Question

With a very precise volumetric measuring device, the volume of a liquid sample is determined to be 6.321 L (liters). Three students are asked to determine the volume of the same liquid sample using a less precise measuring instrument. How do you evaluate the following work of each student with regards to precision, and accuracy
Students
Trials A B C
1 6.35L 6.31L 6.38L
2 6.32L 6.31 L 6.32L
3 6.33L 6.32L 6.36L
4 6.36L 6.35L 6.36L

Answer 1

Answer:

See explanation

Explanation:

Solution:-

- Three students measure the volume of a liquid sample which is 6.321 L.

- Each student measured the liquid sample 4 times. The data is provided for each measurement taken by each student as follows:

                                                 Students

                      Trial          A            B               C

                         1            6.35        6.31          6.38

                        2            6.32        6.31          6.32

                        3            6.33        6.32         6.36

                        4            6.36        6.35         6.36

- We will define the two terms stated in the question " precision " and "accuracy"

- Precision refers to how close the values are to the sample mean. The dense cluster of data is termed to be more precise. We will use the knowledge of statistics and determine the sample standard deviation for each student.

- The mean measurement taken by each student would be as follows:

                       $E ( A ) = \frac{6.35 +6.32+6.33+6.36}{4} \\\\E ( A ) = 6.34\\\\E ( B ) = \frac{6.31 +6.31+6.32+6.35}{4} \\\\E ( B ) = 6.3225\\\\E ( C ) = \frac{6.38 +6.32+6.36+6.36}{4} \\\\E ( C ) = 6.355$

- The precision can be quantize in terms of variance or standard deviation of data. Therefore, we will calculate the variance of each data:

 

                        $Var ( A ) = \frac{6.35^2+6.32^2+6.33^2+6.36^2}{4} - 6.34^2\\\\Var ( A ) = 0.00025\\\\Var ( B ) = \frac{6.31^2+6.31^2+6.32^2+6.35^2}{4} - 6.3225^2\\\\Var ( B ) = 0.00026875\\\\Var ( C ) = \frac{6.38^2+6.32^2+6.36^2+6.36^2}{4} - 6.355^2\\\\Var ( C ) = 0.000475$

- We will rank each student sample data in term sof precision by using the values of variance. The smallest spread or variance corresponds to highest precision. So we have:

                   Var ( A )          <          Var ( B )        <    Var ( C )

                   most precise                                      Least precise

- Accuracy refers to how close the sample mean is to the actual data value. Where the actual volume of the liquid specimen was given to be 6.321 L. We will evaluate the percentage difference of sample values obtained by each student .

                       $P ( A ) = \frac{6.34-6.321}{6.321}*100= 0.30058\\\\P ( B ) = \frac{6.3225-6.321}{6.321}*100= 0.02373\\\\P ( C ) = \frac{6.355-6.321}{6.321}*100= 0.53788$

- Now we will rank the sample means values obtained by each student relative to the actual value of the volume of liquid specimen with the help of percentage difference calculated above. The least percentage difference corresponds to the highest accuracy as follows:

                   P ( B )         <       P ( A )         <      P ( C )

            most accurate                                least accurate