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# Statistics lecture 8 (chapter 7)

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Confidence Intervals

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### Statistics lecture 8 (chapter 7)

1. 1. 1
2. 2. • A PARAMETER is a number that describes a population statistic• A STATISTIC is a number that describes a characteristic in the sample data 2
3. 3. • Inferential statistics – Draw conclusion from data – Sample • Describe data – Use sample statistic to infer population parameter • Estimation • Hypothesis testing 3
4. 4. Data collection Raw data Graphs Information Descriptive statistics Measures • location • spread Estimation Statistical Decision inference making Hypothesis testing 4
5. 5. • Estimation – Numerical values assigned to a population parameter using a sample statistic• Sample mean x used to estimate population mean μ• Sample variance s2 used to estimate population variance σ2• Sample stand dev s used to estimate population stand dev σ ˆ• Sample proportion p used to estimate population proportion p 5
6. 6. • Steps in estimation – Select sample – Get required information from the sample – Calculate sample statistic – Assign values to population parameter 6
7. 7. • Read example 7.1 page 214 7
8. 8. • Sample statistic used to estimate a population parameter is called an ESTIMATOR• An estimator is a rule that tells us how to calculate the estimate and it is generally expressed as a formula 8
9. 9. POPULATION ESTIMATE ESTIMATOR PAPARMETER (VALUE OF (Formula) STATISTIC)MEAN µVARIANCE σ2 s2PROPORTION p 9
10. 10. • Two types of estimate:- –Point estimates –Interval estimates 10
11. 11. • A single number that is calculated from sample data• Resulting number then used to estimate the true value of the corresponding population parameter 11
12. 12. • A random sample of 10 employees reveals the following dental expenses in rands for the preceding year:660; 2172; 1476; 510; 3060; 1248; 1038;2550; 1896 and 1074Determine a point estimate for:-1. The population mean2. The population variance 12
13. 13. • Answer p215 13
14. 14. • If we take another random sample of 10 employees the mean obtained for this random sample will almost certainly differ from the one you have just calculated• Point estimates do not provide information about how close the point estimate is to the population parameter• Point estimates do not consider the sample size or variability of the population from which the sample was taken 14
15. 15. • Sample size and variability of population will affect the accuracy of the estimate so a point estimate is really not very useful• This problem can be overcome by using INTERVAL ESTIMATES 15
16. 16. • No 1 – 6 page 216 16
17. 17. Point Estimates – A single sample statistic used to estimate the population parameter Population distribution Population parameter Sample distribution Point estimator 17
18. 18. Confidence interval – An interval is calculated around the sample statisticPopulation parameterincluded in interval Confidence interval 18
19. 19. Confidence interval – An upper and lower limit within in which the Example: population parameter is expected to lie Meaning of a 90% confidence interval: – Limits will vary from sample to sample – Specify90% of all possible samples taken from the probability that the interval will include population will produce an interval that will the parameter include the population parameter – Typical used 90%, 95%, 99% – Probability denoted by • (1 – α) known as the level of confidence • α is the significance level 19
20. 20. • An interval estimate consists of a range of values with an upper & lower limit• The population parameter is expected to lie within this interval with a certain level of confidence• Limits of an interval vary from sample to sample therefore we must also specify the probability that an interval will contain the parameter• Ideally probability should be as high as possible 20
