Austin Meyer, PhD
MS4
Basic Statistics
Epidemiology
Generalizability
P-value
Confidence Interval
Efficacy
Effectiveness



Investigators are studying prostate specific antigen (PSA) as a predictor for prostate cancer. To make the statistics easier, they are going to assume that PSA is a normally distributed population variable. Which of the following is correct under their assumption?
The normal distribution is unimodal and symmetric.
The important invariant properties (for you) of normal distributions are the following:





Which of the following corresponds to the measures of central tendency on the graph from left to right?
Mode is most common, median is middle, mean is average value.
Always remember that the y-axis on these plots are counts or frequency. Therefore, which line is closest to the peak on the y-axis is the mode. The median is always in the middle. The mean is the most susceptible to outliers so in a skewed distribution it will always be farthest out on the tail.

Two sample: \(H_0\) = There is no difference between the disease and no disease groups
Run the T-test (in this case, in R language)
norm1 <- rnorm(5000, mean = 4.75, sd = 1.2)
norm2 <- rnorm(5000, mean = 5.25, sd = 1.2)
(t.test(norm1, norm2))$p.value
## [1] 4.687206e-88
Have we rejected the null hypothesis?
Yes, we have accepted \(H_A\). There is a difference between disease and no disease groups.
| Healthy | Disease | Total | |
|---|---|---|---|
| Exposed | 40 | 60 | 100 |
| Not Exposed | 500 | 400 | 900 |
| Total | 540 | 460 | 1000 |
Table 1: A 2x2 contingency table
| Exposure Status | Never Sick | Sometimes Sick | Mostly Sick | Total |
|---|---|---|---|---|
| High | 10 | 20 | 180 | 210 |
| Medium | 20 | 100 | 20 | 140 |
| Low | 100 | 40 | 10 | 150 |
| Total | 130 | 160 | 210 | 500 |
Table 2: A 3x3 contingency table
| Exposure Status | Never Sick | Infrequently Sick | Sometimes Sick | Mostly Sick | Always Sick | Total |
|---|---|---|---|---|---|---|
| Super High | 10 | 90 | 34 | 12 | 12 | 158 |
| Very High | 30 | 345 | 54 | 43 | 21 | 493 |
| High | 70 | 57 | 67 | 65 | 32 | 291 |
| Medium | 200 | 33 | 87 | 25 | 42 | 387 |
| Low | 130 | 89 | 58 | 45 | 56 | 378 |
| Very Low | 100 | 54 | 36 | 23 | 78 | 291 |
| Super Low | 90 | 23 | 36 | 63 | 8 | 220 |
| Total | 530 | 691 | 372 | 276 | 249 | 2118 |
Table 3: A 7x5 contingency table
Spearman correlation compares ranked values of two variables

\(r =\) correlation between variables
\(r^2 = \) amount of variance in y that is explained by x



| Clot | No Clot | Total | |
|---|---|---|---|
| OCP Use | 500 | 400 | 900 |
| No OCP Use | 80 | 20 | 100 |
| Total | 580 | 420 | 1000 |
A study was conducted on OCPs and blood clots, and the data is shown. Which of the following is the best method to assess the association between OCP use and blood clots?
What kind of data is this?
The only test available that utilizes categorical data is the Chi-square test. All of the other tests require at least rank or quantitative data.
To test a new biomarker, investigators plan a cross-sectional study comprised of two groups. In one group, the researchers will include men with confirmed prostate cancer. In the other group, researchers will include men with no evidence of prostate cancer. The investigators will assume their biomarker is normally distributed. What is the best test to investigate whether the biomarker can distinguish the two groups?
The number of groups and distribution is all that matters
The two sample T-test is the appropriate test in this case. The two sample Mann-Whitney U-test could work as well, but is slightly less efficient for normally distributed data than the T-test. The Pearson correlation requires two measured variables on the same sample. A chi-squared test requires categorical (i.e. count) data. An analysis of variance is typically used to measure the difference in means of three or more groups.


Attack rate
Incidence
Prevalence

Investigators are studying the association between mesothelioma and asbestos exposure. Due to the relative rarity of the disease, they design a very large case-control study. In the end, they find an \(OR = 20\ (19.54;20.52,\ p < 0.001)\). After assuming that the OR is a good approximation of risk, the authors conclude that the risk of mesothelioma is 20 times higher in those exposed to asbestos compared to control. Why is their assumption reasonable?
Think about the denominators for odds and risks.
The odds ratio is (A / B) / (C / D) and the risk ratio is (A / (A + B)) / (C / (C + D)). In the case where the number of people with the disease is small, the numbers A and C become very small. In that case, B is a good approximation of A + B and D is a good aproximation of C + D. Thus, the RR ~ (A / B) / (C / D).

