# Difference between revisions of "SMHS OR RR"

## Scientific Methods for Health Sciences - Odds Ratio and Relative Risk

### Overview

The relative risk is a measure of dependence that allows us to compare two probabilities in terms of their ratio $\frac{p_1}{p_2}$ rather than their difference $(p_1 – p_2)$. Relative risk is a commonly used measure in public health studies. Another way to compare two probabilities is in terms of the odds. If an event takes place with probability p, then the odds of the event occurring are $\frac{p}{1 - p}$. The odds ratio is the ratio of odds for two complementary probabilities.

### Motivation

Suppose we study brain cancer in the context of cell phone use. The table below illustrates some simulated data. One clear healthcare question in this case-study could be: Is cell phone use associated with a higher incidence of brain cancer? To address this question, we can look at the relative risk of brain cancer in people who use cell phones.

 Brain Cancer (BC) Total Yes No Cell Phone (CP) Yes 18 80 98 (B) No 7 95 102 (C) Total 25 175 200

First, we compute the (conditional!) probabilities (P) of brain cancer (BC) given either cell phone use, P1, or no cell-phone use, P2. We can then form their ratio to determine whether the relative risk of brain cancer (BC) is higher in cell phone users (CP) than in non-users (NCP).

$$P_1 = P(BC|CP) = \dfrac {18}{98} = 0.184$$

$$P_2= P(BC|NCP) = \dfrac {7} {102} = 0.069$$

Therefore, the relative risk of brain cancer in cell phone users is: $$RR= \frac{P(BC|CP)}{P(BC|NCP)} = \frac {0.184}{0.069} = 2.67.$$

The risk of having brain cancer is more than 2.5 times greater among cell phone users compared to non-cell phone users.

For the same example, the odds ratio (OR) of brain cancer relative to cell-phone use is:

$$OR = \frac{\frac{P \left( BC \mid CP \right)}{1 - P \left( BC \mid CP \right)}}{\frac{P \left( BC \mid NCP \right)}{1 - P \left( BC \mid NCP \right)}} = \frac{\frac{\frac{18}{98}}{1 - \frac{18}{98}}} {\frac{\frac{7}{102}}{1 - \frac{7}{102}}} =\frac{\frac{0.184}{0.816}}{\frac{0.069}{0.931}} = 3.04$$

Thus, the odds of having brain cancer is about 3 times greater for cell phone users than it is for non-cell phone users. We could have compared the odds of owning a cell phone given that a patient had brain cancer (i.e., the column-wise probabilities), $P(CP|BC) = 18/25 = 0.72$ versus $P(CP|NBC) = 80/175 = 0.457$. However, this does not seem as important scientifically.

### Theory

 Factor 1 Total Yes No Factor 2 Yes $n_{1,1}$ $n_{1,2}$ $n_{1,1} + n_{1,2}$ No $n_{2,1}$ $n_{2,2}$ $n_{2,1} + n_{2,2}$ Total $n_{1,1} + n_{2,1}$ $n_{2,1} + n_{1,2}$ $N=n_{1,1} + n_{1,2} + n_{2,1} + n_{2,2}$

