Ch-squared tests 6D 1 a H 0 : The data can be modelled by a Po() dstrbuton. H 1 : The data cannot be modelled by Po() dstrbuton. The observed and expected results are shown n the table. The last two columns (for 5 and >5) have been combned to get all values to be greater than 5. x 0 1 3 4 5 Total Observed (O ) 1 3 4 4 1 5 100 xpected ( ) 13.53 7.07 7.07 18.04 9.0 5.7 100 ( O ) 0.1730 0.6119 0.388 1.9690 0.9843 0.0138 4.101 Note that the values found for the test statstc (X ) wll vary accordng to any roundng you do. The test statstc must be calculated wth suffcent accuracy to ensure that the ch-squared test s correct. A range of answers that encompass sensble degrees of accuracy would be accepted n an examnaton. The number of degrees of freedom ν = 6 1= 5 (there are sx cells after combnng the last two columns wth a sngle constrant on the total that the frequences agree). From the tables: χ 5 (5%)=11.070 As 4.101 s less than 11.070, there s nsuffcent evdence to reject H 0 at the 5% level. The data may be modelled by Po(). b If λ s calculated, then ths becomes another constrant and there would be one less degree of the freedom; ν = 6 = 4 H 0 : The data can be modelled by a dscrete unform dstrbuton. H 1 : The data cannot be modelled by a dscrete unform dstrbuton. The number of degrees of freedom ν = 5 (sx data cells wth a sngle constrant on the total) From the tables: χ 5 (5%)=11.070 xpected frequency= ( ) 15+ 3+19+ 0+14+11 = 10 6 6 =17 O = 1 17 (15 17) + (3 17) + (19 17) + (0 17) + (14 17) + (11 17) ( ) = 1 98 (4+ 36+ 4+ 9+9+36)= 17 17 = 5.765 As 5.765 s less than 11.070, there s nsuffcent evdence to reject H 0 at the 5% level. The data may be modelled by a dscrete unform dstrbuton. Pearson ducaton Ltd 018. Copyng permtted for purchasng nsttuton only. Ths materal s not copyrght free. 1
3 a X = 1 13+ 9+3 13 15+13+ 9+13 = 70 50 =1.4 b H 0 : The data can be modelled by Po(1.4). H 1 : The data cannot be modelled by Po(1.4). As λ=1.4 the expected frequences must be calculated usng these equatons: P(X =)= e 1.4 1.4 and! = 50P(X =) as there are 50 observatons n the data Calculate the probablty n the fnal column (for x 5) by summng the other probabltes and subtractng from 1. The observed and expected results are: x 0 1 3 4 5 Total Observed (O ) 15 13 9 13 0 0 50 P(X = ) 0.466 0.345 0.417 0.118 0.0395 0.0143 1 xpected ( ) 1.330 17.6 1.083 5.639 1.974 0.71 50 The last 3 classes must be combned so all the values are 5 or more. Ths gves: x 0 1 3 Total Observed (O ) 15 13 9 13 50 P(X = ) 0.466 0.345 0.417 0.1666 1 xpected ( ) 1.330 17.6 1.083 8.33 50 ( O ) 0.578 1.0514 0.7875.6181 5.035 The number of degrees of freedom ν = (four data cells wth two constrants as λ s estmated by calculaton) From the tables: χ (10%)= 4.605 As 5.035 s greater than 4.605, H 0 should be rejected at the 10% level. The data cannot be modelled by Po(1.4). Pearson ducaton Ltd 018. Copyng permtted for purchasng nsttuton only. Ths materal s not copyrght free.
