![]() |
|
![]() | ||
|
Subject:
Epidemiological Statistics
Category: Health > Conditions and Diseases Asked by: garyresearcher-ga List Price: $50.00 |
Posted:
28 Dec 2002 22:29 PST
Expires: 27 Jan 2003 22:29 PST Question ID: 134523 |
In an initial study the incidence of a specific disease was 145 cases per 100,000 person-years (or 18/12,457 person-years). In a different study, using different methodology five or six years following the initial study, the incidence rate was 307 cases per 100,000 person-years (or 122/39,676 person-years). Is it proper to state the incidence rate ratio is 2.1 (307/145) with 95% C.I., 1.3 to 3.5? What is the proper manner to compare rates based on two different studies? We can assume that ascertainment of cases was 100% in each case. |
![]() | ||
|
Subject:
Re: Epidemiological Statistics
Answered By: answerguru-ga on 29 Dec 2002 01:17 PST Rated: ![]() |
Hi garyresearcher-ga, The short answer to your question is that the method you have used to compare rates based on two different studies is accurate under certain, perhaps unrealistic, conditions. I came across an excellent resource that gives a great outline of epidemiology and its basic practices for measurement. These come in the form the a PowerPoint slideshow but the Google-generated HTML version is also available: PowerPoint version: http://www.iihe.org/education/lectures/epi1.ppt HTML version: ://www.google.ca/search?q=cache:tF4Dzl1lDZMC:www.iihe.org/education/lectures/epi1.ppt+Epidemiology+definition&hl=en&ie=UTF-8 The first piece of key information concerns the definition of epidemiology (slide 2): "Study of the distribution and determinant of diseases and injuries in human populations ": - Concerned with frequencies and types of injuries and illness in groups of people - Concerned with factors that influence the distribution of illness and injuries Next, it is important to consider the fundamental assumptions associated with epidemiology (slide 4): 1. Disease doesn’t occur at random 2. Disease has causal and preventive factors 3. Disease is not randomly distributed throughout a population With the information from both the problem statement and the facts above, the following conclusions can be made regarding the accuracy of a method attempting to compare rates based on two different studies (in addition to the 100% ascertainment assumption): 1. The two groups being considered in each of the studies must be judged as being "similar" demographically and perhaps geographically if it is relevant to the disease being considered. 2. The time differential between when the two studies were conducted must not affect the data that has been collected. For example, if the two groups used are not mutually exclusive, then it is reasonable to believe that the drastic increase could be a result of the disease spreading. This is not desireable, since we are clearly comparing the rates on the basis of the methodologies used in each study. 3. The non-random nature of disease is a threat to any statistical method that assumes randomness in some way form during the course of analysis. On the basis of these conclusions, a "perfect scenario" for obtaining this type of data would be to perform both methods at the same time on the same group (this will satisfy the first two conclusions). Next, calculating the incidence ratio would be fine since it makes no assumptions as to where in the data set each case was likely to occur. In order to calculate the confidence intervals, a specific distribution type (such as the normal distribution) must be specified along with a value for standard deviation. This appears to have been done in the question, though this information seems to have been left out of the problem statement. If you have any problems understanding the information above, please feel free to post a clarification and I will respond to it in a timely manner. Cheers! answerguru-ga | |
| |
| |
| |
| |
| |
|
garyresearcher-ga
rated this answer:![]() Given P(a)=18/12,457=0.00144 and P(b)=122/39676=0.00307. Based on P(a) we would expect 57 cases (computed as 18x39676/12457). The question is what are the chances (the probability of getting 122 cases (or an additional 65 cases) given the expectation is just 57 cases? This cannot be that difficult a computation and would be very meaningful to me. |
![]() | ||
|
Subject:
Re: Epidemiological Statistics
From: krobert-ga on 30 Dec 2002 19:15 PST |
I have to go with the GA researcher on this one. The argument that you would expect 57 more cases in the other set based on computation of ratios would only be valid if the incidence was completely random which, by the definition of epidemiology above, is not true. The answer of "what is the probability of this" depends on the data given. You could play around with a make-believe set of data to try to find this out... get standard deviations and such, or use a gaussian distribution... but that depends on the factor of the set being random... which isn't the case. So, basically other than the "this is different than that" factor, you really can't prove anything by statistically comparing these studies. krobert-ga |
