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The tin and zinc contents of a brass sample are analyzed with the following results: (a) \(\mathrm{Zn}: 33.27,\) 33.37, and 33.34\% and (b) Sn: 0.022, 0.025, and 0.026\%. Calculate the standard deviation and the coefficient of variation for each analysis.

Short Answer

Expert verified
Zn: SD ≈ 0.041, CV ≈ 0.123%; Sn: SD ≈ 0.00245, CV ≈ 10.08%

Step by step solution

01

Calculate the Mean for Zinc

To find the mean (average) of the zinc contents, add up all the measured values and divide by the number of measurements. The zinc measurements are 33.27, 33.37, and 33.34.\[ \text{Mean for Zn} = \frac{(33.27 + 33.37 + 33.34)}{3} = \frac{99.98}{3} = 33.327 \]
02

Calculate the Standard Deviation for Zinc

The standard deviation measures how much variation there is from the average. Use the formula:\[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \]For our zinc values:- \(x_1 = 33.27\), \(x_2 = 33.37\), \(x_3 = 33.34\), and \(\bar{x} = 33.327\)Calculate:\[ s_{Zn} = \sqrt{\frac{(33.27 - 33.327)^2 + (33.37 - 33.327)^2 + (33.34 - 33.327)^2}{3-1}} \]\[ s_{Zn} = \sqrt{\frac{0.00336}{2}} \approx 0.041 \]
03

Calculate the Coefficient of Variation for Zinc

The coefficient of variation (CV) is a standardized measure of dispersion of a probability distribution or frequency distribution, and is calculated as:\[ \text{CV}_{Zn} = \left( \frac{s}{\bar{x}} \right) \times 100"]\]Using our calculated values:\[ \text{CV}_{Zn} = \left( \frac{0.041}{33.327} \right) \times 100 \approx 0.123\% \]
04

Calculate the Mean for Tin

Similarly, find the mean of the tin contents using the same formula.For tin measurements of 0.022, 0.025, and 0.026:\[ \text{Mean for Sn} = \frac{(0.022 + 0.025 + 0.026)}{3} = \frac{0.073}{3} = 0.0243 \]
05

Calculate the Standard Deviation for Tin

Again, use the standard deviation formula for the tin measurements.\(x_1 = 0.022\), \(x_2 = 0.025\), \(x_3 = 0.026\), and \(\bar{x} = 0.0243\)\[ s_{Sn} = \sqrt{\frac{(0.022 - 0.0243)^2 + (0.025 - 0.0243)^2 + (0.026 - 0.0243)^2}{3-1}} \]\[ s_{Sn} = \sqrt{\frac{0.0000189}{2}} \approx 0.00245 \]
06

Calculate the Coefficient of Variation for Tin

Calculate the coefficient of variation using:\[ \text{CV}_{Sn} = \left( \frac{s}{\bar{x}} \right) \times 100 \]For tin:\[ \text{CV}_{Sn} = \left( \frac{0.00245}{0.0243} \right) \times 100 \approx 10.08\% \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Deviation
Standard deviation is a statistical measure that shows how much variation or dispersion exists from the average value (mean) of a set of data. In simpler terms, it tells us how "spread out" the values are. A smaller standard deviation means that the data points are closer to the mean, while a larger standard deviation indicates the data points are more spread out.

To calculate the standard deviation for a set of values, we follow these steps:
  • Compute the mean (average) of the values.
  • Subtract the mean from each value to find the deviation of each point.
  • Square each of these deviations.
  • Find the average of these squared deviations, adjusting for the sample size if necessary (using the formula for an unbiased estimate).
  • Take the square root of that average. This result is the standard deviation.
For example, in the exercise, the standard deviation of the zinc measurements (\(\{33.27, 33.37, 33.34\}\)) was calculated to show us how tightly the values are clustered around the mean value of 33.327.
Coefficient of Variation
The coefficient of variation (CV) is a normalized measure of how much the data varies. It is useful in comparing the degree of variation from one data set to another, even if the means are drastically different. The CV is expressed as a percentage by dividing the standard deviation by the mean and then multiplying by 100.

The formula for calculating the CV is:
\[ \text{CV} = \left( \frac{\text{Standard Deviation}}{\text{Mean}} \right) \times 100 \]
The CV is particularly useful because it provides a sense of relative variability regardless of the magnitude of the mean. For instance, in the given exercise, the tin analysis had a mean of 0.0243% and a standard deviation of approximately 0.00245. This yielded a CV of approximately 10.08%, indicating a relatively high level of dispersion relative to the size of the mean.
  • A low CV indicates the data points are close to the mean, implying less variability.
  • A high CV indicates more variability, which might be less desirable in precise scientific measurements such as analytical chemistry.
Statistical Analysis in Chemistry
In analytical chemistry, statistical analysis is quintessential in ensuring precision and accuracy of measurements. Chemistry often involves working with measurements that might vary slightly due to experimental conditions, equipment sensitivity, or natural variability in samples.

