A physical examination was used to screen for breast cancer in 2,500 women with biopsy-proven adenocarcinoma of the breast and in 5,000 age- and race-matched control women. The results of the physical examination were positive (i. e., a mass was palpated) in 1,800 cases and in 800 control women, all of whom showed no evidence of cancer at biopsy. The sensitivity of the physical examination was:

72%

Explanation:

There are two main caracteristics of the test: sensitivity and specificity.

Sensitivity (Sn) is ability of the test to correctly count those with the disease. Specificity (Sp) is ability of the test to correctly identify those without the disease.

To calculate Sn and Sp, there has to be true information on number of diseased and desease-free patients which in this case, we do have according to biopsy results.

2500 patients has carcinoma

5000 patients are disease-free individuals

According to the physical examination, there are 1800 patients with carcinoma (among 2500 with biopsy-established carcinoma, these are true positive) and 800 patients with carcinoma among those 5000 which do not have a disease, according to biopsy results (these are false positive). True negative will be the rest of the disease-free patients (5000-800) and false negative are remaining patients out of 2500 certainly positive (2500-1800).

A table can be made:

                 Disease

                 +         -

Test  +    TP       FP  

          -    FN       TN

TP - true positive

FP - false positive

FN - false negative

TN - true negative

Sn and Sp are calculated:

Sn=      Sp=

In this case we have:

                                          Carcinoma

                                             +            -

Physical examination  +    1800     800  

                                      -    700       4200

Sn=%

Sp=%

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i wish you would have provided choices to simplify it and allow me to give an exact answer.