Subject:
Re: Epidemiological Statistics
From: answerguru-ga on 31 Dec 2002 01:11 PST |
Hello again Gary, I still do disagree, and the following deductive steps should help you understand where your methodology is flawed: 1. Your ultimate goal is to perform a hypothesis test 2. Hypothesis tests require confidence intervals 3. Confidence intervals can only be obtained when specific distribution information is known (such as distribution type, mean, and standard deviation) 4. Data within a given distribution MUST be random 5. Epidemiology assumes that disease is NOT randomly distributed Since the field of study restricts the assumption of randomness, and the nature is distributions requires randomness, it is impossible to fit this problem into a distribution without making assumptions such as those mentioned earlier. Since a distribution is now no longer possible, confidence intervals cannot be calculated and hypothesis testing cannot occur. Do you now understand why this is not so clear cut? Under the rules of the Google Answers agreement, I cannot have contact with you outside of this forum so I will not be able to provide my email address. You will need to communicate through this board. answerguru-ga |
Subject:
Re: Epidemiological Statistics
From: garyresearcher-ga on 31 Dec 2002 09:31 PST |
I have a Statistics book, "Introduction to Statistics" by Ronald E. Walpole. In a worked example on page 6.16 of the book, there is a problem very similar to the one I have been asking. It states, "Suppose that on the average 1 person in every 1000 is an alcoholic. Find the probability that a random sample of 8000 people will yield fiewer than 7 alcoholics." The worked solution states, "Since p is very close to zero and n is quite large, we shall approximate with the Poisson distribution using mu=(8000)(0.001)=8, Hence if X represents the number of alcoholics, we have Pr(X<7)=0.3134." My problem has almost an exact analogy to this problem. It is common in epidemiological analyses of rare events to also assume a Poisson distribution. This is not a problem and is an acceptable practice. The book uses Chebyshev's theorem to compute confidence intervals. What is your criticism of this approach? |
Subject:
Re: Epidemiological Statistics
From: answerguru-ga on 01 Jan 2003 11:48 PST |
Gary, I believe I now understand why you believe so strongly that using this method is correct. I have no criticism regarding the example you cited because it seems to me that alcoholism can be seen as a random occurance (though some may beg to differ). The reason for this is that alcoholism is not seen as contagious nor hereditary...I believe that this is the assumption that the example has made. In this case, I have no problem with the use of the Poisson distribution nor the use of Chebyshev's theorem for confidence intervals. The disease that you are considering, however, may very well be contagious and/or hereditary (you have not specified the disease so I cannot say for sure). These two components define whether or not randomness can be assumed...if either are true then you cannot assume a random nature. Under this scenario, your example about alcoholism would not be transferrable to the problem at hand. answerguru-ga |
Subject:
Re: Epidemiological Statistics
From: garyresearcher-ga on 01 Jan 2003 13:29 PST |
The disease is Herpes Zoster (shingles), which is regarded as a disease that is not acquired by exposure to others. The onset of herpes zoster occures when a persons cell-mediated immunity (declines), usually in old age (after around 50) and the latent VZV virus (that one acquired as a child due to onset of chickenpox--varicella) finally reactivates and HZ results. My data shows that rather than a gradual age-related decline in immunity which causes increasing HZ incidence with advancing age, it is in reality, the loss of exogenous (outside) exposures to wild-type chickenpox that causes the increasing incidence with age. Exogenous exposures to natural chickenpox in the community and the frequent re-exposures actually boosted individual's immune system to suppress the reactivation of HZ. Now that a varicella vaccine is universally given to children, I am finding that children with natural disease are experiencing shingles at a rate two-fold to three-fold higher than in the prelicensure era. Appreciate your further comments on this scenario. I have three different studies using various methodologies that all consistently yield the hypothesis given above. |
Subject:
Re: Epidemiological Statistics
From: answerguru-ga on 01 Jan 2003 14:36 PST |
