Solution Unit 2 Sem 2 PDF
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This document contains solutions to various unit 2 semester 2 problems. It details statistical tests including z-tests and t-tests.
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2. (Note : College material solution is not correct) Test Applicable : Z Test H0 : 1 = 2 [Method 1 is not better for processing the product] H1 : 1 > 2 [Method 1 is better for processing the product] Given :...
2. (Note : College material solution is not correct) Test Applicable : Z Test H0 : 1 = 2 [Method 1 is not better for processing the product] H1 : 1 > 2 [Method 1 is better for processing the product] Given : Population - 1 Population - 2 (Method - 1) (Method - 2) Mean 1 = 106 2 = 100 S.D. 1 = 12 2 = 10 Sample size n1 = 14 n2 = 14 Now, 1 - 2 S.E. of 12 22 144 100 Z = = + = + = 4.17 S.E. of 1 - 2 1 - 2 n1 n2 14 14 106 - 100 = 4.17 = 1.44 I Z I = 1.44 Since, I Z I = 1.44 < Z(5%, One) = 1.645, H0 is accepted. Conclusion : Method 1 is not better for processing the product. 3. Test Applicable : Z Test H0 : 1 = 2 [There is no significant difference between average life of two makes of bulbs] H1 : 1 2 [There is significant difference between average life of two makes of bulbs] Given : Population - 1 Population - 2 (Philips) (HMT) Mean 1 = 1500 2 = 1512 S.D. s1 = 60 s2 = 80 Sample size n1 = 50 n2 = 40 Now, 1 - 2 S.E. of s2 s2 3600 6400 Z = = 1 + 2 = + = 15.23 S.E. of 1 - 2 1 - 2 n1 n2 50 40 1500 - 1512 = 15.23 = - 0.7879 I Z I = 0.7879 Since, I Z I = 0.7879 < Z(10%, Two) = 1.645, H0 is accepted. Conclusion : There is no significant difference between average life of two makes of bulbs. 4. Shri Balaji Prakashan Test Applicable : Z Test H0 : 1 = 2 [There is no significant difference between average sales of the two salesmen] H1 : 1 2 [There is significant difference between average sales of the two salesmen] Given : Population - 1 Population - 2......................(Salesman......... Publisher..........-. A) of Educational (Salesman - B) Study Materials Mean 1 = 170 2 = 205 S.D. s1 = 20 s2 = 25 Sample size n1 = 200 n2 = 180 1 Now, 1 - 2 S.E. of s12 s22 400 625 Z = = + = + = 2.34 S.E. of 1 - 2 1 - 2 n1 n2 200 180 170 - 205 = 2.34 = 14.96 I Z I = 14.96 Since, I Z I = 14.96 > Z(5%, Two) = 1.96, H0 is rejected. Conclusion : There is significant difference between average sales of the two salesmen. 5. Test Applicable : Z Test H0 : 1 = 2 [There is no significant difference between average sales of the two states] H1 : 1 2 [There is significant difference between average sales of the two states] Given : Population - 1 Population - 2 (State - A) (State - B) Mean 1 = 2500 2 = 2200 S.D. s1 = 400 s2 = 550 Sample size n1 = 400 n2 = 400 Now, 1 - 2 S.E. of s2 s2 160000 302500 Z = = 1 + 2 = + = 34.0036 S.E. of 1 - 2 1 - 2 n1 n2 400 400 2500 - 2200 = 34.0036 = 8.8225 I Z I = 8.8225 Since, I Z I = 8.8225 > Z(1%, Two) = 2.58, H0 is rejected. Conclusion : There is significant difference between average sales of the two states. 6. Test Applicable : Z Test H0 : 1 = 2 [Male students who participate in college athletics are not taller than other male students] H1 : 1 2 [Male students who participate in college athletics are taller than other male students] Given : Population - 1 Population - 2 (Participants) (Nonparticipants) Mean 1 = 68.2 2 = 67.5 S.D. s1 = 2.5 s2 = 2.8 Sample size n1 = 50 n2 = 50 Now, Shri Balaji Prakashan 1 - 2 S.E. of s2 s2 6.25 7.84 Z = = 1 + 2 = + = 0.53 S.E. of 1 - 2 1 - 2 n1 n2 50 50 68.2 - 67.5 = 0.53.............................. Publisher of Educational Study Materials........ =....1.32 I Z I = 1.32 Since, I Z I = 1.32 < Z(5%, One) = 1.645, H0 is accepted. Conclusion : Male students who participate in college athletics are not taller than other male students. 