Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

For each of the following tables, calculate (i) the relative risk and (ii) the odds ratio. $$\begin{aligned}&\text { (a) }\\\&\begin{array}{|rr|}\hline 25 & 23 \\\492 & 614 \\\\\hline\end{array}\end{aligned}$$ $$\begin{aligned}&\text { (b) }\\\&\begin{array}{|cr|}\hline 12 & 8 \\\93 & 84 \\\\\hline\end{array}\end{aligned}$$

Short Answer

Expert verified
For table (a), the relative risk is (25*1106) / (48*492) and the odds ratio is (25*614) / (492*23). For table (b), the relative risk is (12*177) / (20*93) and the odds ratio is (12*84) / (93*8).

Step by step solution

01

Define the Terms

Define the key terms: Relative risk is a ratio of the probability of an event occurring in an exposed group versus a non-exposed group. The odds ratio is a measure of association between an exposure and an outcome; it is the ratio of the odds of an event occurring in an exposed group to the odds of it occurring in a non-exposed group.
02

Calculate the Relative Risk for Table (a)

Use the formula for relative risk: RR = (a/(a+b)) / (c/(c+d)), where a=25, b=23, c=492, and d=614. So, RR = (25/(25+23)) / (492/(492+614)) = (25/48) / (492/1106) = (25*1106) / (48*492).
03

Calculate the Odds Ratio for Table (a)

Use the formula for the odds ratio: OR = (a/c) / (b/d), where a=25, b=23, c=492, d=614. So, OR = (25/492) / (23/614) = (25*614) / (492*23).
04

Calculate the Relative Risk for Table (b)

Using the same formula RR = (a/(a+b)) / (c/(c+d)), for table (b) where a=12, b=8, c=93, and d=84, RR = (12/(12+8)) / (93/(93+84)) = (12/20) / (93/177) = (12*177) / (20*93).
05

Calculate the Odds Ratio for Table (b)

Using the same formula OR = (a/c) / (b/d), for table (b) where a=12, b=8, c=93, d=84, OR = (12/93) / (8/84) = (12*84) / (93*8).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability in Exposed vs Non-Exposed Group
Understanding the probability of an event occurring in groups that have been exposed to a particular factor compared to those that have not is a cornerstone of epidemiological research. In the context of our exercise, the exposed group refers to individuals who have been subject to a specific condition or treatment (represented by the first column of numbers), whereas the non-exposed group consists of those who have not (the second column of numbers).

For instance, in Table (a) with numbers 25 (exposed and event occurred) and 492 (exposed and event did not occur), against 23 (non-exposed but event occurred) and 614 (non-exposed and event did not occur), the probabilities we calculate are essential in determining the potential impact of the exposure on the likelihood of the event. In Table (b), the groups are smaller, yet the method remains the same. To convey the difference in these probabilities more accurately to students, it might be helpful to provide real-life context or examples on how exposure might affect the outcome, like the effect of a vaccine on the likelihood of getting a disease.
Measure of Association
The measure of association is a crucial concept that helps us quantify the relationship between an exposure and an outcome. It tells us how much more (or less) likely an event is to occur in one group versus another. In our exercise, two common measures of association are computed: relative risk (RR) and odds ratio (OR).

The relative risk compares the risk of developing an outcome in the exposed group to the non-exposed group, providing a direct measure of risk increase or decrease due to exposure. On the other hand, the odds ratio compares the odds (not the probability) of an event occurring in both groups, which can be particularly useful when dealing with rare events. It's important to clarify, however, that while OR can approximate RR well when the event is rare, they are not the same and can lead to different interpretations, especially in cases where the event is common.
Statistical Analysis in Life Sciences
Statistical analysis plays a vital role in the life sciences for validating hypotheses and making informed decisions based on data. It applies to a variety of fields, from medicine and pharmacy to environmental and agricultural sciences. In the given exercise, statistical methods are used to discern the strength of association between exposure and outcomes, such as diseases or physiological reactions.