Although this question has clearly gone off on a tangent from its original statistical nature, you last comment leads me to believe that if an external factor is present (in this case exposure to natural chickenpox), randomness cannot be assumed. In order for this assumption to be made validly one must consider the following: 1. The external factor (natural chickenpox) must occur randomly 2. There is no correlation between age and the exposure to this external factor Clearly neither of these are true since, as far as I know, chickenpox spreads in a contagious manner and is not randomly spread through an arbitrary region. We already know that there IS a correlation between age and exposure to natural chickenpox (in fact, the data you have provided states it is a negative correlation). answerguru-ga |
Subject:
Re: Epidemiological Statistics
From: garyresearcher-ga on 01 Jan 2003 20:12 PST |
The incidence rate of HZ is computed only among those individuals that have previously had varicella (chickenpox). While varicella is contagious, HZ is not contagious. Individuals in the community that have had chickenpox receive a boost when exposed to a child with chickenpox; however, those with chickenpox are most infectious 2 to 3 days before the rash breaks out. Therefore, the individuals that provide the exogenous boost or exposure are generally unknown. There is a 2nd mechanism that also provides a boost in immunity--it is called asymptomatic edogenous reactivation--this is what limits the occurrence of HZ to about 500/100,000 person-years in the community, even if there is no boosting from exogenous exposures. Therefore, it is the overall effect of varicella in the community that plays a role in providing exogenous boosting in adults, and the overall effect of varicella in schools that influence the reactivation of herpes zoster in children. While there is a correlation between age and chickenpox (i.e., children usually have onset of checkenpox in Grade K or 1st grade (first exposure in school), or at a younger age due to exposure in pre-school. Herpes zoster, however, among those that have a prior history of chickenpox is due to a decline in cell-mediated immunity which is an individual body process that is accelerated in the absense of varicella (chickenpox) disease in the school or community. I have as yet found no increase in the 10 to 19 year olds (who have a more mature immune system and have sufficient CMI to suppress reactivation of HZ at this time). Children aged <10 years, however, with immature immune systems received repeated exogenous boosts due to exposure to varicella in the community. In view of the above explanation, "the external factor chickenpox" is reduced throughout the entire community with universal vaccination of children. Children previously had more exogenous boosts in the school environment then say 30-40 year-olds who received only occassional exposures when participating in activities with their children or when shopping and coming into close proximity to other children that were to have onset of varicella (but have not as yet broke out), so they would not even recognize such a contact. I am honestly trying to comprehend these assumptions and applicability of statistical analysis myself and do not wish to offend anyone and am trying to fully understand the scope of others comments. In view of the above explaination, since I have stratified my analysis to a specific age group and am not comparing across other age groups, I feel I have not invalidated the statistical approach. All children presently receive less boosts due to fewer varicella cases in the community since the introduction of varicella vaccine. There are actually 70-80% fewer cases of chickenpox today then there were in 1995. Those children that remain that have had natural (wild-type) disease are the ones that are affected by the reduction of exogenous boosts they previously received from other children in the community that had natural chickenpox. In view of the above, could you please site where the assumptions could still be in error. The issue has major consequences since instead of a cost-benefit savings, I have computed a U.S. annual cost of $90 million due to increases in morbidity and mortality of HZ disease in adults for the next 30 years, rather than the $80 million in medical savings due to universal varicella vaccination. These figures assume varicella vaccination are 100% successful in eliminating varicella disease; but this assumption was based on the premise that there was no immunologically-mediated link between varicella incidence prevalence and HZ incidence. |
Subject:
Re: Epidemiological Statistics
From: answerguru-ga on 02 Jan 2003 12:18 PST |
