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Data Set 8: Hodgkin’s
Disease
Angela Meng and Laura Mockensturm
1
Background
● Cancer of the lymphatic system
● Body’s ability to fight the disease is weakened
Histological Types:
● Lymphocyte predominance (LP): males younger than 18, great
survival rates
● Nodular sclerosis (NS): affects the colon
● Mixed cellularity lymphoma (MC): affects white blood cells and
plasma cells
● Lymphocyte depletion (LD): adults with immune-deficient viruses,
fewer white blood cells
2
The Present Study
● 538 patients
● Followed for 3 months during treatment
3
Odds Ratio and Relative Risk
● Positive: LP
● Partial: NS and MC
● None: LD
Relative Risk of Positive Reaction
● LP is...
o 0.5% higher than the NS patients
o 22.9% higher than the MC patients
o 184.6% higher than the LD patients
4
Cumulative Proportions
5
Chi-Squared Test for Independence
● The chi-squared test statistic can be calculated by the formula
below with df=(I-1)(J-1)
● X-squared = 75.8901, df = 6, p-value=2.517e-14
● Strong evidence of dependence
● Standard residuals
6
Cumulative Logit Model
Model:
Model Assumptions:
● Separate intercept, αj, for each cumulative logit
● Same slope, β
7
Model 1
● Explainatory variable (X) - histological type with 4 categories
● Response variable (Y) - response with 3 categories with natural
ordering (positive - partial - none)
8
9
Goodness of fit test
● Residual deviance=2.7283, df=3,p-value=0.4354
● Indicates our current model is adequate for describing the data
● Pearson’s residuals all fall within the range of ±2
10
Model 1.1
● Model with no explanatory variable (with intercept term only)
● Residual deviance=68.2955
11
Likelihood Ratio Test
● Test statistic
65.5672,df=3,p-value=3.793411e-14
● Indicates there is strong association between histological type and
response
12
Log-linear Model
Treats X and Y symmetrically
DF = (I-1)(J-1)
13
Model 2 -log-linear model with homogeneous association
14
Goodness of fit test
● Residual deviance=68.295, df=6,p-value=9.141963e-13
● Indicates our current model is not adequate for describing the data
15
Examine the standard residuals
16
How to improve our model?
● Only option will be including the interaction between histological
type and response, which gives us the saturated model
17
Model 2.1
18
Conclusion: Model
● Best Model: Cumulative logit
i) logit[P(Y≤1)] = -1.3181 + 2.2481LP + 1.6389MC + 2.2222NS
ii) logit[P(Y≤2)] = -0.3679 + 2.2481LP + 1.6389MC + 2.2222NS
where LP, MC and NS are dummy variables
19
Conclusion: Meaning
● Odds Ratio:
o Between LP and LD is 9.4697
o Between MC and LD is 5.1495
o Between NS and LD is 9.2276
● LP is most likely to show positive response
● LD is least likely to show positive response
20
Citations
 "Hodgkin's Lymphoma." Mayo Clinic. Mayo Foundation for Medical
Education and Research, 15 Aug. 2014. Web. 04 Apr. 2015.
 "Childhood Hodgkin Lymphoma Treatment (PDQ®)." National Cancer
Institute. National Institute of Healths, 28 Jan. 2015. Web. 04 Apr. 2015.
21

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Hodgkin's disease final presentation

