For Archmage ONLY

1. Concluding that no relationship exists between two events when, in fact, there is a relationship between them, is referred to as what type of error?

Type II Error: Because we fail to reject the null hypothesis, which is false.

2. A study has the following characteristics:

The subject group is composed of individuals who have a history of having developed the outcome of interest

The control group is composed of individuals who have a history of not having developed the outcome of interest

This type of study would be a experimental study.

3. Statins (cholesterol lowering agents) have been shown to decrease the level of cholesterol in patients with hypercholesterolemia. According to the hierarchy of evidence, type of research study does this statement represent?

Case-control study

4. Why may the reporting of a significant “relative risk” in the conclusions of a research study be misleading? What would be a more meaningful value to report?

Relative risk is used to compare the results of the risks in the two groups in a study. In extreme cases, relativities produce astounding ratios, when in fact they are not really good to look at. For example, without the medicine, 10 out of 100 patients recovered from the illness. With the medicine, 19 out of 100 patients recovered. When we report relative risk, we say that there is an improvement of 90% (10 to19). Truth be told, the medicine only improved the case by 9%. Hence, absolute risk is a more meaningful value to report.

5. True or false: One of the arguments against stopping a study early “for benefit” is that the good results might just be a statistically random high point in the data and not be a reflection of a true effect.

a. True

b. False

6. ____________________ data is data that describes categories that have a hierarchical relationship.

a. Discrete

b. Nominal

c. Ordinal

7. An asymmetric curved graph in which the peak of the curve is toward the left (i.e., toward the lower numbers), is said to be skewed toward the __________________.

a. Right

b. Left

8. Your clinical colleague tells you about a patient whom she treated with an exercise program that she had developed herself. The patient improved and she suggested that you use her program on your next patient who presents with the same problem. According to the hierarchy of evidence, what type of research study is her treatment an example of?

Case report

9. Some pathologists claim that small, localized, non-invasive neoplasms (cancers) may be very slow growing, not prone to invade nor metastasize and therefore may not be clinically significant. If this could be shown to be true, there would be benefits for patients in whom this diagnosis is made (not having to undergo surgery, chemotherapy, nor radiation treatments and decreased emotional strain), and benefits for society (decreased costs of medical care). There have been no randomized controlled trials to determine whether or not treatment of these cancers actually provides any benefit in terms of improved life expectancy or reduced morbidity. Why do you think there have been no randomized controlled trials performed to determine this?

There may be a number of reasons for this. It can be that there is no known way yet to perform trials on this, or it can be that funding is limited. It can also be that the currently known techniques are being considered as the most effective. Or it can be that the economic advantage to hospitals is greatly appreciated that undergoing such trials will reduce or erase this advantage.

10. A new diagnostic test has been developed to Diagnose Disease Y. The new test was compared to the standard diagnostic test (“the gold standard”) on the same set of patients, and the following data were obtained:

Number of patients diagnosed by the new test as having Disease Y, who actually had Disease Y = 3,252

Number of patients in the group who were diagnosed by the standard test as having Disease Y = 4,579

Number of patients diagnosed by the new test as not having Disease Y = 4, 737

Number of patients diagnosed by the new test as having Disease Y, who actually do not have Disease Y= 784

The standard diagnostic test is a tissue biopsy, and therefore, is considered to be 100% sensitive for diagnosing Disease Y.

a. Create the 2x2 table that shows this data (9)

	Gold Standard
Serum Test	Positive	Negative	Total
Positive	3252	784	4036
Negative	1327	3410	4737
Total	4579	4194	8773

b. Calculate:

i. The sensitivity of the new test

= 3252/4579 = 71.02%

ii. The specificity of the new test

= 3410/4194 = 81.31%

iii. The false positive rate of the new test

= 784/4194 = 18.69%

iv. The false negative rate of the new test

= 1327/4579 = 28.98%

v. The positive predictive value of the new test

= 3252/4036 = 80.57%

vi. The negative predictive value of the new test

= 3410/4737 = 71.99%

vii. The likelihood ratio positive for the new test

= 71.02% / (1 – 81.31%) = 3.7992

viii. The likelihood ratio negative for the new test

ix. = (1 – 71.02%) / 81.31% = 0.35643

x.

