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# Business Statistics STAT 2160

WMU

GPA 3.54

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This 29 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 Staff in Fall. Since its upload, it has received 8 views. For similar materials see /class/216959/stat-2160-western-michigan-university in Statistics at Western Michigan University.

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Date Created: 09/30/15

1 Chapter 5 Contents 1 Outline 2 Introduction 21 Overview of Chapter 5 3 Chapter 5 31 Sampling Proportion 32 Law of Large Numbers for Sample Percentages 33 Sampling Distribution of 16 is Approximately Normal 34 Estimating the population proportion p 35 Estimate the Standard Error of proportion p 36 Class exercise 2 Introduction 21 Overview of Chapter 5 Goals and Objectives 0 To learn about the sampling distribution of the proportion Topics 0 Sample Proportion 0 Law of Large Numbers 0 Sampling Distribution of 16 0 Estimate the population proportion p 0 Estimate the Standard Error of proportion p 3 Chapter 5 31 Sampling Proportion Sampling Proportion Let us start with an example 0 Suppose TV World sells 60 extended warranties with 300 TV sets sold The warranty sales rate is 020 o Therefore7 let X denote the number of successes out of a sample of n observations 0 Where X is a binomial random variable with parameters 71 and p Also7 the proportion of successes in a sample is a random variable Sampling Proportion o 16 g number of successes total number of observations in the sample 0 For the binomial7 X is expected to be around np give or take np1 7 p g n o For the proportion7 16 is expected to be take 7V 175717 LIT Sampling Proportion example give or 0 TV World example The number of warranties sold is expected to be around 60 i 7 o The proportion of warranties sold is expected to be around 60 7 E i m or 2 i 002 32 Law of Large Numbers for Sample Percentages Law of Large Numbers for Sample Percentages o The sample percentage tends to get closer to the true percentage as sample size increases Application of the Law of Large Numbers 0 Recall if TV World sold 300 TV sets 16 2 and SD 002 o If TV World sold 1200 TV sets 16 2 and 2 172 now SD 0 00115 33 Sampling Distribution of is Approximately Nor mal Sampling Distribution of is Approximately Normal If TV World sold 100 TV sets last year7 the percentage of set sold with extended warranties is expected to be around 20 give or take 4 Estimate the likelihood that it sold warranties with each TV is more than 25 of those sets7 in other words7 PL39 gt 025 Sampling Distribution of is Approximately Normal Given n 100 and p 2 0 PL gt 25 2 172 normalODF25 9999 2 0 01056 mmde 34 Estimating the population proportion p Estimating the population proportion p o The population proportion p are generally unknown and are estimated from the data 0 Suppose we want to estimate the number of students planning to attend graduate school 0 Will the sample proportion equal the population proportion Yes or No o If not7 by how much will we miss it Estimating the population proportion p Consider this example 71 40 graduating seniors7 X 6 is the number of graduating seniors planning to attend graduate school 0 What is the proportion of graduating seniors planning to attend grad uate school p g 015 o By how much will we miss the true population proportion SE3 w L w73915lt14315gt 005646 iClickeI39 Question Recently7 a random sample of 40 small retail businesses found that 32 had experienced cash ow problems in their rst year of operation What proportion of small retail businesses had cash ow problems in their rst year of operation a 051 b 196 c 25 Cl 125 e 080 Estimating the population proportion p Consider this example 71 40 graduating seniors X 6 is the number of graduating seniors planning to attend graduate school o If we take another sample of size 40 graduating seniors what is the probability that 17 percent or less of the graduating seniors will attend graduate school 0 PL S 17 n0rmODFi99 17 15 05646 06384 o If we take another sample of size 40 graduating seniors what is the probability that 20 percent or more of the graduating seniors will attend graduate school 0 PL 2 2 normODF2 99 15 05646 01879 35 Estimate the Standard Error of proportion p Summarizing the Estimate of the population proportion p o 16 is an estimate of the population proportion ie EH3 p 0 We will miss it by the standard error of the proportion SE13 131 13 Summarizing the Estimate of the population proportion p o The population proportion p is estimated using the sample proportion 13 ie EH3 p This estimate tends to miss