Students should be aware that these calculations are part of a larger process that typically includes data collection, data cleaning, analysis, and interpretation. For clarity, we use relative risk and odds ratio as they are particularly suited to studies where conditions or treatments are compared. The real-life impact of correctly applying these statistical tools can directly influence public health policies and medical practices. Additionally, emphasizing the ethical implications of research and the importance of statistical integrity could enhance students' understanding of the field's significance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The growth factor pleiotrophin is associated with cancer progression in humans. In an attempt to monitor the growth of tumors, doctors measured serum pleiotrophin levels in patients with pancreatic cancer and in a control group of patients. They found that only 2 of 28 control patients had serum levels more than two standard deviations above the control group mean, whereas 20 of 41 cancer patients had serum levels this high. \(^{27}\) Use Fisher's exact test to determine whether a discrepancy this large \((2\) of 28 versus 20 of 41\()\) is likely to happen by chance. Use a directional alternative and let \(\alpha=0.05\)

In an experiment to treat patients with "generalized anxiety disorder," the drug hydroxyzine was given to 71 patients, and 30 of them improved. A group of 70 patients were given a placebo, and 20 of them improved. \(^{51}\) Let \(p_{1}\) and \(p_{2}\) represent the probabilities of improvement using hydroxyzine and the placebo, respectively. Construct a \(95 \%\) confidence interval for \(\left(p_{1}-p_{2}\right)\).

Patients with coronary artery disease were randomly assigned to either receive angioplasty plus medical therapy \((n=1149)\) or medical therapy alone \((n=1138)\) in a clinical trial. Over the next several years 85 angioplasty and 95 medical therapy patients died, with cause of death categorized as cardiac, other, or unknown. The following table shows a cross classification of the data. \(^{39}\) Is there statistically significant evidence, at the 0.10 level, to conclude that there is an association between treatment group (angioplasty versus medical therapy) and outcome? (a) State the null and alternative hypotheses in context. (b) How many degrees of freedom are there for a chi-square test? (c) The \(P\) -value for the chi-square test is \(0.87 .\) If \(\alpha=0.10\), what is your conclusion regarding \(H_{0}\) ? $$ \begin{array}{|lccccc|} \hline & && {\text { Unknown }} \\ & {\text { Cardiac death }} & \text { Other death } & \text { cause of death } & \text { Alive } & \text { Total } \\ \hline \text { Angioplasty } & 23 & 45 & 17 & 1,064 & 1,149 \\ \text { Medical therapy } & 25 & 51 & 19 & 1,043 & 1,138 \\ \text { Total } & 48 & 96 & 36 & 2,107 & 2,287 \\ \hline \end{array} $$

Researchers in Norway found that of 61,042 children born to women who took folic acid while pregnant there were 64 who developed autism. \({ }^{63}\) This compares to 50 out of 24,134 for women who did not take folic acid. (a) Calculate the sample value of the odds ratio. (b) Construct a \(95 \%\) confidence interval for the population value of the odds ratio. (c) Interpret the confidence interval from part (b) in the context of this setting.

To compare the impacts of roads in two adjacent ecoregions in Brazil researchers sampled stretches of roadway over a period of 7 years in each region and recorded the species of animals killed by vehicles. In the Atlantic Forest habitat there were 178 roadkills observed, of which 51 were six-banded armadillos ( Euphractus sexcinctus). In the adjacent Cerrado habitat, there were 318 roadkills observed, of which 66 were six-banded armadillos. \({ }^{15}\) Use a chi-square test to assess the evidence that sixbanded armadillos make up different proportions of species killed for the two regions. (a) State the null hypothesis in words. (b) State the null hypothesis in symbols. (c) Compute the sample proportion of six-banded armadillos killed among roadkills in each ecosystem and display the results in a table similar to Table 10.1 .2 (d) Create a table displaying the observed and expected counts. (e) The value of \(\chi_{s}^{2}=3.948\) and the nondirectional \(P\) -value is \(0.047 .\) If \(\alpha=0.05,\) what are your conclusions? (f) Based on these findings, would it be reasonable to surmise that six-banded armadillos are in more danger of being hit at one location than the other, or is further information needed? If so, what information would be helpful?

See all solutions

Recommended explanations on Biology Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free