Question 11 of 15 Incorrect Map sapling learning Determine the precision and acc

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Question 11 of 15 Incorrect Map sapling learning Determine the precision and accuracy of these data for methylmercury Sample 1 Sample 2 Sample 3 21.8 ng/mL 78.3 ng/mL 233 ng/mL Known concentratio 33.1 18.2 23.5 20.3 20.2 24.4 72.7 58.9 71.5 90.8 70.4 73.5 241 225 203 249 195 221 Experimentally determined values (ng/mL) Sample 2 Precision Rel Std Deviation) Sample 3 Precision Rel Std Deviation) Sample 1 Precision Rel Std Deviation) Number 24.3 Sample 1 Accuracy Number Number 18.3 7.87 Sample 2 Accuracy Relative Error) Number Sample 3 Accuracy Relative Error) Number There is additional feedback available! View this feedback by clicking on the bottom divider bar. Click on the divider bar again to hide the additional feedback -8.7 3.7 Clase Previous Give Up & View Solution # Try Again Next Exit Explanation

Explanation / Answer

I will tell you how to do this with the first sample, and then, you do the same with the remaining sample and you will get the correct result.

For the SDR in sample 1 in precision, you need to take the avg value of the 6 experimental values there. Doing that, you got the following value: 23.3833 ng/mL.

Now to get the SDR = SD/X

The SD obtained is 5.3274 so:
SDR = 5.3274/23.3833 * 100 = 22.78%

To get the relative error in accuracy, you need to substract the concentration value with each of the experimental values, and the difference of them, you have to divide by the value of the concentration value given, and then take an average of all that data. Doing that with the sample 1 we have:

1: 33.1 - 21.8/21.8 * 100 = 1.13%
2. 18.2 - 21.8/21.8 * 100 = -16.51%
3. 23.5 - 21.8/21.8 * 100 = 7.798%
4. 20.3 - 21.8/21.8 * 100 = -6.88%
5. 20.2 - 21.8/21.8 * 100 = -7.34%
6. 24.4 - 21.8/21.8 * 100 = 11.93%

The average is -1.6453%

Do the same thing with the sample 2 and 3. Hope this helps

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