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Revision workshop 17 january 2013

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Revision workshop 17 january 2013

  1. 1. REVISION WORKSHOP NUBE 17 TH JANUARY 2013
  2. 2. Organising and graphing quantitative data in a frequency distribution table. • Frequency table consists of a number of classes and each observation is counted and recorded as the frequency of the class. • If n observations need to be classified into a frequency table, determine: – Number of classes: c  1  3,3log n xmax  xmin – Class width  c 2
  3. 3. Organising and graphing quantitative data in a frequency distribution table. Example: The following data represents the number of telephone calls received for two days at a municipal call centre. The data was measured per hour. 8 11 12 20 18 10 14 18 16 9 5 7 11 12 15 14 16 9 17 11 6 18 9 15 13 12 11 6 10 8 11 13 22 11 11 14 11 10 9 19 14 17 9 3 3 16 8 2 3
  4. 4. Frequency distribution Number of classes  1  3,3log n  1  3,3log 48  6,5  7 xmax  xmin 22  2 Class width    2,86  3 k 7 8 11 12 20 18 10 14 18 16 9 5 7 11 12 15 14 16 9 17 11 6 18 9 15 13 12 11 6 10 8 11 13 22 11 11 14 11 10 9 19 14 17 9 3 3 16 8 2 4
  5. 5. Frequency distribution – first class [ xmin; ; min) class width) 2 5)32x – second class [ 5 ;; 8  3 ) width) 5 5 5 ) class “[“ value is included in class 8 11 12 20 18 10 14 18 16 9 5 7 11 12 15 14 16 9 17 11 6 18 9 15 13 12 11 6 10 8 “)“ value is excluded from class 11 13 22 11 11 14 11 10 9 19 14 17 9 3 3 16 8 2 5
  6. 6. Frequency distribution Classes Count [2;5) │││ 3 8 11 12 20 …. [5;8) |││││ | 4 5 7 11 12 …. [8;11) |│││││││││││ 11 6 18 9 15 …. [11;14) |│││││││││││││ | 13 11 13 22 11 …. [14;17) │││││││││ 9 19 14 17 9 …. [17;20) |││││││ 6 [20;23) ││ 2 6
  7. 7. Frequency distribution Classes Frequency (f) [2;5) 3 [5;8) 4 [8;11) 11 [11;14) 13 [14;17) 9 [17;20) 6 [20;23) 2 Total 48 7
  8. 8. Frequency distribution Classes f % frequency [2;5) 3 3/48×100 = 6,3 [5;8) 4 4/48×100 = 8,3 [8;11) 11 11/48×100 = 22,9 [11;14) 13 27,1 [14;17) 9 18,8 [17;20) 6 12,5 [20;23) 2 4,2 Total 48 100 8
  9. 9. Frequency distribution Classes f %f Cumulative frequency (F) [2;5) 3 6,3 3 [5;8) 4 8,3 3+4=7 [8;11) 11 22,9 7 + 11 = 18 [11;14) 13 27,1 18 + 13 = 31 [14;17) 9 18,8 31 + 9 = 40 [17;20) 6 12,5 40 + 6 = 46 [20;23) 2 4,2 46 + 2 = 48 Total 48 100 9
  10. 10. Frequency distribution Classes f %f F %F [2;5) 3 6,3 3 3/48×100 = 6,3 [5;8) 4 8,3 7 7/48×100 = 14,6 [8;11) 11 22,9 18 18/48×100 = 37,5 [11;14) 13 27,1 31 64,6 [14;17) 9 18,8 40 83,3 [17;20) 6 12,5 46 95,8 [20;23) 2 4,2 48 100 Total 48 100 10
  11. 11. Frequency distribution Classes f F Class mid-points (x) [2;5) 3 3 (2 + 5)/2 = 3,5 [5;8) 4 7 (5 + 8)/2 = 6,5 [8;11) 11 18 (8 + 11)/2 = 9,5 [11;14) 13 31 (11 + 14)/2 = 12,5 [14;17) 9 40 15,5 [17;20) 6 46 18,5 [20;23) 2 48 21,5 Total 48 11
