Standard deviation example

LISTING OP RATE OP VAPORIZATION (G/SBC) AT CONSTANT TEMPERATURE CALCULATION OF AVERAGE VALUES AND STANDARD DEVIATION (EXAMPLE FOR 1 HOUR)... 

Always use, y(ayt i) which gives the sample standard deviation. (Example 2.1a)... 

Count cells and calculate the means, ranges, and standard deviations. Example results from a CCL21 chemotaxis assay (performed as described above) are shown in Fig. 7. In this example, CCL21 prepared at the University of Nebraska Biological Process Development Facility was compared to commercially available CCL21 (Fig. 7). 

Example 12 Suppose Analyst A made five observations and obtained a standard deviation of 0.06, where Analyst B with six observations obtained 5-3 = 0.03. The experimental variance ratio is ... 

To calculate the standard deviation, we obtain the difference between the mean value (3.117 see Example 4.1) and each measurement, square the resulting differences, and add them to determine the sum of the squares (the numerator of equation 4.1)... 

The variance is just the square of the absolute standard deviation. Using the standard deviation found in Example 4.3 gives the variance as... 

Consider, for example, the data in Table 4.1 for the mass of a penny. Reporting only the mean is insufficient because it fails to indicate the uncertainty in measuring a penny s mass. Including the standard deviation, or other measure of spread, provides the necessary information about the uncertainty in measuring mass. Nevertheless, the central tendency and spread together do not provide a definitive statement about a penny s true mass. If you are not convinced that this is true, ask yourself how obtaining the mass of an additional penny will change the mean and standard deviation. 

How we report the result of an experiment is further complicated by the need to compare the results of different experiments. For example. Table 4.10 shows results for a second, independent experiment to determine the mass of a U.S. penny in circulation. Although the results shown in Tables 4.1 and 4.10 are similar, they are not identical thus, we are justified in asking whether the results are in agreement. Unfortunately, a definitive comparison between these two sets of data is not possible based solely on their respective means and standard deviations. 

There is a temptation when analyzing data to plug numbers into an equation, carry out the calculation, and report the result. This is never a good idea, and you should develop the habit of constantly reviewing and evaluating your data. For example, if analyzing five samples gives an analyte s mean concentration as 0.67 ppm with a standard deviation of 0.64 ppm, then the 95% confidence interval is... 

To begin with, we must determine whether the variances for the two analyses are significantly different. This is done using an T-test as outlined in Example 4.18. Since no significant difference was found, a pooled standard deviation with 10 degrees of freedom is calculated... 

Weighted normal calibration curve for the data in Example 5.13. The lines through the data points show the standard deviation of the signal for the standards. These lines have been scaled by a factor of 50 so that they can be seen on the same scale as the calibration curve. 

In Example 7.6 we found that an analysis for the inorganic ash content of a breakfast cereal required a sample of 1.5 g to establish a relative standard deviation for sampling of 2.0%. How many samples are needed to obtain a relative sampling error of no more than 0.80% at the 95% conhdence level ... 

Precision For samples and standards in which the concentration of analyte exceeds the detection limit by at least a factor of 50, the relative standard deviation for both flame and plasma emission is about 1-5%. Perhaps the most important factor affecting precision is the stability of the flame s or plasma s temperature. For example, in a 2500 K flame a temperature fluctuation of +2.5 K gives a relative standard deviation of 1% in emission intensity. Significant improvements in precision may be realized when using internal standards. 

For example, when the activity is determined by counting 10,000 radioactive particles, the relative standard deviation is 1%. The analytical sensitivity of a radiochemical method is inversely proportional to the standard deviation of the measured ac-... 

From Example A6.2 we know that after 100 steps of the countercurrent extraction, solute A is normally distributed about tube 90 with a standard deviation of 3. To determine the fraction of solute in tubes 85-99, we use the single-sided normal distribution in Appendix lA to determine the fraction of solute in tubes 0-84 and in tube 100. The fraction of solute A in tube 100 is determined by calculating the deviation z (see Chapter 4)... 

This result shows that the square root of the amount by which the ratio M /M exceeds unity equals the standard deviation of the distribution relative to the number average molecular weight. Thus if a distribution is characterized by M = 10,000 and a = 3000, then M /M = 1.09. Alternatively, if M / n then the standard deviation is 71% of the value of M. This shows that reporting the mean and standard deviation of a distribution or the values of and Mw/Mn gives equivalent information about the distribution. We shall see in a moment that the second alternative is more easily accomplished for samples of polymers. First, however, consider the following example in which we apply some of the equations of this section to some numerical data. 

Example 3. A centrifugal pump moving a corrosive Hquid is known to have a time-to-failure that is well approximated by a normal distribution with a mean of 1400 h and a standard deviation of 120 h. A particular pump has been in operation for 1080 h. In order to plan maintenance activities the chances of the pump surviving the next 48 h must be deterrnined. 

Since the t distribution relies on the sample standard deviation. s, the resultant distribution will differ according to the sample size n. To designate this difference, the respec tive distributions are classified according to what are called the degrees of freedom and abbreviated as df. In simple problems, the df are just the sample size minus I. In more complicated applications the df can be different. In general, degrees of freedom are the number of quantities minus the number of constraints. For example, four numbers in a square which must have row and column sums equal to zero have only one df, i.e., four numbers minus three constraints (the fourth constraint is redundant). 

The price of flexibility comes in the difficulty of mathematical manipulation of such distributions. For example, the 3-parameter Weibull distribution is intractable mathematically except by numerical estimation when used in probabilistic calculations. However, it is still regarded as a most valuable distribution (Bompas-Smith, 1973). If an improved estimate for the mean and standard deviation of a set of data is the goal, it has been cited that determining the Weibull parameters and then converting to Normal parameters using suitable transformation equations is recommended (Mischke, 1989). Similar estimates for the mean and standard deviation can be found from any initial distribution type by using the equations given in Appendix IX. 

For example, the deterministic value for the yield strength, Sy, for SAE 1018 cold drawn steel for the size range tested is approximately 395 MPa (Green, 1992). Table 4.6 gives the mean and standard deviation as A(540,41) MPa. The lower bound value as used in deterministic design becomes ... 

Values typieally range from 0.077 to 0.33 for the statie eoeffieient of frietion for steel on steel under an interferenee fit with no lubrieation (Kutz, 1986). The interferenee and eoeffieient of frietion are eorrelated in praetiee but for the example here, we assume statistieal independenee. Also, assuming that 6 standard deviations eover the range given, we ean derive that ... 

Once the mean and standard deviation have been determined, the frequency distribution determined from the PDF can be compared to the original histogram, if one was constructed, by using a scaling factor in the PDF equation. For example, the expected frequency for the Normal distribution is given by ... 

The variability or spread of the data does not always take the form of the true Normal distribution of course. There can be skewness in the shape of the distribution curve, this means the distribution is not symmetrical, leading to the distribution appearing lopsided . However, the approach is adequate for distributions which are fairly symmetrical about the tolerance limits. But what about when the distribution mean is not symmetrical about the tolerance limits A second index, Cp, is used to accommodate this shift or drift in the process. It has been estimated that over a very large number of lots produced, the mean could expect to drift about 1.5cr (standard deviations) from the target value or the centre of the tolerance limits and is caused by some problem in the process, for example tooling settings have been altered or a new supplier for the material being processed. 

Solving for the seeond order terms in the varianee equation for both the mean and standard deviation is shown below for the above example ... 

The standard deviation of the sample is defined as the square root of the variance. For the example,... 

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