Whenever chemists measure concentrations or compositions, they rely on statistics like standard deviation and coefficient of variation to understand the reliability of their results. This statistical rigor helps verify that:
  • Measurements are consistent and reproducible across different trials.
  • Variations are within acceptable limits established by research or industry standards.
  • Uncertainty is accounted for, allowing chemists to make informed decisions from their data.
Using statistics allows chemists to:
  • Quantify the uncertainty in their measurements.
  • Improve methodologies by identifying sources of error.
  • Present data that are clear and reliable for further scientific analysis or industrial application.
Thus, tools like standard deviation and coefficient of variation are not just abstract mathematical concepts but are essential in maintaining the credibility of chemical analyses and research outcomes.

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Most popular questions from this chapter

A standard serum sample containing 102 meq/L chloride was analyzed by coulometric titration with silver ion. Duplicate results of 101 and 98 meq/L were obtained. Calculate (a) the mean value, (b) the absolute error of the mean, and (c) the relative error in percent.

Calculate the absolute uncertainty in the answers of the following: (a) \((2.78 \pm 0.04)(0.00506 \pm\) 0.00006 ), (b) \((36.2 \pm 0.4) /(27.1 \pm 0.6)\), (c) \((50.23 \pm 0.07)(27.86 \pm 0.05) /(0.1167 \pm 0.0003)\).

Climate Change and Propagation of Uncertainty. Many factors can cause changes in Earth's climate. These range from well-known greenhouse gases such as carbon dioxide and methane to changes in ozone levels, the effective reflectivity of Earth's surface, and the presence of \(\mathrm{nm}\) - \(\mu \mathrm{m}\) sized aerosol particles that can scatter and absorb sunlight in the atmosphere. Some effects lead to a warming influence on climate, while others may cool the Earth and atmosphere. Climate scientists attempt to keep score and determine the net effect of all competing processes by assigning radiative forcing values \(\left(\mathrm{W} / \mathrm{m}^{2}\right)\) to each effect independently and then summing them. Positive radiative forcing values warm climate, while negative values cool the Earth and atmosphere. This approach is insightful since the net radiative forcing \(\left(\Delta F_{n e t}\right)\) can be linked to expected mean temperature change \(\left(\Delta T_{\text {surface }}\right)\) via the climate sensitivity parameter \((\lambda),\) which is often assigned values of \(0.3-1.1 \mathrm{~K} /\left(\mathrm{W} / \mathrm{m}^{2}\right)\). $$ \Delta T_{\text {surface }}=\lambda \times \Delta F_{\text {net }} $$ The Intergovernmental Panel on Climate Change (IPCC) has studied the work of many scientists to provide current best estimates of radiative forcings and associated uncertainties for each effect. These are described in the figure and table shown below. $$ \begin{array}{lc} \hline \text { Climate Effect } & \text { Best Estimate } \pm \text { Uncertainty }\left(\mathbf{W} / \mathbf{m}^{2}\right) \\ \hline \text { Long-lived Greenhouse Gases } & 2.61 \pm 0.26 \\ \text { Tropospheric and Stratospheric Ozone } & 0.30 \pm 0.22 \\ \text { Surface Albedo } & -0.10 \pm 0.20 \\ \text { Direct Aerosol Effect } & -0.50 \pm 0.36 \\ \text { Indirect Aerosol Effect } & -0.70 \pm 0.7 \\ \hline \end{array} $$ Use the rules for propagation of uncertainty to Ask Yourself: (a) The sum of the radiative forcing terms is \(1.61 \mathrm{~W} / \mathrm{m}^{2}\). What is the uncertainty associated with this estimate? Which of the individual terms seem to dominate the magnitude of the overall uncertainty? (b) If a value of \(0.7 \pm 0.4 \mathrm{~K} /\left(\mathrm{W} / \mathrm{m}^{2}\right)\) is used as the climate sensitivity parameter \((\lambda),\) what is the expected surface temperature change? What is the associated uncertainty in this estimate? (c) Instrumental temperature records from the 1850 s to present day suggest the global mean surface temperature has increased by roughly \(0.8^{\circ} \mathrm{C} .\) Is this estimate consistent with your calculation from b?

(a) You and your friends are visiting the M\&M candy factory at Christmas time. Just as you walk in the door, there is a terrible spill and 200,000 red M\&Ms along with 50,000 green M\&Ms have just spilled on the floor. You madly start grabbing M\&Ms and have managed to collect 1000 of them before management catches you. How many red M\&Ms have you probably picked up? (b) Continued from question above, if you repeat this scenario many times, what will be the absolute standard deviation of green M\&Ms retrieved?

Explain situations when a least-squares linear fit is not appropriate and should not be used. There are at least two common important cases.

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