Prevalence of Breast and Bottle Feeding

Document Sample
Prevalence of Breast and Bottle Feeding Powered By Docstoc
					Design and Analysis of Clinical Study
7. Analysis of Case-control Study

              Dr. Tuan V. Nguyen
       Garvan Institute of Medical Research
                Sydney, Australia
                        Overview

•   Risk
•   Odds ratio (OR)
•   Standard erros of OR
•   95% confidence interval of OR
•   Interpretation
                  Results of a CC study

                                          Risk       Cases   Controls
                 Population               factor
                                          Presence       a      c

                                          Absence        B      d




        Cases                 Controls



Risk+           Risk-   Risk+            Risk-
           a              b          c               d
                                Odds

• Let p be a probability (p is between 0 and 1)
• Odds is defined as:


                               p
                       odds 
                              1 p
• Odds is the likelihood of the occurrence of an event to the non-
  occurrence of the event

• Obviously, when p = 0.5, odds = 1.
                         Odds Ratio (OR)

                Risk factor     Cases        Controls


                Presence          a             c

                Absence           b             d


                                                                     a
• Odds of risk factor presence to absence in cases:       Ocases   
                                                                     b
                                                                     c
• Odds of risk factor presence to absence in controls: Ocontrols 
                                                                     d

                         Ocases     a /b ad
• Odds ratio:       OR                
                         Ocontrols c / d b  c
                 Interpretation of OR

• OR = 1: the risk of disease is the same for exposed
  (presence of the risk factor) and unexposed (absence of
  the risk factor)

• OR > 1: the risk of disease higher in exposed than in
  unexposed. In other words, the risk factor increases the
  probability of getting disease.

• OR < 1: the risk of disease lower in exposed than in
  unexposed.
            Sampling Distribution of OR

• OR is an estimate of a true OR (l)
• Estimate is subject to sampling variation
• Sampling variation is characterized by
   – Mean
   – Variance
• OR is an unbiased mean of l.
• Variance of OR ?
   – Can not be determined directly
   – Have to determined indirectly – log(OR)
                  Variance and 95% of OR

Risk        Cases   Controls   • OR = (ad) / (bc)
   factor                      • L = log(OR)
Presence      a        c       • var(L) = s2 = 1/a + 1/b + 1/c + 1/d
Absence       b        d       • Standard error of L:
                                  s = sqrt(s2)
                               • 95% confidene interval (CI) of L:

                                   L–1.96*s    to    L+1.96*s

                               • 95% CI of OR:
                                 eL–1.96*s to eL–1.96*s
     An Example: Prostate cancer and Agent Orange

•   Giri et al (2004) conducted a pilot case control study to examine the association
    between prostate cancer risk and Agent Orange exposure among American
    soldiers who served in Vietnam during the war time. The investigators identified
    47 Vietnam era veterans with a diagnosis of prostate cancer. They further
    randomly selected 142 veterans who were hospitalized for a reason other than
    cancer from a hospital data base. In each group, the investigators assessed
    whether an individual was exposed to Agent Orange (AO) during the war. The
    results are as follows

     Exposure status               Prostate Cancer                 Controls

     Exposed to AO                         11                         17
     Not exposed to AO                    29                          106
    An Example: Prostate Cancer and Agent Orange

Exposure status          Prostate Cancer            Controls

Exposed to AO                   11                      17
Not exposed to AO               29                      106

•   OR = (11 x 106) / (17 x 29) = 2.37
•   L = log(2.37) = 0.861
•   var(L) = s2 = 1/11 + 1/17 + 1/29 + 1/106 = 0.1936
•   s = sqrt(0.1936) = 0.44
•   95% confidene interval (CI) of L:
    = 0.861–1.96*0.44 to 0.861+1.96*0.44
    = –0.00151 to 1.724
• 95% CI of OR
  = e-0.00151 to e1.724
  = 1.00 to 5.63
                             Distribution of OR
                                     Histogram of OR



                   200
                   150
       Frequency

                   100
                   50
                   0




                         0   2   4      6        8     10   12

                                            OR



Distribution of OR = 2.37 with 95% confidence interval = 1.0 to 5.63
             Matched Case-Control Study

• Confounding is a potential problem of CC study
• Matched CC study: cases and controls are matched for a
  confounding factor.
   – Step 1: Select a case
   – Step 2: Record the confounding factor
   – Step 3: Randomly select a control who has exactly the same
     confounding factor as the case
   – Repeat steps 1-3 until the sample size is adequate.
   – 2x2 Table
                                          Controls

          Cases                 Exposed         Unexposured
          Exposed                  a                 b
          Unexposed                c                 d
    An Example: Hip fracure and Osteoporosis

•   A case control study was under taken to examine the association between
    osteoporosis and risk of hip fracture. 51 cases of hip fracture were identified, and
    were individually age-matched with 51 controls (who had no fracture). The
    results are shown in the following table (“0” is for “non-osteoporosis” and 1 is for
    “osteoporosis”).

                       pair         age         Hip fx       Control
                         1           61            0             0
                         2           62            1             0
                         3           64            0             0
                         4           64            1             0
                         5           64            1             1
                                          ...
                        49          85            1               1
                        50          86            0               0
                        51          87            1               1
         Analysis of Matched CC by the z-test

•   Summary of results

                                      Hip fracture

     Controls              Osteoporosis    Non-osteoporosis
     Osteoporosis               11                   2
     Non-osteoporosis           28                   10

•   Proportion of hip fx who had osteoporosis: P1 = (11+28)/51 = 0.765
•   Proportion of controls who had osteoporosis: P2 = (11+2)/51 = 0.255
•   Difference: D = P1 – P2 = 0.765 – 0.255 = 0.51
•   Variance of D: s2 = (28+2) / (51)2 = 0.0115
•   Standard error of D: s = sqrt(0.0115) = 0.107
•   z = 0.51 / 0.107 = 4.77
•   Highly significant!
                Analysis of Matched CC by OR

•   Summary of results

                                        Hip fracture

     Controls                Osteoporosis    Non-osteoporosis
     Osteoporosis                 11                   2
     Non-osteoporosis             28                   10

           •    OR = 28/2 = 14
           •    L = Log(OR) = log(14) = 2.64
           •    Variance of L: s2 = 1/28 + 1/2 = 0.536
           •    Standard error of L: s = sqrt(0.536) = 0.732
           •    95% CI of L: 2.64 ± 1.96*0.732 = 1.20 to 4.07

           • 95% CI of OR: e1.20 = 2.33 to e4.07 = 58.8
                   Statistical Methods

The relationship between [risk factor] and [outcome, disease] was
assessed by the odds ratio (OR) for case-control study. In this analysis,
the study’s result was summarized into a 2x2 table with two categories of
risk factor and outcome. The exposure OR was then computed from the
table. The statistical significance of OR was assessed by the 95%
confidence interval (CI) of OR. A CI does not include unity was
considered statistically signigicant at the 5% significance level.
                              Exercises

• Preston-Martin et al (1989) conducted a case-control study to investigate
  the etiology of acoustic neuromas. Eighty-six men were diagnosed to
  have acoustic neuromas. The researchers selected another 86 men of
  the same race and within 5 years of age compared to the cases.

• The researchers were interested in the effect of exposure to loud noise.
  They found 58 cases and 46 controls had had some exposure to loud
  noise at work. Carry out an appropriate analysis and interpret the result
  of analysis.
                                Exercises

•   R Doll and B Hill. BMJ 1950; ii:739-748


                                  Lung Cancer   Controls
    Smokers                            647        622
    Non-smokers                         2         27

				
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
views:4
posted:10/12/2011
language:English
pages:18