Nichols-Assignment-E

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University Of Arizona *

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376

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Mechanical Engineering

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Jan 9, 2024

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1 BME 376: Assignment E- Nichols • This assignment is focused on diagnostic tests. All answers should be entered in the space provided below each question. • Most/each sub-question(s) will be graded as 1 (complete) or 0 (incomplete). • Upload a completed word or PDF document to D2L. • Type equations (using in-built Word features) and text. Avoid screenshots or images. You may use the following equation (based on B ayes’ theorem) as the basis for this assignment. 𝑃??𝑖??? 𝑃???𝑖??𝑖?? 𝑉𝑎𝑙?? = 𝑆???𝑖?𝑖?𝑖?𝑦 ∗ 𝑃???𝑎𝑙???? 𝑆???𝑖?𝑖?𝑖?𝑦 ∗ 𝑃???𝑎𝑙???? + (1 − 𝑆???𝑖?𝑖?𝑖?𝑦)(1 − 𝑃???𝑎𝑙????) 1. Cervical cancer is a disease for which the chance of containment is high given that it is detected early. The Pap Smear is a widely accepted screening procedure that can detect cervical cancer; it has been credited with being primarily responsible for the decreasing death rate due to cervical cancer in recent years. The sensitivity and specificity of Pap Smear are 83.75% and 81.36%, respectively. The prevalence of cervical cancer among women is 8.3 cases per 100,000 women. 1.1. [2 points] Compute the positive predictive value of the test. 𝑃??𝑖??? 𝑃???𝑖??𝑖?? 𝑉𝑎𝑙?? = 𝑆???𝑖?𝑖?𝑖?𝑦 ∗ 𝑃???𝑎𝑙???? 𝑆???𝑖?𝑖?𝑖?𝑦 ∗ 𝑃???𝑎𝑙???? + (1 − 𝑆???𝑖?𝑖?𝑖?𝑦)(1 − 𝑃???𝑎𝑙????) 𝑃??𝑖??? 𝑃???𝑖??𝑖?? 𝑉𝑎𝑙?? = 0.8375 ∗ 0.000083 0.8375 ∗ 0.000083 + (1 − 0.8136)(1 − 0.000083) 𝑃??𝑖??? 𝑃???𝑖??𝑖?? 𝑉𝑎𝑙?? = 0.00037 1.2. [2 points] Compute the negative predictive value of the test. 𝑁??𝑎?𝑖?? 𝑃???𝑖??𝑖?? 𝑉𝑎𝑙?? = 𝑆???𝑖?𝑖?𝑖?𝑦 ∗ (1 − 𝑃???𝑎𝑙????) 𝑆???𝑖?𝑖?𝑖?𝑦 ∗ (1 − 𝑃???𝑎𝑙????) + (1 − 𝑆???𝑖?𝑖?𝑖?𝑦)(𝑃???𝑎𝑙????) 𝑁??𝑎?𝑖?? 𝑃???𝑖??𝑖?? 𝑉𝑎𝑙?? = 0.8136 ∗ (1 − 0.000083) 0.8136 ∗ (1 − 0.000083) + (1 − 0.8375)(0.000083) 𝑁??𝑎?𝑖?? 𝑃???𝑖??𝑖?? 𝑉𝑎𝑙?? = 0.99998 1.3. [2 points] Comment on the predictive values. • What would you suggest if the test turns out to be positive? If the test is positive, it is very, very unlikely that they have cervical cancer, but they should take another test to verify. • What would you suggest if the test turns out to be negative? If the test is negative, it is very, very like that the person does not have cervical cancer.