2 8. Test Applicable : Z Test Let, a customer trying the product after seeing the advertisement, be a success H0 : P1 = P2 [Adevertisement A is not more effective than B] H1 : P1 P2 [Adevertisement A is not effective than B] Given : Population - 1 Population - 2 (Advertisement A) (Advertisement B) Sample size n1 = 60 n2 = 100 Proportion p 1 = 18 = 0.3 p 2 = 22 = 0.22 60 100 Z = p1 - p2 S.E. of S.E. of p1 - p2 p1 - p2 = pq 1 + 1 n1 n2 ( = (0.25) (0.75) 60 ) 1. + 1. 100 ( ) 0.3 - 0.22 = 0.07 = 0.07 Where : n1p1 + n2p2 18 + 22 p = = = 0.25 = 1.14 n1 + n2 160 I Z I = 1.14 q = 1 - p = 1 - 0.25 = 0.75 Since, I Z I = 1.14 < Z(5%, One) = 1.645, H0 is accepted. Conclusion : Adevertisement A is not more effective than B. 9. Test Applicable : Z Test Let, a person being recovered from the diseases, be a success H0 : P1 = P2 [The serum does not help to cure the disease] H1 : P1 P2 [The serum helps to cure the disease] Given : Population - 1 Population - 2 (Group A) (Group B) Sample size n1 = 100 n2 = 100 Proportion p 1 = 75 = 0.75 p 2 = 65 = 0.65 100 100 Z = p1 - p2 S.E. of S.E. of p1 - p2 p1 - p2 = pq (1n 1 + 1 n2 )= (0.7) (0.3) (1001. + 1001.) 0.75 - 0.65 = 0.06 = 0.06 Where : n1p1 + n2p2 75 + 65 p = = = 0.7 = 1.67 n1 + n2 200 I Z I = 1.67 q = 1 - p = 1 - 0.7 = 0.3 Since, I Z I = 1.67 > Z(5%, One) = 1.645, H0 is rejected. Conclusion : Shri Balaji Prakashan The serum helps to cure the disease. 10. Test Applicable : Z Test Let, a train arriving on time at the station, be a success.H. 0.. :....P. 1.. =....P. 2.. [There.......is..no.. significant... Publisher..........difference of Educational in the proportions Study arriving Materials on time at the two stations] H1 : P1 P2 [There is significant difference in the proportions arriving on time at the two stations] 3 Given : Population - 1 Population - 2 (Delhi) (New Delhi) Sample size n1 = 157 n2 = 172 Proportion p 1 = 74 = 0.47 p 2 = 65 = 0.38 157 172 Z = p1 - p2 S.E. of p1 - p2 S.E. of p 1 - p2 = pq n1 ( 1 + 1 n2 ) = (0.42) (0.58) 157 ( 1. + 1. 172 ) 0.47 - 0.38 = 0.054 = 0.054 Where : n1p1 + n2p2 74 + 65 p = = = 0.42 = 1.67 n1 + n2 329 I Z I = 1.67 q = 1-p = 1 - 0.42 = 0.58 Since, I Z I = 1.67 < Z(5%, Two) = 1.96, H0 is accepted. Conclusion : There is no significant difference in the proportions arriving on time at the two stations. 11. Test Applicable : Z Test Let, a person being a tea drinker, be a success H0 : P1 = P2 [There is no sufficient decrease in the consumption of tea due to increase in excise duty] H1 : P1 P2 [There is sufficient decrease in the consumption of tea due to increase in excise duty] Given : Population - 1 Population - 2 (Before) (After) Sample size n1 = 500 n2 = 600 Proportion p 1 = 400 = 0.8 p 2 = 400 = 0.67 500 600 Z = p1 - p2 S.E. of S.E. of p1 - p2 p1 - p2 = pq 1 + 1 n1 n2 ( ) = (0.73) (0.27) 1. + 1. 500 600 ( ) 0.8 - 0.67 = 0.027 = 0.027 Where : n1p1 + n2p2 400 + 400 p = = = 0.73 = 4.81 n1 + n2 1100 I Z I = 4.81 q = 1 - p = 1 - 0.73 = 0.27 Since, I Z I = 4.81 > Z(5%, One) = 1.645, H0 is rejected. Conclusion : There is sufficient decrease in the consumption of tea due to increase in excise duty. 12. Test Applicable : Z Test Let, a person being literate, be a success Shri Balaji Prakashan H0 : P1 = P2 [The difference between the proportions will remain hidden] H1 : P1 P2 [The difference between the proportions will not remain hidden] Given : Population - 1 Population - 2 (Town A) (Town B).....Sample....... size............. Publisher.......... n. 1...=... 1200 n2 of= Educational 900 Study Materials Proportion P1 = 0.3 P2 = 0.25 i.e. Q1 = 0.7 Q2 = 0.75 4 p1 - p2 S.E. of P1Q1 + P2Q2 (0.3) (0.7) + (0.25) (0.75) Z = = = S.E. of p1 - p2 p1 - p2 n1 n2 1200 900 0.3 - 0.25 = 0.02 = 0.02 = 2.5 I Z I = 2.5 Since, I Z I = 2.5 > Z(5%, Two) = 1.96, H0 is rejected. Conclusion : The difference between the proportions will not remain hidden. 