Lecture 2Lecture 2

The analysis of cross-tabulations

Cross-tabulationsCross-tabulations

• Tables of countable entities or frequencies • Made to analyze the association,

relationship, or connection between two variables

• This association is difficult to describe statistically

• Null- Hypothesis: “There is no association between the two variables” can be tested

• Analysis of cross-tabulations with larges samples

Delivery and housing tenureDelivery and housing tenure

Housing tenure Preterm Term Total

Owner-occupier 50 849 899

Council tentant 29 229 258

Private tentant 11 164 175

Lives with parents 6 66 72

Other 3 36 39

Total 99 1344 1443

Delivery and housing tenureDelivery and housing tenure

• Expected number without any association between delivery and housing tenure

Housing tenure Pre Term Total

Owner-occupier 899

Council tenant 258

Private tenant 175

Lives with parents 72

Other 39

Total 99 1344 1443

Delivery and housing tenureDelivery and housing tenureIf the null-hypothesis is trueIf the null-hypothesis is true

• 899/1443 = 62.3% are house owners.• 62.3% of the Pre-terms should be house owners:

99*899/1443 = 61.7

Housing tenure Pre Term Total

Owner-occupier 899

Council tenant 258

Private tenant 175

Lives with parents 72

Other 39

Total 99 1344 1443

Delivery and housing tenureDelivery and housing tenureIf the null-hypothesis is trueIf the null-hypothesis is true

• 899/1443 = 62.3% are house owners.• 62.3% of the ‘Term’s should be house owners:

1344*899/1443 = 837.3

Housing tenure Pre Term Total

Owner-occupier 61.7 899

Council tenant 258

Private tenant 175

Lives with parents 72

Other 39

Total 99 1344 1443

Delivery and housing tenureDelivery and housing tenureIf the null-hypothesis is trueIf the null-hypothesis is true

• 258/1443 = 17.9% are council tenant.• 17.9% of the ‘preterm’s should be council tenant:

99*258/1443 = 17.7

Housing tenure Pre Term Total

Owner-occupier 61.7 837.3 899

Council tenant 258

Private tenant 175

Lives with parents 72

Other 39

Total 99 1344 1443

Delivery and housing tenureDelivery and housing tenureIf the null-hypothesis is trueIf the null-hypothesis is true

• In general

Housing tenure Pre Term Total

Owner-occupier 61.7 837.3 899

Council tenant 17.7 240.3 258

Private tenant 12.0 163.0 175

Lives with parents 4.9 67.1 72

Other 2.7 36.3 39

Total 99 1344 1443

row total * column total

grand total

Delivery and housing tenureDelivery and housing tenureIf the null-hypothesis is trueIf the null-hypothesis is true

• In general

Housing tenure Pre Term Total

Owner-occupier 50(61.7) 849(837.3) 899

Council tenant 29(17.7) 229(240.3) 258

Private tenant 11(12.0) 164(163.0) 175

Lives with parents 6(4.9) 66(67.1) 72

Other 3(2.7) 36(36.3) 39

Total 99 1344 1443

row total * column total

grand total

Delivery and housing tenureDelivery and housing tenureIf the null-hypothesis is trueIf the null-hypothesis is true

Housing tenure Pre Term Total

Owner-occupier 50(61.7) 849(837.3) 899

Council tenant 29(17.7) 229(240.3) 258

Private tenant 11(12.0) 164(163.0) 175

Lives with parents 6(4.9) 66(67.1) 72

Other 3(2.7) 36(36.3) 39

Total 99 1344 1443

2

all_cells

10.5O E

E

Delivery and housing tenureDelivery and housing tenuretest for associationtest for association

• If the numbers are large this will be chi-square distributed.

• The degree of freedom is (r-1)(c-1) = 4• From Table 13.3 there is a 1 - 5%

probability that delivery and housing tenure is not associated

2

all_cells

10.5O E

E

Chi Squared TableChi Squared Table

Delivery and housing tenureDelivery and housing tenureIf the null-hypothesis is trueIf the null-hypothesis is true

• It is difficult to say anything about the nature of the association.

