistics mqhmip umlmlmt he Confidence Interval for population proportion Suppose we want to estimate the population proportion using intervals 0 Recall CI is ptest i ME 0 Therefore use 7 success 7 X New 7 sampleSize 7 E 0 ME CVal x SE 0 This CI works well if nxpgt5andnx17pgt5 Note that is it woks well if the expected number of successes and the expected number of Dc failures are both greater than 5 7 W panmcnt of Statistics EAS Sensor Example If a sales clerktails to remove the EAS sensor when an item is purchased it can result in an embarrassing situation for the customer A survey was conducted to study consumer reaction to such false alarms Of 250 customers surveyed 40 said that it they were to set off an EAS alarm because store personnel did not deactivate the merchandise then they would never shop at the store againquot Department of Statistics mvmunnm mumuxa EAS Sensor Example continued 7 4o 7 7 161716 7 ptesti io a SE 7 T7002319 o CVal 042 2025 invNorm1 7 025 196 0 ME 196 X 002319 004544 0 95CI 11456 20544 Depanmcnt of StaKiSUCS warnmtb tmrumm w EAS Sensor Example continued 161 7 16 250 002319 40 ptest 250 7016 SEA 7 p CVal 042 2025 invNorm1 7 025 196 ME 196 x 002319 004544 95CI 11456 20544 Interpret the 95 CI We are 95 confident that the true proportion is between 011 and 021 Depanmcm of Statistics mummymm he EAS Sensor Example continued using Tl calculators a Do this STAT a TESTS l liPropZint l x 40 l 11250 iciLevel 95 l CALCULATE Department of Staxisucs EAS Sensor Example continued using Tl calculators 9 Do this STAT a TESTS l liPropZint l x 40 l 11250 iciLevel 95 l CALCULATE e READOUT Zinterval ll456 20544 P 16 n 250 Department of Staxisucs EAS Sensor Example continued using Tl calculators a Do this STAT a TESTS l lePropz mt l x 40 l 11250 iciLevel 95 l CALCULATE e READOUT Zinterval ll456 20544 P 16 n 250 9 We are 95 confident that the true proportion is between 1 1 and 21 Depanmcm of Sumsucs mm mm w Exercise 17 on page 105 Given x 600 n2000 o 600 7 2000 7 7 31 7 3 i SEp 7 W 7 00102 CVal 2042 2025 invNorm1 7 025 196 ME 196 X 00102 00201 ptest 03 Depanmcnt of StaKiSUCS warnmtb tmrumm he Exercise 17 on page 105 Given x 600 n2000 o 600 7 2000 7 7 31 7 3 i SEp 7 W 7 00102 CVal 2042 2025 invNorm1 7 025 196 ME 196 X 00102 00201 ptest 03 o 95CI 2799 3201 Deparrmcm of Statistics mm mm w Exercise 17 on page 105 Given x 600 n2000 o 600 7 2000 7 7 31 7 3 i SEp 7 W 7 00102 CVal 2042 2025 invNorm1 7 025 196 ME 196 X 00102 00201 ptest 03 o 95CI 2799 3201 o Interpret the 95 CI We are 95 confident that the true proportion is between 028 and 032 Q Department of Scansucs Sample Size Determination based on CI of proportion 0 What is the true proportion of success p 0 Decide which confidence level to use 0 Determine margin of error that you39re willing to accept Department of Su sm EAS Example Suppose that you are a student with a grant to study this EAS issue and you realize that there are not enough funds to gather data on 250 subjects So you want to determine a new sample size by relaxing the confidence level to 90 and use p16 and the ME of 004544 what is the new sample size Department of Statistics mum meann a u m EAS Example continued answer 2 w xmnim Deparrmcm of Staxistics EAS Example continued answer 2 w xmnem O CVal90 217Vg2 23905 invNorm1 7 05 1645 0 ME 04544 0 f1 16 note see discussions on next slide Department of Staxisucs EAS Example continued answer 2 w xmnem O CVal90 217Vg2 23905 invNorm1 7 05 1645 0 ME 04544 0 f1 16 note see discussions on next slide 9 2 122 x 161 7 16 1761 9 177 Deparrmcm of Sta sucs mm mw w Sample Size Determination for Proportion discussions 2 w xi oemi 0 When a rough estimate of p is available such as that from a pilot study or some educated guess use it for 3 above Otherwise use for f a conservative estimate Department of Statistics ruqhnmtb tmrukmt he Sample Size Determination for Proportion discussions 2 w xi oemi 0 When a rough estimate of p is available such as that from a pilot study or some educated guess use it for 3 above Otherwise use for f a conservative estimate 0 It is recommended to always round it up to the next integer as n 177 here which is rounded up from 1761 Depanmcm of Statistics mummymm w Chapter 4 Part 2 Binomial Distribution JC Wang Department of Statistics Western Michigan University Depanmcm of s 39 Outline 0 Computing Binomial Probabilities 0 Properties of a Binomial Distribution 0 Computing Binomial Probabilities by TI Calculators 0 Binomial Random Variables 9 Normal Approximation 0 Example 9 When Approximation is Good 9 Examples 0 Binomial Example 0 Normal Example Department of Starisucs Outline 0 Computing Binomial Probabilities 0 Properties of a Binomial Distribution 0 Computing Binomial Probabilities by TI Calculators O Binomial Random Variables Department of Starisucs Properties of a binomial distribution 0 Only