by an amount called the SE13 0 The SE13 is calculated as w 0 As sample size increases the SE13 decreases like the square root of sample size 36 class exercise class exercise Only ve percent of US families have a net worth in excess of 1 million dollars and thus can be called millionaires However 30 percent of MSs 31000 employees are millionaires Harvard Business Review July August 2000 class exercise If a random samples of 100 MS employees are selected at random what proportion of the samples will have more than 36 millionaires Given p 3 n 100 PH gt 36 o P13gt 36 normODF3699 3 v 00952 class exercise If a random samples of 100 MS employees are selected at random what proportion ofthe samples will have less than 29 millionaires Given p 3 n 100 PH lt 29 o P13lt 29 normODF79929 3 3815 04136 mummy class exercise If a random samples of 100 MS employees are selected at random7 what proportion of the samples will be between 25 and 35 millionaires Given p 37 n 1007 P25 13 g 35 P25 g f g 35 V3 173 normCDF25735737 l 07248 problem 1 page 68 0 Given p g 6 49 on problem 1 page 68 answers Given p 67 n 497 PH gt Pm gt g normODF 99 6 60436 01704 iClicker39 Question Recently7 a random sample of 40 small retail businesses found that 32 had experienced cash ow problems in their rst year of operation what is the probability that between 70 and 85 of small retail businesses had cash ow problems in their rst year of operation a 051 b 020 C 02715 d 07285 e 080 Questions Questions 1 Chapter 7 Contents 1 Outline 1 2 Introduction 1 21 Overview of Chapter 7 1 3 Chapter 7 2 31 Con dence Interval Applications 2 32 De nitions 2 33 Con dence Interval Properties 3 34 One Sample Z con dence interval 4 35 One Sample t con dence interval 6 36 Two Independent Samples t con dence interval 10 37 Two Dependent Samples t con dence interval 13 38 One Sample Proportion z con dence interval 17 2 Introduction 21 Overview of Chapter 7 Goal and Objectives 0 To develop con dence intervals estimates 0 To understand that each interval has two end points lower and upper bound Interpret the con dence interval To compute the con dence interval a point estimate i the margin of error 0 To determine the sample size 3 Chapter 7 31 Con dence Interval Applications Applications of Estimation in Business 0 Store inventory value 0 Manufacture process 0 Distribution process 0 Drug delivery 0 Auditor 32 De nitions De nitions 0 Point estimate ptest is a single sample statistic that estimates the population parameter7 such as7 the mean or proportion o Interval estimate of the true population mean takes into account the sampling distribution of the point estimate where we have an upper bound and a lower bound De nitions 0 Cl con dence interval 0 CV critical value 0 ME margin of error 0 SE standard error 0 SD standard deviation 0 ptest point estimate 0 Z normal distribution critical valueuse invnorm o twl students t distribution critical value with n l degrees of freedom use math solver or the ian 33 Con dence Interval Properties Con dence Interval ptest i ME Where the point estimate is a population mean M or population proportion Em p marginOfError criticalValue gtlt standardError In other words ME CVSE Standard Error Most of the time we will not have the SD of population mean but we can compute sample SE of the mean SE2 SD 7 W Also we will not have the SD of population proportion but we can H3071 n compute sample proportion SE SE13 Critical Value 2 for normal distribution It for students t distribution The students t distribution has n 7 1 degrees of freedom df n 7 1 ZCritical Value percentile invNormpercentile Example 95 con dence interval will give 25 in each tail of the bell shaped curve therefore the CV Z025 invNorm025 7196 Since these curves are symmetric the other CV Z975 196 tCritical Value Tl 83 Tl 84 0 math a solver a tcdfL U D 7 A where L CV U 9999 D n 1 A area under the bell shaped curve of one tail and place the cursor on L and press alpha enter 0 or use ianpercentz ledf 34 One Sample Z con dence interval One Sample Z con dence interval Let7s look at l sample Zcon dence interval 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 0 What are the costs associated with putting too little cereal in a box 0 Let7s construct a 95 con dence interval Example Given n 25 i 365 SD 15 and 95 Cl 0 Suppose SD is a process SD therefore 039 15 grams 0 Suppose we want a 95 con dence interval 0 Therefore the critical values are 2025 7196 and 2975 196 0 Sample size n 25 0 Sample average i 365 Example oMEOVgtltSE o The critical value CV for 95 Cl means that the area under the curve of one tail is 5 2 or 0025 therefore7 zpemmle invNorm025 7196 