Chapter L, Problem 21PS

You and your lab partner are asked to determine the density of an aluminum bar. The mass is known accurately (to four significant figures). You use a simple metric ruler to measure its dimensions and obtain the results for Method A. Your partner uses a precision micrometer and obtains the results for Method B

The accepted density of aluminum is 2.702 g/cm³.

1. (a) Calculate the average density for each method. Should all the experimental results be included in your calculations? If not, justify any omissions.
2. (b) Calculate the percent error for each method's average value.
3. (c) Calculate the standard deviation for each set of data.
4. (d) Which method's average value is more precise? Which method is more accurate?

Interpretation Introduction

Interpretation: The average density for each method and the justifications for omissions has to be given.

Concept introduction:

Precision: Is a measurement indicates how well several determination of the same quantity agree.

Accuracy: Is the agreement of a measurement with the accepted value of the quantity.

Percent error: The difference between measured result and the accepted value.

$\text{Percent}\text{error}\text{=}\frac{\text{Error}\text{in}\text{measurement}}{\text{Accepted}\text{value}}\text{×}\text{100\%}$

$\text{Error in measurement = Experimentally determined value - Accepted value}\text{.}$

Answer to Problem 21PS

Average value for Method A is $\text{2}{\text{.4 g/cm}}^3$

Average value for Method B is $\text{3}{\text{.480 g/cm}}^3$

For B, the $\text{5}{\text{.811 g/cm}}^3$ data point can be excluded because it is more than twice as large as all other points for method B. using only the first three points, average $\text{= 2}{\text{.703 g/cm}}^3$

Explanation of Solution

Average density Value for Method A and B:

$\begin{array}{l}\text{Average density for Method A: }\frac{2.2+\text{ 2}\text{.3+ 2}\text{.7 + 2}\text{.4}}{4}\\ \text{ = 2}{\text{.4 g/cm}}^3\text{.}\\ \text{Average density for Method B: }\frac{2.703+\text{ 2}\text{.701 + 2}\text{.705 + 5}\text{.811}}{4}\\ \text{ = 3}{\text{.48 g/cm}}^3\end{array}$

Hence, the average for method A and B are obtained as shown above.  B, the $\text{5}{\text{.811 g/cm}}^3$ data point can be excluded because it is more than twice as large as all other points for method B. using only the first three points, average $\text{= 2}{\text{.703 g/cm}}^3$

Interpretation Introduction

Interpretation: The percent error for each method’s average value has to be calculated.

Concept introduction:

Precision: Is a measurement indicates how well several determination of the same quantity agree.

Accuracy: Is the agreement of a measurement with the accepted value of the quantity.

Percent error: The difference between measured result and the accepted value.

  $\text{Percent}\text{error}\text{=}\frac{\text{Error}\text{in}\text{measurement}}{\text{Accepted}\text{value}}\text{×}\text{100\%}$

  $\text{Error in measurement = Experimentally determined value - Accepted value}\text{.}$

Answer to Problem 21PS

Method A: error $\text{= 0}{\text{.3 g/cm}}^3$ or about $10\%$

Method B: error $\text{= 0}{\text{.001 g/cm}}^3$ or about $0.04\%$

Explanation of Solution

Percent error for Method A and B:

$\begin{array}{l}\text{Error}\text{in}\text{measurement =}\text{2}{\text{.4 g/cm}}^3\text{ - 2}{\text{.702 g/cm}}^3\\ \text{Percent}\text{error}\text{for}\text{method}\text{A}\text{=}\frac{\text{2}{\text{.4 g/cm}}^3\text{ - 2}{\text{.702 g/cm}}^3}{\text{2}{\text{.702 g/cm}}^3}\text{×}\text{100\%}\\ \text{ = 10\%}\\ \text{Percent}\text{error}\text{for}\text{method}\text{B}\text{= }\frac{\text{2}{\text{.703 g/cm}}^3\text{ - 2}{\text{.702 g/cm}}^3}{\text{2}{\text{.702 g/cm}}^3}\text{×}\text{100\%}\\ \text{ = 0}\text{.04\%}\end{array}$

From the above calculation, the percent errors are obtained as shown above.

Interpretation Introduction

Interpretation: The standard deviation for each set of data.

Concept introduction:

Precision: Is a measurement indicates how well several determination of the same quantity agree.

Accuracy: Is the agreement of a measurement with the accepted value of the quantity.

Percent error: The difference between measured result and the accepted value.

