/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 One important aspect of medical ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

One important aspect of medical diagnosis is its reproducibility. Suppose that two different doctors examine 100 patients for dyspnea in a respiratory- disease clinic and that doctor A diagnosed 15 patients as having dyspnea, doctor B diagnosed 10 patients as having dyspnea, and both doctor \(A\) and doctor \(B\) diagnosed 7 patients as having dyspnea. Compute the Kappa statistic and its standard error regarding reproducibility of the diagnosis of dyspnea in this clinic.

Short Answer

Expert verified
The Kappa statistic is 0.50 with a standard error of approximately 0.1407.

Step by step solution

01

Understand the Kappa Statistic

The Kappa statistic measures the agreement between two raters, adjusted for the agreement occurring by chance. It is calculated as follows: \( \kappa = \frac{P_o - P_e}{1 - P_e} \), where \( P_o \) is the observed agreement, and \( P_e \) is the expected agreement by chance.
02

Construct a Contingency Table

Using the information given, we create a contingency table:\[ \begin{array}{c|cc|c}\text{Doctor B} & \text{Dyspnea} & \text{No Dyspnea} & \text{Total} \\hline\text{Dyspnea} & 7 & 8 & 15 \\text{No Dyspnea} & 3 & 82 & 85 \\hline\text{Total} & 10 & 90 & 100 \\end{array} \]
03

Calculate Observed Agreement \((P_o)\)

The observed agreement \(P_o\) is the proportion of all cases in which the two doctors agree. From the table, agreements occur in 7 (both diagnosed dyspnea) + 82 (both did not diagnose dyspnea) cases, making \(P_o = \frac{7 + 82}{100} = 0.89\).
04

Calculate Expected Agreement \((P_e)\)

The expected agreement \(P_e\) is calculated by taking the product of the marginals for both diagnoses:- Dyspnea agreements: \(\frac{15}{100} \times \frac{10}{100} = 0.015\)- No Dyspnea agreements: \(\frac{85}{100} \times \frac{90}{100} = 0.765\)Thus, \(P_e = 0.015 + 0.765 = 0.78\).
05

Calculate Kappa \((\kappa)\)

Using the values from previous steps, we find \( \kappa = \frac{0.89 - 0.78}{1 - 0.78} = \frac{0.11}{0.22} = 0.50\). This indicates a moderate agreement between the doctors above chance level.
06

Calculate Standard Error of Kappa

The standard error \(SE(\kappa)\) is given by \( \sqrt{\frac{P_o(1 - P_o)}{N(1 - P_e)^2}}\), where \(N\) is the total number of observations:\(SE(\kappa) = \sqrt{\frac{0.89 \times 0.11}{100 \times 0.22^2}} = \sqrt{\frac{0.0979}{4.84}} \approx 0.1407\).
07

Interpret the Kappa Statistic

A Kappa of 0.50 suggests moderate agreement between the doctors. The standard error indicates the precision of this estimate, helping us understand the variability around this agreement measure.

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 91Ó°ÊÓ!

Key Concepts

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

Reproducibility in Medical Diagnosis
Reproducibility in medical diagnosis refers to the consistency of a diagnostic method when used by different observers or over time. It is crucial, as consistent diagnosis can lead to consistent treatment and better patient care. In our example, two doctors examined the same set of 100 patients for the presence of dyspnea. If the examination results are reproducible, we can trust that the observed diagnosis is reliable.

When reproducibility is lacking, it can lead to differences in treatment outcomes due to inconsistent patient diagnosis. Consistency between healthcare providers ensures that patients receive accurate and uniform diagnoses, which is why evaluating reproducibility using statistical measures like the Kappa statistic is vital. It allows us to quantify the level of agreement beyond what would be expected by chance alone, marking the effectiveness and harmony of the diagnostic method used.
Observed and Expected Agreement
The observed agreement between two or more raters is simply how often they concur in their assessments. In the dyspnea example, the two doctors agreed on the diagnosis 89% of the time, which is a high observed agreement. This is calculated by adding up the instances where both agreed on either diagnosis of disease or absence of disease and then dividing by the total observations.

However, agreement can sometimes occur simply by chance, which does not necessarily indicate true agreement. The expected agreement accounts for this by calculating what agreement would look like in a situation where judgments are made randomly based on certain probabilities. In our case, the expected agreement was 78%.