1 6+ 36+3 0+ 4 10+5 6+ 6 4 a r = 6+ 36+ 0+10+ 6+ meanr =np So.4=6p p=0.4 = 40 100 =.4 Ths s one constrant snce a parameter p has been estmated by calculaton. b H 0 : The data can be modelled by B(6, 0.4) H 1 : The data cannot be modelled by B(6, 0.4) Fnd the expected frequences by multplyng the total frequency 100 (ths s a second constrant) by the probablty P(X = ) usng the probablty equaton for a bnomal random varable. 0 6 ( X = 0) = 100 P( X = 0) = 100 0.4 0.6 = 4.666 0 1 5 ( X = 1) = 100 P( X = 1) = 100 0.4 0.6 = 18.66 1 X = = X = = = 3 3 ( X = 3) = 100 P( X = 3) = 100 0.4 0.6 = 7.648 3 4 ( X = 4) = 100 P( X = 4) = 100 0.4 0.6 13.84 4 = 5 1 ( X = 5) = 100 P( X = 5) = 100 0.4 0.6 = 3.6864 5 6 0 ( X = 6) = 100 P( X = 6) = 100 0.4 0.6 = 0.4096 6 4 ( ) 100 P( ) 100 0.4 0.6 31.104 Combne to get 5 Combne to get 5 After combnng the relevant cells, ths gves: x 1 3 4 Total Observed (O ) 6 36 0 18 100 xpected ( ) 3.38 31.104 7.648 17.9 100 ( O ) 0.3061 0.7707.1156 0.0004 3.197 The number of degrees of freedom ν = (four data cells wth two constrants as p s estmated by calculaton) From the tables: χ (5%)=5.991 As 3.19 s less than 5.991, there s nsuffcent evdence to reject H 0 at the 5% level. The data may be modelled by B(6, 0.4) Pearson ducaton Ltd 018. Copyng permtted for purchasng nsttuton only. Ths materal s not copyrght free. 3
5 H 0 : The rate of accdents s constant at the factores. H 1 : The rate of accdents sn t constant at the factores. Total number of accdents= 81 Total number of employees= 15 (thousand) 81 Mean rate of accdents= = 5.4 ( per thousand) 15 The calculaton of the mean rate s one constrant Multply the number of employees n each factory by the mean rate of accdents to get the expected frequences of accdents. The observed and expected results are: Factory A B C D Total Observed (O ) 14 5 8 1 81 xpected ( ) 1.6 16. 7 5.4 10.8 81 ( O ) 0.0074 0.988 0.1481 1.519 0.1333 1.8395 There are 5 cells and 1 constrant, so the number of degrees of freedom s 5 1 = 4 From the tables: χ 4 (5%)=9.488 As 1.84 s less than 9.488, there s nsuffcent evdence to reject H 0 at the 5% level. Ths supports the hypothess that accdents occur at a constant rate at the factores. Pearson ducaton Ltd 018. Copyng permtted for purchasng nsttuton only. Ths materal s not copyrght free. 4
6 H 0 : The data can be modelled by a Posson dstrbuton. H 1 : The data cannot be modelled by Posson dstrbuton. Total frequency = +8+15+18+14+13+7+3= 80 1 8+ 15+ 3 18+ 4 14+5 13+6 7+7 3 Mean=λ= 80 Calculate the expected frequences as follows: 0 = 80 P(X = 0)=80 e 3.45 3.45 0 0! =.540 = 76 80 = 3.45 1 = 80 P(X =1)=80 e 3.45 3.45 1 = 8.76 1! Smlarly =15.114, 3 =17.381, 4 =14.991, 5 =10.344, 6 = 5.948, 7 =.931 = 8 80 ( + 0 + + 1 7) = 1.989 To get values for greater than 5, combne the frst two cells and the last three cells: x 1 3 4 5 6 Total Observed (O ) 10 15 18 14 13 10 80 xpected ( ) 11.30 15.114 17.381 14.991 10.344 10.868 80 ( O ) 0.1500 0.0008 0.00 0.0655 0.068 9 0.0693 0.990 The number of degrees of freedom ν = 4 (sx data cells wth two constrants as λ s estmated by calculaton) From the tables: χ 4 (5%)=9.488 As 0.99 s less than 9.488, there s nsuffcent evdence to reject H 0 at the 5% level. The data may be modelled by a Posson dstrbuton. Pearson ducaton Ltd 018. Copyng permtted for purchasng nsttuton only. Ths materal s not copyrght free. 5
7 a Breakdowns occur sngly, ndependently and at random. They occur at a constant average rate. b H 0 : The data can be modelled by a Posson dstrbuton. H 1 : The data cannot be modelled by Posson dstrbuton. Total frequency = 50+ 4+1+ 9+5=100 1 4+ 1+3 9+ 4 5 Mean=λ= = 95 100 100 = 0.95 Calculate the expected frequences as follows: 0 =100 P(X = 0)=100 e 0.95 0.95 0 = 38.674 0! 