  • 1. Data Set 8: Hodgkin’s Disease Angela Meng and Laura Mockensturm 1
  • 2. Background ● Cancer of the lymphatic system ● Body’s ability to fight the disease is weakened Histological Types: ● Lymphocyte predominance (LP): males younger than 18, great survival rates ● Nodular sclerosis (NS): affects the colon ● Mixed cellularity lymphoma (MC): affects white blood cells and plasma cells ● Lymphocyte depletion (LD): adults with immune-deficient viruses, fewer white blood cells 2
  • 3. The Present Study ● 538 patients ● Followed for 3 months during treatment 3
  • 4. Odds Ratio and Relative Risk ● Positive: LP ● Partial: NS and MC ● None: LD Relative Risk of Positive Reaction ● LP is... o 0.5% higher than the NS patients o 22.9% higher than the MC patients o 184.6% higher than the LD patients 4
  • 6. Chi-Squared Test for Independence ● The chi-squared test statistic can be calculated by the formula below with df=(I-1)(J-1) ● X-squared = 75.8901, df = 6, p-value=2.517e-14 ● Strong evidence of dependence ● Standard residuals 6
  • 7. Cumulative Logit Model Model: Model Assumptions: ● Separate intercept, αj, for each cumulative logit ● Same slope, β 7
  • 8. Model 1 ● Explainatory variable (X) - histological type with 4 categories ● Response variable (Y) - response with 3 categories with natural ordering (positive - partial - none) 8
  • 9. 9
  • 10. Goodness of fit test ● Residual deviance=2.7283, df=3,p-value=0.4354 ● Indicates our current model is adequate for describing the data ● Pearson’s residuals all fall within the range of ±2 10
  • 11. Model 1.1 ● Model with no explanatory variable (with intercept term only) ● Residual deviance=68.2955 11
  • 12. Likelihood Ratio Test ● Test statistic 65.5672,df=3,p-value=3.793411e-14 ● Indicates there is strong association between histological type and response 12
  • 13. Log-linear Model Treats X and Y symmetrically DF = (I-1)(J-1) 13
  • 14. Model 2 -log-linear model with homogeneous association 14
  • 15. Goodness of fit test ● Residual deviance=68.295, df=6,p-value=9.141963e-13 ● Indicates our current model is not adequate for describing the data 15
  • 16. Examine the standard residuals 16
  • 17. How to improve our model? ● Only option will be including the interaction between histological type and response, which gives us the saturated model 17
  • 19. Conclusion: Model ● Best Model: Cumulative logit i) logit[P(Y≤1)] = -1.3181 + 2.2481LP + 1.6389MC + 2.2222NS ii) logit[P(Y≤2)] = -0.3679 + 2.2481LP + 1.6389MC + 2.2222NS where LP, MC and NS are dummy variables 19
  • 20. Conclusion: Meaning ● Odds Ratio: o Between LP and LD is 9.4697 o Between MC and LD is 5.1495 o Between NS and LD is 9.2276 ● LP is most likely to show positive response ● LD is least likely to show positive response 20
  • 21. Citations  "Hodgkin's Lymphoma." Mayo Clinic. Mayo Foundation for Medical Education and Research, 15 Aug. 2014. Web. 04 Apr. 2015.  "Childhood Hodgkin Lymphoma Treatment (PDQ®)." National Cancer Institute. National Institute of Healths, 28 Jan. 2015. Web. 04 Apr. 2015. 21

Editor's Notes

  1. Before fitting a model to our data, we wanted to see if the the variables are dependent. Residuals shows our observed values deviates a lot, further evidence to reject the Ho of independence.
  2. All p-values are sig
  3. Outside the range
  4. Talk about perfect fit, we do not want a saturated model. This makes sense because log-linear model does not consider the natural ordering in the response, some information from the original data is lost when using log-linear model, which results in lack of fit in data.
  5. The estimated odds that patients with histological type LP is likely to show a response to the treatment in the direction of positive response (Y≤j) rather than no response (Y>j) is e^2.2481 = 9.4697 times the estimated odds of that of patients with histological type LD. Similarly, the estimated odds that patients with histological type MC is likely to show a response to the treatment in the direction of positive response (Y≤j) rather than no response (Y>j) is e^1.6389 = 5.1495 times the estimated odds of that of patients with histological type LD. And the estimated odds that patients with histological type NS is likely to show a response to the treatment in the direction of positive response (Y≤j) rather than no response (Y>j) is e^2.2222 = 9.2276 times the estimated odds of that of patients with histological type LD.