21. 21. SO REMEMBER•We can choose the probability•Probability is denoted by (1-α)•Typical values are 0.9 (90%); 0.95 (95%) and 0.99 (99%)•The probability is known as the LEVEL OF CONFIDENCE•α is known as the SIGNIFICANCE LEVEL•α corresponds to an area under a curve•Since we take the confidence level into account when weestimate an interval, the interval is called CONFIDENCEINTERVAL 21
22. 22. Confidence interval for Population Mean, n ≥ 30- population need not be normally distributed- sample will be approximately normal    CI (  )1   x  Z1   , if  is known  2 n  s  CI (  )1   x  Z1   , if  is not known  2 n 22
23. 23.    Example :CI (  )1   x  Z1   , if  is known  2 n 90% confidence interval  s CI (  )1   x  Z1   , if  is not known  2 n 1 –   0,90   0,10 1 90% of all sample  0,10 means fall in this area   0, 05 2 2These 2 areas added Confidence leveltogether = α i.e. 10% 1–α =1-α  1-α   0, 05  0, 05  2 = 0,90 2 2 x Lower conf limit Upper conf limit 23
24. 24. 24
25. 25. See handout 25
26. 26. A random sample of repair costs for 150hotel rooms gave a mean repair cost ofR84.30 and a standard deviation of R37.20.Construct a 95% confidence interval for themean repair cost for a population of 2000hotel rooms 26
27. 27. Example 7.3 p218 27
28. 28. • Four commonly used confidence levels  1-α α 2 z 0,9 0,1 0,05 1,64 0,95 0,05 0,025 1,96 0,98 0,02 0,01 2,33 0,99 0,01 0,005 2,57 28
29. 29. • Confidence interval for Population Mean, n ≥ 30 • Example: – Estimate the population mean with 90%, 95% and 99% confidence, if it is known that – s = 9 and n = 100 – Solution: The confidence intervals are s 990% xz   x  1, 64  x  1, 48 1 n 100 2 s 995% xz   x  1, 96  x  1, 76 1 n 100 2 s 999% xz   x  2, 57  x  2, 31 29 1 n 100 2
30. 30. Confidence level influence width of interval90% x  1, 48  Width of interval = 2 x 1,48 = 2,9695% x  1, 76  Width of interval = 2 x 1,76 = 3,5299% x  2, 31  Width of interval = 2 x 2,31 = 4,62 Margin of error becomes smaller if: • z-value smaller 90% • σ smaller 95% • n larger 99% 30
31. 31. • Example – A survey was conducted amongst 85 childrenmean hours 95% confident the to determine the number of hours they spend in front of the TV every week. children spend watching TV is – The results indicate that thebetween 23,866 and 25,134 24,5 mean for the sample was hours with a standard deviation of 2,98 hours. hours per week – Estimate with 95% confidence the population mean hours that children spend watching TV.  s   2,98   x  z1     24,5  1,96   2 n  85    24,5  0, 634   23,866 ; 25,134 31
32. 32. • Confidence interval for Population Mean, n < 30 – For a small sample from a normal population and σ is known, the normal distribution can be used. – If σ is unknown we use s to estimate σ – We need to replace the normal distribution with the t-distribution ▬ standard normal  s CI (  )1   x  tn 1;1   ▬ t-distribution  2 n 32
33. 33. t Distribution• Refer to handout on how to read the critical value t n-1; 1- 𝛼 2 33
34. 34. • Example – The manager of a small departmental store is concerned about the decline of his weekly sales. 99% confident the mean weekly – He calculated the average and standard deviation of his sales for the past 12 weeks, x =sales will be between R12400 and s = R1346 – Estimate with 99% confidence the populationR13 606,86 R11 193,14 and mean sales of the departmental store. t11;0.995  s   1346   x  tn 1;1    12400  3,106   2 n  12   12400  1206,86  11193,14 ; 13606,86  34
35. 35. EXAMPLE 2• A study of absenteeism among workers at a local mine during the previous year was carried out. A random sample of 25 miners revealed a mean absenteeism of 9.7days with a variance of 16 days. Construct a confidence interval for the average number of days of absence for miners for last year. Assume the population is normally distributed. 35
