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- 1. Inferential Test Procedures Probabilityand Statistics (MA202): Unit V Department of Applied Sciences, DIT University Dehradun
- 2. Contents Z-test (Largesample test) For single mean and difference of means t-test (Small sample test) For single mean and difference of means Chi-Square test For goodness of fit and independence of attributes
- 3. Z-test (Large sampletest n≥30) For single mean H0: µ = µ0 H1 : µ ≠ µ0 𝑧𝑐𝑎𝑙 = 𝑑𝑒𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑟𝑟𝑜𝑟 = ҧ 𝑥− 𝜇0 𝜎 𝑛 If 𝒁𝐜𝐚𝐥 < 𝒁𝐭𝐚𝐛 , null hypothesis is accepted For difference of means H0: µ1 = µ2 H1: µ1 ≠ µ2 𝑧𝑐𝑎𝑙 = ҧ 𝑥− ത 𝑦 𝜎1 2 𝑛1 + 𝜎2 2 𝑛2 If 𝜎1 2 and 𝜎2 2 are unknown, sample variances estimates 𝑠1 2 and 𝑠2 2 are used If 𝜎1 2 = 𝜎2 2 = 𝜎2 (say), for unknown 𝜎2 estimate 𝜎2 is used where 𝜎2 = 𝑛1𝑠1 2+𝑛2𝑠2 2 𝑛1+𝑛2 If 𝒁𝐜𝐚𝐥 < 𝒁𝐭𝐚𝐛 , null hypothesis is accepted
- 4. Example1. A sampleof 900 members has a mean 3.4 and standard deviation 2.61. Is the sample from a large population of mean 3.25 and standard deviation 2.61? For single mean H0: µ = 3.25 ( the sample has been drawn from the population with mean 3.25 and standard deviation 2.61) H1 : µ ≠ 3.25 (Two-tailed) 𝑧𝑐𝑎𝑙 = ҧ 𝑥− 𝜇0 𝜎 𝑛 = 3.4−3.25 2.61 900 = 1.73 𝒁𝐭𝐚𝐛=1.96 at 5% level of significance Since 𝒁𝐜𝐚𝐥 < 𝒁𝐭𝐚𝐛 , null hypothesis is accepted
- 5. Example2. The averagehourly wage of a sample of 150 workers in a plant ‘A’ was Rs. 2.56 with standard deviation of Rs. 1.08. The average hourly wage of a sample of 200 workers in plant ‘B’ was Rs. 2.87 with a standard deviation of Rs. 1.28. Can an applicant safely assume that the hourly wages paid by plant ‘B’ are higher than those paid by plant ‘A’? For difference of means H0: µ1 = µ2 H1: µ1 < µ2 (Left-tailed test) 𝑧𝑐𝑎𝑙 = ҧ 𝑥− ത 𝑦 𝜎1 2 𝑛1 + 𝜎2 2 𝑛2 = 2.56−2.87 1.08 2 150 + 1.28 2 200 = 2.46 𝜎1 2 and 𝜎2 2 are unknown, sample variances estimates 𝑠1 2 and 𝑠2 2 are used 𝒁𝐭𝐚𝐛 = 1.645 at 5% level of significance (one-tailed) If 𝒁𝐜𝐚𝐥 > 𝒁𝐭𝐚𝐛 , null hypothesis is rejected and we can conclude that the average hourly wages paid by plant ‘B’ are certainly higher than ‘A’.
- 6. t-test (Small sampletest n<30) For single mean H0: µ = µ0 H1 : µ ≠ µ0 𝑡𝑐𝑎𝑙 = ҧ 𝑥− 𝜇0 𝑆 𝑛 𝑤ℎ𝑒𝑟𝑒 𝑆 = σ 𝑥𝑖− ҧ 𝑥 2 𝑛−1 𝑡𝑡𝑎𝑏 𝑤𝑖𝑡ℎ 𝑛 − 1 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 If 𝒕𝐜𝐚𝐥 < 𝒕𝐭𝐚𝐛 , null hypothesis is accepted * sample variance 𝑠2 = σ 𝑥𝑖− ҧ 𝑥 2 𝑛 so, 𝑛. 𝑠2 = (𝑛 − 1)𝑆2 For difference of means H0: µ1 = µ2 H1: µ1 ≠ µ2 𝑡𝑐𝑎𝑙 = ҧ 𝑥− ത 𝑦 𝑆. 1 𝑛1 + 1 𝑛2 𝑤ℎ𝑒𝑟𝑒𝑆 = σ 𝑥𝑖− ҧ 𝑥 2+σ 𝑦𝑖− ത 𝑦 2 𝑛1+𝑛2−2 𝑡𝑡𝑎𝑏 𝑤𝑖𝑡ℎ (𝑛1 + 𝑛2 − 2) 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 If 𝒕𝐜𝐚𝐥 < 𝒕𝐭𝐚𝐛 , null hypothesis is accepted
- 7. Example3.The specimen ofcopper wires drawn from a large lot have the following breaking strength (in kg. weight): 578, 572, 570, 568, 572, 578, 570, 572, 596, 544. Test whether the mean breaking strength of the lot may be taken to be 578 kg. weight. For single mean H0: µ = 578 kg. H1 : µ ≠ 578 kg. 𝑡𝑐𝑎𝑙 = ҧ 𝑥− 𝜇0 𝑆 𝑛 = 1.4917 𝑤ℎ𝑒𝑟𝑒 𝑆 = σ 𝑥𝑖− ҧ 𝑥 2 𝑛−1 = 12.719 𝑡𝑡𝑎𝑏 = 2.262 𝑤𝑖𝑡ℎ 10 − 1 = 9 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 and 5% level of significance Since 𝒕𝐜𝐚𝐥 < 𝒕𝐭𝐚𝐛 , null hypothesis is accepted We can conclude that the mean breaking strength of copper wires lot may be taken as 578 kg. 𝒙𝒊 𝒙𝒊 − ഥ 𝒙 𝒙𝒊 − ഥ 𝒙 𝟐 578 6 36 572 0 0 570 -2 4 568 -4 16 572 0 0 578 6 36 570 -2 4 572 0 0 596 24 576 544 -28 784 5720 1456