When considering results, it's important to compare the observed agreement to the expected agreement. The difference between these values forms the basis for calculating the Kappa statistic, which provides insight into how much of the agreement is genuine versus chance-driven.
Contingency Table Analysis
Contingency table analysis is a statistical tool used to show the frequency distribution of variables in a tabular form. It is particularly useful in assessing categorical data and examining the relationship between them.

In the context of the medical diagnosis example, a contingency table was crafted to categorize patients based on the diagnosis from each doctor, highlighting agreements and disagreements. The table comprises four fields: patients diagnosed with dyspnea by both doctors, only by one doctor, neither by both, and totals for each.

This analysis facilitates easy visualization of the data, helping identify patterns of agreement and disagreement. By examining this table, we can further calculate key statistics like the observed and expected agreements. The simplicity of accessing the proportions directly aids in driving analyses in statistical measures, such as identifying how much better than chance the agreement between the doctors is, through the computation of the Kappa statistic.

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

A study group of 576 working women \(30-49\) years of age who took phenacetin- containing analgesics and a control group of 533 comparably aged women without such intake were identified in 1968 and followed for mortality and morbidity outcomes. One hypothesis to be tested was that phenacetin intake may influence renal (kidney) function and hence have an effect on specific indices of renal morbidity and mortality. The mortality status of these women was determined from 1968 to \(1987 .\) The researchers found that 16 of the women in the study group and 1 of the women in the control group died, where at least one cause of death was considered renal [13]. What statistical test should be used to compare the total mortality experience of the study group with that of the control group?

Researchers collected the following data concerning comparability of diagnoses of schizophrenia obtained from primary-care physician report as compared with proxy report (from spouses). Data were collected concerning 953 people (referred to as index subjects). The researchers found that schizophrenia was identified as present on 115 physician reports and 124 proxy reports. Both physician and proxy informants identified 34 people as positive, and they are included among the 115 and 124 individuals described. Suppose there is no difference in the percentage of subjects identified as schizophrenic by physician and by proxy informants. Does this mean the two sources of information are the same for each individual? Why or why not?

Aminoglycosides are powerful broad-spectrum antibiotics used for gram-negative infections often in seriously ill patients. For example, the drugs are often prescribed for drug-resistant tuberculosis as recommended by the World Health Organization. However, these drugs have serious side effects, including irreversible hearing loss referred to as ototoxicity. The most commonly prescribed aminoglycoside is gentamicin. A clinical trial was set up in China to assess whether the addition of aspirin to a standard regimen of gentamicin would have an effect on the incidence of ototoxicity [28]. There were 195 patients enrolled in a prospective, randomized, double-blind clinical trial. Of these, 106 patients were randomized to a twice daily regimen of \(80-160\) mg of gentamicin plus placebo and 89 patients were randomized to receive the same regimen of gentamicin \(+3 \mathrm{g}\) of aspirin (ASA) daily. (i) What is a prospective study? What are its advantages? (ii) What is a randomized study? What are its advantages? (iii) What is a double-blind study? Are there advantages vs. other approaches? After 2 weeks of treatment, 14 of the placebo patients and 3 of the ASA patients developed clinically significant hearing loss.

It is fairly well known that perception of weight by adolescents does not always agree with actual weight. What is less clear is whether perception of weight differs by gender. For this purpose, a study was performed among students in a local high school, where students provided their actual height and weight by self-report. The following data were obtained from 286 students \((143 \text { boys and } 143\) girls). (The data for this problem were provided by Laura Rosner.) The students were classified as underweight if their bodymass index (BMI) \(\left(\mathrm{kg} / \mathrm{m}^{2}\right)\) was less than \(18.0 \mathrm{kg} / \mathrm{m}^{2}\), as normal if their BMI was \(\geq 18.0\) and \(<25.0,\) and overweight if their BMI was \(\geq 25.0\). Based on these criteria, 17 of the girls were underweight, 113 were of normal weight, and 13 were overweight. For the boys, 7 were underweight, 115 were of normal weight, and 21 were overweight. What test procedure can be used to assess whether the weight status of boys significantly differs from girls?

A case-control study was performed among 145 subjects with macular degeneration and 34 controls, all of whom were \(70-\) to 79 -year-old women. A genetic risk score was developed to help differentiate the cases from the controls. The risk score was categorized into six groups \((1,2,3,4,\) 5, 6), with 6 being the highest risk and 1 being the lowest risk. The data in Table 10.48 were obtained relating the risk score to case/control status. What is the relative risk for AMD among women in group 6 compared with women in group \(1 ?\) What does the relative risk mean in words?

See all solutions

Recommended explanations on Math 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.