Statistics lecture 9 (chapter 8)

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Hypothesis Testing

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Statistics lecture 9 (chapter 8)

  1. 1. 1
  2. 2. Data collection Raw data Graphs Information Descriptive statistics Measures • location • spread Estimation Statistical Decision inference making Hypothesis testing 2
  3. 3. • Hypothesis testing is a procedure for making inferences about a population• A hypothesis test gives the opportunity to test whether a change has occurred or a real difference exists
  4. 4. • A hypothesis is a statement or claim that something is true H0 always contains = sign – Null hypothesis H0 • No effect or no difference • Must be declared true/false – Alternative hypothesis H1 • True if H0 found to be false – If H0 is false – reject H0 and accept H1 – If H0 is true – accept H0 and reject H1
  5. 5. • State the null and alternative hypotheses that would be used to test each of the following statements:-• A manufacturer claims that the average life of a transistor is at least 1000 hours (h) H0: µ = 1000 H1: µ > 1000• A pharmaceutical firm maintains that the average time for a certain drug to take effect is 15 mins H0: µ = 15 H1: µ ≠ 15• The mean starting salary of graduates is higher than R50000 per annum H0: µ = 50000 H1: µ > 50000
  6. 6. • Hypothesis testing consists of 6 steps: 1. State the hypothesis 2. State the value of α 3. Calculate the test statistic 4. Determine the critical value 5. Make a decision 6. Draw conclusion 6
  7. 7. Hypothesis Testing Step 1 One/Two populations State the hypothesis Two tailed H0: parameter = ? H1: parameter ≠ ?Ho – Null hypothesis Right tailedH1 – Alternative Hypothesis H0: parameter = ? H1: parameter > ? Left tailed H0: parameter = ? H1: parameter < ? 7
  8. 8. Hypothesis Testing Step 2State the value of α Decision Actual situation H0 is true H0 is false Do not Correct reject H0 decision Correct Reject H0 decision 8
  9. 9. Hypothesis Testing Step 2 State the value of α Decision Actual situation H0 is true H0 is false Two types of errors: Do not Correct Type II error = reject H0 decision  Type I error Correct Reject H0 = decisionα - Probability of Type I error Level of significance 1%, 5%, 10% Determine critical value/s 9
  10. 10. LEVEL OF SIGNIFICANCE • Probability (α) of committing a type I error is called the LEVEL OF SIGNIFICANCE • α is specified before the test is performed • You can control the Type I error by deciding, before the test is performed, what risk level you are willing to take in rejecting H0 when it is in fact true • Researchers usually select α levels of 0.05 or smaller 10
  11. 11. Hypothesis Testing Step 3Calculate value of test statisticThere are different test statistics for testing:• Single population – Mean, proportion, variance• Difference between two population – Means, proportions, variances 11
  12. 12. REJECTION AND NON-REJECTION REGIONS• To decide whether H0 will be rejected or not , a value, called the TEST STATISTIC has to be calculated by using certain sample results• Distribution of test statistic often follows a normal or t distribution.• Distribution can be divided into 2 regions:- – A region of rejection – A region of non- rejection 12
  13. 13. Hypothesis Testing Step 4Determine the critical value/values Right tailed Two tailed Left tailed H1: parameter > ? H1: parameter ≠ ? H1: parameter < ? H0 H0 H0 α α/2 α/2 α - Critical value – from tables 13
  14. 14. Hypothesis Testing Step 5Make decision Right tailed Two tailed Left tailed H1: parameter > ? H1: parameter ≠ ? H1: parameter < ? H0 H0 H0 α α/2 α/2 α Accept Rej Rej Accept Rej Rej Accept H0 H0 H0 H0 H0 H0 H0 14
  15. 15. Hypothesis Testing Step 6Draw conclusion Accept H0 ? H0 Reject H0 ?? ? Test statistic 15
  16. 16. STEPS OF A HYPOTHESIS TESTStep 1 • State the null and alternative hypothesesStep 2 • State the values of αStep 3 • Calculate the value of the test statisticStep 4 • Determine the critical valueStep 5 • Make a decision using decision rule or graphStep 6 • Draw a conclusion 16
  17. 17. Concept Questions• 1 – 12 page 261 textbook 17
  18. 18. Hypothesis test for Population Mean, n ≥ 30- population need not be normally distributed- sample will be approximately normal Testing H0: μ = μ0 for n ≥ 30 Alternative Decision rule: Test statistic hypothesis Reject H0 if H1: μ ≠ μ0 |z| ≥ Z1- α/2 x  0 z H1: μ > μ0 z ≥ Z1- α  s    H1: μ < μ0 z ≤ -Z1- α  n 18 Use σ if known
  19. 19. • Example – It will be cost effective to employee an additional staff member at a well known take away restaurant if the average sales for a day is more than R11 000 per day. – A sample of 60 days were selected and the average sales for the 60 days were R11 841 with a standard deviation of R1 630. – Test if it will be cost effective to employ an additional staff member. – Assume a normal distributed population. Use α = 0,05.