13. Test Applicable : Z Test Let, a person being smoker, be a success H0 : P1 = P2 [There is no significant difference in smoking habit among men between two cities] H1 : P1 P2 [There is significant difference in smoking habit among men between two cities] Given : Population - 1 Population - 2 Sample size n1 = 1000 n2 = 1200 Proportion p 1 = 750 = 0.75 p 2 = 1000 = 0.83 1000 1200 Z = p1 - p2 S.E. of S.E. of p1 - p2 p1 - p2 = pq 1 + 1 n1 (n2 ) = (0.79) (0.21) 1. + 1. 1000 1200 ( ) 0.75 - 0.83 = 0.017 = 0.017 Where : n1p1 + n2p2 750 + 1000 p = = = 0.79 = - 4.71 n1 + n2 2200 I Z I = 4.71 q = 1 - p = 1 - 0.79 = 0.21 Since, I Z I = 4.71 > Z(5%, Two) = 1.96, H0 is rejected. Conclusion : There is significant difference in smoking habit among men between two cities. 15. Test Applicable : t Test H0 : 1 = 2 [The packets delivered at the end of the month are not heavier than that of delivered at the beginning of the month] H1 : 1 < 2 [The packets delivered at the end of the month are heavier than that of delivered at the beginning of the month] Given : Population - 1 Population - 2 (Beginning of the month) (End of the month) Mean 1 = 5.25 2 = 4.96 S.D. s1 = 1.2 s2 = 1.15 Sample size n1 = 20 n2 = 10 Now, 1- Shri Balaji Prakashan 2 (n1 - 1) s12 + (n2 - 1) s22 (19) 1.44 + (9) 1.32 t = s2 = = = 1.4014 s /n + /n 1 1 2 1 n 1 + n2 - 2 20 + 10 - 2 5.25 - 4.96. i.e. s = 1.184 = 1.184 /20 + /10 1 1 d.f. = n1 + n2 - 2 = 20 + 10 - 2 = 28............................... Publisher of Educational Study Materials........=... 0.632 I t I = 0.632 Since, I t I = 0.632 < t(5%, One, 28) = 1.071, H0 is accepted. Conclusion : The packets delivered at the end of the month are not heavier than that of delivered at the beginning of the month. 5 16. Test Applicable : t Test H0 : 1 = 2 [Sensitivity achieved by the formal program is not higher than that achieved under the informal program] H1 : 1 2 [Sensitivity achieved by the formal program is higher than that achieved under the informal program] Given : Population - 1 Population - 2 (Fromal) (Informal) Mean 1 = 92 2 = 84 S.D. s1 = 15 s2 = 19 Sample size n1 = 12 n2 = 15 Now, 1- 2 (n1 - 1) s12 + (n2 - 1) s22 (11) 225 + (14) 361 t = s2 = = = 301.16 s /n + /n 1 1 1 2 n 1 + n2 - 2 12 + 15 - 2 92 - 84. i.e. s = 17.35 = 17.35 /12 + /15 1 1 d.f. = n1 + n2 - 2 = 12 + 15 - 2 = 25 = 1.19 I t I = 1.19 Since, I t I= 1.19 < t(5%, One, 25) = 1.708, H0 is accepted. Conclusion : Sensitivity achieved by the formal program is not higher than that achieved under the informal program. 17. Test Applicable : t Test H0 : 1 = 2 [There is no significant difference between two population means] H1 : 1 2 [There is significant difference between two population means] Given : Population - 1 Population - 2 (Type A) (Type B) Mean 1 = 600 2 = 640 S.D. s1 = 11 s2 = 12 Sample size n1 = 9 n2 = 8 Now, 1- 2 (n1 - 1) s12 + (n2 - 1) s22 (8) 121 + (7) 144 t = s2 = = = 131.73 s /n + /n 1 1 1 2 n 1 + n2 - 2 9+8-2 600 - 640. i.e. s = 11.48 = 11.48 /9 + /8 1 1 d.f. = n1 + n2 - 2 = 9 + 8 - 2 = 15 = - 7.17 I t I = 7.17 Since, I t I = 7.17 > t(5%, Two, 15) = 2.131, H0 is rejected. Conclusion : There is significant difference between two population means. Shri Balaji Prakashan 18. Test Applicable : t Test H0 : 1 = 2 [The male customers do not spend less money on an average than the female....................customers]...................... Publisher of Educational Study Materials H1 : 1 2 [The male customers spend less money on an average than the female customers] Given : Population - 1 Population - 2 (Male) (Female) Mean 1 = 80 2 = 96 S.D. s1 = 17.50 s2 = 14.40 Sample size n1 = 25 n2 = 20 6 Now, 1 - 2 (n1 - 1) s12 + (n2 - 1) s22 (24)306.25 + (19)207.36 t = s2 = = = 262.55 s /n + 1/n 1 1 2 n 1 + n2 - 2 25 + 20 - 2 80 - 96. i.e. s = 16.20 = 16.2 1/25 + 1/20 d.f. = n1 + n2 - 2 = 25 + 20 - 2 = 43 = - 3.29 I t I = 3.29 Since, I t I = 3.29 > t(5%, One, 43) = 1.682, H0 is rejected. Conclusion : The male customers spend less money on an average than the female customers. 