2. Suppose a new test to detect canine parvovirus shows the following result: Parvo Present Parvo Absent Positive Test Result Negative Test Result 1000 30 20 1200 a. Compute the sensitivity and specificity of the test and interpret them.
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- Interpreting p-values, however, are not foolproof and may lead to certain errors. Differentiate between types 1 and 2 errors; discuss these errors in the context of your study. Assume the following: 1a. You committed a type 1 error. What could be the practical consequences arising from those errors? What steps would you take to avoid making these errors? 1b. You committed a type 2 error. What could be the practical consequences arising from those errors? What opportunities did you end up missing because of such an error? What steps would you take to avoid making these errors?In which case is a type II error committed?Examus cdn.studentuae.examus.netridban3D1&sessi... Choose the correct answer for the following question: A simple random sample of front-seat occupants involved in car crashes is obtained. Among 1550 occupants not wearing seat belts, 12 were killed. Among 6500 occupants wearing seat belts, 12 were killed. Using this data with 0.05 significance level to test the claim that the fatality rate is higher for those not wearing seat belts, the alternative hypothesis is given as ct912 10 a. 362 b. H:P1 SP2 C. Hj: p P2 cf917 d. Hj: p>P2 ce362dcf917 ce362dct917 ce3624ci917 362dcf917 ce362acf917 ce362dcf917 0% ce362dcf917 MacBook Pro 7. e362dcf917 C0 P.
- A new virus has taken root in a country. Government officials are reporting that 9.4% of the population is currently infected with the virus. However, epidemiologists across the country claim to be observing a much higher infection rate. It was found that 171 people out of a randomly selected sample of 1490 people from around the country were infected with the virus. Use the critical value method to determine if the sample data support the epidemiologists' supposition that the true rate of infection in the country is higher than 9.4%. Use a significance level of 5%. State the null and alternative hypothesis for this test. Ho: ? V H₁: ? ✓ Determine if this test is left-tailed, right-tailed, or two-tailed. Otwo-tailed Oright-tailed Oleft-tailed Should the standard normal (2) distribution or Student's (t) distribution be used for this test? O The standard normal (2) distribution should be used The Student's t distribution should be used Determine the critical value(s) for this hypothesis…A researcher wants to test the effectiveness of a new drug to treat asthma. One hundred patients with chronic, stable asthma were randomly assigned to one of three groups to receive a placebo (0mg), 1mg of drug, or 5 mg of drug. After treatment, a doctor tested the patient’s forced expiratory volume for 1 second (FEV1). Which statistical test would be best to determine if there is a significant difference between groups? one-sample t-test dependent samples t-test independent samples t-test ANOVAA researcher wants to test the effectiveness of a new drug to treat asthma. One hundred patients with chronic, stable asthma were randomly assigned to one of three groups to receive a placebo (0mg), 1mg of drug, or 5 mg of drug. After treatment, a doctor tested the patient’s forced expiratory volume for 1 second (FEV1). Which statistical test would be best to determine if there is a significant difference between groups?
- A new noninvasive screening test is proposed that is claimed to be able to identify patients with impaired glucose tolerance based on a battery of questions related to health behaviors. The new test is given to 75 patients. Based on each patient's response to the questions they are classified as positive or negative for impaired glucose tolerance. Each patient also submits a blood sample and their glucose tolerance status id determined and tabulated below. What is the sensitivity of the screening test? screening test impaired glucose tolerance not impaired positive 17 13 negative 8 37 please clearly explain how you got this so I can try to figure it out on my ownA psychiatrist devised a short screening test for depression. An independent blind comparison was made with a gold standard for diagnosis of depression among 215 psychiatric outpatients. Among the 69 outpatients found to be depressed according to the gold standard, 44 patients were positive for the test. Among 146 patients found not to be depressed according to the gold standard, 39 patients were found to be positive for the test. 1. False-negative error rate 2. False-positive error rate 3. Positive predictive value 4. Negative predictive value 5. Positive likelihood ratio (Use the already rounded-off answers when 6. Negative likelihood ratio (Use the already rounded-off answers when computing) computing)Select all the answers that are associated with the p value Type I error alpha significance level Critical value Type II error
- 3. Activated-Protein-C (APC) resistance is a serum marker that has been associated with hrombosis (the formation of blood clots often leading to heart attacks) among adults. A dy assessed this risk factor among adolescents. To assess the reproducibility of the assay, lit-sample technique was used in which a blood sample was provided by 10 people; each ople was split into two aliquots (sub-samples), and each aliquot was assessed separately. 1. To assess the reliability, compute the Standard Error of Mean TABLE 2.19 APC resistance split-samples data Sample number A А -В 2. To assess the variability of the assay, the investigators need coefficient of variation. Compute the coefficient of variation (CV) for each subject by obtaining the mean and standard deviation over the 2 replicates for each subject, and give your interpretation based on the evidence. 1 2.22 1.88 0.34 to compute the 3.42 3.59 -0.17 3 3.68 3.01 0.67 4 2.64 2.37 0.27 2.68 2.26 0.42 6. 3.29 3.04 0.25 7 3.85 3.57 0.28 8.…Define and explain Type I and Type II errors and relate each to the selection of an alpha level.For a given disease gene, specified mode of inheritance and observed data set, if the p-value for a chi-square goodness of fit analysis is 0.98, and the significance level is 0.001, then we: a) make no decision b) accept the null hypothesis that the disease almost certainly follows the specified mode of inheritance at the given gene. c) reject the null hypothesis at the 0.0001 level. d) reject the null hypothesis that the disease follows the specified mode of inheritance at the given gene. e) reject the null hypothesis that the disease does not follow the specified mode of inheritance at the given gene.