SMHS OR RR - Revision history

Dinov at 12:27, 18 June 2020

2020-06-18T12:27:38Z

Dinov: /* Inference */

2017-03-15T21:05:45Z

‎Inference

Glenbrau: /* Problems */

2015-03-27T17:11:10Z

‎Problems

Glenbrau: /* Applications */

2015-03-27T17:10:34Z

‎Applications

Glenbrau: /* Interpretation */

2015-03-27T15:34:51Z

‎Interpretation

Glenbrau: /* Interpretation */

2015-03-27T15:31:24Z

‎Interpretation

Glenbrau: /* Applications */

2015-03-24T19:47:51Z

‎Applications

Glenbrau: /* Inference */

2015-03-24T19:09:15Z

‎Inference

Glenbrau: /* Interpretation */

2015-03-24T19:06:31Z

‎Interpretation

Dinov: /* Overview */

2015-03-17T13:07:43Z

‎Overview

@@ Line 2: / Line 2: @@
 ===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
+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.
+\((p_1 – p_2)\). 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===
@@ Line 40: / Line 40: @@
 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.
+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===
@@ Line 50: / Line 50: @@
 |Yes||No
 |-
-| rowspan=2|Factor 2||Yes||$n_{1,1}$||$n_{1,2}$||$n_{1,1} + n_{1,2}$
+| rowspan=2|Factor 2||Yes|| \(n_{1,1} \)|| \)n_{1,2} \)|| \(n_{1,1} + n_{1,2} \)
 |-
-|No||$n_{2,1}$||$n_{2,2}$||$n_{2,1} + n_{2,2}$
+|No|| \(n_{2,1} \)|| \(n_{2,2} \) || \(n_{2,1} + n_{2,2} \)
 |-
-| colspan=2|Total||$n_{1,1} + n_{2,1}$||$n_{2,1} + n_{1,2}$||$N=n_{1,1} + n_{1,2} + n_{2,1} + n_{2,2}$
+| colspan=2|Total|| \(n_{1,1} + n_{2,1} \) || \(n_{2,1} + n_{1,2} \) || \(N=n_{1,1} + n_{1,2} + n_{2,1} + n_{2,2} \)
 |}
 </center>
@@ Line 69: / Line 69: @@
 * '''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 \(A|B\) 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.
+**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.
+**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.
+**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}$.
+**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 [https://en.wikipedia.org/wiki/Odds_ratio#Statistical_inference standard error of $ ln(OR)$ may be approximated by]:
+* ''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 [https://en.wikipedia.org/wiki/Odds_ratio#Statistical_inference 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:
+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., $SE(ln(OR))$).
+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 [http://socr.ucla.edu/htmls/dist/StudentT_Distribution.html SOCR Student’s T-distribution calculator] to compute the value of the standard normal Z statistics for a given false-positive rate, $\alpha$.
+You can use the [http://socr.ucla.edu/htmls/dist/StudentT_Distribution.html 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 [[AP_Statistics_Curriculum_2007_Normal_Std|critical value for the normal distribution]], $z_{\frac{\alpha}{2}}$, can be computed using the [http://socr.ucla.edu/htmls/dist/Normal_Distribution.html SOCR Normal distribution calculator], or the [http://socr.umich.edu/Applets/Z-table.html Normal Distribution Table].
+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 [[AP_Statistics_Curriculum_2007_Normal_Std|critical value for the normal distribution]], \(z_{\frac{\alpha}{2}}\), can be computed using the [http://socr.ucla.edu/htmls/dist/Normal_Distribution.html SOCR Normal distribution calculator], or the [http://socr.umich.edu/Applets/Z-table.html Normal Distribution Table].
 ===Applications===
 * [http://dx.doi.org/10.1016/j.ijnurstu.2012.11.014 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.
+: 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.
 <center>Predictors of adverse events as shown in final logistic regression analysis.</center>
@@ Line 210: / Line 210: @@
 ===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 $OR$.
+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 \(OR\).
 <center>

@@ Line 80: / Line 80: @@
 ====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)$  is:
+* ''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 [https://en.wikipedia.org/wiki/Odds_ratio#Statistical_inference standard error of $ ln(OR)$ may be approximated by]:
 $$ SE(ln(OR))= \sqrt{\frac {1} {n_{1,1}}+ \frac {1} {n_{1,2}} + \frac {1} {n_{2,1}} + \frac{1} {n_{2,2}}}.$$

@@ Line 210: / Line 210: @@
 ===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.
+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 $OR$.
 <center>

@@ Line 96: / Line 96: @@
 * [http://dx.doi.org/10.1016/j.ijnurstu.2012.11.014 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.
+: 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.
 <center>Predictors of adverse events as shown in final logistic regression analysis.</center>

@@ Line 74: / Line 74: @@
 *'''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.
+**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.

@@ Line 118: / Line 118: @@
 * [http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3308724/ 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 included 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).
+: 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 [[SMHS_OR_RR#References|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.