  12. 12. Frequency distribution Classes f %f F %F (x) [2;5) 3 6,3 3 6,3 3,5 [5;8) 4 8,3 7 14,6 6,5 [8;11) 11 22,9 18 37,5 9,5 [11;14) 13 27,1 31 64,6 12,5 [14;17) 9 18,8 40 83,3 15,5 [17;20) 6 12,5 46 95,8 18,5 [20;23) 2 4,2 48 100 21,5 Total 48 100 12
  13. 13. Histograms Classes f %f [2;5) 3 6,3 [5;8) 4 8,3 [8;11) 11 22,9 y-axis [11;14) 13 27,1 [14;17) 9 18,8 [17;20) 6 12,5 [20;23) 2 4,2 x-axis 13
  14. 14. Histograms Number of telephone calls per hour at a municipal call centre 14 Number of hours 12 10 8 6 4 2 0 2 5 8 11 14 17 20 23 Number of calls 14
  15. 15. Definitions Frequency Polygon A line graph of a frequency distribution and offers a useful alternative to a histogram. Frequency polygon is useful in conveying the shape of the distribution Ogive A graphic representation of the cumulative frequency distribution. Used for approximating the number of values less than or equal to a specified value 15
  16. 16. Frequency polygons Class mid-points (x) f %f 3,5 3 6,3 6,5 4 8,3 9,5 11 22,9 y-axis 12,5 13 27,1 15,5 9 18,8 18,5 6 12,5 21,5 2 4,2 x-axis 16
  17. 17. Frequency polygons Number of telephone calls per hour at a municipal call centre (x) 14 3,5 Number of hours 12 6,5 10 8 9,5 6 12,5 4 2 15,5 0 18,5 0.5 3.5 6.5 9.5 12.5 15.5 18.5 21.5 24.5 21,5 Arbitrary mid-points to Number of calls close the polygon. 17
  18. 18. Ogives Classes F %F [2;5) 3 6,3 [5;8) 7 14,6 [8;11) 18 37,5 y-axis [11;14) 31 64,6 [14;17) 40 83,3 [17;20) 46 95,8 [20;23) 48 100 x-axis 18
  19. 19. Ogives Ogive of number of call received at a call centre per hour 100 number of hours 90 % Cumulative 80 70 60 50 40 30 20 10 0 2 5 8 11 14 17 20 23 Number of calls None of the hours had less than 2 calls. 19
  20. 20. Ogives Ogive of number of call received 20% of the hours had at a call centre per hour more than 17 calls 100 number of hours per hour. 90 % Cumulative 80 70 80% of the 60 hours had 50 less than 40 30 17 calls 20 per hour. 10 0 2 5 8 11 14 17 20 23 50% of Number ofhad less the hours calls than 12 calls per hour. 20
  21. 21. Exam question 2 A garbage removal company would like to start charging by the weight of a customers bin rather than by the number of bins put out. They select a sample of 25 customers and weigh their garbage bins. The weights in kg are given below:- 14.5 5.2 16.0 14.7 15.6 18.9 13.5 24.6 24.5 7.4 13.2 23.4 13.9 12.0 22.5 31.4 16.1 10.9 25.1 22.1 14.8 15.1 4.9 17.0 10.3 1. Construct a frequency table to describe the data. Include a frequency and relative (%) frequency column. (Hint: start the class intervals with the whole number just smaller than the lowest value in the dataset)
  22. 22. Procedure 1. Calculate the range of the dataset 2. Calculate the no of classes 3. Calculate the class width 4. Construct table showing the intervals calculated in 1 to 3 5. Put in the tally for each interval and then show as frequency 6. Calculate the relative (%) frequency 13 marks