two possible outcomes for each observation Depanmcm of Staxisucs Properties of a binomial distribution 0 Only two possible outcomes for each observation 9 Probability of success is p probability of failure is q 1 7 p Department of Starisucs Properties of a binomial distribution 0 Only two possible outcomes for each observation 9 Probability of success is p probability of failure is q 1 7 p 0 Observations are independent Department of Staxisucs m npmm Properties of a binomial distribution 0 Only two possible outcomes for each observation 9 Probability of success is p probability of failure is q 1 7 p 0 Observations are independent Note 0 Sample size is fixed 0 Number of successes X is of interest a De m panmcm of Staxisucs mummww Example of binomial distribution Stat 2160 multiple choice quiz has 5 questions with 4 choices for each question def success answering a question correctly 0 probability of success Department of Swims Example of binomial distribution Stat 2160 multiple choice quiz has 5 questions with 4 choices for each question def success answering a question correctly 0 probability of success o probability of failure Department of Statistics Example of binomial distribution Stat 2160 multiple choice quiz has 5 questions with 4 choices for each question def success answering a question correctly 0 probability of success o probability of failure 0 sample size Department of Statistics Example of binomial distribution Stat 2160 multiple choice quiz has 5 questions with 4 choices for each question def success answering a question correctly 0 probability of success o probability of failure 0 sample size o What is probability that a student will answer exactly 3 questions correctly in other words PX Sin 5p 025 q 075 7 Department of Statistics um m u m Computing Binomial Probabilities PX Sin 5p 025q 075 7 Department of Statistics Computing Binomial Probabilities PX Sin 5p 025 7 0 Formula 5 3 73 3573p q5 00879 Note 0 1 and p0 1 and n1x2xxn Department of Staxisucs Computing Binomial Probabilities PX Sin 5p 025 7 0 Formula 5 3 73 3573p q5 00879 Note 0 1 and p0 1 and n1x2xxn 0 Using T Calculator lt2ndgtltDISTRgtibinomiaIPDF5 25 3ltENTERgt and get 00879 Depanmcnt of StaKiSUCS warnmtb tmrumm w Computing Binomial Probabilities using Tl calculator keyword exactly 20 of TV buyers at TVMore purchase the store39s extended warranty Say that 10 TV sets were sold in one day What is the probabilities that exactly 3 extended warranties were sold in other words PX3ln 10p 2 7 Depanmcnr of Staxisucs mqhmip umlmlnm he Computing Binomial Probabilities using Tl calculator keyword exactly 20 of TV buyers at TVMore purchase the store39s extended warranty Say that 10 TV sets were sold in one day What is the probabilities that exactly 3 extended warranties were sold in other words PX3ln 10p 2 7 lt2ndgtltDlSTRgtlbinomPDF1023ltENTERgt and get 02013 Depanmcm of Statistics mummymm w Cumulative Probabilities cumulative distribution function keyword at most ie lefttail probability PX lnp PXOor10rorjlnp PXOPX1PXj 20 of TV buyers at TVMore purchase the store39s extended warranty Say that 10 TV sets were sold in one day What is the probability that at most 3 extended warranties were sold in other words PX Sin 10p 2 7 lt2ndgtltDISTRgtlbinomCDF1023ltENTERgt and get 08791 E Dwanmmfsmms Cumulative Probabilities right tail probability keyword at least ie righttail probability PXjlnp PXjorj1orornlnp PXjPXj1PXn 1PX 1lnlp AA 20 of TV buyers at TVMore purchase the stores extended warranty Say that 10 TV sets were sold in one day What is the probability that at least 3 extended warranties were sold ie 17PX 371l10 2 17PX 2l10 2 7 Dcpanmcnr of Stansucs PX3ln10p2 17 lt2ndgtltDlSTRgtlbinomCDF1022 03222 Binomial Random Variables 0 Expected value EX np ie expected number of successes Department of Staxisucs Binomial Random Variables 0 Expected value EX np ie expected number of successes 6 Standard deviation sdX np1 7 p 1npq Department of Staxisucs Binomial Random Variables TVMore Example Suppose that 20 of TV buyers at TVMore purchase the store39s extended warranty If 26 TVs were sold last week the expected number of extended warranties should be around np 26 x 02 52 give or take lnp1 7 p 26 x 021 7 02 20396 Say that the extended warranty cost is 100 how much revenue will be generated 100 x 52 i 100 X 20396 520 i 20396 Note Multiplication rule of location and spread Department of Statistics Outline 9 Normal Approximation 0 Example When Approximation is Good Department of Statistics Graphical Representation of binomial probabilities bar graph and probability histogram Binomiain15po5 Binomiain14po5 H 2B 7 Pgtltgtlt narncigm n 2n 7 mar Q Uu r Unnr i i 7 i I I n 5 in none rrl I we X i probabiiity i D Pgtltgtlt narncigm bararea u 15 r 3 me it prubabiiity cf 3 success is u 5 5 U i the shape is symmetric about nZ EL n ma 7 u