Chemistry: In comparing groups, the first or the second, which group was more precise or was the precision the same

In comparing groups, the first or the second, - id14103931, jacksonyodell8601

Ответ:

houtchhaytang

04.11.2021

Chemistry

Comparing the data obtained online, the first group is more precise

The reason the above selection is correct is as follows;

Question: Parts of the question appear missing from a similar question online and included here;

First Group:

$\begin{array}{|c|c|c|c|c|c|c|} Team \ a &Team \ b&Team \ c&Team \ d&Team \ e&Team \ f&Team \ g\\2.77 \ cm&2.60 \ cm&2.80 \ cm&2.65 \ cm&2.75 \ cm&2.65 \ cm& 2.68 \ cm\end{array}\right]$

Second Group

$\begin{array}{|c|c|c|c|c|c|c|} Team \ a &Team \ b&Team \ c&Team \ d&Team \ e&Team \ f&Team \ g\\2.70 \ cm&2.78 \ cm&2.62 \ cm&2.65 \ cm&2.75 \ cm&2.80 \ cm& 2.60 \ cm\end{array}\right]$

Calculations for the first group:

The average = (2.77 + 2.60 + 2.80 + 2.65 + 2.75 + 2.65 + 2.68)/7 = 2.7

The range = 2.80 - 2.60 = 0.20

The approximate ± range of the average = ±0.2/2 = ±0.1

The precision of the first measurement is 2.7 ± 0.1 cm

Calculations for the second group:

The average = (2.70 + 2.78 + 2.62 + 2.65 + 2.75 + 2.80 + 2.60)/7 = 2.7

The range = 2.80 - 2.60 = 0.20

The approximate ± range from the average = ±0.2/2 = ±0.1

The range of values from the average is approximately ±0.1 cm

The precision of the first measurement is 2.7 ± 0.1 cm

Method to determine precision:

Precision is given by the finding the average deviation and the standard deviation as follows;

$\mathbf{Average \ deviation} = \dfrac{\sum \left | x - \mu \right |}{n}$

$\mathbf{Standard\ deviation} =\sqrt{\dfrac{\sum \left ( x - \mu \right )^2}{n}}$

Where;

μ = The mean or average = 2.7

n = The number of items (count) of the data = 7

For the first group, we have;

$\mathbf{\dfrac{\sum \left | x - \mu \right |}{n}} = \dfrac{0.07 + 0.1 + 0.1 + 0.05 + 0.05 + 0.05 + 0.02 }{7} = \mathbf{0.062857}$

The average deviation of the first group = 0.062857

∑(x - μ)² ≈ 0.0328

$\mathbf{The \ standard\ deviation} =\sqrt{\dfrac{0.0328}{7}} \approx \mathbf{0.068452}$

For the second group, we have;

$\mathbf{\dfrac{\sum \left | x - \mu \right |}{n}} = \dfrac{0.0+ 0.08 + 0.08 + 0.05 + 0.05 + 0.1 + 0.1 }{7} = \mathbf{0.06571428571}$

∑(x - μ)² ≈ 0.0378

$\mathbf{The \ standard\ deviation} =\sqrt{\dfrac{0.0378}{7}} \approx \mathbf{0.073485}$

By using the average deviation, and standard deviation values, the deviation of the second group is more than the first group and therefore the first group is more precise

Learn more about accuracy and precision here:

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In comparing groups, the first or the second, which group was more precise or was the precision the same

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