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Testing Mean and Proportion Differences Worksheet Answers 1. Two populations have mound – shaped, symmetric distributions. A random sample of 16 measurements from the first population had a sample mean of 20, with sample standard deviation 2. An independent random sample of 9 measurements from the second population had a sample mean of 19, with a sample standard deviation of 3. Test the claim that the population mean of the first population exceeds that of the second. Use 𝛼 = 0.05 𝐻0 ∶ 𝜇1 = 𝜇2 𝑧= 𝑥̅1 −𝑥̅2 𝜎2 𝜎2 √ 1+ 1 𝑛1 𝑛2 𝐻1 ∶ 𝜇1 > 𝜇2 = 20−19 = 2 2 √2 +3 20 19 𝛼 = 0.05 1 √ = 0.2+0.474 1 √0.674 = 1.218 𝑃 − 𝑣𝑎𝑙𝑢𝑒(𝑧 > 1.218) = 1 − 0.8997 = 0.1003 Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 > 0.05, we do not reject 𝐻0 2. A random sample of 49 measurements from a population with a population standard deviation 3 had a sample mean of 10. An independent random sample of 64 measurements from a second population with population standard deviation 4 had a sample mean of 12. Test the claim that the population means are different. Use a 1% level of significance 𝐻0 ∶ 𝜇1 = 𝜇2 𝑧= 𝑥̅ 1 −𝑥̅ 2 𝜎2 𝜎2 √ 1+ 1 𝑛1 𝑛2 = 𝐻1 ∶ 𝜇1 ≠ 𝜇2 20−19 2 2 √2 +3 20 19 = 1 √0.2+0.474 𝛼 = 0.01 = 1 √0.674 = 1.218 ** two tailed 𝑃 − 𝑣𝑎𝑙𝑢𝑒(𝑧 > 1.218) = 2(1 − 0.8997) = 2(0.1003) = 0.2006 Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 > 0.01, we do not reject 𝐻0 3. For one binomial experiment, 75 binomial trials produced 45 successes. For a second independent binomial experiment, 100 binomial trials produced 70 successes. At the 5% level of significance, test the claim that the probability of success for the two binomial experiments differ. 𝑟1 = 45 , 𝑛1 = 75 , 𝑟2 = 70 , 𝑛2 = 100 𝐻0 ∶ 𝑝1 − 𝑝2 = 0 𝑝̅ = 𝑟1 +𝑟2 𝑛1 +𝑛2 𝑝̂1 = 𝑧= 𝑟1 𝑛1 = 𝐻1 ∶ 𝑝1 ≠ 𝑝2 45+70 = 75+100 45 = 115 175 = 0.60 75 ̅𝑞 ̅ 𝑝 = ̅𝑞 ̅ 𝑝 √𝑛 +𝑛 1 2 = 0.657 𝑝̂2 = 𝑝̂1 −𝑝̂2 𝛼 = 0.05 𝑟2 𝑛2 = 𝑞̅ = 1 − 𝑝̅ = 1 − 0.657 = 0.343 70 100 = 0.70 −0.10 √ (0.657)(0.343) (0.657)(0.343) + 75 100 = 𝑝̂1 − 𝑝̂2 = 0.60 − 0.70 = −0.10 −0.10 √0.003+0.002 = −0.10 √0.005 = −1.414 𝑃 − 𝑣𝑎𝑙𝑢𝑒 = 2𝑃(𝑧 < −1.414) = 2(0.0793) = 0.1586 Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 > 0.05, we fail to reject 𝐻0 4. REM ( rapid eye movement ) sleep is sleep during which most dreams occur. It is thought that children have more REM sleep than adults. Assume REM sleep time is normally distributed for both children and adults. Children : 𝑛1 = 10 , 𝑥̅1 = 2.8 hours , 𝜎 = 0.5 hours Adults : 𝑛2 = 10 , 𝑥̅2 = 2.1 hours , 𝜎 = 0.7 hours Do these data indicate that, on average, children tend to have more REM sleep time than adults ? Use 𝛼 = 0.01 𝐻0 ∶ 𝜇1 = 𝜇2 𝑧= 𝑥̅1 −𝑥̅2 𝜎2 𝜎2 √ 1+ 1 𝑛1 𝑛2 𝐻1 ∶ 𝜇1 > 𝜇2 = 2.8−2.1 = 2 2 √0.5 +0.7 10 10 𝛼 = 0.01 0.7 √ = 0.025+0.049 𝑃 − 𝑣𝑎𝑙𝑢𝑒(𝑧 > 2.56) = 1 − 0.9948 = 0.0052 Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 ≤ 0.01, we reject 𝐻0 0.7 √0.074 = 2.56 5. A random sample of 𝑛1 = 10 regions in New England gave the following violent crime rates ( per million population ) 𝒙𝟏 ∶ New England Crime Rate 3.5 3.7 4.0 3.9 3.3 4.1 1.8 4.8 2.9 3.1 Another random sample of 𝑛2 = 12 regions in the Rocky Mountain states gave the following violent crime rates ( per million population ) 𝒙𝟐 ∶ Rocky Mountain States 3.7 4.3 4.5 5.3 3.3 4.8 3.5 2.4 3.1 3.5 5.2 2.8 Do the data indicate that the violent crime rate in the Rocky Mountain region is higher than that in New England? Use 𝛼 = 0.01 Using a calculator 𝑥̅1 = 3.51 , 𝑠1 = 0.77 , 𝑛1 = 10 𝑥̅2 = 3.87 , 𝑠2 = 0.94 , 𝑛2 = 12 𝐻0 ∶ 𝜇1 = 𝜇2 𝑡= 𝑥̅1 −𝑥̅ 2 𝑠2 𝑠2 √ 1+ 1 𝑛1 𝑛2 𝐻1 ∶ 𝜇1 < 𝜇2 = 3.51−3.87 = 2 2 √0.77 +0.94 10 12 𝛼 = 0.01 −0.36 √ = 0.059+0.074 −0.36 √0.133 = −0.99 𝑃 − 𝑣𝑎𝑙𝑢𝑒 𝑡 = −0.99 with 𝑑. 