36. 36. EXAMPLE 2 - ANSWER• Example 7.6, page 222 textbook 36
37. 37. CLASSWORK• Do concept questions 7 – 19, page 223 textbook 37
38. 38. • Confidence interval for Population proportion – Each element in the population can be classified as a success or failure number of successes x ˆ Sample proportion p = = sample size n – Proportion always between 0 and 1 – For large samples the sample proportion ˆ p is approximately normal  p (1  p )  ˆ ˆ CI ( p )1   p  z1  ˆ   2 n  38
39. 39. • Example – A sales manager needs to determine the proportion of defective radio returns that is made on a monthly basis. 95% confident the mean monthly – In December 65 new radios werewill be will in January returns sold and be between 13 were returned for rework. 10,3% and 29,7% – Estimate with 95% confidence the population proportion of returns for December. 13 p ˆ  0, 2 65  p (1  p )   ˆ ˆ 0, 2(1  0, 2)   p  z1  ˆ    0, 2  1,96   2 n   65    0, 2  0, 097    0,103 ; 0, 297  39
40. 40. EXAMPLE 2• A cellphone retailer is experiencing problems with a high % of returns. The quality control manager wants to estimate the % of all sales that result in returns. A sample of 40 sales showed that 8 cellphones were returned. Construct a 99% confidence interval for the % of all sales that result in returns 40
41. 41. EXAMPLE 2• Answer – example 7.9 page 225, textbook 41
42. 42. • Confidence interval for Population Variance – Population variance very often important – Very often required for quality control – Sample drawn from a normal population – Sample variance is based on a random sample of size n – Distribution of s2 resulted from repeated sampling is a χ2 (chi-square) distribution 42
43. 43. • Confidence interval for Population Variance – χ2 (chi-square) distribution • Skewed to the right distribution • Shape varies in relation to the degrees of freedom • Critical values from the χ2-table A4(read same way as t distribution) • Critical value of χ21 - α specifies an area to the left • Critical value of χ2α specifies an area to the right 43
44. 44. • Confidence interval for Population Variance  2 (n  1) s (n  1) s  2 CI ( )1 2  2 ; 2   n 1;1   n 1;   2 2  44
45. 45. • Example – For a binding machine to work on its optimum capacity 90% confident the variation in the variation in the temperature of the room is vital. temperature will be will be between – The temperature for 30 consecutive hours were measured and sample standard and 0,757 were found to 0,315 deviation degrees be 0,68 degrees. – What will be a 90% confidence interval for σ2?   (n  1) s 2 (n  1) s 2   29(0, 682 ) 29(0, 682 )  CI ( 2 )1   2 ; 2  ;    n 1;1   n 1;    29;0,95  2  29;0,05  2   2 2   29(0, 682 ) 29(0, 682 )  n= 30; s = 0.68; α = 0.1  ;   42,56 17, 71    0,315;0, 757  45
46. 46. The total revenue for a sample of 10hardware stores in a well-known chain wasrecorded for a particular week. The results(in R1000) were as follows: 129.78;130.11;129.83;130.02;129.67;129.87;129.88;129.86130.18 and 129.91. Construct a 90%confidence interval for the standarddeviation of the total weekly revenue for allhardware stores in this chain 46
47. 47. Answer example 2 47
48. 48. Answer example 2 contd   n 1s 2 n 1s 2 CI( )0,9 =  2 2 ; 2   n1;1   n1;   2 2  90,0234 90,0234 =  ;   16,92 3,32   = [0,0124;0,0634]CI()0,9 = [ 0,0124 ; 0,0634 ]  = [0,1114;0,2518] 48
49. 49. CONCEPT QUESTIONS• Nos 20 – 28, page 228 textbook 49
50. 50. Where are we?• So far we have looked at interval estimation procedures for µ, p and σ2 for a SINGLE POPULATION• We are now going to look at interval estimation procedures for:- – The difference between two population means – The difference between two population proportions – The ratio of two population variances 50