on r i i 7 5 i i Deparrmcm o ma Normal Approximation of binomial probabilities n2n7 r Pgtltx gtltEmumaln ZE P U7 m7 Px70 5ltYltx05 mummdnaae Einumial Prubabiliw u in 7 VNurmalw 5 2 U 2 U355 PmbabillW n ma 7 D actual pmbabiliW u appruximalE prubabliW i i i I 5 Xl 15 2D 25 n In 7 r PM S x Pgtlt 2 x PYltx05 m7 PYgtx705 u in 7 n ma 7 u an 7 i i i i f Deparrmcm or s Normal Approximation TV more example Say n 26 and p 2 Using the Normal Curve make sure npgt5and nqgt5 Here26x 252gt5and 26 x 1 7 2 208 gt 5 Therefore conditions hold and the standard deviation is m 20396 Let39s say X g 5 PX g 5 z PY lt 55 normaICDF79999555220396 05585 Department of Statistics Normal Approximation TVmore example continued 020 7 015 7 010 7 005 7 Normal Approximation TVmore example comparison with exact binomial probability Sayn26andp 2anng5 Using the Binomial Probability PX Sin 26 p 2 binomCDF26 2 5 05775 Note Normal approximation gives a value close to the precise binomial method Department of Su sm Conditions for Good Approximation Normal curve gives a reasonable approximation for the binomial probabilities whenever both np gt 5 and nq gt 5 Note 0 np expected number of successes o nq expected number of failures Depanmcm of Statistics mm mm w Outline 9 Examples 0 Binomial Example 0 Normal Example Binomial Example travel agents example The rate of commission that commercial airlines pay travel agents has been declining for several years In an attempt by travel agents to raise revenue many agents are now charging their customers a ticket fee typically between 10 and 15 dollars According to the ASTA about 90 percent of travel agents charge customer fees when purchasing an airline ticket Department of Statistics mvmunnm mumuxa Travel Agents Example continued Suppose that a random sample of 55 travel agents is selected Assume that the number of the 55 travel agents charging a ticket is distributed as a binomial random variable 0 What are the mean and sd of this distribution 0 What assumptions are necessary 0 What is the probability that none of the travel agents will charge a tee in other words PX Oln 55p 9 7 Department of Stansucs u um Travel Agents Example continued 0 What is the probability that at least 46 of our sample of travel agents will charge a tee in other words PX 2 46in 55p 9 7 o What is the probability that at most 50 of our sample of travel agents will charge a tee in other words PX g 50in 55p 9 7 Department of Swims Normal Example packaging bags example Plastic bags used for packaging produce are manufactured so that the breaking strength of the bag is normally distributed with a mean of 5 pounds per square inch and a sd of 15 pounds per square inch Given u 5 a 15 o What is the proportion of bags produced having a breaking strength of less than 317 pounds per square inch in other words PX lt 317 7 o What is the proportion of bags produced having a breaking strength between 32 and 42 pounds per square inch in other words P32 g X g 42 7 Department of Statistics WWWWMW a Packaging Bags Example conitinued o What is the proportion of bags produced having a breaking strength of at least 36 pounds per square inch in other words PX 2 36 7 0 Between what two values symmetrically distributed around about the mean will 95 of the breaking strength tall in other words Pa g X g b 7 Department of Swims Chapter 1 Data Presentation Statistics and Data Section 20269 JC Wang Department of Statistics Western Michigan University Department of Statistics w m mu Goal and Objectives 1 To understand the difference between types of data categorical and numerical 2 To learn how to analyze each type gt To explore the categorical levels of data gt To explore the numerical levels of data Department of Statistics w m mu Outline Statistics and Data Statistics and Data Types of Data Levels of Measurement Example of Data Types and Levels of Measurement Summarizing Categorical Data Relative Frequency Table Bar Chart Pie Chart Summarizing Numerical Data StemandLeaf Plot Relative Frequency Table and Histogram FiveNumber Summary and Boxplots Dotplot Data Shape Department of Statistics uwmqm Statistics and Data gt Statistics gt broad sense collection of methodsprocedures for collecting data and analyzing data gt narrower sense numbers arise out of data analysis eg average gt Data gt collection of measurements made on a number of subjects gt example CLASS DATA set page 2 of textbook has six different measurements on ten students Department of Statistics uwmqm Two Types of Data gt Categorical such as color of student eyes gt Numerical such as temperature of this room in Fahrenheit Department of Statistics w mtmmm Levels of Measurement Categorical Data 1 Nominal ordering does not exist eg gender SSN eye color 2 Ordinal ordering does exist eg military rank class levels rating scales Department of Statistics w mhmwmur Levels of