Chi-squared tests 6E. So degrees of freedom=(3 1) (3 1)=4. Critical value is χ 4 2 (5%)=9.488

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1 Ch-squared tests 6 1 A 3 table has 3 rows (h) and columns (k) so: Degrees of freedom ν =(3 1)( 1)= Crtcal value s χ (5%)=5.991 H 0 : Ownershp s not related to the localty. H 1 : Ownershp s related to the localty. Degrees of freedom ν =(3 1)( 1)= Crtcal value s χ (5%)=5.991 Test statstc X = 13.1, whch s greater than so reject H 0 at 5% sgnfcance level and conclude that there s evdence to suggest that ownershp of a televson s related to the localty. 3 a There are three groups of students and so three rows of data, and three exam classfcatons so three columns of data. When calculatng the expected values, t s not necessary to calculate the last value n each row because t has to equal the row total. Smlarly, t s not necessary to calculate the last value n each column because t has to equal the column total. So degrees of freedom=(3 1) (3 1)=4 b H 0 : There s no assocaton between the student group and the results. H 1 : There s an assocaton between the student group and the results. Crtcal value s χ 4 (5%)=9.488 Test statstc X = 10.8, whch s greater than so reject H 0 at 5% sgnfcance level and conclude that there s evdence to suggest an assocaton between the student groups and the exam results they acheve. Pearson ducaton Ltd 018. Copyng permtted for purchasng nsttuton only. Ths materal s not copyrght free. 1

2 4 H 0 : There s no assocaton between Mathematcs and nglsh results. H 1 : There s an assocaton between Mathematcs and nglsh results. These are the observed frequences (O ) wth totals for each row and column: Maths grades nglsh grades A B C Total A B C Total Calculate the expected frequences ( ) for each cell. For example: xpected frequency Mathematcs A and nglsh A= =1.105 The expected frequences ( ) are: Maths grades nglsh grades A B C A B C The test statstc (X ) calculatons are: O ( O ( O X = = The number of degrees of freedom ν =(3 1)(3 1)=4; from the tables: χ 4 (5%)=9.488 As s less than 9.488, there s nsuffcent evdence to reject H 0 at the 5% level. Ths suggest that the Mathematcs and nglsh results are ndependent. Pearson ducaton Ltd 018. Copyng permtted for purchasng nsttuton only. Ths materal s not copyrght free.

3 5 H 0 : There s no assocaton between staton and lateness. H 1 : There s an assocaton between staton and lateness. These are the observed frequences (O ) wth totals for each row and column: Staton On tme Late Total A B C Total Calculate the expected frequences ( ) for each cell. For example: xpected frequency On tme and Staton A = = The expected frequences ( ) are: Staton On tme Late A B C The test statstc (X ) calculatons are: O ( O X = ( O = The number of degrees of freedom ν =(3 1)( 1)=; from the tables: χ (5%)=5.991 As s less than 5.991, there s nsuffcent evdence to reject H 0 at the 5% level. There s no reason to beleve there s an assocaton between staton and lateness. 6 H 0 : There s no assocaton between gender and grades. H 1 : There s an assocaton between gender and grades. The number of degrees of freedom ν =(5 1)( 1)=4; from the tables: χ 4 (1%)=13.77 As X (= 14.7) s greater than 13.77, reject H 0 at the 1% level. There s evdence to suggest an assocaton between gender and grade. Pearson ducaton Ltd 018. Copyng permtted for purchasng nsttuton only. Ths materal s not copyrght free. 3

4 7 a The contngency table showng observed frequences s: Factory A B Total OK Defectve Total b H 0 : There s no assocaton between factory and qualty. H 1 : There s an assocaton between factory and qualty. Calculate the expected frequences ( ) for each cell. For example: xpected frequency OK and Factory A = =45 The expected frequences ( ) are: Factory A B OK Defectve The test statstc (X ) calculatons are: O ( O ( O X = = The number of degrees of freedom ν =( 1)( 1)=1; from the tables: χ 1 (5%)=3.841 As s less than 3.841, there s nsuffcent evdence to reject H 0 at the 5% level. There s no reason to beleve there s an assocaton between a factory and garment qualty. Pearson ducaton Ltd 018. Copyng permtted for purchasng nsttuton only. Ths materal s not copyrght free. 4

