1.1.5 Basic Principles and Techniques (Reference interval studies, Method Evaluation) Flashcards by Miguel Acosta | Brainscape

Q

Reference interval derives from?

a. Confidence interval
b. Sensitivity
c. Precision
d. Accuracy

A

a

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Q

Done to confirm the validity of an existing or published Reference interval for an analyte

a. Verifying a reference interval
b. Establishing a reference interval

A

a

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Q

Also known as Transference study

a. Verifying a reference interval
b. Establishing a reference interval

A

a

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Q

In Verifying a reference interval, It requires at least ______ study individuals and the Reference interval is adopted if ___________ of the subjects fall outside the range

A

20

<10%

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Q

In Establishing a reference interval, requires at least _______ study individuals and the reference interval is set base on the ______________

A

120

Confidence interval = mean +_ 2s

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Q

In establishing a reference interval, n=120
Mean = 5.0 mmol/L
s = 0.5

What is the reference interval?

A

4-6 mmol/L

Solution:

0.5 x 2 = 1

5 - 1 = 4
5 + 1 = 6

= 4 - 6 mmol/L

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Q

First step in method evaluation; usually done by running two control materials twice a day over 10 day period (2x2x10)

a. Precision study
b. Recovery study

A

a

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Q

Invoves spiking a sample with a known amount of a analyte and determining how much of it can be detected by the method in the presence of other compounds in the matrix

a. Precision study
b. Recovery study

A

b

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Q

In t-test, it compares the ________ of two groups of data or the ________ of two methods

A

mean

Accuracy

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Q

F test compares the ________ of two groups of data or the _________ of two procedures

A

Standard deviation

Precision

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Q

Used to compare two methods using the best fit line through the data points

A

Linear regression

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Q

Linear regression

x= ________________

A

Independent variable
(Reference method)

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Q

Linear regression

y= ________________

A

Dependent variable
(New method)

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Q

ability of a method to detect only the analyte of interest

a. Analytical sensitivity
b. Analytical Specificity

A

b

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Q

Ability of a method to detect the smallest concentration of an analyte

a. Analytical sensitivity
b. Analytical Specificity

A

a

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Q

Not prone to False negative

a. Analytical sensitivity
b. Analytical Specificity

A

a

Q

Used for screening test

a. Analytical sensitivity
b. Analytical Specificity

A

a

Q

Not prone to False positive

a. Analytical sensitivity
b. Analytical Specificity

A

b

Q

Desired in confirmatory test

a. Analytical sensitivity
b. Analytical Specificity

A

b

Q

Ability of a test to detect absence of a given disease or condition; proportion of individuals with NO DISEASE who have a negative test result

a. Diagnostic sensitivity
b. Diagnostic specificity
c. Positive predictive value
d. Negative predictive value

A

b

Q

The probability that a negative test result indicates absence of disease proportion of individuals with a NEGATIVE RESULT who truly do not have the disease

a. Diagnostic sensitivity
b. Diagnostic specificity
c. Positive predictive value
d. Negative predictive value

A

d

Q

Ability of a test to detect a given disease or condition; proportion of individuals with the DISEASE who have a positive test result

a. Diagnostic sensitivity
b. Diagnostic specificity
c. Positive predictive value
d. Negative predictive value

A

a

Q

Probability that a poritive test result indicates disease; proportion of individual with a POSITIVE RESULT who truly have the disease

a. Diagnostic sensitivity
b. Diagnostic specificity
c. Positive predictive value
d. Negative predictive value

A

c

Q

2 x 2 Table

Disease ; No disease

(+) TP = 48 ; 12 = FP
(-) FN = 2 ; 188 = TN

What is the Percentage Sensitivity?

A

96%

Solution:

%Se = (TP / (TP + FN)) x 100
= (48 / (48 + 2)) x 100
= 0.96 x 100
= 96%

Q

2 x 2 Table

Disease ; No disease

(+) TP = 48 ; 12 = FP
(-) FN = 2 ; 188 = TN

What is the Percentage Specificity?

A

94%

Solution:

%Se = (TN / (TN + FP)) x 100
= (188 / (188 + 12)) x 100
= 0.94 x 100
= 94%

Q

2 x 2 Table

Disease ; No disease

(+) TP = 48 ; 12 = FP
(-) FN = 2 ; 188 = TN

What is the Percentage PPV?

A

80%

Solution:

%PPV = (TP / (TP + FP)) x 100
= (48 / (48 + 12)) x 100
= 0.80 x 100
= 80%

Q

2 x 2 Table

Disease ; No disease

(+) TP = 48 ; 12 = FP
(-) FN = 2 ; 188 = TN

What is the Percentage NPV?

A

99%

Solution:

%NPV = (TN / (TN + FN)) x 100
= (188 / (188 + 2)) x 100
= 0.99 x 100
= 99%