Measurement Numerical Data gt Interval gt distance exists but no ratios gt zero is arbitrary and not an indication of absence of the variable eg s temperature scale IQ scores GPA gt Ratio ratios exists and zero indicates an absence of the variable Department of Statistics m ugmannhnh d nxmmn Interval Measurement Example Temperature gt 600F i SOOF 300F gt 60 F is not twice as warm as 30 F gt 0on not absence of temperature Department of Statistics m ugmannhnh d nxmmn Ratio Measurement Two types gt Discrete result of a counting process whole numbers ie integers eg result of question how many children in your family gt Continuous result of a measuring process eg s age money time mph height weight Department of Statistics mm mm an Example CLASS DATA Variables Sex and Transport are nominal Variable Class Level is ordinal Variable GPA is an interval measurement VVVV Variables Hours taken and Sleep hours last night are ratio measurements Note The column student is used as data label It39s not used in data analysis However it could be used to label observations in graphical presentation or tabular outcomes Other data labels include students names SSNs Ple etc V Department of Statistics mm mm m Relative Frequency Table for categorical data gt Frequency table lists categories of a variable and their respective frequencies gt Frequency is a count of subjects who fall into a particular category gt Relative frequency is the frequency divided by total count Independent of size of data set Used to compare datasets of different sizes Department of Statistics IlaJumnnmp umam Example Graduation Rate Data on page Department of Statistics IlaJumnnmp umam Relative Frequency Table for nominal data Arrange the rows of relative frequency table in decreasing order of frequencies of the categories Redo previous example Note it is recommended not to apply this to ordinal data in such case the rows should be kept in the order of categories Department of Statistics i mmnyauIu umam Bar Chart is used to present frequencies or relative frequencies graphically draw bars at the same base with heights representing frequencies or relative frequencies Note use only relative 3 7 frequency for comparison of multiple data sets having the same variable of same categories Frequency 1 l Department of Statistics i mmnyauIu umam Pareto Chart for nominal data A Pareto chart is a bar chart used for nominal data only recommended in which bars are arranged in decreasing order left to right of heights Redo previous example Frequency 1 l Department of Statistics 1mmq mam Pie Chart gt Picture illustrating the distribution of data gt Circle divided into slices number of slices corresponds to the number of categories gt Size of slice is proportional gt Relative frequency percents make it easier to create a proportional pie chart Department of Statistics 1mmq mam Example Summer Quiz Data MKT 71 Department of Statistics 1mmq mam Drawback of Pie Chart It human eyes not good at comparing areas of slices l poor choice of colors or filled patterns makes judging area even more difficult OD human eyes are good at judging positions of bar tops hence a bar chart works better than a pie chart pie chart with proportions of categories shown is no good bar chart with reference grid lines works better or even a relative frequency table is enough to convey the comparison 4s Department of Statistics 1mmq mam Use of StemandLeaf Plot to identify the following typical value spred about the typical value data gaps shape distribution of data number amp location of peaks outlying values V VVVVV Department of Statistics ummmw Construction of StemandLeaf Plot sort data recommended select leading digits for stems list all possible stems record leaf for each observation beside the corresponding stem round off if necessary ODN t O1 indicate units for stems amp leaves note if isolated extreme values low or high exist write them out instead 05 epartment of Statistics uvnywunqnmnw Example of StemandLeaf Plot Summer Quiz Data sorted 8 ll l3 19 21 2 25 25 25 23 31 35 39 27 Stemplot of Summer Quiz Data stem width 10 0 8 1 139 Ltlf5359 3 159 Department of Statistics t m Stemplot of Quiz Data different set of stems Each stem in the previous example split in two stem width 5 558 JgtJgtLOLONNHHO U lt U IHKOHGD 0 Department of Statistics t m ReIFreq Table for Numerical Data Increased sample sizes required data to be condensed gt Advantage summary table of data arranged into numerically ordered class groupings gt Disadvantage information lost individual values lost through grouping Department of Statistics uvymnnuu Jmam Calculations of RelFreq Class RF Class freq sample size gt Expressed as a proportion fraction or percent gt Independent of sample size gt Used when comparing multiple data sets of different size a right way to do epartment of Statistics uvymnnuu Jmam