A firm producing plate glass has developed a less expensive tempering process to allow for fireplaces to rise to a higher temperature without breaking. To test it, five different plates of glass were drawn randomly from a production run, then cut in half, with one-half tempered by the new process, and the other half by the old. The two halves then were heated until they broke, yielding the following data: Breaking Temperature New Old 475 485 446 435 495 493 493 496 416 423 Does the new process shows significant improvement over the old process? Assume 5% significance level. (Please specify the null and alternative hypotheses carefully) 2. In a large American university in 1969, the male and female professors were sample independently, yielding the following annual salaries (in thousand of dollars): Male Female 10 8 10 11 17 16 14 18 21 15 Test if there was wage inequality using the following methods: Assume 5% level of significance. 1. ANOVA 2. Two population t-test. 3. Are the two methods yield different conclusion? Why or Why not? 3. To see what difference class attendance made, a professor sampled grades from his large statistics class of 550 students. From the 230 students who attended class less than half the time (the \"irregulars\"), he took a random sample of 5 grades. From the remaining 320 students who attended at least half the time (the \"regulars\"), he took an independent random sample of 5 other grades: Irregulars Regulars 41 69 81 66 56 93 64 75 62 92 Does the data support the contention that \"it is worth 13 marks to come to class regularly? Assume 5% level of significance.(Please specify the null and alternative hypotheses carefully). 4. Why the F-distribution is positively skewed? Explain. . A firm producing plate glass has developed a less expensive tempering process to allow for fireplaces to rise to a higher temperature without breaking. To test it, five different plates of glass were drawn randomly from a production run, then cut in half, with one-half tempered by the new process, and the other half by the old. The two halves then were heated until they broke, yielding the following data: Breaking Temperature New Old 475 485 446 435 495 493 493 496 416 423 Does the new process shows significant improvement over the old process? Assume 5% significance level. (Please specify the null and alternative hypotheses carefully) H0: 1= 2 H A: 1 > 2 =0.05 SUMMARY Coun t 5 5 Groups New Old Test statistics= D.f=n1+n2-2=8 Sum 2325 2332 465466.4 1136.5 1199.8 + 5 5 Averag e 465 466.4 = -0.0647662 Varianc e 1136.5 1199.8 P-value=0.475 Critical value =t 0.05, 8=1.8595 Since test statistics is less than critical value, H0 is NOT rejected we conclude that there is no sufficient evidence to support the claim that the new process shows significant improvement over the old process In other words we conclude that new process shows NO significant improvement over the old process 2. In a large American university in 1969, the male and female professors were sample independently, yielding the following annual salaries (in thousand of dollars): Male Female 10 8 10 11 17 16 14 18 21 15 Test if there was wage inequality using the following methods: Assume 5% level of significance. H0: 1= 2 H A: 1 2 =0.05 Male X1 X21 Female X2 10 10 17 14 21 (x1) = 72 (x1)2 =5184 8 11 16 18 15 X22 100 100 289 196 441 64 121 256 324 225 (x2)= 68 x12 =1126 x22 =990 (x2)2 = 4624 SST= ( 1126+ 990 ) ( 72+ 68 )2 =156 10 5184 4624 ( 72+ 68 ) + 5 4 10 SSBetween= 2 =1.6 SSE=156-1.6=154.4 ANOVA Source of SS Variation 1.6 Between Groups Within Groups 154.4 156 Total df MS F P-value F crit 1 1.6 0.082902 0.7807 5.3177 8 19.3 9 Since the tests statistics F=0.0829 (with a p-value of 0.7807) is less than critical value F 0.05, 1, 8=5.318 H0 is not rejected. We conclude that there is no significant wage difference between males and female at 5% level of significant Two population t-test. H0: 1= 2 H A: 1 2 =0.05 Groups Male Female T= Count Sum 5 5 14.413.6 22.3 16.3 + 5 5 72 68 Averag e 14.4 13.6 Varianc e 22.3 16.3 = 0.2879 Critical value=2.306 Decision: Since tests statistics (0.2879) is less than critical value (2.306) H0 is NOT rejected We conclude that there is no significant wage difference between males and female at 5% level of significant Are the two methods yield different conclusion? Why or Why not? No, Since in both tests we have conclude that there is no significant in salaries between the two genders. 3. To see what difference class attendance made, a professor sampled grades from his large statistics class of 550 students. From the 230 students who attended class less than half the time (the \"irregulars\"), he took a random sample of 5 grades. From the remaining 320 students who attended at least half the time (the \"regulars\"), he took an independent random sample of 5 other grades: Irregulars Regulars 41 69 81 66 56 93 64 75 62 92 Does the data support the contention that \"it is worth 13 marks to come to class regularly? Assume 5% level of significance.