  20. 20. • Solution – The population of interest is the daily sales – We want to show that the average sales is more than R11 000 per day. – H1 : μ > 11 000 – The null hypothesis must specify a single value of the parameter  – H0 : μ = 11 000 – Need to test if R11 841( x ) is significant more than R11 000(μ)
  21. 21. • Solution At α = 0,05 if it will be cost effective to – H0 : μ = 11 000 employ an additional staff member – the average monthly income is more – H1 : μ > 11 000 than R11 000 – α = 0,05 x  0 11841  11000 α = 0,05z   3,99 – s  1630   0 z1-α 1,65  n 60 Accept H0 Reject H0 – Reject H0 Critical value Z 1-α = Z 0.95 = 1.65
  22. 22. • 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 22
  23. 23. Concept QuestionsQuestions 13 – 19, page 267, textbook 23
  24. 24. • Example – Health care is a major issue world wide. One of the concerns is the waiting time for patients at clinics. – Government claims that patients will wait less than 30 minutes on average to see a doctor. – A random sample of 25 patients revealed that their average waiting time was 28 minutes with a standard deviation of 8 minutes. – On a 1% level of significance can we say that the claim from government is correct? 24
  25. 25. • Solution – The population of interest is the waiting time at clinics – Want to test the claim that the waiting time is less than 30 minutes – H1 : μ < 30 – The null hypothesis must specify a single value of the parameter  – H0 : μ = 30 – Need to test if 28( x ) is significant less than 30(μ) 25
  26. 26. • Solution At α = 0,01 we can not say that – H0 : μ = 30 the average waiting time at – H1 : μ < 30 clinics is less than 30 minutes – α = 0,01 α = 0,01 x  0 28  30 t   1, 25 –  s  8   -t1-α -2,492 0  n 25 Reject H0 Accept H0 – Accept H0 tn-1;1-α = t24;0.99= -2.492 26
  27. 27. • 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 27
  28. 28. • Example – A market research company investigates the claim of a supplier that 35% of potential buyers are preferring their brand of milk – A survey was done in several supermarkets and it was found that 61 of the 145 shoppers indicated that they will buy the specific brand of milk – Assist the research company with the claim of the supplier on a 10% level of significance. 28
  29. 29. • Solution – The population of interest is the proportion of buyers – Want to test the claim that the proportion is 35% = 0,35 – H0 : p = 0,35 – The alternative hypothesis must specify that the proportion is not 35% – H1 : p ≠ 0,35 – Sample proportion p = number of successes = x  61  0, 42 ˆ sample size n 145 – Need to test if 0,42 p  is significant different from 0,35(p) ˆ 29
  30. 30. • Solution /2  0,05At – = 0 : p = 0,35 not say α H 0,10 we canthat 35%p ≠the clients will – H1 : of 0,35 -z-1,65 z1-/2 1-/2 +1,65 prefer the brand of milk – α = 0,10 Reject H0 Accept H0 Reject H0 p  p0 ˆ 0, 42  0,35 z   1, 76 p0 (1  p0 ) 0,35(1  0,35) n 145 Z1-α/2 = Z0.95 = +/- 1.65 – Reject H0 30