Construction of ReIFreq Table at Determine number of classes k 5 to 15 recommended Determine class interval width range k 7 1 Establish classes39 boundaries and use proper inclusion rule left or right so that classes not overlap Must include entire range of data ION 4 Department of Statistics ummmw Construction of ReIFreq Table cont d Choose classes to facilitate interpretation shape etc Tally data into appropriate classes Total frequency of each class Calculate rel freq39s OWVO JO39I Calculate cumulative rel freq39s if needed epartment of Statistics uvlywuhmmnw Example of RelFreq Table Summer Quiz Data sorted 8 ll l3 19 21 232132132529 31 35 39 47 range47i 839andsetk 4 then Width 39 4 7 1 13 rightinclusion 2 75 8 8 21 21 34 34 47 Class Freq RelFreq Cum RelFreq 75 8 1 0071 0071 8 21 4 0286 0357 21 34 6 0429 0786 34 47 3 0214 1000 Total 14 10 class interval includes right boundary Department of Statistics woumnnmp uma Construction of Histogram Vertical bar chart constructed at boundaries of each class Bars of same width and joined at boundaries Height represents class freq or relfreq Classes listed and evenly spaced along horizontal axis Rel Freq listed along vertical axis scaled to account for highest rel freq VVVVV gt Comparison histogram use rel freq only epartment of Statistics I urmnnmp umar Freq Histogram Example Summer Quiz Data 4 classes Frequency Distribution Summer Quiz Data to L0 v 3 5 3 lt0 u w e N o 5 8 21 34 47 quiz scores cl 39 ht boundary Freq Histogram Example Summer Quiz Data 5 classes Frequency Histogram Summer Quiz Data to L0 v 5 z w lt0 u w E N o i i i i i i 0 10 20 30 40 50 Quiz Scores k5 rangeke13941o Department of Statistics ltx amtnyyv rman Department of Statistics ltx amtnyyv rman Fivenumber Summary gt MIN Minimum data value gt C21 first quartile upper boundary of first quarter of data values gt MED middle value of data second quartile gt GS third quartile upper boundary of third quarter of data values gt MAX Maximum data value Department of Statistics mmqm Fivenumber Summary Summer Quiz Example data 8 ll l3 19 21 23 25 25 25 28 31 35 39 47 orderl 2 3 4 5 6 7 8 9 10 ll l2 l3 14 1714 25nl375 5nl75 75nlll25 Use textbook formula pages 12 amp 13 MIN 8 C21 16 3rd obs 4th obs2 13192 MED 25 7th obs 8th obs 2 25252 as 33 11th obs 12th obs2 31352 MAX 47 epartmcnt of Statistics mmqm Construction of Boxplot Skeletel Boxplot Draw 5Iines that includes range of data and quartiles Draw box extending from C21 to C23 Draw ine inside box at median Extend horizontal lines whiskers from box to minimum and maximum ODN k of Statistics mw iman Skeletel Boxplot Summer Quiz Example Skeletel Boxplot Summer ll Quiz Example 10 2o 30 40 em Scores Note methods used for quartiies Department of Statistics m mw mm Schematic Boxplot Box and median line remained V V Outliers outlying values displayed with symbol such as circle Whiskers extended to smallest and largest values that are normal IQR InterQuartile Range C23 7 C21 values lt C21 7 151QR and values gt C23 151QR are outliers Example as seen in the comparison boxplots of HS GPA amp GPA2 page 12 of textbook V V V Department of Statistics 1mm mm Construction of Dotplot Simple graph that displays along an axis line Each observation is shown as a dot above the line Illustrates the pattern of variation in data VVVV Provides information similar to that found in a stemandleaf plot gt Example as seen in the comparison dotplots of HS GPA amp GPA2 page 13 of textbook epartment of Statistics IzJmnnun umam Symmetry and Skewness based on histogram stemplot and dotplot gt Histogram draw a smooth curve over the top of the bars gt Stemplot firt turn 90 degrees counterclockwise draw a smooth curve over the ends of the leaves gt Dotplot draw a smooth curve over the top of the stacked dots gt Examine the tails of the curve examples on page 14 of textbook gt Left tail longer than right leftskewed data gt Right tail longer than left rightskewed data gt Both tails approximately the same symmetric data Department of Statistics 1mmq mam Symmetry and Skewness based on boxplot gt Use only skeletel boxplot in horizontal position to examine data shape gt Examine the length of the whiskers example on page 14 of textbook gt Left whisker longer than right leftskewed data gt Right whisker longer than left rightskewed data gt Both whiskers approximately the same symmetric data Department of Statistics 1mmq mam Chapter 8 Testing Hypotheses Tests for Proportions and Two Sample Problems JC Wang Department of Statistics Western Michigan University Outline OneProportion Z Test OneProportion Z Test OneProportion Z Test Wald s Method OneProportion Z Test Scores Method TwoSample Problems TwoSample t Test PairedSamples t Test PairedSamples vs Two Independent Samples Types of Errors Two Types of Decision Errors An Example partment of Statistics mm 5 H i 7 Prescription