- 8. Example4.Sample of salesin similar shops in two towns are taken for a new product with the following results: Is there any evidence of difference in sales in the towns? For difference of means H0: µ1 = µ2 H1: µ1 ≠ µ2 𝑡𝑐𝑎𝑙 = ҧ 𝑥− ത 𝑦 𝑆. 1 𝑛1 + 1 𝑛2 = 57−61 2.236. 1 5 + 1 7 =3.055 𝑡𝑡𝑎𝑏 = 2.228 𝑤𝑖𝑡ℎ (𝑛1 + 𝑛2 − 2) = 10 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 and 5% level of significance Since 𝒕𝐜𝐚𝐥 > 𝒕𝐭𝐚𝐛 , null hypothesis is rejected We can conclude that the difference in sales in two towns is significant at 5% level of significance. Town Mean Sales Variance size of sample A 57 5.3 5 B 61 4.8 7
- 9. Chi-Square ( 𝝌𝟐 )Test For goodness of fit H0: Fit is good H1: Fit is bad Expected value 𝑬𝒊 is the average value Oi is the observed value 𝝌𝒄𝒂𝒍 𝟐 = σ 𝑶𝒊 −𝑬𝒊 𝟐 𝑬𝒊 𝜒𝑡𝑎𝑏 2 𝑤𝑖𝑡ℎ 𝑛 − 1 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 If 𝜒𝑐𝑎𝑙 2 < 𝜒𝑡𝑎𝑏 2 we accept the null hypothesis For independence of attributes H0: Attributes are independent H1: Attributes are not independent Expected value 𝑬𝒊 is the average value Oi is the observed value 𝝌𝒄𝒂𝒍 𝟐 = σ 𝑶𝒊 −𝑬𝒊 𝟐 𝑬𝒊 𝜒𝑡𝑎𝑏 2 𝑤𝑖𝑡ℎ 𝒓 − 𝟏 𝒄 − 𝟏 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 If 𝜒𝑐𝑎𝑙 2 < 𝜒𝑡𝑎𝑏 2 we accept the null hypothesis
- 10. Example5. A dieis thrown 132 times with following results: Is the die unbiased? Solution: It is a case of goodness of fit H0: Fit is good or die is unbiased H1: Fit is bad or die is biased Expected value 𝑬𝒊 is average = 22 𝝌𝒄𝒂𝒍 𝟐 = σ 𝑶𝒊 −𝑬𝒊 𝟐 𝑬𝒊 = 9 𝜒𝑡𝑎𝑏 2 = 11.071 𝑤𝑖𝑡ℎ 6 − 1 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 at 5% level of significance Since 𝜒𝑐𝑎𝑙 2 < 𝜒𝑡𝑎𝑏 2 so we may accept the null hypothesis So we can conclude that the die is unbiased. X 𝑶𝒊 𝑬𝒊 𝑶𝒊 −𝑬𝒊 𝑶𝒊 −𝑬𝒊 𝟐 𝑶𝒊 −𝑬𝒊 𝟐 𝑬𝒊 1 16 22 -6 36 1.636364 2 20 22 -2 4 0.181818 3 25 22 3 9 0.409091 4 14 22 -8 64 2.909091 5 29 22 7 49 2.227273 6 28 22 6 36 1.636364 132 9 Number turned up 1 2 3 4 5 6 Frequency 16 20 25 14 29 28
- 11. Example6. The tablegiven below shows the data obtained during outbreak of smallpox: Test the effectiveness of vaccination in preventing the attack from smallpox? Solution: It is a case of independence of attributes H0: Vaccination and attack are independent H1: Vaccination and attack are not independent 𝑬𝒊 for AB = (500*216)/2000= 54 𝑬𝒊 for Ab = (500*1784)/2000=446 𝑬𝒊 for aB = (1500*216)/2000=162 𝑬𝒊 for ab = (1500*1784)/2000=1338 𝝌𝒄𝒂𝒍 𝟐 = σ 𝑶𝒊 −𝑬𝒊 𝟐 𝑬𝒊 = 14.6431 𝜒𝑡𝑎𝑏 2 = 3.841 𝑤𝑖𝑡ℎ 2 − 1 ∗ 2 − 1 = 1 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 at 5% level of significance Since 𝜒𝑐𝑎𝑙 2 > 𝜒𝑡𝑎𝑏 2 so we reject the null hypothesis So we can conclude that the vaccination is effective in preventing the attack from smallpox. Groups 𝑶𝒊 𝑬𝒊 𝑶𝒊 −𝑬𝒊 𝑶𝒊 −𝑬𝒊 𝟐 𝑶𝒊 −𝑬𝒊 𝟐 𝑬𝒊 AB 31 54 -23 529 9.796296 Ab 469 446 23 529 1.186099 aB 185 162 23 529 3.265432 ab 1315 1338 -23 529 0.395366 2000 2000 14.64319 Attacked (B) Not Attacked (b) Total Vaccinated (A) 31 469 500 Not Vaccinated (a) 185 1315 1500 Total 216 1784 2000
- 12.