or invNorm975 196 7 S 7 15 7 0 SE 7 W i V i 3 0 ME i196 gtlt 3 i588 Example 0 Since the con dence interval is the ptest i ME 0 Cl 365 i 588 359127 37088 0 Therefore7 we are 95 con dent that the population mean is between 359 and 371 0 Since 3687 the value that is printed on the box indicates the manufac turing process is working properly7 is within the interval7 there is no reason to conclude that anything is wrong with the process Example zCI Let7s use the Tl 83 Tl 84 o STAT a TESTS a Zintemal a STATS a 039 15 a i 365 a n 25 a Clevel 95 a CALCULATE o READOUT o Zinterval a 3591237088 a i 365 a n 25 0 Since 3687 the target of the package7 is within the interval production should continue Note 0 The value of Z selected for constructing such a con dence interval is called the critical value for the distribution 0 There are different critical values for each level of con dence7 1 7 04 where 04 signi cance level SL Note 0 This leads us to an interesting challenge There is a trade off between the width of the con dence interval and the level of con dence 0 We have been dealing with NM0 where a population or process SD7 is known What happens when standard deviation 0 is not from a population or process SD ls this requirement rigid Can we compute standard deviation from the sample Let us review some history rst 35 One Sample t con dence interval One Sample t con dence interval Let7s look at 1 sample t con dence interval History of the Student7s t William Gosset7 an employee of Guinness Breweries in lreland7 had a pre occupation with making statistical inferences about the mean when SD was unknown Since the employees of the company were not allowed to publish their scienti c work under their own name He chose the pseudonym 77Stu dent77 Therefore7 his contribution is still known as Students t Distribution t based CI of the Mean Summer II Quiz Data ThisistheresultsofaSummerIIQuiz 8 11 13 19 21 23 25 25 25 28 31 35 39 47 0 Construct a 95 CI for Summer 11 Quiz Data of 14 students 0 Given i 257 SD 10777771 14 want a 95 CI t based CI of the Mean Example Construct a 95 CI for Summer 11 Quiz Data of 14 students Given i 257 SD 10777771 14 want a 95 CI 0 SE SD 7 10777 7 o W 7 W 7 28803 0 CV o z 7wT0257 14 71 721604 t based CI of the Mean Construct a 95 CI for Summer 11 Quiz Data of 14 students Given i 257 SD 10777771 14 SE 288037 CV i21604 want 95 CI 0 ME CV gtlt SE where CV t 0 ME i21604 gtlt 28803 i62225 o ptest 25 o 95 CI ptest i ME 0 95 CI a 18777731223 Example t CI Let7s use the Tl 83 Tl 84 o STAT a TESTS a tIntemal a STATS a i 25 a 51 10777 a n 14 a Clevel 95 a CALCULATE READOUT tIntemal a 1877731223 a i 25 a n 14 Interpret the con dence interval We are 95 con dent that the true mean quiz score is between 188 and 312 Determine Sample Size 0 What sample size should we use for the average Quiz Score determina tion if we want 95 con dence7 ME 5 and s10777 7 02 7 1962gtlt107772 7 N oniME2i 52 7178m18 page 100 1 class work given Slow Wave Sleep Data 21 20 22 7 9 14 23 9 10 25 15 17 11 a i and S a i 156154 and S 61310 b What is the population average and SD C7 VVVVVV Not possible c The sample average will miss the population average by the c The sample average will miss the population average by the SE page 100 1 class work Given n 137 SD 613 S SEEW 613 M13 MECVgtltSE SEE 170 95 CV t02512 inoT02512 i21788 ME CV gtlt SE i21788 gtlt170 i3704 What is a 95 Cl 95 01119171932 use ptesti ME or STAT a EDIT enter data from page 100 STAT a TESTS a tInterial a DATA a List L1 a Freg 1 a C 7 Level 95 a Calculate page 100 1 g g FF If the con dence level to reduced to 907 The new interval will be shorter Calculate a 90 con dence interval for the population average 90 Cl a 12585718646 lnterpret the 90 Cl We are 90 percent con dent that the true population average is be tween 13 and 19 page 100 1 u Wu VV k Does the 90 Cl suggest that elderly men over 60 spend 20 of their sleep in REM No since 20 is not in the 90 C a 1319 What sample size should we use if we change the ME to 25 with 95 con dence CV2gtltSD2 7 1962X61312 n ME2 252 i 24 iClickeI39 Question A random sample of twenty ve factory workers were asked how many vacation days they take per year The average of the sample was 2285 days7 and the standard deviation of the sample was 580 day Construct a 95 con dence interval for the average number of vacation days per year taken by factory workers a 209 248 b 205 252 c 210 250 d 224 233 e 225 232 iClickeI39 Question A random sample of twenty ve factory workers were asked how many vacation days they take per year The average of the sample was 2285 days7 and the standard deviation of the sample was 580 day