As is less than , there is insufficient evidence to reject H 0 at the 5% level. The data may be modelled by Po(2).

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1 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 Total Observed (O ) xpected ( ) ( O ) 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%)= As s less than , 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%)= xpected frequency= ( ) = =17 O = 1 17 (15 17) + (3 17) + (19 17) + (0 17) + (14 17) + (11 17) ( ) = 1 98 ( )= = As s less than , 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

2 3 a X = = =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 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 Total Observed (O ) P(X = ) xpected ( ) The last 3 classes must be combned so all the values are 5 or more. Ths gves: x Total Observed (O ) P(X = ) xpected ( ) ( O ) The number of degrees of freedom ν = (four data cells wth two constrants as λ s estmated by calculaton) From the tables: χ (10%)= As 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.

3 a r = meanr =np So.4=6p p=0.4 = =.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) = = ( X = 1) = 100 P( X = 1) = = X = = X = = = 3 3 ( X = 3) = 100 P( X = 3) = = ( X = 4) = 100 P( X = 4) = = 5 1 ( X = 5) = 100 P( X = 5) = = ( X = 6) = 100 P( X = 6) = = ( ) 100 P( ) Combne to get 5 Combne to get 5 After combnng the relevant cells, ths gves: x Total Observed (O ) xpected ( ) ( O ) 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

4 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 ) xpected ( ) ( O ) 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

5 6 H 0 : The data can be modelled by a Posson dstrbuton. H 1 : The data cannot be modelled by Posson dstrbuton. Total frequency = = Mean=λ= 80 Calculate the expected frequences as follows: 0 = 80 P(X = 0)=80 e ! =.540 = = = 80 P(X =1)=80 e = ! Smlarly =15.114, 3 =17.381, 4 =14.991, 5 =10.344, 6 = 5.948, 7 =.931 = 8 80 ( ) = To get values for greater than 5, combne the frst two cells and the last three cells: x Total Observed (O ) xpected ( ) ( O ) 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

6 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 = = Mean=λ= = = 0.95 Calculate the expected frequences as follows: 0 =100 P(X = 0)=100 e = ! 1 =100 P(X =1)=100 e = ! 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 Total Observed (O ) xpected ( ) ( O ) 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 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= = 50.5 ( ) ( ) ( ) ( O ) Test statstc ( X ) = = = The number of degrees of freedom ν = 9 (ten data cells wth a sngle constrant on the total) From the tables: χ 9 (5%)= As s less than , 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

7 9 a The expected number of ltters s modelled by B(8, 0.5) As total frequency = R= 00 P( X = 3) = = S = X = = = P( 4) T = X = = = P( 5) 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 Totals Observed (O ) xpected ( ) ( O ) 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%)= 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= Varance= fx n fx n = = = 300 = =.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

8 10 c s= 0 = 300 P(X = 0)= 300 e ! 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 ( ) = 300 ( )=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%)= As 15.7 s greater than , H 0 should be rejected at the 5% level. The data cannot be modelled by Po(.4) 11 a Mean= = 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 = 37.4 ( d.p.)! t= 7 =148 ( )=.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 Total Observed (O ) xpected ( ) ( O ) 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%)= As s less than , 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

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