Drug Dosage Example gt Suppose you are concerned if doctors were consistently adjusting dosages for weight of elderly patients You studied 2000 prescriptions and found that 626 prescriptions the doctors failed to adjust the dosage gt Let s consider the given information to answer the above question gt Is there evidence that the proportion of doctors who failed to adjust dosage is different from g of the written prescription gt Given X 626 n 2000 p0 partmcnt of Statistics mmn ammlmmn rm Prescription Drug Dosage Example gt How many tails is this test A Two Note the keyword different gt What are the hypotheses Hopvs Hapy gt What is the significance level a 005 gt What is the sample size n 2000 gt Now there are two methods to determine the SE D partmentof Statistics rm 7 Prescription Drug Dosage Example Wald s method gt What is the SE using the Wald s Method p 2000 gt What is the test statistic 626 0313 SE 001037 031371 2 W 7 3 7196046 SE 001037 gt What is the pvalue ofTails x normCDF TSl9999 2 X normCDF1 96046 9999 004994 partmcnt of Sta marlsn Prescription Drug Dosage Example Wald s method continued gt Conclusion Is the p value lt significance level ie ls 049946 lt 005 Yes Reject Ho there is statistically significant evidence that the proportion of doctors who failed to adjust dosage is different than of the written prescriptions D partmentof Statistics noquot Prescription Drug Dosage Example Scores method gt What is the SE using Scores method P01P0 1 i SE 7 n i 2000 7 0010541 gt What is the test statistic i ptest 7 Ho Value 7 0313 7 i i Z SE 0010541 1929 gt What is the pvalue ofTails x normCDF TSl9999 2 X normCDF19299999 00537 partmcnt of Statistics mmn mmmmmn mi Prescription Drug Dosage Example Scores method continued gt Conclusion Is the p value lt significance level ie Is 0537 lt 005 No Do not reject Ho there is not enough evidence to conclude that the proportion of doctors who failed to adjust dosage differs than of the written prescriptions Note Wald s method is statistically significant while the Scores method is not significant D partmentof Statistics nun 39 Prescription Drug Dosage Example Scores method using Tl calculators Note The TI calculators use scores method gt Do this STAT gt TESTS i l PropZTest i p0 33333i X 626 l I7 2000 l prop y p0 l CALCULATE gt Readout l PropZTest prop y 33333 2 7192898937 P 0537320441 P 313 n 2000 gt Conclusion Do not reject Ho partmcnt of Statistics mmwp aum nmmn Gasoline Example An independent testing agency has been contracted to determine whether there is any difference in the gasoline mileage output of two different gasoline on the same model car Gas A is tested on 25 cars and produced a sample mean of 185 mpg with a sample SD of 46 mpg Gas B is tested on 24 cars and produced a sample mean of 193 mpg with a sample SD of 52 mpg At the 005 level of significance Is there evidence of a significant difference between the averages of the two gasoline performances Department of Statistics quot9439 r Gasoline Example con nued S I7 gt How many tails in this test A two Note keyword is difference gt What are the hypotheses HoiMAMBVSH1 1MA MB partmcnt of Statistics wmmun v v mmmn Gasoline Example con nued gt What is the significance level a 005 gt What are the sample sizes nA 25 and n3 24 gt Should we pool the variances SD Yes if larger SD smaller SD lt 2 Since 113 is less than 2 we should indicate Pooled YES on our Tl83 calculator Department of Statistics rm 39 Gasoline Example continued using TI calculators gt Do this STAT gt TESTS l 2 SampTTEST ix 185l 8X146ln125lY2193lSX252ln224l M1 M2 l pooled yes i CALCULATE gt Readout 2 SampTTest 1 M2 1 75709832971 P 5707311948 df 47 SXp 490279989 partmcnt of Statistics mmn mmmmnm mi Gasoline Example con nued gt What is the conclusion ls pvalue lt a ie is 05707 lt 005 No do not reject Ho there is not enough evidence of a significant performance difference between the averages of the two gasolines Department of Statistics nun 39 Gasoline Example con nued Suppose in addition to the above information as obtained from TI calculators you are interested in finding the SE Remember this equation 80 14011 Y1Y2M1M2 08 i 0 SE 1 75710 This is much easier than finding the pooled SD and using SESE2SE Department of Statistics mtllnuwmumrrp agm nmmn PairedSamples t Test blood pressure gauge example A digital blood pressure gauge manufactured for home use was tested recently in the following manner Eight individuals had their blood pressure taken first by a highly respected physician then by the home device The data are shown below partment of Statistics mm Blood Pressure Gauge Example con nued gt What are the hypotheses HoiMD0VS H1 1MD 0 gt What is the significance level a 005 gt What is the sample size partmcnt of Sta mm1n Blood Pressure Gauge Example con nued gt Data entryediting STAT gt EDIT and enter pairs into