$$RR=\frac{\frac{n_{1,1}}{n_{1,1}+ n_{1,2}}}{\frac{n_{2,1}}{n_{2,1}+n_{2,2}}}.$$

$$OR = \frac{n_{1,1} × n_{2,2}}{n_{1,2}× n_{2,1}}.$$

#### Interpretation

• RR: In general, the measure relative risk (RR) is interpreted as follows:
• RR = 1 indicates that the probabilities of two events are the same.
• RR > 1 implies that there is an increased risk.
• RR < 1 implies that there is a decreased risk.
• OR
• If event $A|B$ has probability $p = 1/2$, then the odds are $\frac{1/2}{1/2}=1$, or $1:1$, or 1 to 1. This means the probability that event $A|B$ occurs is equal to the probability that it does not occur.
• If event $A|C$ has probability $p = 3/4$, then the odds are $\frac{3/4}{1/4}= 3$, or 3 to 1. The probability that the event $A|C$ occurs is three times as large as the probability that it does not occur.
• Similarly, if $A|D$ has probability $p = 1/4$, then the odds are $\frac {1/4}{3/4}=\frac {1}{3}$, or 1 to 3. The probability that the event $A|D$ occurs is three times smaller than the probability that it does not occur.
• RR vs. OR
• The formula and reasoning for the relative risk is a little bit easier to follow. In most cases the $OR$ and $RR$ measures are roughly equal to each other.
• Odds ratios have an advantage over relative risk because they can be calculated no matter the row or column comparison.
• Relative risk runs into problems when the study design is a cohort study or a case-control design.
• Odds ratios are an approximation of relative risk: $OR = RR \times \frac{1-P_2} {1-P_1}$.

#### Inference

• Inference on OR: In practice, we commonly report ORs along with their confidence intervals (CIs). It turns out that the distribution of ORs is not normal; however, the log-transformed OR is approximately normally distributed and the standard error of $ln(OR)$ may be approximated by:

$$SE(ln(OR))= \sqrt{\frac {1} {n_{1,1}}+ \frac {1} {n_{1,2}} + \frac {1} {n_{2,1}} + \frac{1} {n_{2,2}}}.$$

Thus, if $\alpha$ is the false-positive (i.e., Type I) error rate, the $(1-\alpha)100\%$ CI of the log-transformed OR can be computed by: $$ln(OR)±z_{\frac{\alpha}{2}}SE(ln(OR)),$$

where the OR point estimate is $OR = \frac {n_{1,1}×n_{2,2}}{n_{1,2}×n_{2,1}}$, and the standard error of the log-transformed OR is listed above (i.e., $SE(ln(OR))$).

You can use the SOCR Student’s T-distribution calculator to compute the value of the standard normal Z statistics for a given false-positive rate, $\alpha$.

NOTE: Remember that the lower $(L=ln(OR)-z_{\frac{\alpha}{2}}SE(ln(OR))$ and upper $(U=ln(OR)+z_{\frac{\alpha}{2}}SE(ln(OR))$ limits of the $ln(OR)$ confidence interval represent log-transformed data. To convert these confidence limits into real OR terms, you need to invert the log transform (i.e., using the exponential function). Thus, the $CI(OR)$ would be: $(e^L,e^R)$. Also remember that the critical value for the normal distribution, $z_{\frac{\alpha}{2}}$, can be computed using the SOCR Normal distribution calculator, or the Normal Distribution Table.