5 8 H 0 : There s no assocaton between gender and susceptblty to nfluenza. H 1 : There s an assocaton between gender and susceptblty to nfluenza. The contngency table showng observed frequences s: Boys Grls Total Flu No flu Total 8 50 Calculate the expected frequences ( ) for each cell. For example: xpected frequency Boy and Flu = 3 = The expected frequences ( ) are: Boys Grls Flu No flu The test statstc (X ) calculatons are: O ( O ( O X = = The number of degrees of freedom ν =( 1)( 1)=1; from the tables: χ 1 (5%)=3.841 As s greater than 3.841, reject H 0 at the 5% level. There s evdence of an assocaton between the gender and susceptblty to nfluenza. Pearson ducaton Ltd 018. Copyng permtted for purchasng nsttuton only. Ths materal s not copyrght free. 5

6 9 H 0 : There s no assocaton between choce of beach and the gender of the organsms. H 1 : There s an assocaton between choce of beach and the gender of the organsms. Observed (O ) Gender Beach A B C Total Male Female Total Calculate the expected value for each cell by multplyng column and row totals and dvdng by the grand total, for example: AM, = = xpected ( ) Gender Beach A B C Male Female The test statstc (X ) calculatons are: O ( O ( O X = = 7.71 The number of degrees of freedom ν =(3 1)( 1)=; from the tables: χ (5%)=5.991 As 7.71 s greater than 5.991, reject H 0 at the 5% level. There s evdence of an assocaton between the gender of an organsm and the beach on whch t s found. Pearson ducaton Ltd 018. Copyng permtted for purchasng nsttuton only. Ths materal s not copyrght free. 6

7 10 H 0 : There s no assocaton between age and number of credt cards. H 1 : There s an assocaton between age and number of credt cards. These are the observed frequences (O ) wth totals for each row and column: Age Number of cards 3 > 3 Total > Total Calculate the expected frequences ( ) for each cell. For example: xpected frequency 30 and 3 = = The expected frequences ( ) are: Age Number of cards 3 > > The test statstc (X ) calculatons are: O ( O ( O X = = The number of degrees of freedom ν =( 1)( 1)=1; from the tables: χ 1 (5%)=3.841 As s greater than 3.841, reject H 0 at the 5% level. There s evdence of an assocaton between the age and number of credt cards possessed. Pearson ducaton Ltd 018. Copyng permtted for purchasng nsttuton only. Ths materal s not copyrght free. 7

8 11 a H 0 : There s no assocaton between njury rate and choce of gym. H 1 : There s an assocaton between njury rate and choce of gym. b Calculate the expected value for each cell by multplyng column and row totals and dvdng by the grand total: C,I = =6.88 ( d.p.) c The test statstc (X ) calculatons are: O ( O ( O X = = 7.7 The number of degrees of freedom ν =(4 1)( 1)=3; from the tables: χ 3 (5%)=7.815 As 7.7 s less than 7.815, there s nsuffcent evdence to reject H 0 at the 5% level. There s no reason to beleve there s an assocaton between njury rate and choce of gym. Pearson ducaton Ltd 018. Copyng permtted for purchasng nsttuton only. Ths materal s not copyrght free. 8

9 1 a H 0 : There s no assocaton between scence studed and salary. H 1 : There s an assocaton between scence studed and salary. b Calculate the expected value for each cell: B,0 0 = =.90 The expected frequences ( ) are: Salary 0 0k 0k 40k 40 60k 60k 80k > 80k Bology Chemstry Physcs Requre each cell of the expected table to have a value at least 5; merge the frst two columns (so create a category 0 40k) and the last two columns (for a category > 60k). Salary k 60k > 60k Bology Chemstry Physcs The test statstc (X ) calculatons are: O ( O ( O X = =.030 The number of degrees of freedom ν =(3 1)(3 1)=4; from the tables: χ 4 (5%)=9.488 As.030 s less than 9.488, there s nsuffcent evdence to reject H 0 at the 5% level. There s no reason to beleve there s an assocaton between scence studed and salary. Pearson ducaton Ltd 018. Copyng permtted for purchasng nsttuton only. Ths materal s not copyrght free. 9

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

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