20. Test Applicable : Paired t Test H0 : 1 = 2 [Campaign cannot be judged to be a success] H1 : 1 2 [Campaign can be judged to be a success] _ d ; d.f. () = n - 1 = 6 - 1 = 5 t = s ( ) n = 3.5 (3.08/ 6 ) = 2.78 I t I = 2.78 Working : Sales before campaign Sales after campaign d d2 Shops xi yi y i - xi A 53 55 5 25 B 28 29 1 1 C 31 30 -1 1 D 48 55 7 49 E 50 56 6 36 F 42 45 3 9 _ 21 121 d 21 d = = = 3.5 n 6 _2 d2 - n d 121 - 6(12.25) sd2 = = = 9.5 s = 3.08 n-1 6 -1 Since, I t I = 2.78 > t(5%, One, 5) = 2.015, H0 is rejected. Conclusion : Campaign can be judged to be a success. Shri Balaji Prakashan 21. Test Applicable : Paired t Test H0 : 1 = 2 [Attending the crash course does not increase the writing speed] H1 : 1 2 [Attending the crash course increases the writing speed].......................................... Publisher of Educational Study Materials _ d ; d.f. () = n - 1 = 8 - 1 = 7 t = s ( ) n = 10.13 ( /8) 9.86 = 2.91 I t I = 2.91 7 Working : Before course After course d d2 xi yi y i - xi 81 97 16 256 75 72 -3 9 89 93 4 16 91 110 19 361 65 78 13 169 70 69 -1 1 90 115 25 625 64 72 8 64 _ 81 1501 d 81 d = = = 10.13 n 8 _2 d 2 - n d 1501 - 8(102.62) sd2 = = = 97.15 s = 9.86 n-1 8 -1 Since, I t I = 2.91 > t(5%, One, 5) = 2.015, H0 is rejected. Conclusion : Attending the crash course increases the writing speed. 22. Test Applicable : Paired t Test H0 : 1 = 2 [The use of gasoline additive does not increase the gasoline mileage] H1 : 1 2 [The use of gasoline additive increases the gasoline mileage] _ d ; d.f. () = n - 1 = 5 - 1 = 4 t = s ( ) n = 1.78 (1.59/ 5 ) = 2.50 I t I = 2.50 Working : Without With d d2 xi yi y i - xi 24.6 26.3 1.7 2.89 28.3 31.7 3.4 11.56 18.9 18.2 - 0.7 0.49 23.7 25.3 1.6 2.56 15.4 18.3 2.9 8.41 Shri Balaji Prakashan _ 8.9 25.91 d 8.9 d = = = 1.78 n 5 _2 d2 - n d 25.91 - 5(3.17).....s.d.. =...... n..-. 1......=........5. -1 = 2.515 ofEducational s = Study 1.59 Materials 2.......... Publisher Since, I t I = 2.50 < t(1%, One, 4) = 3.747, H0 is accepted. Conclusion : The use of gasoline additive does not increase the gasoline mileage. 8 24. (Note : College material solution is not correct) Test Applicable : F Test H0 : 1 = 2 [Women do not have greater variation in attitude on political issues than men] H1 : 1 > 2 [Women have greater variation in attitude on political issues than men] Given : Population - 1 Population - 2 (Women) (Men) Variance s12 = 120 s22 = 80 Sample size n1 = 41 n2 = 31 (n2 - 1) n1s1 ; d.f. () = [(n - 1), (n - 1)] = [(41 - 1), (31 - 1)] = (40, 30) 2 F = 1 2 (n1 - 1) n2s22 (31 - 1) (41) (120) = (41 - 1) (31) (80) = 1.49 Since, F = 1.49 < FTab = 1.79, H0 is accepted. Conclusion : Women do not have greater variation in attitude on political issues than men. 25. Test Applicable : F Test H0 : 1 = 2 [Pipes produced by two processes A and B have same variability] H1 : 1 2 [Pipes produced by two processes A and B have different variability] Given : Population - 1 Population - 2 (Process - B) (Process - A) Variance s1 = (3.8) = 14.44 2 2 s22 = (2.9)2 = 8.41 Sample size n1 = 16 n2 = 23 (n2 - 1) n1s1 ; d.f. () = [(n - 1), (n - 1)] = [(16 - 1), (23 - 1)] = (15, 22) 2 F = 1 2 (n1 - 1) n2s22 (23 - 1) (16) (14.44) = (16 - 1) (23) (8.41) = 1.75 Since, F = 1.75 < FTab = 2.498, H0 is accepted. Conclusion : Pipes produced by two processes A and B have same variability. 