  23. 23. Range 31.4 - 4.9 = 26.5 No of classes K or c= 1+3.3logn n = 25 K or c= 3.3 log (25) = 5.61 ≈ 6 Class Width xmax  xmin = 26.5/6 = 4.41 ≈ 5 Class width  c
  24. 24. No of classes = 6 Class width = 5 INTERVALS TALLY FREQUENCY (f) RELATIVE FREQUENCY (%f) 4-<9 111 3 12 9 - < 14 1111 1 6 24 14 - < 19 1111 1111 9 36 19 - < 24 111 3 12 24 - < 29 111 3 12 29 - < 34 1 1 4 25 100
  25. 25. Exam question 2 2. Comment on the interval 4% of bins weighed between containing the lowest 29 & 34 kg percentage 3. In which interval do the data Largest no. of bins weighed tend to cluster? Which between 14 & 19kg. We descriptive statistics measure, assume mode will fall in this can we assume, would be interval (highest frequency) found in this interval? 4. Comment on the shape of +ve skewed as more the distribution without values located in lower drawing a graph . Give reasons intervals 7 MARKS
  26. 26. Quartiles & Box & Whisker Plots
  27. 27. • Quartiles • Percentiles • Interquartile range 27
  28. 28. QUARTILES 28
  29. 29. • QUARTILES – Order data in ascending order. – Divide data set into four quarters. 25% 25% 25% 25% Min Q1 Q2 Q3 Max 29
  30. 30. Example – Given the following data set: 2 5 8 −3 5 2 6 5 −4 Determine Q1 for the sample of nine measurements: •Order the measurements −4 −3 2 2 5 5 5 6 8 1 2 3 4 5 6 7 8 9 Q1 is the  n  1  1 4   9  1  1 4  2,5th value Find difference between data for 2 & 3 2-(-3)=5 and multiply by the decimal portion of value : 5 x 0.5 = 2.5 30 Add to smallest figure: -3 + 2.5: Q1 = 0.5
  31. 31. Example – Given the following data set: 2 5 8 −3 5 2 6 5 −4 Determine Q3 for the sample of nine measurements: −4 −3 2 2 5 5 5 6 8 1 2 3 4 5 6 7 8 9 Q3 is the  n  1  3 4   9  1  3 4  7,5th value Q3 = 5 + 0,5(6 − 5) = 5,5 31
  32. 32. Example – Given the following data set: 2 5 8 −3 5 2 6 5 −4 Interquartile range = Q3 – Q1 Q3 = 5,5 Q1 = −0,5 Interquartile range = 5,5 – (−0,5) =6 32
  33. 33. INTERQUARTILE RANGE (IQR) • Difference between the third and first quartiles • Indicates how far apart the first and third quartiles are IQR = Q3 – Q1 33
  34. 34. BOX & WHISKER PLOT • Provides a graphical summary of data based on 5 summary measures or values – First quartile, median, third quartile ,lower limit, upper limit • Box and whisker plot detects outliers in a data set LL = Q1 – 1,5 (IQR) UL = Q3 + 1,5 (IQR) 34
  35. 35. BOX-AND-WISKER PLOT Me = 12,38 LL = Q1 – 1,5(IQR) = 9,36 – 1,5(6,31) = –0,11 Q3 = 15,67 Q1 = 9,36 UL = Q3 + 1,5(IQR) = 15,67 – 1,5(6,31) = 25,14 IRR = 6,31 1,5(IQR) IQR 1,5(IQR) 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 • Any value smaller than −0,11 will be an outlier. • Any value larger than 25,14 will be an outlier. 35
  36. 36. Exam question 3 The Tubeka brothers spent the following amounts in Rand on groceries over the last 8 weeks:- 54 56 89 67 74 57 43 51 1. Calculate a five number summary table 2. Construct a box and whisker plot for the data 3. Determine whether there are any outliers. Show calculations 20 MARKS PROCEDURE 1. Reorder the data set 2. Identify maximum and minimum values in dataset 3. Calculate median 4. Calculate Q1 & Q3 5. Construct plot 6. Calculate upper & lower limits for dataset to determine if outliers present
