1.1.5 Basic Principles and Techniques (Reference interval studies, Method Evaluation) Flashcards
Reference interval derives from?
a. Confidence interval
b. Sensitivity
c. Precision
d. Accuracy
a
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
Also known as Transference study
a. Verifying a reference interval
b. Establishing a reference interval
a
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
20
<10%
In Establishing a reference interval, requires at least _______ study individuals and the reference interval is set base on the ______________
120
Confidence interval = mean +_ 2s
In establishing a reference interval, n=120
Mean = 5.0 mmol/L
s = 0.5
What is the reference interval?
4-6 mmol/L
Solution:
0.5 x 2 = 1
5 - 1 = 4
5 + 1 = 6
= 4 - 6 mmol/L
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
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
b
In t-test, it compares the ________ of two groups of data or the ________ of two methods
mean
Accuracy
F test compares the ________ of two groups of data or the _________ of two procedures
Standard deviation
Precision
Used to compare two methods using the best fit line through the data points
Linear regression
Linear regression
x= ________________
Independent variable
(Reference method)
Linear regression
y= ________________
Dependent variable
(New method)
ability of a method to detect only the analyte of interest
a. Analytical sensitivity
b. Analytical Specificity
b
Ability of a method to detect the smallest concentration of an analyte
a. Analytical sensitivity
b. Analytical Specificity
a
Not prone to False negative
a. Analytical sensitivity
b. Analytical Specificity
a
Used for screening test
a. Analytical sensitivity
b. Analytical Specificity
a
Not prone to False positive
a. Analytical sensitivity
b. Analytical Specificity
b
Desired in confirmatory test
a. Analytical sensitivity
b. Analytical Specificity
b
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
b
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
d
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
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
c
2 x 2 Table
Disease ; No disease
(+) TP = 48 ; 12 = FP
(-) FN = 2 ; 188 = TN
What is the Percentage Sensitivity?
96%
Solution:
%Se = (TP / (TP + FN)) x 100
= (48 / (48 + 2)) x 100
= 0.96 x 100
= 96%
2 x 2 Table
Disease ; No disease
(+) TP = 48 ; 12 = FP
(-) FN = 2 ; 188 = TN
What is the Percentage Specificity?
94%
Solution:
%Se = (TN / (TN + FP)) x 100
= (188 / (188 + 12)) x 100
= 0.94 x 100
= 94%
2 x 2 Table
Disease ; No disease
(+) TP = 48 ; 12 = FP
(-) FN = 2 ; 188 = TN
What is the Percentage PPV?
80%
Solution:
%PPV = (TP / (TP + FP)) x 100
= (48 / (48 + 12)) x 100
= 0.80 x 100
= 80%
2 x 2 Table
Disease ; No disease
(+) TP = 48 ; 12 = FP
(-) FN = 2 ; 188 = TN
What is the Percentage NPV?
99%
Solution:
%NPV = (TN / (TN + FN)) x 100
= (188 / (188 + 2)) x 100
= 0.99 x 100
= 99%