Question

In: Statistics and Probability

4.12. miRNA Identifies NSCLS. Tremendous efforts have been made to de- velop cancer biomarkers by detecting...

4.12. miRNA Identifies NSCLS. Tremendous efforts have been made to de- velop cancer biomarkers by detecting circulating extracellular miRNAs directly released from tumors. Yet, until recently none of the cell-free biomarkers has been accepted to be used for early detection of non-small cell lung cancer (NSCLC).
Ma et al. (2015) investigated whether analysis of miRNA expressions of peripheral blood mononuclear cells (PBMC) has diagnostic value for NSCLC. They identified several PBMC miRNAs with a significantly al- tered expression level in subjects with NSCLC. In a training set of 84 patients with confirmed NSCLC and 69 cancer-free smokers, a panel of two miRNAs (miRs-19b-3p and -29b-3p) produced 72.62% sensitivity and 82.61% specificity in identifying NSCLC.
(a) From the data supplied, recover the table with TP, FP, TN, and FN. (b) If Mr. Molina is one of the participants in the study and tested posi- tive for NSCLC, what is the probability that he has the disease?
(c) Find F-measure and Youden Index for the test.
(d) The prevalence of undiagnosed NSCLC in the population of smokers aged over 60 is 0.3%. Mr. Ramussen and Ms. Antonetti are randomly selected from that population and tested for NSCLC. What is the prob- ability that Mr. Ramussen has disease if he tested positive? What is the probability that Ms. Antonetti does not have the disease if she tested negative.
(e) Based on pretest symptoms there was 1:3 odds that Mr. Bhwana has disease. He tested positive. What are his posttest odds?

Solutions

Expert Solution

(a) The confusion matrix is shown below:

>> Sensitivity = 72.62% = TP / TP + FN

We have TP + FN = 84

Therefore, TP = 72.62% * 84 = 61

FN = 84-61 = 23

>> Specificity = 82.61% = TN / TN + FP

We have TN + FP = 69

Therefore, TN = 82.61% * 69 = 57

FP = 69-57 = 12

Hence we have the confusion matrix as:

(b) Probability of Mr Molina having the disease = TP / TP + FP = 61/ (61 + 12) = 0.835

(c) F- measure = 2TP / 2TP + FP + FN = 2* 61 / (2*61 + 12 + 23) = 0.77

Youden Index = Sensitivity + Specificity - 1

= 72.62 % + 82.61% - 1

= 55.23%

(d) Probability of Mr Ramussen having the disease = TP / TP + FP = 61/ (61 + 12) = 0.8356

Probability of Ms Antonetti not having the disease = TN / TN + FN = 57/ (57 + 23) = 0.7125

(e) Probability for Mr. Bhwana having the disease = 1/4 * 0.8356 = 0.2089

  = 1:4 odds


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