  31. 31. • Hypothesis testing for Population Variance – Draw conclusions about variability in population – Χ2 –distribution with (n - 1) degrees of freedom Testing H0: σ2 = σ20 Alternative Decision rule: Test statistic hypothesis Reject H0 if Χ2 ≤ Χ2n-1;α/2 or H1: σ2 ≠ σ20 (n  1) s 2 Χ2 ≥ Χ2n-1;1- α/2 2  H1: σ2 > σ20 Χ2 ≥ Χ2n-1;1- α  02 H1: σ2 < σ20 Χ2 ≤ Χ2n-1;α 31
  32. 32. Concept Questions• Questions 20 – 24 page 272 32
  33. 33. • Example – The variation in the content of a 340ml can of beer should not be more than 10ml2. – To test the validity of this, 25 cans of beers revealed a variance of 12ml2. – On a 5% level of significance can we say the variation in the content of the cans is too large? 33
  34. 34. • Solution – The population of interest is the variation in content – Want to test the claim that the variance is more than 10ml2 – H1 : σ2 > 10 – The null hypothesis must specify that the variance is 10ml2 – H0 : σ2 = 10 – Need to test if 12(s2) is significant more than 10(σ2) 34
  35. 35. • Solution – H0 : σ2 = 10   0,05 – H1 : σ2 > 10 – α = 0,05 Χ2n – 1; 1-α +36,42 (n  1) s 2 (25  1)12   2   28,8 Accept H0 Reject H0  02 10 At α = 0,05 we can not say – Accept H0 that the variation in the content of the cans is moreΧ n-1; 1-α = Χ 24; 0.95 = 36.42 2 2 than 10ml2 35
  36. 36. • Hypothesis tests for comparing two Populations Drawn from different – Difference between two means samples, samples have no relation • Independent samples – Large samples – Small samples Samples are • Dependent samples related – Difference between two proportions – Difference between two variances 36
  37. 37. • Hypothesis tests for comparing two Populations – H0: Population 1 parameter = Population 2 parameter – H1: Population 1 parameter ≠ Population 2 parameter – H1: Population 1 parameter > Population 2 parameter – H1: Population 1 parameter < Population 2 parameter μ1 / σ21 / p1 μ2 / σ22 / p2
  38. 38. Difference between two Population Means,independent samples, n1 ≥ 30 and n2 ≥ 30 Testing H0: μ1 = μ2 for n1 ≥ 30 and n2 ≥ 30 Alternative Decision rule: Test statistic hypothesis Reject H0 if H1: μ1 ≠ μ2 |z| ≥ Z1- α/2 x1  x2 z H1: μ1 > μ2 z ≥ Z1- α s12 s2 2  H1: μ1 < μ2 z ≤ -Z1- α n1 n2 38 Use σ12 and σ22 if known
  39. 39. Example A leading television manufacturer purchasescathode ray tubes from 2 businesses (A and B). Arandom sample of 36 cathode ray tubes frombusiness A showed a mean lifetime of 7.2 yearsand a std dev of 0.8 years, while a random sampleof 40 cathode ray tubes from business B showed amean lifetime of 6.7 years and a std dev of 0.7 yrs.Test at a 1% significance level whether the meanlifetime of the cathode ray tubes from business A islonger than the mean lifetime of business B 39
  40. 40. Answer Company A: n1 = 36, x1 = 7,2 and s1 = 0,8. Company B: n2 = 40, x 2 = 6,7 and s2 = 0,7. H0: 1 = 2  H1: 1 > 2   = 0,01 x1  x 2 z= s12 s2 2  n1 n 2 7,2  6,7 = 0,8 0,7 2 2  36 40 0,5 = 0,1733 = 2,8852 Z1 = Z0,99 = 2,33 Therefore, reject H0. There is enough evidence to say that the mean lifetime of the tubes from Company A is longer than that of Company B. 40