Hi again Gary, As is often the case with these types of analyses, the devil is truly in the detail! After considering the details of the disease as well as those pertaining to your study, I can conclude the following about the validity of the assumptions you would need to make: 1. With the new information you provided regarding HZ, I can see validity in the assumption that occurances within the population you are studying are truly random. 2. Since it is clear that chickenpox are a prerequisite for getting HZ, your population must consist entirely of inidividuals who have already had chickenpox. Furthermore, you must disprove that there is any correlation between age and HZ (this only needs to be done for the age ranges being considered in your study). Since I can validate that these two elements have been conclusively resolved (based on the sum of the information you have provided throughout our correspondence), your use of an analysis similar to the example you provided is now justified. We really needed to dig deep to accurately answer this question, but I think you now have a true understanding of the validity of your analysis. This should help you when you face the group who will be evaluating your methodology. I believe I have provided you with all the information you required. If you now feel satisfied that your question has been answered sufficiently, I would appreciate if you could revise your rating/tip decision. If you are unable to do this on the website, you can contact the GA Editors at: answer-editors@google.com Please cite Question ID #134523 and provide the information you would like to update. I thank you for you patience, and I hope you are satisfied with your Google Answers experience. Best Regards, answerguru-ga Google Answers Researcher |
Subject:
Re: Epidemiological Statistics
From: garyresearcher-ga on 02 Jan 2003 13:08 PST |
My only comment would be, that some statistical expert out there should have taken my original data, and onsidered a binomial trail with n=39676 and probability of success p=0.001445 and computed that the probability of observing 122 cases would be (combination 39676 and 122)x0.001445 raised to the 122 power x 9.999855 raisted to the 39,554 power which is approx. equal to e (2.718...) raised to the negative 5.75302 power x 5.75302 raised to the 122 power, divided by 122 factorial = 1.6377738710x10 raised to the -113 which is approximately zero. Thus, the interpretation is that in the 1st study the chance of opersving the case is p=0.001445. If we assume that the chance of observing the case in the new study remains the same, then the probability of observing 122 cases in 39,676 person-years would be virtually zero. But the fact is that we have observed 122 cases in the new study. Logically, this means that the chance of observing cases in the new study is significantly better than in the old study. In other words, we can conclude that the cases have increased significantly. To further argue about the significance of the new study, we can formulate the following statistical hypotheses: Null Hypothesis: There is no difference between the two studies in terms fo the chance of observing cases, i.e., H0: p=0.00145 Alternative Hypothesis: The chance of observing cases is greater in the new study than in the old study, i.e., H1: p>0.00145. Let X be the random variable that follows a binomial distribution with 39676 trials and the probability of success is some p (0<p<1). Under the null hypothesis, p=0.001445. P-value=P(X>=122)+P((X-np)/sqrt(np(1-p))>=(122-39676*0.001445)/sqrt(39676*0.001445(1-0.001445)) = P(Z>=4.84691) which is approx. 0.0000. Where Z follows the standard normal distribution. We have applied the Central Limit Theorem for the computation above. Since p-value is extremely small, we can reject the null hypothesis by concluding that p>0.00144. That is to say, we can conclude that the chance of observing cases is greater inthe new study than in the old study. This is essentially what I was looking for which is not much different from the original problem statement. |
Subject:
Re: Epidemiological Statistics
From: answerguru-ga on 02 Jan 2003 13:38 PST |
I agree that the formulation you provided in your last comment is accurate, though it is worth noting that your specific study is an exception within the field of epidemiology...the details necessary to undertake a meaningful analysis came later. Any statistician would follow a similar protocol - a full understanding of the problem is always needed before a statistical method can be applied. answerguru-ga |
Subject:
Re: Epidemiological Statistics
From: garyresearcher-ga on 02 Jan 2003 14:01 PST |
I am updating my rating from 1 Start to 5 Stars. Lastly, does anyone have a computer program listing of the software to do such similar calculations as the one I provided above? Thanks again. |
If you feel that you have found inappropriate content, please let us know by emailing us at answers-support@google.com with the question ID listed above. Thank you. |
Search Google Answers for |
Google Home - Answers FAQ - Terms of Service - Privacy Policy |