Question

1.Two students are calibrating a P-1000. They want to check how accurate and precise the P-1000 is at 500 µL.  

1. What should the dial be set to on the P-1000 micropipette? (2 pts.)

B. The students raw data are provided

Student 1	Student 2
Weight (g)	Weight (g)
0.5001	0.4999
0.5034	0.5009
0.4982	0.5027

1. a. What was the % error and mean deviation for each student? ( 4 points)

b. What student was more accurate? Why?

c. Which student was more precise? Why?

Question has been updated.

Answer 1

Answer 1a)

Student 1:

mean value of all three measuremetnt is 0.5005

Standard deviation is = 0.002631

Standard error = standard deviation / sqrt(n) = 0.002631 /sqrt(3)= 0.001519

% error = |Predicted mean - Actual mean| * 100/ Actual mean

= |0.5005 -0.5|*100 / 0.500

=  0.0005 *100/0.5

= 0.1%

Student 2:

mean value of all three measuremetnt is 0.5012

Standard deviation is = 0.001418

Standard error = standard deviation / sqrt(n) = 0.002631 /sqrt(3)= 0.0008192

% error = |Predicted mean - Actual mean| * 100/ Actual mean

= |0.5012 -0.5|*100 / 0.500

=  0.0012 *100/0.5

= 0.24%

Answer 1b)

student 1 is more accurate than student 2 as % error of student 1 (0.1%) is less than %error of student 2(0.24%)

Answer 1c)

Student 2 is more precise than student 1 as the standard error of student 2 (0.0008192) is less than standard error of student 1(0.001519)