26. Test Applicable : F Test H0 : 1 = 2 [There is no significant difference between two population variances] H1 : 1 2 [There is significant difference between two population variances] Given : Population - 1 Population - 2 Variance s1 = (2) = 4 2 2 s22 = (1.9)2 = 3.61 Shri Balaji Prakashan Sample size n1 = 9 n2 = 13 (n2 - 1) n1s12 ; d.f. () = [(n - 1), (n - 1)] = [(9 - 1), (13 - 1)] = (8, 12) F = 1 2 (n1 - 1) n2s22 (13 - 1) (9) (4).........=... (9...-.1)..(13)..................... Publisher of Educational Study Materials....(3.61) = 1.75 Since, F = 1.15 < FTab = 3.51, H0 is accepted. Conclusion : There is no significant difference between two population variances. 9 28. Test Applicable : ANOVA F Test (One Way Classification) H0 : 1 = 2 = 3 [Average life time of three brands of tyres are equal] H1 : atleast one i j [Average life time of three brands of tyres are not equal] : ANOVA Table : Source d.f. Sum of Squares (SS) Mean Sum of Square (MSS) = SS Fcal d.f. Between 3 - 1 = 2 BSS 13.9 BSS = 13.9 MBSS = = = 6.95 k-1 2 6.95 Fcal = = 2.12 Error ESS 36.1 3.28 14 - 3 = 11 ESS = 36.1 MESS = = = 3.28 (Within) n-k 11 d.f = (2, 11) Total 14 - 1 = 13 TSS = 50 Since, F = 2.12 < F[0.05, (2, 11)] = 3.98, H0 is accepted. Conclusion : Average life time of three brands of tyres are equal. Working : Calculation for various SS Brand Lif-time ’000 miles (Yij) ni Total (Ti) Ti2 Ti2 /ni A 45 42 43 44 42 5 216 46656 9331.2 B 41 40 42 43 4 166 27566 6891.5 C 44 42 38 43 39 5 206 42436 8487.2 Total n = 14 G = 588 24709.9 = G = (588) = 24696 2 2 CF n 14 TSS = Yij2 - CF = 452 + 422 + 432 +......... + 392 - 24696 = 50 BSS = (Ti /ni) - CF = 24709.9 - 24696 2 = 13.9 ESS = TSS - BSS = 50 - 13.9 = 36.1 29. Test Applicable : ANOVA F Test (One Way Classification) H0 : 1 = 2 = 3 = 4 [Effect of fertilizers does not differ significantly] H1 : atleast one i j [Effect of fertilizers differs significantly] : ANOVA Table : Source d.f. Sum of Squares (SS) Mean Sum of Square (MSS) = SS Fcal d.f. Fertilizerz 4 - 1 = 3 BSS 2940 BSS = 2940 MBSS = = = 980 k-1 3 980 Fcal = = 5.99 within ESS 3272 163.6 24 - 4 = 20 ESS = 3272 MESS = = = 163.6 groups n-k 20 d.f = (3, 20) Shri Balaji Prakashan Total 24 - 1 = 23 TSS = 6212 Since, F = 5.99 > F[0.05, (3, 20)] = 3.10, H0 is rejected. Conclusion : Effect of fertilizers differs significantly........................................... Publisher of Educational Study Materials 30. Test Applicable : ANOVA F Test (One Way Classification) H0 : 1 = 2 = 3 [Three salesmen do not differ in their selling ability significantly] H1 : atleast one i j [Three salesmen differ in their selling ability significantly] 10 : ANOVA Table : Source d.f. Sum of Squares (SS) Mean Sum of Square (MSS) = SS Fcal d.f. Between 3 - 1 = 2 BSS 1000 500. BSS = 1000 MBSS = = = 500 Fcal = k-1 2 272.73 Error ESS 3000 = 1.83 14 - 3 = 11 ESS = 3000 MESS = = = 272.73 (Within) n-k 11 d.f = (2, 11) Total 14 - 1 = 13 TSS = 4000 Since, F = 1.83 < F[0.05, (2, 11)] = 3.98, H0 is accepted. Conclusion : Three salesmen do not differ in their selling ability significantly. Working : Calculation for various SS Salesman Sales (Yij) ni Total (Ti) Ti2 Ti2 /ni A 30 40 30 50 0 5 150 22500 4500 B 60 30 30 40 4 160 25600 6400 C 70 30 40 60 50 5 250 62500 12500 Total n = 14 G = 560 23400 = G = (560) = 22400 2 2 CF n 14 TSS = Yij2 - CF = 302 + 402 + 302 +......... + 502 - 22400 = 4000 BSS = (Ti2 /ni) - CF = 23400 - 22400 = 1000 ESS = TSS - BSS = 4000 - 1000 = 3000 31. Test Applicable : ANOVA F Test (One Way Classification) H0 : 1 = 2 = 3 = 4 [There is no significant difference in performance of the four machines] H1 : atleast one i j [There is significant difference in performance of the four machines] : ANOVA Table : Source d.f. Sum of Squares (SS) Mean Sum of Square (MSS) = SS Fcal d.f. Between 4 - 1 = 3 BSS 540.69 180.23 BSS = 540.69 MBSS = = = 