Biostats - Desperately need help - Exam this Thursday!

[acab]

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Please help. I'm taking the beast on Thursday and really suck at biostats. Please help me set up these problems correctly (these come from NBME 1 and/or 2):

3. What is the odds ratio for an older woman (relative to that of a younger woman) to have an infant with a low birth weight?
A) (21 / 425) / (54 / 799) = 0.73
B) (54 / 799) / (21 / 425) = 1.37
C) (54 / 745) / (21 / 404) = 1.39
D) (54 / 799) x 100 = 6.8
E) (54 / 75) / (745 / 1149) = 11.1

C?- Since I don't understand how to even think about these problems, I just try to set them up in the 2x2 grid and follow the formulas given in FA. Here's what I did:
Age=Disease and Test=Birth Weight - a=54, b=745, c=21, d=404.

Am I wrong to set it up this way?


9. Serum LDL-cholesterol concentrations are measured in blood samples collected from 100 healthy volunteers. The data follow a normal distribution. The mean and standard deviation for this group is 130 mg/dL and 20 mg/dL, respectively. The standard error of the mean is 5.0. The 95% confidence interval about the mean is 120 to 140 mg/dL. If blood samples are collected from 25 healthy volunteers instead of 100, which of the following best expresses the expected impact on the standard error and the 95% confidence interval about the mean?

Increased std error since n decreases. Would CI increase also (because std dev would increase)?

5. The prevalence of a disease is half as great in town A than it is in town B, but the incidence of the disease is no different in town A than it is in town B. Which of the following best explains these findings?

A) The case fatality rate is twice as high in town A.
B) The duration of the disease is twice as long in town B.
C) The number of new cases in town A is twice as many as those in town B.
D) People in town A use medical care facilities half as often as those in town B.
E) The proportion of asymptomatic cases is twice as much in town B than in town A

B?

28. The prevalence of breast cancer is compared in two groups of women based on parity. The following data are obtained at age 70 in both groups:

Children Yes and Breast Cancer Yes = 120
Children No and Breast Cancer Yes = 180
Children Yes and Breast Cancer No = 1380
Children No and Breast Cancer No = 820

Based on these data, what is the relative risk (risk ratio) for development of breast cancer in childless women compared with women who have children?

A) 0.67
B) 1.2
C) 1.5
D) 1.8
E) 2.25

E? - breast Ca with children = 120:1500, and breast Ca in childless = 180:1000. Is this right?

Thanks for all the help!

 

[stochastic]

Well, I think you did this incorrectly, but I agree that C is the correct answer. So… you might be in that population of “lucky guessers”, but maybe it is me that is incorrect.

The stem for this question (which was left out of the original post) states that:

A case-control study is conducted to determine if women who delay childbearing are at increased risk for having children with low birth weight. A group of women who were at least 35 years old during their first pregnancy are compared with a group of women who were between the ages of 20 and 24 years during their first pregnancy.

Normally the purpose of a “test” is to indicate presence of disease. From this stem, it seems to me that time of childbearing is the “test”. The disease (negative outcome) would be low birth weight, making normal birth weight the absence of disease. Therefore, you should have it set up as disease = birth weight and age = test. With a = 54 = TP; b = 745 = FP; c = 21 = FN; d = 404 = TN.

OR = (TP/FP)/(FN/TN) = C

The reason you got it correct was because, you had everything in the same order as FA did. But if you were assuming age = presence of disease, I would think >34 would be positive and 20-24 would be negative. Since you made two errors in series, they cancelled each other out and you arrived at the correct answer on accident.

Maybe, I am looking at this wrong. I hope not, because my test is Friday. I hope this helped. My advice (if it is any good) would be to try to replace the a,b,c,d with TP, FP, FN, TN and always write these in next to the numbers in the table, because evidently they aren’t always arranged the same way FA is.

Good luck.

 

[stochastic]

I think it's E. SEM = standard deviation/sqrt👎 and CI = mean + or - 1.96*(SEM). I am going to assume that the mean and standard deviation stay the same. Therefore the denominator is just getting smaller in SEM which is increasing SEM and thus CI too.

 

[stochastic]

Yes, prevalence = incidence x duration

 

[acab]

Thanks a lot, Stochastic!
Hopefully I won't lose too many points on biostats.

 

[stochastic]

Yep.

TP = 180
FP = 820
FN = 120
TN = 1380

RR = (TP/(TP+FP))/(FN/(FN+TN)) = (180/180+820)/(120/120+1380) = 2.25