Question

Consider a sample that contains analyte at the detection limit defined in Figure. Explain the following statements: There is approximately a $\mathbf{1}\mathbf{\%}$chance of falsely concluding that a sample containing no analyte contains analyte above the detection limit. There is a$\mathbf{50}\mathbf{\%}$chance of concluding that a sample that really contains analyte at the detection limit does not contain analyte above the detection limit.

Short answer

The following statements was explained.

1. Above the detection limit approximately

$1\%$

chance of falsely concluding that the sample does contain no analyte.

2. Above the detection limit approximately$50\%$chance of concluding that the sample doesn't contains analyte.

Step-by-step solution

The detection limit

The concentration of an analyte that produces a signal with a standard deviation three times that of a blank signal.

The minimum detectable concentration $\mathbf{=}\frac{\mathbf{3}\mathbf{s}}{\mathbf{m}}$

Where,

$\mathit{S}\mathbf{=}$Standard deviation.

$\mathbf{m}\mathbf{=}$Slope of linear calibration curve.

Explanation of statements

The first statement means that since there is approximately $1\%$chance of false negative results showing up, there will also be approximately $1\%$of samples (that do NOT contain the analytes) that will respond with a signal above the detection limit.

For the second statement -- which is a hypothetical situation for false positive results -- means that$50\%$ of the samples containing the analyte with a signal going below the limit of detection.