180.23 Fcal = k-1 3 7.15 Error ESS 85.75 = 25.21 16 - 4 = 12 ESS = 85.75 MESS = = = 7.15 (Within) n-k 20 d.f = (3, 12) Total 16 - 1 = 15 TSS = 626.44 Since, F = 25.21 < F[0.01, (3, 112] = 7.59, H0 is rejected. Conclusion : There is significant difference in performance of the four machines. Shri Balaji Prakashan Working : Calculation for various SS Machine No. of flowers per 100 sq. met. (Yij) ni Total (Ti) Ti2 Ti2 /ni A 9 8 11 12 4 40 1600 400.......................................... Publisher of Educational Study Materials B 6 8 10 4 4 28 784 196 C 14 12 18 9 4 53 2809 702.25 D 20 22 25 23 4 90 8100 2025 Total n = 16 G = 211 3323.25 CF = G 2 = (211) 2 = 2782.56 n 16 11 TSS = Yij2 - CF = 92 + 82 + 112 +......... + 232 - 2782.56 = 626.44 BSS = (Ti2 /ni) - CF = 3323.25 - 2728.56 = 540.69 ESS = TSS - BSS = 626.44 - 540.69 = 85.75 32. Test Applicable : Z Test H0 : 1 = 2 [It cannot be concluded that girls are more intelligent than boys] H1 : 1 2 [It canbe concluded that girls are more intelligent than boys] Given : Population - 1 Population - 2 (Girls) (Boys) Mean 1 = 84 2 = 81 S.D. s1 = 10 s2 = 12 Sample size n1 = 121 n2 = 81 Now, 1 - 2 S.E. of s2 s2 100 144 Z = = 1 + 2 = + = 1.602 S.E. of 1 - 2 1 - 2 n1 n2 121 81 84 - 81 = 1.602 = 1.87 I Z I = 1.87 Since, I Z I = 1.87 > Z(5%, One) = 1.645, H0 is rejected. Conclusion : It canbe concluded that girls are more intelligent than boys. 33. Test Applicable : Z Test Let, a person being recovered from the diseases, be a success H0 : P1 = P2 [The serum does not help to cure the disease] H1 : P1 P2 [The serum helps to cure the disease] Given : Population - 1 Population - 2 (Group A) (Group B) Sample size n1 = 100 n2 = 100 Proportion p1 = 75 = 0.75 p 2 = 65 = 0.65 100 100 Z = p1 - p2 S.E. of S.E. of p1 - p2 p1 - p2 = pq 1 + 1 n1 ( n2 )= (0.7) (0.3) (1001. + 1001.) 0.75 - 0.65 = 0.06 = 0.06 Where : n1p1 + n2p2 75 + 65 p = = = 0.7 = 1.67 n1 + n2 200 I Z I = 1.67 q = 1-p = 1 - 0.7 = 0.3 Shri Balaji Prakashan Since, I Z I = 1.67 < Z(1%, One) = 2.33, H0 is accepted. Conclusion : The serum does not help to cure the disease. 34........................................... Publisher of Educational Study Materials Test Applicable : Z Test Let, an animal responding to the drug, be a success H0 : P1 = P2 [There is no significant difference between efffectiveness of the two drugs] H1 : P1 P2 [There is significant difference between efffectiveness of the two drugs] 12 Given : Population - 1 Population - 2 (Group 1) (Group 2) Sample size n1 = 100 n2 = 90 Proportion p 1 = 71 = 0.71 p 2 = 58 = 0.64 100 90 Z = p1 - p2 S.E. of S.E. of p1 - p2 p1 - p2 = pq 1 + 1 n1 ( n2 ) = (0.68) (0.32) 1. + 1. 100 90 ( ) 0.71 - 0.64 = 0.068 = 0.068 Where : n1p1 + n2p2 71 + 58 p = = = 0.68 = 1.03 n1 + n2 190 I Z I = 1.03 q = 1 - p = 1 - 0.68 = 0.32 Since, I Z I = 1.03 < Z(5%, Two) = 1.96, H0 is accepted. Conclusion : There is no significant difference between efffectiveness of the two drugs. 35. Test Applicable : Paired t Test H0 : 1 = 2 [There is no change in ability of trainee after training] H1 : 1 2 [There is change in ability of trainee after training] _ d ; d.f. () = n - 1 = 9 - 1 = 8 t = s ( )n = 5 ( /9) 9.206 = 1.63 I t I = 1.63 Working : Score before training Score after training d d2 Trainee xi yi y i - xi 1 75 70 -5 25 2 70 77 7 49 3 46 57 11 121 4 68 60 -8 64 5 68 79 11 121 6 43 64 21 441 7 55 55 0 0 8 68 77 9 81 9 77 76 -1 1 Shri Balaji Prakashan _ 45 903 d 45 d = = = 5 n 9 _2 d - n d 2 903 - 9(25).... s. d...=......n. -..1.....=......9. -1.......... =... 84.75 s = 9.206Study Materials of Educational 2 Publisher Since, I t I = 1.63 < t(5%, Two, 8) = 2.306, H0 is accepted. Conclusion : There is no change in ability of trainee after training. 