If true infections are low, denominator \(A+B \approx B\) and \(C+D \approx D\)
Two studies were conducted on different samples from the same population to assess the relationship between oral contraceptive use and the risk of deep venous thrombosis (DVT). Study A showed an increased risk of DVT among oral contraceptive users, with a relative risk of 2.0 and a 95% CI of 1.2-2.8. Study B showed a relative risk of 2.05 and a 95% CI of 0.8-3.1. Which of the following statements is most likely to be true regarding these 2 studies?
What gives a narrower confidence interval?









This is the real prevalence of HIV... Where would you put the cutoff?

Assume a steady-state population that is not changing in any way. Which of the following statements is true for people who test positive regarding moving the cutoff for a positive test from the solid to the dotted line?
Question prefaces a positive test result


The highest yield cutoff is the x-value that maximizes the distance from the diagonal to the curve

Again, the best cutoff is the x-value that maximizes the distance from the diagonal to the curve

Closer to the top means better evidence


This post hoc analysis is overly simplified for real life
This understanding is sufficient for step 1
Confounders reduced because a patient can serve as their own control




A study was conducted to evaluate the efficacy of a new antiviral drug for the treatment of the common cold in young children. The study population consisted of 100 children between the age of 2 to 8 years. These children were diagnosed with rhinovirus infection and subsequently given the particular antiviral drug. One week later, it was observed that 92 of the 100 patient were asymptomatic. Which of the following is the true conclusion of this study?
A treatment is tested without a control
Researchers are studying the relationship between mutations in HMG-CoA reductase and CAD. The study population is selected at random. Tissue samples are obtained for genotyping and stress echos are performed to assess CHD. In the subsequent paper, the authors conclude that there is an association between mutations in HMG-CoA reductase and CHD. Which of the following study designs did the authors utilize?
What does the timeline look like?
A study was conducted to evaluate the efficacy of a new antiviral drug. The study population consisted of 100 rhinovirus-infected children. The treatment arm was given an antiviral drug and the control arm was given a placebo. One week later, researchers found that 42 out of 50 treatment patients were asymptomatic and 30 of 50 control patients were asymptomatic. On average, how many people need to be treated with this drug to cure one infected person?
\(ARR = Risk_{Control} - Risk_{Treatment}\)
\(NNT = \frac{1}{ARR}\)
\(ARR = \frac{42}{50} - \frac{30}{50} = \frac{12}{50}\)
\(NNT = \frac{1}{12/50} = \frac{50}{12}\)



Assuming that mortality is simply the incidence of death per 100 patients, after controlling for physician characteristics, what is the relative risk of death within 30 days of discharge for patients with a male physician versus a female physician?
\(Incidence_{Male} = \frac{a}{a + b}\)
\(Incidence_{Female} = \frac{c}{c + d}\)
Thus, mortality is the risk of death.
Mortality is the risk of death. Then, the relative risk is:
\(Risk_{Male} = 11.49\)
\(Risk_{Female} = 11.07\)
\(RR = \frac{11.49}{11.07}\)

Controlling for physician characteristics as before, what is the absolute risk reduction of having a female physician?
Mortality is the risk of death.
\(ARR = Risk_{Male} - Risk_{Female}\)
Mortality is the risk of death. Since this is subtraction it is important to have the units correct.
\(ARR = 0.1149 - 0.1107 = 0.0042\)

Given the information from the previous slide, on average how many patients would need to be treated by a female physician to save a life?
Mortality is the risk of death.
\(ARR = Risk_{Male} - Risk_{Female}\)
\(NNT = \frac{1}{ARR}\)
Thus, mortality is the risk of death. Since this is subtraction it is important to have the units correct.
\(ARR = 0.1149 - 0.1107 = 0.0042\)
\(NNT = \frac{1}{ARR} = \frac{1}{0.0042} = 238.1\)

Given the information from the previous slide, if we no longer allowed men to treat general medicine patients approximately how long on average would it take for a female physician to save a patient that otherwise would have died under the previous treatment system?
Approximately how many individual study patients are being seen each year by the physicians in this study?
In this study, between 130 and 180 patients are being seen annually by female and male physicians respectively. Thus, the time to save a patient in years is:
\(\frac{238.1}{131.9} = 1.81\)
\(\frac{238.1}{180.5} = 1.32\)