  37. 37. 43 51 54 56 57 67 74 89 xmin = 43 xmax = 89 median = (56+57)/2 = 56.5 Q1 = 51.75 Q3 = 72.25 Q1 = (n+1) (1/4) = (8+1) x ¼ = 2.25 value Between 51 & 54 54-51 = 3 multiply by decimal portion of value 3x 0.25 = 0.75 and add the lower value Q1 = 51 + 0.75 = 51.75 Q3 = (n+1) (¾) = (8+1) x ¾ = 6.75 value Between 67 & 74 74 – 67 = 7 multiply by decimal portion of value 7 x 0.75 = 5.25 and add lower value Q3 = 67 + 5.25 = 72.25
  38. 38. 43 51 54 56 57 67 74 89 xmin = 43 xmax = 89 median = (56+57)/2 = 56.5 Q1 = 51.75 Q3 = 72.25 OUTLIERS 1. Calculate upper & lower limits LL = Q1 – 1,5 (IQR) UL = Q3 + 1,5 (IQR) IQR = 72.25 – 51.75 = 20.5 LL = 51.75 – 1,5(20.5) = 21 UL = 72.25 + 1.5(20.5) = 103 No values smaller than 21 or greater than 103 therefore no outliers present
  39. 39. MEASURES OF LOCATION
  40. 40. • ARITHMETIC MEAN – Data is given in a frequency table – Only an approximate value of the mean x fx i i f i where f i  frequency of the i th class interval xi = class midpoint of the i th class interval 40
  41. 41. • MEDIAN – Data is given in a frequency table. – First cumulative frequency ≥ n/2 will indicate the median class interval. – Median can also be determined from the ogive.  ui  li   n  Fi 1  M e  li  2 fi where li = lower boundary of the median interval ui = upper boundary of the median interval Fi -1 = cumulative frequency of interval foregoing median interval fi = frequency of the median interval 41
  42. 42. • MODE – Class interval that has the largest frequency value will contain the mode. – Mode is the class midpoint of this class. – Mode must be determined from the histogram. 42
  43. 43. Example – The following data represents the number of telephone calls received for two days at a municipal call centre. The data was measured per hour. To calculate the Number of Number of mean for the sample calls hours fi xi of the 48 hours: [2–under 5) 3 3,5 determine the class [5–under 8) 4 6,5 midpoints [8–under 11) 11 9,5 [11–under 14) 13 12,5 [14–under 17) 9 15,5 [17–under 20) 6 18,5 [20–under 23) 2 21,5 n = 48 43
  44. 44. Example – The following data represents the number of telephone calls received for two days at a municipal call centre. The data was measured per hour. x  fi xi Number of Number of calls hours fi xi  fi [2–under 5) 3 3,5 597  [5–under 8) 4 6,5 48 [8–under 11) 11 9,5  12, 44 [11–under 14) 13 12,5 Average number [14–under 17) 9 15,5 of calls per hour [17–under 20) 6 18,5 is 12,44. [20–under 23) 2 21,5 n = 48 44
  45. 45. Exam question 3 The number of overtime hours worked by 40 part-time employees of a security company in 1 week is shown in the following frequency distribution:- Hours per Frequency (f) week 2.1 - < 2.8 12 2.8 - < 3.5 13 3.5 - < 4.2 7 4.2 - < 4.9 5 4.9 - < 5.6 2 5.6 - < 6.3 1 1. Estimate the mean number of overtime hours worked 2. What % of employees worked at least 4.2 hours overtime? 8 marks