13 36. Test Applicable : t Test H0 : 1 = 2 [There is no reduction in mean total cholesterol in patients taking the new drug] H1 : 1 < 2 [There is reduction in mean total cholesterol in patients taking the new drug] Given : Population - 1 Population - 2 (New Drug) (Placebo) Mean 1 = 195.9 2 = 227.4 S.D. s1 = 28.7 s2 = 30.3 Sample size n1 = 15 n2 = 15 Now, 1- 2 (n1 - 1) s12 + (n2 - 1) s22 (14) 823.69 + (14) 918.09 t = s2 = = = 870.89 s /n + /n 1 1 1 2 n 1 + n2 - 2 15 + 15 - 2 195.9 - 227.4. i.e. s = 29.51 = 29.51 /15 + /15 1 1 d.f. = n1 + n2 - 2 = 15 + 15 - 2 = 28 = - 2.92 ItI = 2.92 Since, I t I = 2.92 > t(1%, One, 28) = 2.467, H0 is rejected. Conclusion : There is reduction in mean total cholesterol in patients taking the new drug. 37. Test Applicable : Z Test H0 : 1 = 2 [Intensive tutoring is not more effective than paced tutoring] H1 : 1 > 2 [Intensive tutoring is more effective than paced tutoring] Given : Population - 1 Population - 2 (Intensive) (Paced) Mean 1 = 46.31 2 = 42.79 S.D. s1 = 6.44 s2 = 7.52 Sample size n1 = 32 n2 = 38 Now, 1 - 2 S.E. of s2 s2 41.47 56.55 Z = = 1 + 2 = + = 1.67 S.E. of 1 - 2 1 - 2 n1 n2 32 38 46.31 - 42.79 = 1.67 = 2.11 I Z I = 2.11 Since, I Z I = 2.11 > Z(5%, One) = 1.645, H0 is rejected. Conclusion : Intensive tutoring is more effective than paced tutoring. 38. Shri Balaji Prakashan Test Applicable : Paired t Test H0 : 1 = 2 [There is no evidence that radial tyres produce better fuel economy] H1 : 1 2 [There is evidence that radial tyres produce better fuel economy].........._................................ Publisher of Educational Study Materials d ; d.f. () = n - 1 = 12 - 1 = 11 t = s ( ) n = - 0.142 (0.2/ 12 ) = - 2.46 I t I = 2.46 14 Working : Radial Belted d d2 Gas eco xi yi y i - xi 1 4.2 4.1 - 0.1 0.01 2 4.7 4.9 0.2 0.04 3 6.6 6.2 - 0.4 0.16 4 7.0 6.9 - 0.1 0.01 5 6.7 6.8 0.1 0.01 6 4.5 4.4 - 0.1 0.01 7 5.7 5.7 0 0 8 6.0 5.8 - 0.2 0.04 9 7.4 6.9 - 0.5 0.25 10 4.9 4.7 - 0.2 0.04 11 6.1 6.0 0.1 0.01 12 5.2 4.9 - 0.3 0.09 _ - 1.7 0.67 d - 1.7 d = = = - 0.142 n 12 _2 d2 - n d 0.67 - 12(0.02) sd2 = = = 0.04 s = 0.2 n-1 12 -1 Since, I t I = 2.46 > t(5%, One, 11) = 1.796, H0 is rejected. Conclusion : There is evidence that radial tyres produce better fuel economy. 39. Test Applicable : Paired t Test H0 : 1 = 2 [Hypnotism is not effective in reducing pain] H1 : 1 2 [Hypnotism is effective in reducing pain] _ d ; d.f. () = n - 1 = 8 - 1 = 7 t = s ( )n = - 3.1 (2.91/ 8 ) = - 3.01 I t I = 3.01 Working : Score before hypnotism Score after hypnotism d d2 Subject xi yi y i - xi A 6.6 6.8 0.2 0.04 Shri Balaji Prakashan B 6.5 2.4 - 4.1 16.81 C 9.0 7.4 - 1.6 2.56 D 10.3 8.5 - 1.8 3.24 E 11.3 8.1 - 3.2 10.24.................. Publisher....F....................8.1 6.1 of Educational - 2.0 Study Materials 4.0 G 6.3 3.4 - 2.7 7.29 H 11.6 2.0 - 9.6 92.16. _ - 24.8 136.34 d - 24.8 d = = = - 3.1 n 8 _2 d2 - n d 136.34 - 8(9.61) sd2 = = = 8.49 s = 2.91 n-1 8 -1 15 Since, I t I = 3.01 > t(5%, One, 7) = 1.895, H0 is rejected. Conclusion : Hypnotism is effective in reducing pain. 40. Test Applicable : Z Test Let, a person saying ‘yes’ to a survey question, be a success H0 : P1 = P2 [Smokers and non-smokers do not differ significantly in their opinion] H1 : P1 P2 [Smokers and non-smokers differ significantly in their opinion] Given : Population - 1 Population - 2 (Smokers) (Non-smokers) Sample size n1 = 605 n2 = 195 Proportion p 1 = 300 = 0.496 p 2 = 41 = 0.21 605 195 Z = p1 - p2 S.E. of S.E. of p1 - p2 p1 - p2 = pq 1 + 1 n1 (n2 ) = (0.43) (0.57) 1. + 1. 605 195 ( ) 0.496 - 0.21 = 0.04 = 0.04 Where : n1p1 + n2p2 300 + 41 p = = = 0.43 = 7.15 n1 + n2 800 I Z I = 7.15 q = 1 - p = 1 - 0.426 = 0.57 Since, I Z I = 7.15 > Z(5%, Two) = 1.96, H0 is rejected. Conclusion : Smokers and non-smokers differ significantly in their opinion. 