Housing tenure Pre Term Total

Owner-occupier 50(61.7) 849(837.3) 899

Council tenant 29(17.7) 229(240.3) 258

Private tenant 11(12.0) 164(163.0) 175

Lives with parents 6(4.9) 66(67.1) 72

Other 3(2.7) 36(36.3) 39

Total 99 1344 1443

2 by 2 tables2 by 2 tables

Bronchitis No bronchitis Total

Cough 26 44 70

No Cough 247 1002 1249

Total 273 1046 1319

2 by 2 tables2 by 2 tables

Bronchitis No bronchitis Total

Cough 26 (14.49) 44 (55.51) 70

No Cough 247 (258.51) 1002 (990.49) 1249

Total 273 1046 1319

2

all_cells

12.2O E

E

Chi Squared TableChi Squared Table

Chi-squared test for small samplesChi-squared test for small samples

• Expected valued– > 80% >5– All >1

Streptomycin Control Total

Improvement 13 (8.4) 5 (9.6) 18

Deterioration 2 (4.2) 7 (4.8) 9

Death 0 (2.3) 5 (2.7) 5

Total 15 17 32

Chi-squared test for small samplesChi-squared test for small samples

• Expected valued– > 80% >5– All >1

Streptomycin Control Total

Improvement 13 (8.4) 5 (9.6) 18

Deterioration and death

2 (6.6) 12 (7.4) 14

Total 15 17 32

2

all_cells

10.8O E

E

Fisher’s exact testFisher’s exact test

• An example

S D T

A 3 1 4

B 2 2 4

5 3 8

S D T

A 4 0 4

B 1 3 4

5 3 8

S D T

A 1 3 4

B 4 0 4

5 3 8

S D T

A 2 2 4

B 3 1 4

5 3 8

Fisher’s exact testFisher’s exact test

• Survivers: – a, b, c, d, e

• Deaths: – f, g, h

• Table 1 can be made in 5 ways

• Table 2: 30• Table 3: 30• Table 4: 5• 70 ways in total

S D T

A 3 1 4

B 2 2 4

5 3 8

S D T

A 4 0 4

B 1 3 4

5 3 8

S D T

A 1 3 4

B 4 0 4

5 3 8

S D T

A 2 2 4

B 3 1 4

5 3 8

Fisher’s exact testFisher’s exact test

• Survivers: – a, b, c, d, e

• Deaths: – f, g, h

• Table 1 can be made in 5 ways

• Table 2: 30

• Table 3: 30

• Table 4: 5

• 70 ways in total

5 30 1

70 70 2

• The properties of finding table 2 or a more extreme is:

Fisher’s exact testFisher’s exact test

S D T

A f11 f12 r1

B f21 f22 r2

c1 c2 n

S D T

A 3 1 4

B 2 2 4

5 3 8

1 2 1 2

11 12 21 22

! ! ! !

! ! ! ! !

4!4!5!3!0.4286

8!3!1!2!2!

r r c cp

n f f f f

S D T

A f11 f12 r1

B f21 f22 r2

c1 c2 n

S D T

A 4 0 4

B 1 3 4

5 3 8

1 2 1 2

11 12 21 22

! ! ! !

! ! ! ! !

4!4!5!3!0.0714

8!4!0!1!3!

r r c cp

n f f f f

Yates’ correction for 2x2 Yates’ correction for 2x2

• Yates correction: 212

all_cells

O E

E

Streptomycin Control Total

Improvement 13 (8.4) 5 (9.6) 18

Deterioration and death

2 (6.6) 12 (7.4) 14

Total 15 17 32

212

all_cells

8.6O E

E

2

all_cells

10.8O E

E

Chi Squared TableChi Squared Table

Yates’ correction for 2x2 Yates’ correction for 2x2

• Table 13.7– Fisher: p = 0.001455384362148– ‘Two-sided’p = 0.0029– χ2: p = 0.001121814118023– Yates’ p = 0.0037

Odds and odds ratiosOdds and odds ratios

• Odds, p is the probability of an event

• Log odds / logit

1

po

p

ln( ) ln1

po

p

OddsOdds

• The probability of coughs in kids with history of bronchitis.p = 26/273 = 0.095o = 26/247 = 0.105The probability of coughs in kids with

history without bronchitis.p = 44/1046 = 0.042o = 44/1002 = 0.044

Bronchitis No bronchitis Total

Cough 26 (a) 44 (b) 70

No Cough 247 (c) 1002 (d) 1249

Total 273 1046 1319

1

po

p

Odds ratioOdds ratio

• The odds ratio; the ratio of odds for experiencing coughs in kids with and kids without a history of bronchitis.

Bronchitis No bronchitis Total

Cough 26; 0.105 (a) 44; 0.0439 (b) 70

No Cough 247; 9.50 (c) 1002; 22.8 (d) 1249

Total 273 1046 1319

ac

bd

ador

bc

ab

cd

ador

bc

26247

441002

26*10022.40

247*44or

Is the odds ratio different form 1?Is the odds ratio different form 1?

Bronchitis No bronchitis Total

Cough 26 (a) 44 (b) 70

No Cough 247 (c) 1002 (d) 1249

Total 273 1046 1319

1 1 1 1 1 1 1 126 44 247 1002SE ln 0.257a b c dor

0.874 1.96 0.257 _ to_0.874 1.96 0.257 0.37 _ _1.38to

ln( ) ln(2.40) 0.874or

• We could take ln to the odds ratio. Is ln(or) different from zero?

• 95% confidence (assumuing normailty)

Confidence interval of the Odds ratio Confidence interval of the Odds ratio

• ln (or) ± 1.96*SE(ln(or)) = 0.37 to 1.38

• Returning to the odds ratio itself:

• e0.370 to e1.379 = 1.45 to 3.97

• The interval does not contain 1, indicating a statistically significant difference

Bronchitis No bronchitis Total

Cough 26 (a) 44 (b) 70

No Cough 247 (c) 1002 (d) 1249

Total 273 1046 1319

Chi-square for goodness of fitChi-square for goodness of fit

• df = 4-1-1 = 2