L1 physicians and L2 home device then place cursor on L3 and do 2nd2 i 2W1 ie L2 7 L1 gt Do this STAT gt TESTS l T TEST gt DATA l p0 Ol LISTILS l FREQ1 lu y yo i CALCULATE gt Readout T Test y 0 t 1865367247 P 857314815 Y 1125 Sxl70581987 D partmentof Statistics noquot Blood Pressure Gauge Example con nued 5 1125 sD 170582 SE sD 170582 6031 gt What is the test statistic i ptest 7 Ho value 7 1125 7 0 i ti SE 7 6031 701865 gt How many tails in this test A two Note keyword is difference partmcnt of Statistics mmn mmmmmn ml Blood Pressure Gauge Example con nued gt What is the pvalue ofTails x tCDFl TSl 9999 df 2 X tCDF186599997 08573 What is the critical value V CV3 05277 1102577 tINV1 7 V Conclusion ls pvalue lt a ie ls 08573 lt 005 No so do not reject H0 There is no evidence of a significant difference between the physician s readings and the home device readings D partmentof Statistics nun 39 Paired vs Two Independent Samples gt So why would we design a study with paired data if we could just use two independent samples gt The paired analysis will give us a better result because the between subject variation is controlled Department of Statistics mthlnuwmnmwp aum nmmn Type and Type II Errors Occur when wrong decision reached gt Type Error Reject Ho the decision when in fact the true state of nature H0 is true The significance level a is the maximum allowable risk probability of committing type decision error gt Type II Error Do not reject Ho the decision when in fact the true state of nature H0 is false 5 PType Error Stat Dec Watch works Watch broken do not fix it w min prob 1 7 a Type II error the confidence level w prob 5 fix it Type lerror w prob 1 7 5 w max prob a the power partment of Statistics pm Type and Type II Errors cereal packaging example A supervisor is attempting to determine if the cerealfillingprocess is in control ie the average fill per box is the hypothesized 368 grams Letting x denote the sample average of 25 boxes in which the cerealfillingprocess was found to be in control With the help of a statistician the hypotheses Ho p 368 vs H1 p y 368 are formed If H0 is not rejected it is determined that the cerealfillingprocess is in control partmcnt of Statistics mmwp aambnmmn Type and Type II Errors cereal packaging examplecontinued What occurs if the supervisor commits a Type Error The risk of a Type I error involves concluding that the average fill per box has changed from the hypothesized 368 grams when in fact it has not changed Department of Statistics quot9439 r Type and Type II Errors cereal packaging examplecontinued What occurs if the supervisor commits a Type II Error The risk of a Type II error involves concluding that the average fill per box has not changed from the hypothesized 368 grams when in fact it has changed mi partmcnt of Statistics mmn mmmmmn Chapter 4 Part 2 Binomial Distribution JC Wang Department of Statistics Western Michigan University Outline Computing Binomial Probabilities Properties of a Binomial Distribution Computing Binomial Probabilities by TI Calculators Binomial Random Variables Normal Approximation Example When Approximation is Good Examples Binomial Example Normal Example Department of Statistics mum M mm m Properties of a binomial distribution 1 Only two possible outcomes for each observation 2 Probability of success is p probability of failure is q 1 7 p 3 Observations are independent Note gt Sample size is fixed gt Number of successes X is of interest cpartmcntof Stati 39 s wman mmm Example of binomial distribution Stat 2160 multiple choice quiz has 5 questions with 4 choices for each question d r success 2 answering a question correctly gt probability of success gt probability of failure gt sample size gt What is probability that a student will answer exactly 3 questions correctly in other words PX Sin 5 p 025 q 075 7 Department of Statistics quot9439 r Computing Binomial Probabilities PX Sin 5 p 025 q 075 7 gt Formula 5 3 573 35 i 3p q 00879 Note 0 1 and p0 1 and n1x2xxn gt Using Tl Calculator lt2ndgtltDISTRgtlbinomialPDF5 25 3ltENTERgt and get 00879 partmcnt of Statistics l nnn mrnmmw mm Computing Binomial Probabilities using TI calculator keyword exactly 20 of TV buyers at TVMore purchase the store s extended warranty Say that 10 TV sets were sold in one day What is the probabilities that exactly 3 extended warranties were sold in other words PX3ln 10p 2 7 lt2ndgtltDlSTRgtlbinomPDF1023ltENTERgt and get 02013 partment of Statistics mm mmllmniIJ1 Cumulative Probabilities cumulative distribution function keyword at most ie lefttail probability PXSlnp PXOor1ororjlnp e PX0PX1PXj 20 of TV buyers at TVMore purchase the store s extended warranty Say that 10 TV sets were sold in one day What is the probability that at most 3 extended warranties were sold in other words PX g Sin 10p 2 7 lt2ndgtltDISTRgtlbinomCDF1023ltENTERgt