### Applications

• This article retrospectively studies the relationship between surveillance, staffing, and serious adverse events in children on general care postoperative units. The paper investigates these hypotheses: (1) the relationship between patient factors and surveillance is moderated by staffing (i.e., registered nurse hours per patient per shift), and (2) the relationship between staffing and serious adverse events is mediated by surveillance.
The study demonstrates that one additional full-time registered nurse equivalent per day reduced the odds of in-hospital mortality, respiratory failure, pneumonia, and failure to rescue; the greatest cost-benefit was found in adult surgical patients. Table 4 of the results shows the $OR$ and $CI(OR)$. Interpret the findings.
Predictors of adverse events as shown in final logistic regression analysis.
 Factors β (S.E.) p-Value Odds ratio [95% CI) Staffing −0.41 (0.33) 0.219 0.66 [0.35, 1.28] American Society of Anesthesiologists Physical Status 0.94 (0.39) 0.017 2.57 [1.88, 5.55] Comorbidity 0.57 (0.43) 0.189 1.76 [0.76, 4.12] Perioperative complication 0.64 (0.22) 0.003 1.90 [1.24, 2.92] Interaction staffing × surveillance −1.04 (0.42) 0.012 0.354 [0.157, 0.798]
• This article investigates whether hospitals with well-organized care (e.g., improved nurse staffing and work environments) provide better patient care and nurse workforce stability in European countries and the United States. It uses data from 488 clinics in 12 European countries and 617 in the US. It is based on 33,659 nurses and 11,318 patients in Europe and 27,509 nurses and more than 120,000 patients in the US.
Some of the authors’ findings are that nurses in hospitals with better work environments were approximately half as likely to (a) report poor or fair quality of care (Europe, adjusted odds ratio 0.56, 95% confidence interval 0.51 to 0.61; US, 0.54, 0.51 to 0.58) and (b) give their hospitals poor or failing grades on patient safety (0.50, 0.44 to 0.56 EU; 0.55, 0.50 to 0.61 US).
Interpret the results in the table below. Note that in this nurse outcomes study, the authors adjusted the regression estimates (odds ratios) at the hospital level for differences in the composition of nurses between hospitals and between countries (i.e., age, sex, full time employment status, and specialty) using a multilevel model structure in which nurses were nested within hospitals and countries.
Effects of nurse staffing and practice environment on nurse outcomes in study countries.
 Nurse Outcome Europe US Unadjusted odds ratio (95% CI) Adjusted odds ratio (95% CI) Unadjusted odds ratio (95% CI) Adjusted odds ratio (95% CI) Poor or fair quality of care in ward Practice environment 0.58 0.56 0.52 0.54 (0.53 to 0.63) (0.51 to 0.61) (0.49 to 0.56) (0.51 to 0.58) Staffing 1.11 1.11 1.2 1.06 (1.08 to 1.13) (1.07 to 1.15) (1.16 to 1.25) (1.03 to 1.1) Poor or fair quality of care in ward Practice environment 0.5 0.5 0.53 0.55 (0.43 to 0.57) (0.44 to 0.56) (0.48 to 0.59) (0.5 to 0.61) Staffing 1.04 1.1 1.18 1.05 (1.01 to 1.08) (1.05 to 1.16) (1.12 to 1.23) (1 to 1.1) Burnout Practice environment 0.69 0.67 0.69 0.71 (0.63 to 0.76) (0.61 to 0.73) (0.66 to 0.73) (0.68 to 0.75) Staffing 1.06 1.05 1.12 1.03 (1.04 to 1.08) (1.02 to 1.09) (1.08 to 1.15) (1 to 1.06) Job dissatisfaction Practice environment 0.63 0.52 0.58 0.6 (0.57 to 0.69) (0.47 to 0.57) (0.55 to 0.61) (0.57 to 0.64) Staffing 1.1 1.07 1.17 1.06 (1.08 to 1.12) (1.04 to 1.11) (1.13 to 1.21) (1.03 to 1.09) Intention to leave in the next year Practice environment 0.72 0.61 0.7 0.69 (0.66 to 0.79) (0.56 to 0.67) (0.65 to 0.76) (0.64 to 0.75) Staffing 1.04 1.05 1.1 1.03 (1.01 to 1.06) (1.02 to 1.09) (1.05 to 1.15) (0.98 to 1.08) Not confident that patients can manage own care after hospital discharge Practice environment 0.62 0.73 0.71 0.72 (0.56 to 0.69) (0.69 to 0.78) (0.67 to 0.75) (0.68 to 0.77) Staffing 1.08 1.03 1.1 1.04 (1.05 to 1.11) (1 to 1.05) (1.06 to 1.13) (1.01 to 1.07) Not confident that hospital management would resolve patients’ problems Practice environment 0.5 0.53 0.56 0.56 (0.46 to 0.54) (0.48 to 0.58) (0.53 to 0.59) (0.54 to 0.59) Staffing 1.04 1.02 1.12 1.01 (1.01 to 1.07) (0.98 to 1.06) (1.09 to 1.17) (0.98 to 1.03)

### Problems

Formulate some clinically relevant questions in terms of the $OR$ and $RR$, and try to answer them in the following situations. Interpret the results. E.g., the estimate of the relative risk of a heart attack is approximately ____ for those who smoke versus those who do not smoke. Compute the $CI$ of the $OR$.

 Heart Attack (HA) Total Yes No Smoking (S) Yes 33 18 51 No 167 182 349 Total 200 200 400