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Review Questions

  1. Cohort studies follow a group of individuals forward in time. Case-control studies study a portion of individuals, selected based on case status.
  2. Because you do not know sizes of populations at risk.
  3. Parameter:=  either OR or Estimator =  either or
  4. natural logarithmic.
  5. 1
  6. Fisher's test is used when an expected frequency is less than 5.
  7. In the 2-by-2 table used to summarize matched-pair case-control data, table cells t and w contain counts of concordant pairs, while cells u and v contain counts of discordant pairs.
  8. True.
  9. The name of the chi-square statistic used to test matched-pair data is McNemar's test.
  10. Matching can be used to control for confounding factors.
  11. T
  12. F
  13. The odds ratio is significant at alpha = 0.05 if it excludes 1.
  14. (a) and (c) are significant at alpha = 0.05.

Exercises

Part A: Independent Samples

17A.1 Cell phone use and brain tumors, Study 1.  Do these results support or fail to support the theory that recent hand-held cellular telephones use causes brain tumors? This study fails to support the theory that cell phones cause brain tumors. Explain your response. The OR estimates for glioma and meningioma show negative associations. The estimate for acoustic neuroma shows a small positive association, but the confidence interval is consistent with both positive and negative associations. The estimate for all tumor types combined is 1.0, indicating no association between recent cell phone use and intracranial tumors. However, the relatively broad confidence intervals for the parameter (e.g., from 0.6 to 1.5 for all tumor types combined) leaves open possible association. This leads to a very common epidemiologic recommendation--"more study is needed."

17A.3 Esophageal cancer and tobacco consumption (dichotomized exposure) - BD1 dataset

(A) Calculate the odds ratio for the data and a 95% confidence interval for the odds ratio parameter. = 1.96 (95% CI: 1.39 to 2.77) Interpret these statistics. This allows us to say with 95% confidence that the OR parameter is between 1.39 and 2.77. (The meaning of 95% confidence needs to be carefully scrutinized. Beware of careless interpretations of confidence intervals.)  

(B) Calculate a P value for the problem. There are several ways to calculate this statistic. All derive P 0.000. [Calculation of chi-square statistics shown below] Interpret this statistic. The evidence against the null hypothesis is highly significant; the association is said to be highly significant.

X2stat, Pearson's =  (64 - 43.897)2 / 43.897    + (150 - 170.103)2 / 170.103  + (136 - 156.103)2 / 156.103 + (625 - 604.897)2 / 604.897 
                 =  9.206 +  2.376  + 2.589  + 0.668 =  14.838

X2stat, Yates'=  (|64 - 43.897|-0.5)2 / 43.897    + (|150 - 170.103|-0.5)2 / 170.103  + (|136 - 156.103|-0.5)2 / 156.103 + (|625 - 604.897|-0.5)2 / 604.897 =  14.109

(C) Download dataset; print SPSS codebook

(D) SPSS output

17A.5 Doll & Hill 1950 .

(A) Cross-tabulation

Smoke

Cases

Non-cases

 Total

  +

  647

622  

1269  

 

  2

27  

 29

Total

  649

649  

 1298

 

(B) = (647)(27) / (622)(2) = 14.04. 
To calculate the confidence interval: ln(14.04) = 2.6419 and SE = sqrt(647-1 + 622-1 + 2-1 + 27-1) = 0.7350; 95% CI for OR = e(2.6419 1.96 0.7350) = e(1.2013, 4.0825) = 3.3 to 59.3.
(C) This study found a very strong association between smoking and lung cancer. The 95% confidence means that we can be 95% confident the odds ratio parameter falls between 3.3 and 59.3. (These statements assume data are free of systematic error/bias.)

17A.7 Vasectomy and prostate cancer. Calculate the odds ratio = = 0.9493 =  0.95 and its 95% confidence interval. ln() = -0.0520 and SE = 0.2049; 95% CI = e-0.0520 (1.96)(0.2049) = e(-0.4536, 0.3496) = (0.64, 1.42) Interpret your findings. Data are compatible with no increase in risk. [Optional: X2stat = 0.064 with 1 df,  P = 0.800; with continuity correction, X2stat, c = 0.02, P = 0.88]

17A.9 Esophageal cancer and alcohol recorded at four levels (BD1). Calculate the odds ratio associated with each level of alcohol consumption. 1 =  (29)(386) / (29)(386) = 1.00 (reference); 2 = (75)(386) / (280)(29) = 3.57; 3 = (51)(386) / (87)(29) = 7.80; 4 = (45)(386) / (22)(29) = 27.23.  Is there evidence of a dose-response relationship? But of course. 

17A.11 Baldness and myocardial infarction 

(A) Calculate odds ratios associated with each level of baldness using baldness level 1 as the reference category. 1 = 1.00 (reference), 2 = 0.98, 3 = 1.39, 4 = 1.94, 5 = 2.64 (very small sample).  . Interpret these results: There is a positive trend from baldness level 3 to 4 to 5.
(B) X2stat = 14.570, df = 4, P = 0.0057. The association is highly significant.
(C) zstat,trend = 3.39, P = 0.00070. The trend is highly significant.
(D) 
Explain why this is important. Age is related to baldness and is also an independent risk factor for myocardial infraction. The association between baldness and myocardial infarction may merely reflect age differences (confounding).
(E)
Does this materially effect you interpretation of results? The adjustments had a small influence on the odds ratio estimates.  Overall, however, our impression is unchanged: there is no evidence of an increased risk with baldness level 2, a small increase with baldness level 3, and a doubling of risk with baldness level 4. There were too few men with level 5 baldness to make precise statement about this level of exposure.

Part B: Matched-pairs

17B.1 Fruits, vegetables, and adenomatous polyps  

(A) Calculate the odds ratio associated with low fruit/veggie consumption. = 45 / 24 = 1.875 = 1.88. Interpret this result.  Low fruit/veggie consumption was associated with an 88% increase in risk. 
(B) Calculate a 95% confidence interval for the odds ratio.  e0.6286 (1.96)(0.2528) = e(0.1331, 1.1241)= (1.14, 3.07). Interpret this result. This places the odds ratio parameter between 1.1 and 3.1 with 95% confidence.
(C) Calculate a P value for testing H0: OR = 1. zMcN = sqrt[(45 - 24)2 / (45 + 24)] = 2.528; P = 0.011 [This provides significant evidence against the null hypothesis.]
(D) Do data support a connection between low fruit/veggie consumption and colon cancer? Yes.

17B.3 Thrombotic stoke

(A) Odds ratio = 44 / 5 = 8.8
(B)  Unmatched cross-tabulation shown below. The unmatched odds ratio = (46)(99)/(7)(60) = 10.8. This is too big (i.e., overestimates risk by about 200%).

Match broken

Case

Control

Total

  Exposed

46

7

53

  Non-exposed 

60

99

159

Total

106

106

212

17B.5 Estrogen and cervical cancer.  

(A) Calculate the odds ratio...  =  43 / 7 = 6.143 ... and its 95% confidence interval... SElnOR = 0.3887; 95% CI for OR = e1.8153(1.96)(0.3887)e(1.0534,2.5772) = (2.87, 13.16). Interpret your results. There is a large elevation in risk associated with unopposed estrogens, the risk is about 6 times higher in the exposed group.  
[Advanced students: Fisher's exact 95% confidence CI via WinPepi: 2.74 to 16.19]
(B) Calculate a P-value for the problem.  zMcN = sqrt[(43 - 7)2 / (43 + 7)] = 5.09, P 0