51. 51. • Interval estimation for two populations – There is different procedures for the differences in means, proportions and variances. Population Sample Population Sample 1 1 2 2Mean μ1 x1 μ2 x2Variance σ 21 s21 σ 22 s22Std dev σ1 s1 σ2 s2Size N1 n1 N2 n2Proportion P1 ˆ p1 P2 ˆ p2 51
52. 52. • Confidence interval difference in means – Large independent samples   12  2 2  CI ( 1   2 )1   x1  x2   Z1      2 n1 n2   if  12 and  2 is known 2  s12 s2  2 CI ( 1   2 )1   x1  x2   Z1      2 n1 n2  if  12 and  2 not is known 2 NOTE: If 0 is not included in the interval it means that 0 does not occur between the lower and upper 52 boundaries of the interval
53. 53. Example 1Independent random samples of male and femaleemployees selected from a large industrial plantyielded the following hourly wage results:- MALE FEMALE n1 = 45 n2 = 32 𝑥 = 6.00 𝑥 = 5.75 s1 = 0.95 s1 = 0.75Construct a 99% confidence interval for thedifference between the hourly wages for all malesand females and interpret the results 53
54. 54. Example 1- Answer  0,011   0,99 1  1 2 2   0,01  0,995Z 0,995= 2,57  2 2 CI(1 – 2 =  )0,99  x1  x 2   Z  s1  s2    1 2 n1 n 2    2   6  5,75  2,57 0,95  0,75  2  =   45 32     = [–0,2486;0,7486] 54
55. 55. Example 1- answerInterpretation:-At a 99% level of confidence, the differencebetween the hourly wages of males and females isbetween -0.2486 and 0.7486 rand. The value 0 isincluded in the interval which tells us that there is apossibility that there is no difference between thetwo population means. To make sure whetherthere is a difference or not, a hypothesis test (nextchapter!!!!) has to be performed. 55
56. 56. • Confidence interval difference in means – Small independent samples – When sample sizes are small, n1 & n2< 30 we use the t distributionNOTE: If both the limits of the confidence interval arenegative you should suspect that the mean of firstpopulation is smaller than mean of second population56
57. 57. ExampleA plant that operates two shifts per week would like toconsider the difference in productivity for the two shifts. Thenumber of units that each shift produces on each of the 5working days is recorded in the following table:- Monday Tuesday Wednesday Thursday FridayShift 1 263 288 290 275 255Shift 2 265 278 277 268 244Assuming that the number of units produced by each shiftis normally distributed and that the population standarddeviations for the two shifts are equal construct a 99%confidence interval for the difference in mean productivityfor the two shifts and comment on the result. 57
58. 58. x Example 1 - answer  x2 1 x1  x2  n1 n2 1 371 1 332   5 5  274, 2  266, 4       1 2 1 2 2 2 x1  x1 x2  x2 2 n1 2 n2 s1  s2  n1 1  n 2 1  233,7  188,3 sp = n1 1s12  n 2 1s2 2 n1  n 2  2  = 51(233,7)  51(188,3) 5 5 2  = 14,5258  0,01 1   0,99 1  1  2 2   0,01  0,995 t 8; 0,995 = 3,355  1 1  CI(1 – 2 =  x1  x 2   t )0,99   sp     n1 n 2 2;1 2 n1 n 2   1 1 = [(274,2 – 266,4)  3,355(14,5258)  ] 5 5  = [–23,0221;38,6221] At the 99% confidence level, because zero is included in the interval, it is possible that there 58  is no significant difference between the two shifts with respect to productivity.
59. 59. CONCEPT QUESTIONS• Nos 29 -39, p 235 – 237, textbook 59
60. 60. • Confidence interval difference in proportions – Large independent samples  p1 1  p1  p2 1  p2   ˆ ˆ ˆ ˆCI ( p1  p2 )1   p1  p2   z1  ˆ ˆ     2 n1 n2   x1 x2with p1  ˆ and p2  ˆ n1 n2 60
61. 61. Example 1Two groups of males are polled concerningtheir interest in a new electric razor that hasfour cutting edges. A sample of 64 malesunder the age of 40 indicated that only 12were interested while in a sample of 36males over the age of 40, only 8 indicatedan interest. Construct a 95% confidenceinterval for the difference between age frouppopulations 61