(Please specify the null and alternative hypotheses carefully). Mean Variance Observations H0: 1- 2=0 Regulars 79 162.5 5 Irregulars 60.8 208.7 5 HA: 1 - 2=18 Test statistic N 1n 1 N 2n2 2 2 S 1 S 2 N 11 N 21 + n1 n2 T= ( X 1X 2)(u 1u 2) (7960.8)18 T= 3205 162.5 2305 208.7 + ( ) 3201 5 2301 5 ( ) = 0.02339 Critical value=2.30 Since 0.02339 2 =0.05 SUMMARY Coun t 5 5 Groups New Old Test statistics= D.f=n1+n2-2=8 Sum 2325 2332 465466.4 1136.5 1199.8 + 5 5 Averag e 465 466.4 = -0.0647662 Varianc e 1136.5 1199.8 P-value=0.475 Critical value =t 0.05, 8=1.8595 Since test statistics is less than critical value, H0 is NOT rejected we conclude that there is no sufficient evidence to support the claim that the new process shows significant improvement over the old process In other words we conclude that new process shows NO significant improvement over the old process 2. In a large American university in 1969, the male and female professors were sample independently, yielding the following annual salaries (in thousand of dollars): Male Female 10 8 10 11 17 16 14 18 21 15 Test if there was wage inequality using the following methods: Assume 5% level of significance. H0: 1= 2 H A: 1 2 =0.05 Male X1 X21 Female X2 10 10 17 14 21 (x1) = 72 (x1)2 =5184 8 11 16 18 15 X22 100 100 289 196 441 64 121 256 324 225 (x2)= 68 x12 =1126 x22 =990 (x2)2 = 4624 SST= ( 1126+ 990 ) ( 72+ 68 )2 =156 10 5184 4624 ( 72+ 68 ) + 5 4 10 SSBetween= 2 =1.6 SSE=156-1.6=154.4 ANOVA Source of SS Variation 1.6 Between Groups Within Groups 154.4 156 Total df MS F P-value F crit 1 1.6 0.082902 0.7807 5.3177 8 19.3 9 Since the tests statistics F=0.0829 (with a p-value of 0.7807) is less than critical value F 0.05, 1, 8=5.318 H0 is not rejected. We conclude that there is no significant wage difference between males and female at 5% level of significant Two population t-test. H0: 1= 2 H A: 1 2 =0.05 Groups Male Female T= Count Sum 5 5 14.413.6 22.3 16.3 + 5 5 72 68 Averag e 14.4 13.6 Varianc e 22.3 16.3 = 0.2879 Critical value=2.306 Decision: Since tests statistics (0.2879) is less than critical value (2.306) H0 is NOT rejected We conclude that there is no significant wage difference between males and female at 5% level of significant Are the two methods yield different conclusion? Why or Why not? No, Since in both tests we have conclude that there is no significant in salaries between the two genders. 3. To see what difference class attendance made, a professor sampled grades from his large statistics class of 550 students. From the 230 students who attended class less than half the time (the \"irregulars\"), he took a random sample of 5 grades. From the remaining 320 students who attended at least half the time (the \"regulars\"), he took an independent random sample of 5 other grades: Irregulars Regulars 41 69 81 66 56 93 64 75 62 92 Does the data support the contention that \"it is worth 13 marks to come to class regularly? Assume 5% level of significance.(Please specify the null and alternative hypotheses carefully). Mean Variance Observations H0: 1- 2=0 Regulars 79 162.5 5 Irregulars 60.8 208.7 5 HA: 1 - 2=18 Test statistic N 1n 1 N 2n2 2 2 S 1 S 2 N 11 N 21 + n1 n2 T= ( X 1X 2)(u 1u 2) (7960.8)18 T= 3205 162.5 2305 208.7 + ( ) 3201 5 2301 5 ( ) = 0.02339 Critical value=2.30 Since 0.02339 2 =0.05 SUMMARY Coun t 5 5 Groups New Old Test statistics= D.f=n1+n2-2=8 Sum 2325 2332 465466.4 1136.5 1199.8 + 5 5 Averag e 465 466.4 = -0.0647662 Varianc e 1136.5 1199.8 P-value=0.475 Critical value =t 0.05, 8=1.8595 Since test statistics is less than critical value, H0 is NOT rejected we conclude that there is no sufficient evidence to support the claim that the new process shows significant improvement over the old process In other words we conclude that new process shows NO significant improvement over the old process 2. In a large American university in 1969, the male and female professors were sample independently, yielding the following annual salaries (in thousand of dollars): Male Female 10 8 10 11 17 16 14 18 21 15 Test if there was wage inequality using the following methods: Assume 5% level of significance. H0: 1= 2 H A: 1 2 =0.05 Male X1 X21 Female X2 10 10 17 14 21 (x1) = 72 (x1)2 =5184 8 11 16 18 15 X22 100 100 289 196 441 64 121 256 324 225 (x2)= 68 x12 =1126 x22 =990 (x2)2 = 4624 SST= ( 1126+ 990 ) ( 72+ 68 )2 =156 10 5184 4624 ( 72+ 68 ) + 5 4 10 SSBetween= 2 =1.6 SSE=156-1.6=154.4 ANOVA Source of SS Variation 1.6 Between Groups Within