What sample size is needed to obtain a margin of error of 2 days per year with a 99 con dence interval a55 0 C7 00 00 CL 3 C40 56 e32 36 Two Independent Samples t con dence interval Two Independent Samples t con dence interval Let7s look at 2 independent samples t con dence interval Comparing Mean of two Independent Samples 0 We are not limited to comparing an average to a constant Suppose we want to compare the means of two independent samples This often happens in the workplace 0 Remember Cl is ptest i ME ptest d f1 7 f2 ME CV gtlt SE where CV tn1n2g SE WSE12 SE Example diff of 2 indep groups A statistics student designed a study to see if there was any real difference in battery life between brand name AA batteries and generic AA batteries He used six pairs of AA alkaline batteries from two major battery manu factures 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 minutes when the sound stopped AA batteries Example diff of 2 indep groups Given want a 95 Cl a What is the ptest 7 a ptest f1 7 f2 206 71874 186 b What is the standard error b pooled SD 126 SE 727 c What is the 95 CV c CV tmn z mT025 10 22281 11 Example diff of 2 indep groups Given want a 95 Cl d What is the ME d ME CV gtlt SE 22281gtlt 727 161983 e What is the 95 Cl e 95 Cl a ptesti ME a 23534185 f Does this con dence interval suggest that generic AA batteries will last longer than brand name AA batteries f Yes because zero is not within the interval Example diff of 2 indep groups Given want a 95 Cl g lnterpret the 95 Cl g We are 95 con dent that the true average di erence is between 2 and 35 Example 2sample tCI Let7s use the Tl 83 Tl 84 o STAT a TESTS a 2 7 SampTint a STATS ail2060aS1103 n16af21874aS2146an2 6 a Cleoel 95 a Pooled Yes a CALCULATE o READOUT o 2 7 sampTint a 23573485 a Smp 126343 0 Since zero is not within this interval7 we can conclude that there is a di erence between the two mean iClicker Question The director of training for a company manufacturing widgets is inter ested in determining whether different training methods have an effect on the productivity of assembly line employees Assume that the variances in the populations of training methods are equal Construct a 95 con dence interval for the difference of two indepen dent averages a 4082 0518 b 3833 0767 c 4159 0441 d 3845 0755 e 41410459 37 Two Dependent Samples t con dence interval Two Dependent Samples t con dence interval Let7s look at 2 dependent sample t con dence interval Comparing Means of 2 dependent groups 0 We are not limited to comparing two averages of independent samples Suppose we want to compare the means of two related samples 0 Remember Cl is ptest i ME7 where ptest f1 7 f2 0 ME CV gtlt SE where CV tn1 239an71 71 and SE Example Comparing Means of 2 related groups o If we compare January 2002 prices vs January 2003 prices of computer companies7 see page 92 Example Comparing Means of 2 related groups Given ptest 79467 SD 61426 D What is Standard Error D 5diff 7 61426 7 7 7 7 7 27471 w C7 What is 95 Critical Value C7 VVVVVV tam ian0254 i27764 2 0 What is 95 Margin of Error CV gtlt SE 27764 gtlt 27471 i76271 0 Example Comparing Means of 2 related groups d What is a 95 Con dence Interval CL 9501 a ptest i ME 5 0319 15573 D Does this con dence interval suggest a difference between Jan 2002 and Jan 2003 D Yes be zero is NOT within 0 i h V lnterpret the 95 Cl f We are 95 con dent that the true di erenee is between 3 and Z6 Example 2related means t CI Let7s use the Tl 83 Tl 84 the data is on page 92 o STAT a EDIT Enter data into L1 and L2 and place cursor on L3 do 2211 i 2212 a 25 L1 7 L2 STAT a TESTS a tIntemal a DATA a List a L3 Freq 1 a Clevel 95 a CALCULATE READOUT tIntemal a 0318915573 a i 7946 a 5m 61426 a n 5 0 Since zero is not within this interval we can conclude that there is a dz erehee between the two mean page 104 13 Comparing Means of 2 related groups Some stock market analysts have speculated that parts of West Michigan Telecorn might be worth more than the whole For example the company7s communication systems in Ann Arbor and Detroit can be sold to other com munications companies Suppose that a stock market analyst chose nine 9 acquisition experts and asked each to predict the return in percent on in vestment ROI in the company held to the year 2003 if it does business as usual or ii if it breaks up its communication system and sells all its parts Their predictions follow page 104 13 Comparing Means of 2 related groups ROI in percent Expert 1 2 3 4 5 6 7 8 9 Not Break 12 21 8 20 16 5 18 21 10 Break Up 15 25 12 17 17 10 21 28 15 Difference 3 4 4 3 1 5 3 7 5 Given 2 322 and s 28626 What is a 95 Cl 7amp728626 