Question

You were asked to include both your standard error and your percent difference when formally reporting your experimental results. What do they represent? Do you think they should be included in scientific reporting?

1. Student A performs a similar experiment with a different object and obtains the following direct measurements of the mass of their object: [21.06g, 20.98g, 21.12g, 21.21g, 21.10g]
   a. Calculate and report the average, standard deviation, and standard error of Student A’s mass

   measurements using the 1D Stats macro.
   b. Add one gram to each of the five mass values in part (a). Repeat your 1D Stats calculation and

   report the average, standard deviation, and standard error. Which values have changed and which have not? Be sure to show your work, including the new 1D Stats results. The one gram difference in each measurement represents a systematic error in the data.

   c. Change the masses back to their original values and now add a 10 gram systematic error to the first two values only. Repeat your 1D Stats and note what has and has not changed.

   d. If the only data you saw was the data table generated in part (d), do you feel you would have been able to catch the systematic errors in part (d)? Explain in one to three sentences.

Answer 1

(a) The average mean of Student A's mass measurement which will be given by -

_avg = (sum of all data points) / (total number of points)

_avg = [(21.06 g) + (20.98 g) + (21.12 g) + (21.21 g) + (21.10 g)] / (5)

_avg = 21.094 g

The standard deviation of Student A's mass measurement which will be given by -

= (1/N) (m_i - _avg)²

where, (m_i - _avg) = [(21.06 g) - (21.094 g)]² = 0.001156 g²

[(20.98 g) - (21.094 g)]² = 0.012996 g²

[(21.12 g) - (21.094 g)]² = 0.000676 g²

[(21.21 g) - (21.094 g)]² = 0.013456 g²

[(21.10 g) - (21.094 g)]² = 0.000036 g²

then, we get

= (1/5) [(0.001156 g²) + (0.012996 g²) + (0.000676 g²) + (0.013456 g²) + (0.000036 g²)]

= 0.005664 g²

= 0.0752 g

The standard error of Student A's mass measurement which will be given by -

SE = / N

SE = (0.0752 g) / 5

SE = [(0.0752 g) / (2.236)]

SE = 0.0336 g

(b) Add one gram to each of the five mass values in Part (a).

Then, the average mean of Student A's mass measurement which will be given by -

'_avg = (sum of all data points) / (total number of points)

'_avg = [(22.06 g) + (21.98 g) + (22.12 g) + (22.21 g) + (22.10 g)] / (5)

'_avg = 22.094 g

Then, the standard deviation of Student A's mass measurement which will be given by -

' = (1/N) (m_i - _avg)²

where, (m_i - _avg) = [(22.06 g) - (22.094 g)]² = 0.001156 g²

[(21.98 g) - (22.094 g)]² = 0.012996 g²

[(22.12 g) - (22.094 g)]² = 0.000676 g²

[(22.21 g) - (22.094 g)]² = 0.013456 g²

[(22.10 g) - (22.094 g)]² = 0.000036 g²

then, we get

' = (1/5) [(0.001156 g²) + (0.012996 g²) + (0.000676 g²) + (0.013456 g²) + (0.000036 g²)]

' = 0.005664 g²

' = 0.0752 g

Then, the standard error of Student A's mass measurement which will be given by -

SE' = / N

SE' = (0.0752 g) / 5

SE' = [(0.0752 g) / (2.236)]

SE' = 0.0336 g

The value of the average mean of Student A's mass measurement will be changed.

The value of standard deviation & standard error of Student A's mass measurement will not be changed.