  46. 46. Exam question 3 Procedure 1. Calculate the midpoint x for each interval (lower limit + upper limit/2) 2. Multiply f by the midpoint x 3. Total the fx and f columns 4. Divide ∑fx by ∑f
  47. 47. Exam question 3 Hours per week Frequency (f) Mid point (x) fx 2.1 - < 2.8 12 (2.1 + 2.8)/2= 29.4 2.45 2.8 - < 3.5 13 3.15 40.95 3.5 - < 4.2 7 3.85 26.95 4.2 - < 4.9 5 4.55 22.75 4.9 - < 5.6 2 5.25 10.5 5.6 - < 6.3 1 5.95 5.95 40 136.5 Mean = 136.5/40 = 3.41hrs Employees at least 4.2 hrs = 8 8/40 *100 = 20%
  48. 48. PERCENTILES 48
  49. 49. • PERCENTILES – Order data in ascending order. – Divide data set into hundred parts. 10% 90% Min P10 Max 80% 20% Min P80 Max 50% 50% Min P50 = Q2 Max 49
  50. 50. Example – Given the following data set: 2 5 8 −3 5 2 6 5 −4 Determine P20 for the sample of nine measurements: −4 −3 2 2 5 5 5 6 8 1 2 3 4 5 6 7 8 9 P20 is the  n  1    9  1    2 p 100 20 100 nd value P20 = −3 50
  51. 51. Example – The following data represents the number of telephone calls received for two days at a municipal call centre. The data was measured per hour. Number of Number of P60 calls hours fi F = np/100 [2–under 5) 3 3 = 48(60)/100 [5–under 8) 4 7 = 28,8 [8–under 11) 11 18 The first cumulative [11–under 14) 13 31 frequency ≥ 28,8 [14–under 17) 9 40 [17–under 20) 6 46 [20–under 23) 2 48 n = 48 51
  52. 52. Example – The following data represents the number of telephone calls received for two days at a municipal call centre. The data was measured per hour. P60 Number of Number of  u p  l p   100  Fp1  np calls hours fi F  lp  fp [2–under 5) 3 3  11  14  11 28,8  18  [5–under 8) 4 7  13, 49 13 [8–under 11) 11 18 [11–under 14) 13 31 60% of the time less [14–under 17) 9 40 than 13,49 or 40% of [17–under 20) 6 46 the time more than 13,49 calls per hour. [20–under 23) 2 48 n = 48 52
  53. 53. Exam question 3 1. John, one of the part-time workers was told he falls on the 70th percentile. Calculate the value and explain what it means. PROCEDURE 1. Calculate the cumulative frequencies 2. Calculate which class the required percentile falls into by using P =np/100 3. Once you have identified the class use the percentile formula given in the tables book to calculate the value. Take CARE to order the calculation correctly. 4 MARKS
  54. 54. Exam question 3 P = np/100 = 40*70/100 Hours per Frequency Cumulative =28 week (f) F 2.1 - < 2.8 12 12 P70 = 3.5 + [ (4.2-3.5)(28-25)]/7 2.8 - < 3.5 13 25 = 3.5 + 0.8 3.5 - < 4.2 7 32 =3.8 4.2 - < 4.9 5 37 4.9 - < 5.6 2 39 70% of the workers worked fewer hours overtime than John. 70% of 5.6 - < 6.3 1 40 the workers worked fewer than 3.8 hrs. 30% of the workers worked 40 more overtime hours than John. 30% of the employees worked more than 3.8hrs.