11. Two new tests have been developed to diagnose Condition Z. One test (Test A) measures a metabolite “Metabolite A” in the serum. The other test (Test B) measures a different metabolite, “Metabolite B” in the urine. After appropriate preliminary testing, human trials are conducted. Following informed consent, the two tests were run on a group of patients with each patient getting both tests, as well as the “gold standard” test for diagnosing Condition Z. The following data were obtained:

Important!!!! Since some answers depend on the results you obtained in previous sections of the problem, it is important for you that you show *all* work, since I will give partial credit if your analysis and set up are correct, even if your numerical result is incorrect.

Serum level (metabolite A) # of patients Disease positive Disease negative

(mg / ml) (by gold standard) (by gold standard)

0 – 1.99 25 22 3

2.0 – 3.99 273 201 72

4.0 – 5.99 584 572 12

6.0 – 7.99 538 529 9

8.0 – 9.99 175 170 5

>10.0 14 14 0

Urine level (metabolite B)

(ng/ml)

0 – 2.99 30 5 25

3.0 – 5.99 245 46 199

6.0 – 8.99 423 400 23

9.0 – 11.99 628 620 8

12.0 – 14.99 175 169 6

15.0 – 17.99 99 95 4

>18.0 9 9 0

a) For test A, a cutoff value of 4.0 mg / ml of metabolite A is chosen as the boundary between a negative and a positive test result (i.e., values below 4.0 mg / ml are considered “negative” and values of 4.0 mg / ml or greater are considered “positive”).

For test B, a cutoff value of 6.0 ng / ml for metabolite B is chosen as the boundary between a negative and a positive test result (i.e., values below 6.0 ng / ml are considered “negative” and values of 6.0 ng / ml or greater are considered “positive”).

i. Create a fully labeled 2 x 2 table for test A

	Gold Standard
Serum Test	Positive	Negative	Total
Positive	True Positive = 1285	False Positive = 26	1584
Negative	False Negative = 223	True negative = 75	25
Total	1508	101	1609

ii. Create a fully labeled 2 x 2 table for test B

	Gold Standard
Serum Test	Positive	Negative	Total
Positive	True Positive = 1293	False Positive = 41	1570
Negative	False Negative = 51	True negative = 224	30
Total	1344	265	1600

iii. Which test is the more sensitive at diagnosing Condition Z?

Sensitivity of A = 1285 / 1508 = 85.21%

Sensitivity of B = 1293 / 1344 = 96.21%

Test B is more sensitive at diagnosing Condition Z.

iv. Which test is the more specific at diagnosing Condition Z?

Specificity of A = 75 / 101 = 74.26%

Specificity of B = 224 / 265 = 84.53%

Test B is more specific at diagnosing Condition Z.

b) What is the post-test probability of disease being present in a patient with a positive test result for test A if the pre-test probability of Condition Z is 8%, and the cutoff value of 4.0 mg/ml is chosen?

Likelihood ratio positive = 85.21% / (1 – 74.26%) = 3.3102

Post-test probability = 8% * 3.3102 / (8% * 3.3102 + 1 – 8%) = 22.35%

Using the odd-ratio form of Bayes’ Theorem and likelihood ratio, we have:

Post-test odds = pre-test odds x LR+

The pre-test odds is calculated from the pre-test probability as follows:

Odds = probability / (1 – probability)

So, pre-test odds = pre-test probability / (1 – pre-test probability)

Pre-test odds = 0.12 / (1 – 0.12)

= 0.12 / 0.88

Pre-test odds = 0.13636

Then, you multiply the pre-test odds times the LR+ to get the post-test odds:

The LR+ = TPR/FPR, where TPR = sensitivity and FPR = FP / (FP + TN) = (1 – specificity)

LR+ = sensitivity / (1 – specificity)

= 0.8903 / (1 – 0.5589)

= 0.8903 / 0.4411

LR+ = 2.01836

Post-test odds = pre-test odds x LR+

= 0.13636 x 2.01836

Post-test odds = 0.2752

Now, you have to convert post-test odds into post-test probability as follows:

Probability = odds / (1 + odds)

Post-test probability = post-test odds / (1 + post-test odds)

= 0.2752/ (1 + 0.2752)

= 0.2752/ 1. 2752

Post-test probability = 0.2158 = 0.22 or 22%