$\text{Percent}\text{error}\text{=}\frac{\text{Error}\text{in}\text{measurement}}{\text{Accepted}\text{value}}\text{×}\text{100\%}$

$\text{Error in measurement = Experimentally determined value - Accepted value}\text{.}$

Answer to Problem 21PS

Method A: standard deviation $\text{= 0}{\text{.2 g/cm}}^3$

Method B: (including all data points) standard deviation $\text{= 1}{\text{.345 g/cm}}^3$

Method B: (excluding the $\text{5}{\text{.811 g/cm}}^3$ data point) standard deviation $\text{= 0}{\text{.002 g/cm}}^3$

Explanation of Solution

Standard deviation for method A:

$\begin{array}{l}SD=\sqrt{\frac{\sum |x-\mu {|}^2}{N}}\\ \text{where},x\text{is values in the data set,}\\ \mu is\text{average of the data set (mean value),}\\ \text{ N is no}\text{.of data points}\text{.}\\ \\ \text{Given : Method A = 2}\text{.2, 2}\text{.3, 2}\text{.7, 2}\text{.4}\\ \text{Find mean value (}\mu ):\\ \mu =\frac{\text{2}\text{.2+2}\text{.3+ 2}\text{.7+ 2}\text{.4}}{4}\\ \text{ = 2}\text{.4}\\ \begin{array}{cccc}\text{Determination}& \text{Measured values}& \begin{array}{l}\text{Difference}\text{between}\\ \text{Measurement}\text{and}\text{Average}\\ (x-\mu )\end{array}& \begin{array}{l}\text{Square of Difference}\\ |x-\mu {|}^2\end{array}\\ 1& 2.2& -0.2& 0.04\\ 2& 2.3& -0.1& 0.01\\ 3& 2.7& 0.3& 0.09\\ 4& 2.4& 0.0& 0.0\end{array}\\ \sum |x-\mu {|}^2=0.04+0.01+0.09+0.0\\ \text{ = 0}\text{.14}\\ SD=\sqrt{\frac{\text{0}\text{.14}}{4}}=0.2g/c{m}^3\end{array}$

Standard deviation for method B:

$\begin{array}{l}SD=\sqrt{\frac{\sum |x-\mu {|}^2}{N}}\\ \text{where},x\text{is values in the data set,}\\ \mu is\text{average of the data set (mean value),}\\ \text{ N is no}\text{.of data points}\text{.}\\ \\ \text{Given : Method B = 2}\text{.703, 2}\text{.701, 2}\text{.705, 5}\text{.811}\text{.}\\ \text{Find mean value (}\mu ):\\ \mu =\frac{\text{2}\text{.703+ 2}\text{.701+ 2}\text{.705+ 5}\text{.811}}{4}\\ \text{ = 3}\text{.48}\\ \begin{array}{cccc}\text{Determination}& \text{Measured values}& \begin{array}{l}\text{Difference}\text{between}\\ \text{Measurement}\text{and}\text{Average}\\ (x-\mu )\end{array}& \begin{array}{l}\text{Square of Difference}\\ |x-\mu {|}^2\end{array}\\ 1& 2.703& -0.777& 0.6037\\ 2& 2.701& -0.779& 0.6068\\ 3& 2.705& -0.775& 0.6006\\ 4& 5.811& 2.331& 5.4335\end{array}\\ \sum |x-\mu {|}^2=0.6037+0.6068+0.6006+5.4335\\ \text{ = 7}\text{.2446}\\ SD=\sqrt{\frac{\text{7}\text{.25}}{4}}=1.345g/c{m}^3\end{array}$

Standard deviation excluding the $\text{5}{\text{.811 g/cm}}^{\text{3}}$ data point is $\text{0}{\text{.002 g/cm}}^{\text{3}}.$

Interpretation Introduction

Interpretation:

The method that has the more precise value and accurate has to be given.

Concept introduction:

Precision: Is a measurement indicates how well several determination of the same quantity agree.

Accuracy: Is the agreement of a measurement with the accepted value of the quantity.

Percent error: The difference between measured result and the accepted value.

$\text{Percent}\text{error}\text{=}\frac{\text{Error}\text{in}\text{measurement}}{\text{Accepted}\text{value}}\text{×}\text{100\%}$

$\text{Error in measurement = Experimentally determined value - Accepted value}\text{.}$

Explanation of Solution

Method B average is both more precise and more accurate so long as the $\text{5}{\text{.811 g/cm}}^3$ data point is excluded.