  55. 55. CONFIDENCE INTERVALS
  56. 56. Confidence interval – An interval is calculated around the sample statistic Population parameter included in interval Confidence interval 56
  57. 57. 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 – Specify the probability thatsamples taken from 90% of all possible the interval will include the parameter produce an interval that will population will 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 57
  58. 58. • 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 58
  59. 59. 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 we estimate an interval, the interval is called CONFIDENCE INTERVAL 59
  60. 60. 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 60
  61. 61.    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 2 These 2 areas added Confidence level together = 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 61
  62. 62. 62
  63. 63. • 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 63
  64. 64. t Distribution 64
  65. 65. • 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 R11 193,14 and R13 606,86 – Estimate with 99% confidence the population 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  65
  66. 66. • 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 = – Proportion always between 0 and 1 size = sample n – For large samples the sample proportion is approximately normal ˆ p  p (1  p )  ˆ ˆ CI ( p )1   p  z1  ˆ   2 n  66
  67. 67. Exam question 7 1. In a sample of 200 residents of Johannesburg, 120 reported they believed the property taxes were too high. Develop a 95% confidence interval for the proportion of the residents who believe the tax rate is too high. Interpret your answer 2. The time it takes a mechanic to tune an engine in a sample of 20 tune ups is known to be normally distributed with a sample mean of 45 minutes and a sample standard deviation of 14 minutes. Develop a 95% confidence interval estimate for the mean time it will take the mechanic for all engine tune ups. Interpret your answer 15 MARKS
  68. 68. Exam question 7 PROCEDURE 1. Determine what measure your are looking at: mean, proportion or standard deviation 2. Select appropriate formula based on 1. and sample size (t for small sample sizes <30; z for larger sample sizes) 3. Put the numbers into the formula and calculate the confidence intervals
  69. 69. Exam question 7 1. ˆ Sample proportion p = number of successes = x In a sample of 200 residents of sample size n Johannesburg, 120 reported they believed the property  p (1  p )  ˆ ˆ taxes were too high. Develop a CI ( p )1   p  z1  ˆ   2 n  95% confidence interval for 𝑝 = 120/200 = 0.6 the proportion of the Z 1-α = 1.96 residents who believe the tax 2 rate is too high. Interpret your CI = 0.6 +/_1.96 √( 0.6 0.4 )/200 answer CI = 0.6 +/- 0.07 0.53<CI<0.67 At CL of 95% between 53% and 67% of residents believe tax rate is too high
  70. 70. Exam question 7 The time it takes a mechanic  s  CI (  )1   x  t n 1;1   to tune an engine in a  2 n  sample of 20 tune ups is known to be normally 14 = 45 +/- 2.093 √20 distributed with a sample mean of 45 minutes and a sample standard deviation = 45 +/- 6.55 of 14 minutes. Develop a 95% confidence interval 38.45< µ < 51.55 estimate for the mean time At a confidence level of 95% the it will take the mechanic for population average time to complete a all engine tune ups. tune up is between 38.45 and 51.55 Interpret your answer minutes
  71. 71. HYPOTHESIS TESTING
  72. 72. STEPS OF A HYPOTHESIS TEST Step 1 • State the null and alternative hypotheses Step 2 • State the values of α Step 3 • Calculate the value of the test statistic Step 4 • Determine the critical value Step 5 • Make a decision using decision rule or graph Step 6 • Draw a conclusion 72
  73. 73. • Hypothesis test for Population Mean, n < 30 – If σ is unknown we use s to estimate σ – We need to replace the normal distribution with the t-distribution with (n - 1) degrees of freedom Testing H0: μ = μ0 for n < 30 Alternative Decision rule: Test statistic hypothesis Reject H0 if H1: μ ≠ μ0 |t| ≥ tn - 1;1- α/2 x  0 t H1: μ > μ0 t ≥ tn-1;1- α  s     n H1: μ < μ0 t ≤ -tn-1;1- α 73