and get 08791 partmcnt of Statistics m wmmun ammlmmu Cumulative Probabilities right tail probability keyword at least ie righttail probability PX Ell710 PX jor1orornlnp PXjPXj1PXn 1PX1gt1 in P 20 of TV buyers at TVMore purchase the stores extended warranty Say that 10 TV sets were sold in one day What is the probability that at least 3 extended warranties were sold ie PX 2 Sin 1op 2 17PX 371l10 2 17PX 2l10 2 7 1i lt2ndgtltDlSTRgtlbinomCDF1022 03222 partment of Statistics mm 7 Binomial Random Variables gt Expected value EX np ie expected number of successes gt Standard deviation sdX xnp1 7 p W partmcnt of Statistics mm mmmmmn ml Binomial Random Variables TVMore Example Suppose that 20 of TV buyers at TVMore purchase the store s extended warranty If 26 TVs were sold last week the expected number of extended warranties should be around np 26 x 02 52 give or take np1 i p 26 X 021 i 02 20396 Say that the extended warranty cost is 100 how much revenue will be generated 100 X 52 it 100 gtlt 20396 520 in 20396 Note Multiplication rule of location and spread D partmentof Statistics nun 39 Graphical Representation of binomial probabilities bar graph and probability histogram Binomialn15p05 Binomialn14p05 0 20 Pgtltx barhelght o 20 r 015 7 A 0 is r F gtlt 0 i0 7 g E B 010 0 05 r o a o 00 r o 05 e i i 7 i I I 0 5 lo 0 00 e r I I r X 0 20 Pgtltx barhelght bar area 0 is r 3 Note if probability of a success is o 5 X 010 i the shape is symmetric about riZ o 05 r o 00 r l l 75 l l o 5 i0 is x Normal Approximation of binomial probabilities nzne PXx XEl mlal 26 p 2 N Polo 5ltYltX0 5 u i5 7 Hp5 2 I7an 2 mas Elnumlal Prubablllty u in r YNurmalp 5 2i 7 2 mass Pmbablllty n ma 7 D acmal pmbablllty D apprmlmale prubablllty D m i i i i i i U 5 XlU 15 2D 25 U 2U PX 2 x U 15 i PYgtx 05 El lEI muse artmcnr of Statistics ruin mmmmiuiim Normal Approximation TVmore example Say n 26 and p 2 Using the Normal Curve make sure npgt5andnqgt5 Here26x 252gt5and 26 x 1 7 2 208 gt 5 Therefore conditions hold and the standard deviation is W 20396 Let39s say X g 5 PX g 5 z PY lt 55 normalCDF79999555220396 05585 Normal Approximation TVmore example continued 020 7 015 7 010 7 005 7 D partmenr of Statistics Normal Approximation TVmore example comparison with exact binomial probability Sayn26andp2andX 5 Using the Binomial Probability PX 3 Sin 26 p 2 binomCDF26 2 5 05775 Note Normal approximation gives a value close to the precise binomial method partmcnt of Sta marltn Conditions for Good Approximation Normal curve gives a reasonable approximation for the binomial probabilities whenever both np gt 5 and nq gt 5 Note gt np expected number of successes gt nq expected number of failures D partmentof Statistics mm m m mu Binomial Example travel agents example The rate of commission that commercial airlines pay travel agents has been declining for several years In an attempt by travel agents to raise revenue many agents are now charging their customers a ticket fee typically between 10 and 15 dollars According to the ASTA about 90 percent of travel agents charge customer fees when purchasing an airline ticket cpartmcntof Stati 39 s wmma mmquot Travel Agents Example con nued Suppose that a random sample of 55 travel agents is selected Assume that the number of the 55 travel agents charging a ticket is distributed as a binomial random variable gt What are the mean and sd of this distribution gt What assumptions are necessary gt What is the probability that none of the travel agents will charge a fee in other words PX Oln 55p 9 7 Department of Statistics mr Travel Agents Example con nued gt What is the probability that at least 46 of our sample of travel agents will charge a fee in other words PX 2 46in 55 p 9 7 gt What is the probability that at most 50 of our sample of travel agents will charge a fee in other words PX g 50in 55 p 9 7 Department of Statistics mmtm agm nmmn Normal Example packaging bags example Plastic bags used for packaging produce are manufactured so that the breaking strength of the bag is normally distributed with a mean of 5 pounds per square inch and a sd of 15 pounds per square inch Given p 5 a 15 gt What is the proportion of bags produced having a breaking strength of less than 317 pounds per square inch in other words PX lt 317 7 gt What is the proportion of bags produced having a breaking strength between 32 and 42 pounds per square inch in other words P32 g X g 42 7 D partmentof Statistics nmmmun Packaging Bags Example conitinued gt What is the proportion of bags produced having a breaking strength of at least 36 pounds per square inch in other words PX 2 36 7 gt Between what two values symmetrically distributed around about the mean will 95 of the breaking strength fall in other words Pa g X g b 7 partmcnt of Statistics rum wmmun v v mmmn

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