Groups 154.4 156 Total df MS F P-value F crit 1 1.6 0.082902 0.7807 5.3177 8 19.3 9 Since the tests statistics F=0.0829 (with a p-value of 0.7807) is less than critical value F 0.05, 1, 8=5.318 H0 is not rejected. We conclude that there is no significant wage difference between males and female at 5% level of significant Two population t-test. H0: 1= 2 H A: 1 2 =0.05 Groups Male Female T= Count Sum 5 5 14.413.6 22.3 16.3 + 5 5 72 68 Averag e 14.4 13.6 Varianc e 22.3 16.3 = 0.2879 Critical value=2.306 Decision: Since tests statistics (0.2879) is less than critical value (2.306) H0 is NOT rejected We conclude that there is no significant wage difference between males and female at 5% level of significant Are the two methods yield different conclusion? Why or Why not? No, Since in both tests we have conclude that there is no significant in salaries between the two genders. 3. To see what difference class attendance made, a professor sampled grades from his large statistics class of 550 students. From the 230 students who attended class less than half the time (the \"irregulars\"), he took a random sample of 5 grades. From the remaining 320 students who attended at least half the time (the \"regulars\"), he took an independent random sample of 5 other grades: Irregulars Regulars 41 69 81 66 56 93 64 75 62 92 Does the data support the contention that \"it is worth 13 marks to come to class regularly? Assume 5% level of significance.(Please specify the null and alternative hypotheses carefully). Mean Variance Observations H0: 1- 2=0 Regulars 79 162.5 5 Irregulars 60.8 208.7 5 HA: 1 - 2=18 Test statistic N 1n 1 N 2n2 2 2 S 1 S 2 N 11 N 21 + n1 n2 T= ( X 1X 2)(u 1u 2) (7960.8)18 T= 3205 162.5 2305 208.7 + ( ) 3201 5 2301 5 ( ) = 0.02339 Critical value=2.30 Since 0.02339 2 =0.05 SUMMARY Coun t 5 5 Groups New Old Test statistics= D.f=n1+n2-2=8 Sum 2325 2332 465466.4 1136.5 1199.8 + 5 5 Averag e 465 466.4 = -0.0647662 Varianc e 1136.5 1199.8 P-value=0.475 Critical value =t 0.05, 8=1.8595 Since test statistics is less than critical value, H0 is NOT rejected we conclude that there is no sufficient evidence to support the claim that the new process shows significant improvement over the old process In other words we conclude that new process shows NO significant improvement over the old process 2. In a large American university in 1969, the male and female professors were sample independently, yielding the following annual salaries (in thousand of dollars): Male Female 10 8 10 11 17 16 14 18 21 15 Test if there was wage inequality using the following methods: Assume 5% level of significance. H0: 1= 2 H A: 1 2 =0.05 Male X1 X21 Female X2 10 10 17 14 21 (x1) = 72 (x1)2 =5184 8 11 16 18 15 X22 100 100 289 196 441 64 121 256 324 225 (x2)= 68 x12 =1126 x22 =990 (x2)2 = 4624 SST= ( 1126+ 990 ) ( 72+ 68 )2 =156 10 5184 4624 ( 72+ 68 ) + 5 4 10 SSBetween= 2 =1.6 SSE=156-1.6=154.4 ANOVA Source of SS Variation 1.6 Between Groups Within Groups 154.4 156 Total df MS F P-value F crit 1 1.6 0.082902 0.7807 5.3177 8 19.3 9 Since the tests statistics F=0.0829 (with a p-value of 0.7807) is less than critical value F 0.05, 1, 8=5.318 H0 is not rejected. We conclude that there is no significant wage difference between males and female at 5% level of significant Two population t-test. H0: 1= 2 H A: 1 2 =0.05 Groups Male Female T= Count Sum 5 5 14.413.6 22.3 16.3 + 5 5 72 68 Averag e 14.4 13.6 Varianc e 22.3 16.3 = 0.2879 Critical value=2.306 Decision: Since tests statistics (0.2879) is less than critical value (2.306) H0 is NOT rejected We conclude that there is no significant wage difference between males and female at 5% level of significant Are the two methods yield different conclusion? Why or Why not? No, Since in both tests we have conclude that there is no significant in salaries between the two genders. 3. To see what difference class attendance made, a professor sampled grades from his large statistics class of 550 students. From the 230 students who attended class less than half the time (the \"irregulars\"), he took a random sample of 5 grades. From the remaining 320 students who attended at least half the time (the \"regulars\"), he took an independent random sample of 5 other grades: Irregulars Regulars 41 69 81 66 56 93 64 75 62 92 Does the data support the contention that \"it is worth 13 marks to come to class regularly? Assume 5% level of significance.(Please specify the null and alternative hypotheses carefully). Mean Variance Observations H0: 1- 2=0 Regulars 79 162.5 5 Irregulars 60.8 208.7 5 HA: 1 - 2=18 Test statistic N 1n 1 N 2n2 2 2 S 1 S 2 N 11 N 21 + n1 n2 T= ( X 1X 2)(u 1u 2) (7960.8)18 T= 3205 162.5 2305 208.7 + ( ) 3201 5 2301 5 ( ) = 0.02339 Critical value=2.30 Since 0.02339