oSEi W 7 3 709542 CV tn1 239an025 8 23060 ME 2306 gtlt 9542 22004 9501 a ptest 1 ME 4 10218 54226 15 page 104 13 Comparing Means of 2 related groups Does this con dence interval suggest a difference between breaking up the company or not Yes bc zero is NOT within 0 Interpret the 95 Cl We are 95 con dent that the true di erence among the eccperts is between 10 and 54 Example 2I39elated means t CI Let7s use the Tl 83 Tl 84 STAT a EDIT Enter data into L1 and L2 and place cursor on L3 do 2W2 i 2W1 a 215 L2 7 L1 STAT a TESTS a tInterial a DATA a List L3 a Freq 1 a Cleoel 95 a CALCULATE READOUT tInterial a 102542 a i 322 a 5128626 a n 9 Since zero is not within this interval we can conclude that there is a di erence between the two mean iClicker39 Question In order to measure the effect of a storewide sales campaign on non sale items the Director of Sales of a regional supermarket chain took a random sample 12 pairs of stores that were matched according to average weekly sales volume One store of each pair the experimental group was exposed to the sales campaign and the other member of the pair the control group was not The average difference between the two groups and the standard deviation are 456 and 4297 respectively What is the 95 margin of error a 44482 b 27257 c 54514 d 22241 e 22010 38 One Sample Proportion z con dence interval One Sample Proportion z con dence interval Let7s look at 1 sample proportion z con dence interval Estimating the Population Proportion using Intervals 0 Suppose we want to estimate the proportion of a population using intervals Remember Cl is ptest i ME success 7 g sampleSize n Therefore7 ptest 0 ME CV gtlt SE This test works well if n gtlt p gt 5 and n gtlt 17p gt 5 Example CI for proportion If a sales clerk fails to remove the EAS sensor when an item is purchased7 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 surveyed7 40 said that if they were to set off an EAS alarm because store personnel did not deactivate the merchandise7 then 77they would never shop at the store again77 Example CI for proportion Given X 40 n 250 What is a 95 Cl a ptest am 7016 b SE1 514 7 161716 7 b 1 71T 7 002319 C Olg5 Z 2035 C 2 tnoNorm025 7196 Example CI for proportion Given X 40 n 250 p 016 ME CV gtlt SE 196 X 002319 004544 950 a e ptesti ME 0114567 020544 Interpret the 95 Cl We are 95 con dent that the true proportion is between 11 and 2 Example 1Prosznt Let7s use the Tl 83 Tl 84 o STAT a TESTS a 1 PropZint a z 40 a n 250 a Cleoel 95 a CALCULATE o READOUT o Ztnteroal a 0114602054 a p 16 a n 250 0 We are 95 con dent that the true proportion is between 11 and 21 Determine Sample Size 0 What is the true proportion of success p 0 Decide which con dence level to use 0 Determine margin of error that you are willing to accept Example Sample Size Suppose that you are a student with a grant to study this EAS issue7 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 con dence level to 90 and use 16 016 and the ME of 0045447 what is the new sample size Example Sample Size Given n 25013 016 ME 0045447 and CL 90 o n 230 723 o Ol90 z invNorm05 1645 ME 04544 p 16 n V34654f2161716 1761177 page 105 17 Given X 600 n2000 a ptest a 600 i 03 2000 7 b SE1 3 173 b 0 00102 c Olg5 z c 2 invNorm025 196 page 105 17 Given f 037 20025 i1967 SE 00102 ME CV gtlt SE 196 X 00102 00201 950 a ptesti ME a 2799 3201 Interpret the 95 Cl We are 95 con dent that the true proportion is between 28 and 32 iClickeI39 Question The owner of a restaurant serving continental food wants to study charac teristics of customers of her restaurant ln particular7 she decides to focus on two variables the amount of money spent by customers and whether or not customers order dessert The results from a random sample of 40 customers are as follows 0 Amount spent i 48537 S 815 0 15 customers purchased dessert What is a 90 con dence interval estimate of the population proportion of customers who purchase dessert a 0249 0501 b 0265 0485 c 0178 0572 d 0197 0553 e 0225 0525 20 iClickeI39 Question The owner of a restaurant serving continental food wants to study charac teristics of customers of her restaurant ln particular7 she decides to focus on two variables the amount of money spent by customers and whether or not customers order dessert The results from a random sample of 40 customers are as follows 0 Amount spent i 48537 S 815 0 15 customers purchased dessert If she wants to have 95 con dence of estimating the true proportion of customers who purchase dessert to within i00757 what is sample size is needed a 160 b 68 c 113 d 161 e 112 Question Questions 21

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