  41. 41. Difference between two Population Means, independentsamples, n1 < 30 and n2 < 30 Testing H0: μ1 = μ2 for n1 < 30 and n2 < 30 Alternative Decision rule: Test statistic hypothesis Reject H0 if x1  x2 H1: μ1 ≠ μ2 |t| ≥ tn1 + n2 – 2 ; 1- α/2 t with 1 2 sp  n1 n2 H1: μ1 > μ2 t ≥ tn1 + n2 – 2 ; 1- α sp   n1  1 s12   n2  1 s22 H1: μ1 < μ2 t ≤ -tn1 + n2 – 2 ;1- α n1  n2  2 41
  42. 42. Difference between two Population Means, dependentsamples – pairs of observations Observation 1 2 3 ---------- n Sample 1 X11 X12 X13 - - - - - - - - - - X1n Sample 2 X21 X22 X23 - - - - - - - - - - X2n d1 d2 d3 dn Difference (d) (X11 - X21) (X12 - X22) (X13 – X23) (X1n - X2n) d  d  2 1 2  1 d   d and sd  n n n 1 42
  43. 43. Difference between two Population Means, dependentsamples Testing H0: μ1 = μ2 Alternative Decision rule: Test statistic hypothesis Reject H0 if H1: μ1 ≠ μ2 |t| ≥ tn – 1 ; 1- α/2 d t H1: μ1 > μ2 t ≥ tn – 1 ; 1- α  sd     n H1: μ1 < μ2 t ≤ -tn – 1 ;1- α 43
  44. 44. Concept Questions Questions 25 – 29 p 283, textbook 44
  45. 45. Difference between two Population Proportions, largeindependent samples Testing H0: p1 = p2 for n1 ≥ 30 and n2 ≥ 30 Alternative Decision rule: Test statistic hypothesis Reject H0 if H1: p1 ≠ p2 |z| ≥ Z1- α/2 p1  p2 ˆ ˆ z 1 1 p(1  p)    ˆ ˆ H1: p1 > p2 z ≥ Z1- α  n1 n2  n1 p1  n2 p2 ˆ ˆ H1: p1 < p2 z ≤ -Z1- α where p  ˆ 45 n1  n2
  46. 46. Difference between two Population Variances, largeindependent samples Testing H0: σ21 = σ22 Alternative Decision rule: Test statistic hypothesis Reject H0 if H1: σ21 ≠ σ22 F ≥ Fn1-1 ; n2-1 ; α/2 s12 F 2 H1: σ21 > σ22 F ≥ Fn1-1 ; n2-1 ; α s2Assume population 1 has the larger 46variance. Thus always: s21 > s22
  47. 47. • Example• There is a belief that people staying in Cape Town travel less than people staying in Johannesburg. – Random samples of 43 people in Cape Town and 39 in Johannesburg were drawn. – For each person the distance travelled during October were recorded. – Test the belief on a 5% level of significance 47
  48. 48. • Solution – The population of interest is the km travelled – Samples are large and independent – We want to show that Cape Town travel less than Johannesburg – H1 : μ1 < μ2 – The null hypothesis must specify a single value of the parameter  – H0 : μ1 = μ2 48
  49. 49. The belief that people staying in• From the data: Cape Town travel less than people – Cape Town Johanesburg in Johannesburg is not true staying x1 = 604 x 2 = 633a 5% level of significance on n1 = 43 n 2 = 39 s1 = 64 s 2 = 103 -1,65 0 Reject H0 Accept H0 – H0: μ1 = μ2 – H1: μ1 < μ2 – z  x1  x2  604  633  1,51 2 2 2 2 s1 s2 64 103   n1 n2 43 39 – Accept H0 49
  50. 50. Example• Pathological laboratories have a problem with time it takes a blood sample to be analyzed. They hope by introducing some new equipment, the time taken will be reduced. – Blood samples for 10 different types of test were analyzed by the traditional laboratories and by the newly equipped laboratories. – The time, in minutes, were captured for each test. – Did the time reduce? α = 0,01 50
  51. 51. • Solution – The population of interest is the test times – The samples are dependent – We want to show that new times is less than the old times – H1 : μ1 > μ2 – The null hypothesis must specify a single value of the parameter – H0 : μ1 = μ2 51
