Sentence examples for mean peak area from inspiring English sources

The absolute matrix effect and extraction recovery were calculated as follows: Absolute matrix effect = (Mean peak area)set 2 ⁄ (Mean peak area)set 1. Extraction efficiency = (Mean peak area)set 3 ⁄ (Mean peak area)set 2. This post-extraction spike method proposed by Matuszewski et al. [11] provides a quantitative understanding of the level of matrix effect observed for specific analytes.

For estimation of matrix effect, mean peak area ratio of set 1/set 2 × 100 was calculated.

The mean peak area of the tablets was found to be equivalent to that of the pure drug used as a positive control at the same concentration level.

In ice-cold water bath and at pH 6.0, there was no change in mean peak area response for both analytes with respect to time.

HPLC water (N = 6) 97.3 Tap7 97.3 Tap water (N = 6) 88.8 96.2 RiverRiver water (N = 6) 77.7 81.6 95.4 ME1 is matrix effect expressed as the ratio of mean peak area of analyte spiked postextraction to the mean peak area of the same analyte standard multiplied by 100.

Protocol-B: Samples were kept at room temperature for 6.0 h A1 mean peak area observed at RT of LS (m/z: 422.1 → 285.4), A2 mean peak area observed at RT of LHA (m/z:423.3 → 303.7) both in positive ion mode.

Alternatively, the corresponding regression equation was derived using mean peak area-concentration data, and the concentration of the unknown analyte can be computed from the regression equation.

Matrix effects were performed by comparing the mean peak areas of the analyte spiked post-treatment to those of the analyte solutions at corresponding concentrations.

The mean peak areas of the selected estrogens in HPLC water for a 1 mL injection were compared with the same volume injection of those of urine samples for a dilution factor of ten.

The mean peak areas (20 and 40 µg L−1, n = 5) of the selected estrogens of a direct injection (25 µL) were compared with those of the on-line 1 mL volume injection.

The independent T-test suggested no significant difference in the mean peak areas with a p-value = 0.798, whereas Levene's test displayed equal variances with a p-value = 0.072 for both data sets.