62. 62. Example 1 - answer 12 ˆUnder 40: n1 = 64 and p1 = = 0,1875. 64 8 ˆOver 40: n2 = 36 and p2 = = 0,2222. 361   0,95  1   1 0,05  2 2   0,05    0,975Z 0,975 = 1,96  p11 p1  p2 1 p2   ˆ ˆ ˆ ˆCI( p1  p2 ) =  0,9  p1  p2   Z  ˆ ˆ +    1 2 n1 n2    0,18750,8125 0,22220,7778  =  0,1875 0,2222 1,96      64 36    = [–0,2008;0,1314] 62
63. 63. • Confidence interval for the ratio of two population variances• We use the f distribution, table A5. See handout  2  2      2CI   1  s1   2 1  ; s1 F   s2  F     1 2   s2 2 n2 1; n1 1; 2   n1 1; n2 1;   2 2NOTE: If 1 does not lie in the confidence interval, thereis some evidence that the population variances are notequal 63
64. 64. EXAMPLE 1A criminologist is interested in comparing theconsistency of the lengths of sentences given topeople convicted of robbery by two judges. Arandom sample of 17 people convicted of robberyby judge 1 showed a standard deviation of 2.53years, while a random sample of 21 peopleconvicted by judge 2 showed a standard deviationof 1.34 years. Construct a 95% confidence intervalfor the ratio of the two populations variances. Doesthe data suggest that the variances of the lengthsof sentences by the two judges differ? Motivateyour answer. 64
65. 65. Example 1 - answer Judge 1: n = 17 and s = 2,53. 1 1 Judge 2: n2 = 21 and s2 = 1,34.  0,05 1   0,95  2 2   0,05  0,025 F  = F16; 20; 0,025 n1 1;n 2 1; 2  = 2,55 F   = F20; 16; 0,025 n 2 1;n1 1; 2 = 2,68      s 2   s 2     12 1 1    1 CI( 2 )0,95 =  2   2 F ;  2 F n 2 1; n1 1;  s2  n 1;n 1;  s2  2    1 2 2    2,53 2    1  2,53 2,68  2 2  =  2  ;  1,34 2,55  1,34     = [1,3979;9,5536] Yes, at the 95% level of confidence it is possible that the variances differ because 1 is not 65  included in the interval.
66. 66. CONCEPT QUESTIONS• Concept questions 40 – 47, p 241, textbook 66
67. 67. DETERMINING SAMPLE SIZES FOR ESTIMATES• Everything we have done so far has assumed that a sample has ALREADY been taken• We often need to know how large a sample should we take to construct the confidence interval• Many factors can affect sample size such as budget, time and ease of selection• We will now look at how to determine the proper sample size (from a statistical perspective) 67
68. 68. • Sample size for estimating means – Confidence level (1 – α) – Accepted sampling error - e – Need to know σ, else use s  z1   2 n   e NOTE: Sample size, n, is required to be a wholenumber. Therefore always round UP to the nextlargest integer 68
69. 69. EXAMPLE 1A pharmaceutical company is considering arequest to pay for the continuing educationof its research scientists. It would like toestimate the average amount spent by thesescientists for professional memberships.Base on a pilot study the standard deviationis estimated to be R35. If a 95% confidenceof being correct to within +/- R20 is desired,what sample size is necessary? 69
70. 70. Example 1 - answer  = 35 e = 20  0,05 1   0,95 1  1 2 2   0,05  0,975 Z 0,975 = 1,96 2 Z     1 2   n =   e    2 1,9635   =    20  = 11,7649  12 At least 12 scientists should be selected. 70
71. 71. • Sample size for estimating proportions – Confidence level (1 – α) – Accepted sampling error - e – Need to know p, else use p ˆ 2  z1  n  p (1  p )  e  71
72. 72. Example 1An audit test to establish the % ofoccurrence of failures to follow a specificinternal control procedure is to beundertaken. The auditor decides that themaximum tolerable error rate that ispermissible is 5%. What sample size isnecessary to achieve a sample precision of+/- 0.02 with 99% confidence? 72
73. 73. Example 1 - answer p = 0,05 e = 0,02  0,01 1   0,99 1  1 2 2   0,01  0,995 Z 0,995 = 2,57 2 Z    1 2  p 1 p n =      e    2 2,57  =   0,050,95 0,02   = 784,3319  785 A sample size of at least 785 is required. 73 
74. 74. Classwork• Questions 48 – 52, pages 244 – 245 , textbook• Self review test, p245, text book• Izimvo Exchange 1 and 2• Activity 1,2,3• Revision Exercise 1,2,3 and 4 74
75. 75. HOMEWORK• Supplementary questions, p249 – 253, textbook 75