𝑓. = 9 is between 0.250 and 0.125 Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 > 0.01, we do not reject 𝐻0 6. A study of fox rabies in southern Germany gave the following information about different regions and the occurrence of rabies in each region. A random sample of 𝑛1 = 16 locations in region I gave the following information about the number of cases of fox rabies near that location. 𝒙𝟏 Region I Data 1 8 8 8 7 8 8 1 3 3 3 2 5 1 4 6 A second random sample of 𝑛2 = 15 locations in region II gave the following information about the number of cases of fox rabies near the location. 𝒙𝟐 Region II Data 1 1 3 1 4 8 5 4 4 2 2 5 6 9 4 Does this information indicate that there is a difference ( either way ) in the mean number of cases of fox rabies between the two regions ? Use 𝛼 = 0.05 Using a calculator 𝑥̅1 = 4.75 , 𝑠1 = 2.82 , 𝑛1 = 16 𝑥̅2 = 3.93 , 𝑠2 = 2.43 , 𝑛2 = 15 𝐻0 ∶ 𝜇1 = 𝜇2 𝑡= 𝑥̅1 −𝑥̅ 2 𝑠2 𝑠2 √ 1+ 1 𝑛1 𝑛2 𝐻1 ∶ 𝜇1 ≠ 𝜇2 = 4.75−3.93 = 2 2 √2.82 +2.43 16 15 𝛼 = 0.05 0.82 √ = 0.497+0.394 0.82 √0.891 = 0.89 𝑃 − 𝑣𝑎𝑙𝑢𝑒 𝑡 = 0.89 with 𝑑. 𝑓. = 14 and two – tailed is between 0.500 and 0.250 Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 > 0.05, we do not reject 𝐻0 7. Would you prefer spending more federal tax money on the arts? Of a random sample of 𝑛1 = 220 women, 𝑟1 = 59 responded yes. Another random sample of 𝑛2 = 175 men showed that 𝑟2 = 56 responded yes. Does this information indicate a difference ( either way ) between the population proportion of women and the population proportion of men who favor spending more tax dollars on the arts ? Use 𝛼 = 0.05 𝑟1 = 59 , 𝑛1 = 220 , 𝑟2 = 56 , 𝑛2 = 175 𝐻0 ∶ 𝑝1 − 𝑝2 = 0 𝐻1 ∶ 𝑝1 ≠ 𝑝2 𝛼 = 0.05 𝑝̅ = 𝑟1 +𝑟2 𝑛1 +𝑛2 220+175 𝑟1 59 𝑝̂1 = 𝑧= 59+56 = 𝑛1 = 220 𝑝̂1 − 𝑝̂2 𝑝̅ 𝑞̅ 𝑝̅ 𝑞̅ √ 𝑛1 + 𝑛2 = = 0.27 = 115 395 = 0.291 𝑝̂2 = 𝑟2 𝑛2 = 𝑞̅ = 1 − 𝑝̅ = 1 − 0.291 = 0.709 56 175 = 0.32 −0.05 √(0.291)(0.32) + (0.291)(0.32) 220 175 = 𝑝̂1 − 𝑝̂2 = 0.27 − 0.32 = −0.05 −0.05 √0.0004 + 0.0005 = −0.05 √0.0009 = −1.67 𝑃 − 𝑣𝑎𝑙𝑢𝑒 = 2𝑃(𝑧 < −1.67) = 2(0.0475) = 0.095 Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 > 0.05, we fail to reject 𝐻0 8. A random sample of 𝑛1 = 228 voters registered in the state of California showed that 141 voted in the last general election. A random sample of 𝑛2 = 216 registered voters in the state of Colorado showed that 125 voted in the most recent general election. Do these data indicate that the population proportion of voter turnout in Colorado is higher than that in California? Use a 5% level of significance. 𝛼 = 0.05 𝑟1 = 114 , 𝑛1 = 228 , 𝑟2 = 125 , 𝑛2 = 216 𝐻0 ∶ 𝑝1 − 𝑝2 = 0 𝐻1 ∶ 𝑝1 < 𝑝2 𝛼 = 0.05 𝑝̅ = 𝑟1 +𝑟2 = 114+125 𝑛1 +𝑛2 228+216 𝑟1 114 𝑝̂1 = 𝑛1 = 228 𝑝̂1 − 𝑝̂2 𝑧= 𝑝̅ 𝑞̅ 𝑝̅ 𝑞̅ 𝑛1 + 𝑛2 √ = = 0.50 = 239 444 = 0.538 𝑝̂2 = 𝑟2 𝑛2 = 𝑞̅ = 1 − 𝑝̅ = 1 − 0.538 = 0.462 125 216 = 0.58 −0.08 √(0.538)(0.462) + (0.538)(0.462) 228 216 𝑃 − 𝑣𝑎𝑙𝑢𝑒 = 2𝑃(𝑧 < −1.67) = 0.0475 Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 ≤ 0.05, we reject 𝐻0 𝑝̂1 − 𝑝̂2 = 0.50 − 0.58 = −0.08 = −0.08 √0.0011 + 0.0012 = −0.08 = −1.67 0.048