Dixon's Q-test Calculator - Detection of a single outlier

 

Dixon’s test (or the Q-test) has been described in a previous post entitled “Detection of a Single Outlier|Statistical Analysis|Quantitative Data ”. The test is popular because the calculations involved are simple. A solved example is given in the above post.

Are there any limitations to Dixon’s Q-test?

The data excluding  the possible outlier must be normally distributed (use the Kolmogorov-Smirnov test to check if data is normally distributed)
The Q-test is valid for the detection of a single outlier (it cannot be used for a second time on the same set of data). Other forms of Dixon’s Q-test can be applied to the detection of multiple outliers.
The Q-test should be applied with caution – the same applies to all statistical tests used for rejecting data - since there is a probability, equal to the significance level a (a =0.05 at the 95% confidence level) that an outlier identified by the Q-test actually is not an outlier.

Moreover, if two suspect values occur, both of them might be at high end of the measurement range, both at the low end, or one at the high end and one at the low end. In situations like these the test may give erroneous results. As an example consider the following data tested for outliers:

4.0, 4.1, 4.2, 4.3, 4.3, 4.9, 5.1

Two of the above values (4.9 and 5.1) are suspiciously high compared with the mean of the data, yet if Q were calculated (at the 95% confidence level) would give that the tested value 5.1 is not an outlier at the 95% confidence level. Clearly, the possible outlier 5.1 has been masked by the other possible outlier 4.9 giving a low value for Q compared to Qcrit.

An online calculator is given below that can identify outliers in a data set at  six different confidence levels (80%, 90%, 95%, 96%, 98%, 99%). To test a data set for possible outliers follow the steps below:

Check that data is normally distributed (Kolmogorov-Smirnov test, Q-Q plot)
Type data in the yellow-labeled cells
Select the confidence level from the drop-down list
See the tested value and the results

 

 

 

 

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Relevant Posts

Detection of a Single Outlier|Statistical Analysis|Quantitative Data

Detection of Outliers in Analytical Data – The Grubb’s Test

Calibration and Outliers - Statistical Analysis

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References

1. D. B. Rorabacher,  Anal. Chem., 63, 139–146, (1991)
2. D. Harvey,  “Modern Analytical Chemistry”, McGraw-Hill Companies Inc., 2000
3. R.D. Brown, “Introduction to Chemical Analysis”, McGraw-Hill Companies Inc., 1982
4. J. N. Miller, J. C. Miller, "Statistics and Chemometrics for Analytical Chemistry", 6th Edition, Pearson, 2010

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Key Terms

Dixon's q test calculator, calculator online, detecting outlier online, normal distribution, detection of a single outlier

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Home > Weak Acids and Bases – Calculate the pH of a weak acid (a general equation) > Solving Weak Acid and Weak Base pH problems > Weak Acid Weak Base pH calculation solved example > Weak Acid Weak Base pH calculation solved example > Cubic Equation Calculator for Weak Acid-Base Equilibria

 

Cubic Equation Calculator for Weak Acid-Base Equilibria

The mathematical (algebraic) exact method for solving weak acid or weak base equilibrium problems has traditionally been less popular than the alternative approximate method, probably because of the inconveniences related to solving cubic equations. However, modern mathematics software or even spreadsheets like Excel 2010 handle such equations with ease, making the algebraic method more attractive than in the past.

The mathematical approach to solving weak acid-base equilibria has been presented in a previous post entitled  “Weak Acids and Bases – Calculate the pH of a weak acid (a general equation)”.

There are two equilibria present: i) the dissociation of water and ii) the dissociation of the weak acid ka

[H⁺] [OH^-] = k_w        (1)

[H⁺] [A^-] = k_a [HA]        (2)

If the concentration of the acid in solution is C (M, moles/l), a mass balance on the anion A gives:

C = [HA] + [A^-]            (3)

and a charge balance gives:

[H⁺] = [OH^-] + [A^-]             (4)

Let us assume that ka, kw  and C are known. Then there are four equations and four unknowns (shown in red in equations (1) to (4)).

A general expression for [H⁺ ] (or for the pH of a weak acid) in this case would be an expression in terms of the known k_a, k_w  and C. Therefore, the unknowns have to be eliminated starting from [HA] or [OH^-] that are contained in the least number of equations. Let us eliminate first [OH-]. Solving equation (1) for [OH^-] and substituting it in equation (4):

[H⁺] [OH^-] = k_w    and  [OH^-] = k_w / [H⁺]   (1’)

By substituting (1’) to (4) eliminates [OH^-]:

[H⁺] = [OH^-] + [A^-] = k_w / [H⁺] + [A^-]           (5)

Next let us eliminate [HA] by solving equation (3) for [HA] and substituting in equation (2):

C = [HA] + [A^-]   and [HA] = C - [A^-]     (3’)

Substitute (3’) in equation (2) and eliminate [HA]:

[H⁺] [A^-] = k_a [HA]  = k_a (C - [A^-])     (6)

Now let us eliminate [A-] by solving equation (5) for [A-] and substitute in equation (6):

[H⁺] = k_w  / [H⁺] + [A^-]  and  [A^-] = [H⁺] - k_w / [H⁺]     (7)

Substituting (7) in equation (6):

[H⁺] [A^-] =k_a [HA]  = k_a (C - [A^-]) and

[H⁺] ([H⁺] - k_w / [H⁺])  = k_a (C -[H⁺] - k_w / [H⁺]) )      (8)

Rearranging (8) we get a cubic equation in [H⁺]:

[H⁺]³ + k_a[H⁺]² – ( k_w + k_aC) [H⁺] - k_wk_a= 0       (9)

Using the known values of C, k_a, k_w equation (9) can be solved for [H⁺] and the pH can be calculated. In our days, solution of equations like (9) has become easier by using on line cubic calculators such as the one shown below.

For acids use the weak acids calculator. Select the weak acid dissociation constant from the table and copy and paste it in the k_a box. Type the initial acid concentration C in moles/liter (M) in the box named C (molar). The [H⁺] concentration and the pH of the solution are shown at the bottom.

For acids use the weak acids calculator shown below.

For bases use the weak bases calculator shown below. Follow the instructions given above.