  74. 74. • Hypothesis testing for Population proportion number of successes x – Sample proportion p = ˆ = sample size n – Proportion always between 0 and 1 Testing H0: p = p0 for n ≥ 30 Alternative Decision rule: Test statistic hypothesis Reject H0 if H1: p ≠ p0 |z| ≥ Z1- α/2 p  p0 ˆ z H1: p > p0 z ≥ Z1- α p0 (1  p0 ) H1: p < p0 z ≤ -Z1- α n 74
  75. 75. Exam question 8 1. Oliver Tambo airport wants to test the claim that on average cars remain in the short term car park area longer than 42.5 minutes. The research team drew a random sample of 24 cars and found that the average time that these cars remained in the short term parking area was 40 minutes with a sample standard deviation of 2 minutes. Test the claim at 10% level of significance and interpret. 2. The Gautrain Authority add a bus route if more than 55% of commuters indicate they would use the route. A sample of 70 commuters revealed that 42 would use a route from Sandton to Auckland Park. Does this route meet the Gautrain criteria. Use 0.05 significance level 16 MARKS
  76. 76. Exam question 8 Procedure 1. State H0 and Ha 2. Determine the critical value from the appropriate test table using α, and n 3. Compute test statistic (t or z value??) 4. Draw conclusion
  77. 77. Exam question 8 State hypothesis Oliver Tambo airport wants H0: µ = 42.5 to test the claim that on Ha: µ > 42.5 average cars remain in the Determine critical value short term car park area tn-1; 1- α = t 23; 0.9 = 1.319 longer than 42.5 minutes. Reject H0 if the test statistic is > The research team drew a 1.319 random sample of 24 cars Calculate test statistic and found that the average x  0 time that these cars t  s    remained in the short term  n parking area was 40 minutes T= 40-42.5 = -6.12 with a sample standard 2 deviation of 2 minutes. Test √24 the claim at 10% level of Do not reject H0 significance and interpret.
  78. 78. Exam question 8 State hypothesis The Gautrain Authority H0: p = 0.55 add a bus route if more Ha: p > 0.55 than 55% of commuters Determine critical value indicate they would use α = 0.05 Z = 1.64 the route. A sample of 70 Reject H0 if Z test > 1.64 commuters revealed that Calculate test statistic 42 would use a route from number of successes x Sandton to Auckland Park. ˆ Sample proportion p = = sample size n Does this route meet the p  p0 ˆ z Gautrain criteria. Use 0.05 p0 (1  p0 ) n significance level 0.6−0.55 Z= = 0.84 √((0.55)(0.45)/70 Do not reject H0
  79. 79. CORRELATION COEFFICIENT
  80. 80. Coefficient of correlation • The coefficient of correlation is used to measure the strength of association between two variables. • The coefficient values range between -1 and 1. – If r = -1 (negative association) or r = +1 (positive association) every point falls on the regression line. – If r = 0 there is no linear pattern. • The coefficient can be used to test for linear relationship between two variables. 80
  81. 81. Perfect positive High positive Low positive r = +1 r = +0,9 r = +0,3 Y Y Y X X X Perfect negative High negative No Correlation r = -1 r = -0,8 r=0 Y Y Y X X X 81
  82. 82. Exam question 10 The cost of repairing cars that were involved in accidents is one reason that insurance premiums are so high. In an experiment 5 cars were driven into a wall. The speeds were varied between 20km/hr and 80km/hr (X). The costs of repair (Y) were estimated and listed below:- SPEED (Km/h) (X) COST OF REPAIR (R’000) (Y) 20 3 30 5 40 8 60 24 80 34 1. Use calculator to calculate coefficient of correlation. Interpret your answer 2. Calculate and interpret the coefficient of determination for this data 3. Use your calculator to construct regression line equation and predict repair cost at 50km/h 10 MARKS
  83. 83. Exam question 10 1. Put data into calculator 2. Select regression function and select r 3. Calculate coefficient of determination = r2 x100% 4. Interpret results 5. Using Y = A + BX select regression function on calculator and determine values for A & B 6. Put x = 50 into formula and calculate result
  84. 84. Exam question 10 1. r = 0.98 There is a very strong relationship between the repair cost and speed. 2. r2 x 100% = 0.982 x 100 = 96% 96% of the variation in the cost of repair is explained by the variation in the speed at which the car crashed 3. Y = -10.7 +0.55x X = 50 Y = 16.8
  • KIDADROITRIVASARIVAL

    Nov. 14, 2019

NUBE Revision Workshop

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