2 2. A “Gold Standard” procedure is available to detect whether or not a person has tuberculosis (TB). Chest X-rays are routinely used as an alternative simple inexpensive procedure to detect the presence or absence of the disease in individuals. We would like to evaluate how good the X-rays are as diagnostic tools. A sample of 30 people, who are known to have tuberculosis, are given a chest x-ray. Another sample of 1,790, who are known not to have tuberculosis, are also given a chest x-ray. The results are summarized below. X-ray Tuberculosis Yes No Negative 8 1739 Positive 22 51 (a) [4 points] Estimate the sensitivity and specificity of chest X-rays in detecting TB. 𝑆???𝑖?𝑖?𝑖?𝑦 = ?(? ∩ ? 1 ) ?(? 1 ) = 22 1820 30 1820 = 22 30 = 0.73 𝑆???𝑖?𝑖?𝑖?𝑦 = ?(? 𝑐 ∩ ? 2 ) ?(? 2 ) = 1739 1820 1790 1820 = 1739 1790 = 0.97 [4 points] In USA, 9.3 people out of 100,000 have tuberculosis. Compute the predictive value of a positive Test and predictive value of a negative test. 𝑃???𝑎𝑙???? = 9.3 100000 = 0.01648 𝑃??𝑖??? 𝑃???𝑖??𝑖?? 𝑉𝑎𝑙?? = 𝑆???𝑖?𝑖?𝑖?𝑦 ∗ 𝑃???𝑎𝑙???? 𝑆???𝑖?𝑖?𝑖?𝑦 ∗ 𝑃???𝑎𝑙???? + (1 − 𝑆???𝑖?𝑖?𝑖?𝑦)(1 − 𝑃???𝑎𝑙????) 𝑃??𝑖??? 𝑃???𝑖??𝑖?? 𝑉𝑎𝑙?? = 0.73 ∗ 0.000093 0.73 ∗ 0.000093 + (1 − 0.97)(1 − 0.000093) = 0.0023 𝑁??𝑎?𝑖?? 𝑃???𝑖??𝑖?? 𝑉𝑎𝑙?? = 𝑆???𝑖?𝑖?𝑖?𝑦 ∗ (1 − 𝑃???𝑎𝑙????) 𝑆???𝑖?𝑖?𝑖?𝑦 ∗ (1 − 𝑃???𝑎𝑙????) + (1 − 𝑆???𝑖?𝑖?𝑖?𝑦)(𝑃???𝑎𝑙????) 𝑁??𝑎?𝑖?? 𝑃???𝑖??𝑖?? 𝑉𝑎𝑙?? = 0.97 ∗ (1 − 0.000093 ) 0.97 ∗ (1 − 0.000093) + (1 − 0.73)( 0.000093 ) = 0.99997 (b) [2 points] Comment on the usefulness of chest X-rays as a diagnostic tool i.e., what would you suggest if the test turns out to be positive? What would you suggest if the test turns out to be negative? If the test turns out to be positive, there is a small chance that the person has TB, but they should retest to make sure their results were accurate. If the test turns out to be negative, there is a very high chance that the patient does not have TB.

3 3. The following data are from a study comparing self-reported smoking status with measured serum cotinine level. In this study, cotinine level was used as a screening tool for smoking status. For various cut-off points (for cotinine level), the observed sensitives and specificities are given below. Cotinine level (ng/mL) Sensitivity Specificity 5 0.971 0.898 7 0.964 0.931 9 0.960 0.946 11 0.954 0.951 13 0.950 0.954 14 0.949 0.956 15 0.945 0.960 17 0.939 0.963 19 0.932 0.965 a) [3 points] Determine the cut-off point that will yield the most optimal sensitivity and specificity for the cotinine levels as a screening test. 𝐷 2 = (1 − 𝑆???𝑖?𝑖?𝑖?𝑦) 2 + (1 − 𝑆???𝑖?𝑖?𝑖?𝑦) 2 Cotinine Level 5: 𝐷 2 = (1 − 0.971) 2 + (1 − 0.898) 2 = 0.011 Cotinine Level 7: 𝐷 2 = (1 − 0.964) 2 + (1 − 0.931) 2 = 0.006 Cotinine Level 9: 𝐷 2 = (1 − 0.960) 2 + (1 − 0.946) 2 = 0.004516 Cotinine Level 11: 𝐷 2 = (1 − 0.954) 2 + (1 − 0.951) 2 = 0.004517 Cotinine Level 13: 𝐷 2 = (1 − 0.950) 2 + (1 − 0.954) 2 = 0.004616 Cotinine Level 14: 𝐷 2 = (1 − 0.949) 2 + (1 − 0.956) 2 = 0.004537 Cotinine Level 15: 𝐷 2 = (1 − 0.945) 2 + (1 − 0.960) 2 = 0.004625 Cotinine Level 17: 𝐷 2 = (1 − 0.939) 2 + (1 − 0.963) 2 = 0.00509 Cotinine Level 19: 𝐷 2 = (1 − 0.932) 2 + (1 − 0.965) 2 = 0.005849 Cotinine levels of 9 ng/mL are the most optimal because they have the lowest D-squared value. b) [1 point] Cotinine levels can be measured in both saliva and urine. A study determined the AUC - area under the receiver operating characteristic (ROC) curve – was 0.958 and 0.973 for saliva and urine-based cotinine test, respectively. Based on the AUC, which test is more reliable? Based on the AUC, the urine-based cotinine test is more reliable. This is because this test has a higher AUC value which means that its diagnostic accuracy was better, making the test more reliable.

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