Disease Detectives B/C

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Re: Disease Detectives B/C

Post by Nerd_Bunny »

UTF-8 U+6211 U+662F wrote:
Nerd_Bunny wrote:I'm going to try to do "hard" questions. ;)

Info:
You perform a test. You learn that you got 22 false positives, 10 false negatives, 75 true positives, and 38 true negatives.
1. Calculate the sensitivity and interpret your results.
2. Calculate the specificity and interpret your results.
(The following questions do not pertain to the info.)
3. Which organisms contain two types of nucleic acid? (Bacteria, virus, animal, etc.)
4. Define antigen.
5. Define animal in an epidemiology context.
6. Which case studies are used to calculate prevalence?
7. Define and give an example of Simpson's paradox.
8. Define and give an example of Late-Look bias.
1. 75/(75+10) = 75/85 = 15/17; The test detects 15/17 of positive cases.
2. 38/(38+22) = 38/60 = 19/30; The test correctly rules out 19/30 of negative cases.
3. I'm not so sure about this, but I would guess all eukaryotes
4. A foreign substance in response to which antibodies are produced
5. Not sure what you mean here, but arthropods often act as vectors to disease, and vertebrates often act as natural reservoirs to disease. People can contract disease from such vertebrates by zoonosis.
6. You can calculate the prevalence of a disease using a descriptive study, e.g. time series analyses, ecological studies, and surveys (cross-sectional studies).
7. When data has one correlation when separated into groups but the opposite (positive vs negative) correlation when combined, e.g. if the prevalence of the disease increases with increasing age in the age groups of 20-30 and 70-80, but the group of 70-80 has a much smaller prevalence of disease, the prevalence of the disease actually has a negative correlation with age.
8. No idea.
1 + 2: Correct, although sensitivity and specificity are expressed as percents.
3. Bacteria, protozoa, fungi, and animals. Not all eukaryotes.
4. Correct.
5. I wasn't really looking for a really specific answer, but you got it right. :)
6. Correct.
7. Correct.
8. Late-look bias is when the late diagnosis of a disease is falsely perceived as a longer survival rate. An example would be a case-control study performed on a disease where half of those affected die within 2 weeks, while the other half have an average survival of 8 years. Most patients included in the study are those who survive beyond two weeks.
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Re: Disease Detectives B/C

Post by UTF-8 U+6211 U+662F »

Why is relative risk not an appropriate measure of exposure for case-control studies? What measure of exposure is appropriate?
Why is relative risk not an appropriate measure of exposure for cross-sectional studies? What measure of exposure is appropriate?
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Re: Disease Detectives B/C

Post by IcsTam »

UTF-8 U+6211 U+662F wrote:Why is relative risk not an appropriate measure of exposure for case-control studies? What measure of exposure is appropriate?
Why is relative risk not an appropriate measure of exposure for cross-sectional studies? What measure of exposure is appropriate?
Relative Risk is not appropriate for case-control studies because you start with disease status, which means you cannot find incidence. Odds Ratio is appropriate. Cross-sectional studies utilize prevalence rates. Since cross-sectional studies are not time-dependent, they do not have incidence, which means we cannot use relative risk either.
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Re: Disease Detectives B/C

Post by UTF-8 U+6211 U+662F »

IcsTam wrote:
UTF-8 U+6211 U+662F wrote:Why is relative risk not an appropriate measure of exposure for case-control studies? What measure of exposure is appropriate?
Why is relative risk not an appropriate measure of exposure for cross-sectional studies? What measure of exposure is appropriate?
Relative Risk is not appropriate for case-control studies because you start with disease status, which means you cannot find incidence. Odds Ratio is appropriate. Cross-sectional studies utilize prevalence rates. Since cross-sectional studies are not time-dependent, they do not have incidence, which means we cannot use relative risk either.
Oops, I meant measure of association.
Relative risk does not use incidence rates. It uses attack rates, which are time-independent. The reason that we cannot use relative risk for a case-control study or cross-sectional study is that these types of studies are not sufficiently controlled. We cannot determine that the exposure caused the disease or some outside factor did. With a cohort study, however, efforts are taken to control the characteristics of the subjects. Odds ratios are used for both relative risk and cross-sectional studies.
Feel free to ask a question.
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Re: Disease Detectives B/C

Post by IcsTam »

What statistical model is used to measure the difference between three or more sample means for statistical significance?
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Re: Disease Detectives B/C

Post by UTF-8 U+6211 U+662F »

IcsTam wrote:What statistical model is used to measure the difference between three or more sample means for statistical significance?
The standard error of the mean?
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Re: Disease Detectives B/C

Post by IcsTam »

UTF-8 U+6211 U+662F wrote:
IcsTam wrote:What statistical model is used to measure the difference between three or more sample means for statistical significance?
The standard error of the mean?
I was looking for ANOVA (Analysis of Variance), but seeing that you're Division B, I really doubt that'll ever appear on a test for you. Anyway, ask away.
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Re: Disease Detectives B/C

Post by UTF-8 U+6211 U+662F »

There are 100 people at a party. 50 people drink the milk and get sick. 75 people overall drank the milk. 60 people got sick overall at the party. Find and interpret the attributable risk of the milk.
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Re: Disease Detectives B/C

Post by IcsTam »

UTF-8 U+6211 U+662F wrote:There are 100 people at a party. 50 people drink the milk and get sick. 75 people overall drank the milk. 60 people got sick overall at the party. Find and interpret the attributable risk of the milk.
Attributable Risk = Risk(exposed)-Risk(unexposed), so 50/75 - 10/25 = .267. This means that .267 of the incidence in the exposed would be eliminated if the exposure was eliminated.
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Re: Disease Detectives B/C

Post by UTF-8 U+6211 U+662F »

IcsTam wrote:
UTF-8 U+6211 U+662F wrote:There are 100 people at a party. 50 people drink the milk and get sick. 75 people overall drank the milk. 60 people got sick overall at the party. Find and interpret the attributable risk of the milk.
Attributable Risk = Risk(exposed)-Risk(unexposed), so 50/75 - 10/25 = .267. This means that .267 of the incidence in the exposed would be eliminated if the exposure was eliminated.
Good job (although incidence has a different statistical meaning in epidemiology). Your turn!

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