1 =100 P(X =1)=100 e 0.95 0.95 1 = 36.740 1! Smlarly =17.45, 3 = 5.56, 4 =1.315 There s no need to go further, as further terms are extremely small. Fnd 3 to get all the values to be 5 or more. x 0 1 3 Total Observed (O ) 50 4 1 14 100 xpected ( ) 38.674 36.740 17.45 7.134 100 ( O ) 3.317 4.4188 1.703 6.608 16.05 The number of degrees of freedom ν = (four data cells wth two constrants as λ s estmated by calculaton) From the tables: χ (5%)=5.991 As 16.05 s greater than 5.991, reject H 0 at the 5% level. The data cannot be modelled by Po(0.95) 8 H 0 : The przes are unformly dstrbuted. H 1 : The przes are not unformly dstrbuted. Total frequency= 505, so expected frequency for each class= 505 10 = 50.5 ( ) ( ) ( ) ( O ) 56 50.5 49 50.5 50 50.5 Test statstc ( X ) = = + +... + = 10.74 50.5 50.5 50.5 The number of degrees of freedom ν = 9 (ten data cells wth a sngle constrant on the total) From the tables: χ 9 (5%)=16.919 As 10.74 s less than 16.919, there s nsuffcent evdence to reject H 0 at the 5% level. There s no reason to doubt that the przes are dstrbuted unformly. Pearson ducaton Ltd 018. Copyng permtted for purchasng nsttuton only. Ths materal s not copyrght free. 6
9 a The expected number of ltters s modelled by B(8, 0.5) As total frequency = 00 8 3 5 R= 00 P( X = 3) = 00 0.5 0.5 = 43.75 3 8 S = X = = = 4 4 4 00 P( 4) 00 0.5 0.5 54.69 8 T = X = = = 5 5 3 00 P( 5) 00 0.5 0.5 43.75 b H 0 : The data can be modelled by B(8, 0.5) H 1 : The data cannot be modelled by B(8, 0.5) To get values for greater than 5, combne the frst two cells and the last two cells: No of females 1 3 4 5 6 7 Totals Observed (O ) 10 7 46 49 35 6 7 00 xpected ( ) 7.03 1.88 43.75 54.69 43.75 1.88 7.03 00 ( O ) 1.55 1.198 0.116 0.59 1.75 0.776 0.0001 5.69 The number of degrees of freedom ν = 6 (seven data cells wth one constrant; note that p s not estmated by calculaton but gven n the queston) From the tables: χ 6 (5%)=1.59 As 5.69 s less than 1.59, there s nsuffcent evdence to reject H 0 at the 5% level. The data may be modelled by B(8, 0.5) c If p s estmated by calculaton, ths would gve an extra constrant. Ths would reduce the degrees of freedom by 1 so ν = 5 The crtcal value would become χ 5 (5%)=11.070. However, the test statstc (X = 5.69) would stll be less than ths crtcal value so the concluson would reman the same: there s nsuffcent evdence to reject H 0 10 a Mean= 0 33+1 55+ 80+3 56+ 4 56+5 11+ 6 5+ 7 4 33+55+80+56+56+11+5+ 4 Varance= fx n fx n = 718 300 =.4 0 33+1 55+ 4 80+9 56+16 56+ 5 11+ 36 5+ 49 4 = 300 = 46 5.76=.33 ( d.p.) 300 (.4) b The fact that the sample mean s close to the varance supports the use of a Posson dstrbuton. Pearson ducaton Ltd 018. Copyng permtted for purchasng nsttuton only. Ths materal s not copyrght free. 7
10 c s= 0 = 300 P(X = 0)= 300 e.4.4 0 0! t= = 300 P(X = )=300 e.4.4! = 7. (1 d.p.) = 78.4 (1 d.p.) d H 0 : The data can be modelled by a Po(.4) dstrbuton. H 1 : The data cannot be modelled by Po(.4) dstrbuton. e xpected frequency for 7 or more goals 7 = 300 ( 0 + 1 + + 3 + 4 + 5 + 6 ) = 300 (7.+65.3+ 78.4+6.7+ 37.6+18.1+ 7.)=3.5 f As 7 combne wth 6 to gve the data cell 6 or more goals. There are now 7 data cells after combnng these two values and two constrants as the mean has been calculated n part a, so there are 7 = 5 degrees of freedom. g Test statstc (X ) = 15.7; crtcal value s χ 5 (5%)=11.070 As 15.7 s greater than 11.070, H 0 should be rejected at the 5% level. The data cannot be modelled by Po(.4) 11 a Mean= 0 9+1 4+ 43+ 3 34+ 4 1+5 15+ 6 9+ 4+ 43+ 34+ 1+15+ = 383 =.59 ( d.p.) 148 b It could be assumed that the plants occur at a constant average rate and occur ndependently and at random n the meadow. c s= =148 P(X = )=148 e.59.59 = 37.4 ( d.p.)! t= 7 =148 (11.10+8.76+ 37.4+3.15+ 0.8+10.78+ 4.65)=.50 ( d.p.) d H 0 : The data can be modelled by a Po(.59) dstrbuton. H 1 : The data cannot be modelled by Po(.59) dstrbuton. To get values for greater than 5, combne the last two cells: Number of plants 0 1 3 4 5 6 Total Observed (O ) 9 4 43 34 1 15 148 xpected ( ) 11.10 8.76 37.4 3.15 0.8 10.78 7.15 148 ( O ) 0.397 0.788 0.891 0.106 0.00 1.65 3.709 7.545 The number of degrees of freedom ν = 5 (seven data cells wth two constrants as λ has been estmated by calculaton) From the tables: χ 5 (5%)=11.070 As 7.545 s less than 11.070, there s nsuffcent evdence to reject H 0 at the 5% level. The data may be modelled by Po(.59) Pearson ducaton Ltd 018. Copyng permtted for purchasng nsttuton only. Ths materal s not copyrght free. 8