41. Test Applicable : Z Test Let, a patient getting reduction in pain of 3+ scale point, be a success H0 : P1 = P2 [New pain reliever is not more effective than standard pain reliever] H1 : P1 P2 [New pain reliever is more effective than standard pain reliever] Given : Population - 1 Population - 2 (New) (Standard) Sample size n1 = 50 n2 = 50 Proportion p 1 = 23 = 0.46 p 2 = 17 = 0.34 50 50 Z = p1 - p2 S.E. of p1 - p2 p1 - p2 S.E. of ( = pq ) 1 + 1 n1 n2 ( = (0.4) (0.6) 50) 1. + 1. 50 0.46 - 0.34 = 0.1 = 0.1 Where : n1p1 + n2p2 23 + 17 p = = = 0.4 = 1.2 n1 + n2 100 Shri Balaji Prakashan I Z I = 1.2 q = 1 - p = 1 - 0.4 = 0.6 Since, I Z I = 1.2 < Z(1%, One) = 2.33, H0 is accepted. Conclusion : New pain reliever is more effective than standard pain reliever. 42.......................................... Publisher of Educational Study Materials. Test Applicable : Z Test H0 : 1 = 2 [Engine 2 has not higher RPM than Engine 1] H1 : 1 < 2 [Engine 2 has higher RPM than Engine 1] 16 Given : Population - 1 Population - 2 (Engine - 1) (Engine - 2) Mean 1 = 1500 2 = 1600 S.D. 1 = 50 2 = 60 Sample size n1 = 14 n2 = 16 Now, 1 - 2 S.E. of 12 22 2500 3600 Z = = + = + = 20.09 S.E. of 1 - 2 1 - 2 n1 n2 14 16 1500 - 1600 = 20.09 = - 4.98 I Z I = 4.98 Since, I Z I = 4.98 > Z(5%, One) = 1.645, H0 is rejected. Conclusion : Engine 2 has higher RPM than Engine 1. 43. Test Applicable : F Test H0 : 1 = 2 [1st instructor’s variance is not larger than that of 2nd] H1 : 1 > 2 [1st instructor’s variance is larger than that of 2nd] Given : Population - 1 Population - 2 (1st instructor) (2nd instructor) Variance s12 = 62.3 s22 = 59.9 Sample size n1 = 30 n2 = 30 (n2 - 1) n1s1 ; d.f. () = [(n - 1), (n - 1)] = [(30 - 1), (30 - 1)] = (29, 29) 2 F = 1 2 (n1 - 1) n2s22 (30 - 1) (30) (62.3) = (30 - 1) (30) (59.9) = 1.04 Since, F = 1.04 < FTab = 1.6199, H0 is accepted. Conclusion : 1st instructor’s variance is not larger than that of 2nd. 44. Test Applicable : ANOVA F Test (One Way Classification) H0 : 1 = 2 = 3 [Three groups do not differ in their level of intelligence] H1 : atleast one i j [Three groups differ in their level of intelligence] : ANOVA Table : Source d.f. Sum of Squares (SS) Mean Sum of Square (MSS) = SS Fcal d.f. BSS 0.71 Shri Balaji Prakashan Between 3 - 1 = 2 BSS = 0.71 MBSS = = = 0.355 k-1 2 0.355 Fcal = = 0.16 Error ESS 25 2.27 14 - 3 = 11 ESS = 25 MESS = = = 2.27 (Within) n-k 11 d.f = (2, 11) Total........14...-.1. =..13.......TSS.... =...25.71............. Publisher of Educational Study Materials Since, F = 0.16 < F[0.05, (2, 11)] = 3.98, H0 is accepted. Conclusion : Three groups do not differ in their level of intelligence. 17 Working : Calculation for various SS Stream Intelligence Score (Yij) ni Total (Ti) Ti2 Ti2 /ni Arts 15 14 13 12 11 5 65 4225 845 Commerce 12 14 11 13 4 50 2500 625 Science 12 15 14 13 11 5 65 4225 845 Total n = 14 G = 180 2315 = G = (180) = 2314.29 2 2 CF n 14 TSS = Yij2 - CF = 152 + 142 + 132 +......... + 112 - 2314.29 = 25.71 BSS = (Ti /ni) - CF = 2315 - 2314.29 = 0.71 2 ESS = TSS - BSS = 25.71 - 0.71 = 25 45. Test Applicable : ANOVA F Test (One Way Classification) H0 : 1 = 2 = 3 = 4 [Mean effect of all the treatment is same] H1 : atleast one i j [Mean effect of all the treatment is not same] : ANOVA Table : Source d.f. Sum of Squares (SS) Mean Sum of Square (MSS) = SS Fcal d.f. Between 4 - 1 = 3 BSS 82.4 BSS = 82.4 MBSS = = = 27.47 k-1 3 27.47 Fcal = = 1.56 Error ESS 281.5 17.59 20 - 4 = 16 ESS = 281.5 MESS = = = 17.59 n-k 16 d.f = (3, 16) Total 20 - 1 = 19 TSS = 364 Since, F = 1.56 < F[0.05, (3, 16)] = 3.239, H0 is accepted. Conclusion : Mean effect of all the treatment is same. Shri Balaji Prakashan.......................................... Publisher of Educational Study Materials SHRI BALAJI # 9924040043 18