  52. 52. Blood Existing New Difference - Calculate thesample lab lab 1 47 70 -23 difference for each xi 2 65 83 -18 - Calculate the average 3 59 78 -19 differences and the 4 61 46 15 standard deviation of 5 75 74 1 the differences 6 65 56 9 7 73 74 -1 x d  2, 2 8 85 52 33 9 97 99 -2 sd  19,14 10 84 57 27 52
  53. 53. - The hypotheses test for this problem is H0: 1 = 2 H1: 1 > 2 α =0,10- The statistic is 0 1,383 t0.90,9 d t Accept H0 Reject H0  sd     n 2, 2 Using α = 0,10, introducing  some new equipment the 19,14 10 time taken did not reduce. 53  0, 36
  54. 54. • Example – A clothing manufacturer introduced two new swim suit ranges on the market. – Of the 266 clients asked if they will wear range A, 85 indicated they will. – Of the 192 clients asked if they will wear range B, 50 indicated they will. – Can we say there is a difference in the preferences of the two ranges. Use α = 0,05. 54
  55. 55. • Solution – The population of interest is the proportion of clients who will wear the clothing – We want to determine if the proportion of range A differ form the proportion of range B – H1: p1 ≠ p2 – The null hypothesis must specify a single value of the parameter – H0: p1 = p2 55
  56. 56. • From the data: – Range A : p1  266  0,32 ˆ 85 Range B: p2  192  0, 26 ˆ 50 266(0,32)  192(0, 26) – p ˆ  0, 29 266  192 +1,96 -1,96 – H0: p1 = p2 Reject H0 Accept H0 Reject H0 – H1: p1 ≠ p2 – p1  p2 ˆ ˆ 0,32  0, 26 z   1,93 1 1 p(1  p)    ˆ ˆ  0, 29(1  0, 29) 266  192 1 1   n1 n2  – Accept H0 There is no difference in the preferences of the two ranges if α = 0,05. 56
  57. 57. • Example Remember: – An important measure to determine 1 has the largerin the Population service delivery variance. banking sector is the variability in the service 2times. Thus always: s21 > s 2 – An experiment was conducted to compare the service times of two bank tellers. – The results from the experiment: • Teller A: nA = 18 and s2A = 4,03 • Teller B: nB = 26 and s2B = 9,49 – Can we say that the variance in service time of teller A is less than that variance of teller B on a 5% level of significance. – We will then test if the variance in service time of teller B is more that the variation of teller A: s2B > s2A 57
  58. 58. • Solution – The population of interest is the variation in the service time of the two bank tellers. – We want to determine if the variation of teller B is more than the variation of teller A. – H1: S21 > S22 – The null hypothesis must specify a single value of the parameter – H0: S21 = S22 58
  59. 59. • From the data: – The results from the experiment: • Teller A: nA = 18 and s2A = 4,03 • Teller B: nB = 26 and s2B = 9,49 F26-1;18-1;0,05 = 2,18 Accept H0 Reject H0 – H0: 1 = S22 S2 – H1: S21 > S22 S 2 9, 49 Variation of teller B is more F 1 2  2,35 than the variation of teller A S 4, 03 2 on a 5% level of – Reject H0 significance. 59
  60. 60. Concept Questions• Questions 30 -31, p 289, textbook 60
  61. 61. HOMEWORK• Self review test, p289• Izmivo• Revision test• Supplementary Questions p292 – 298 61

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