Interpolation

The Spearman’s Rho Table shows a critical value of .447 for alpha = .05 and n = 20 and .522 for alpha = .02 and n = 20, so what value should we use when alpha = .025 and n = 20?

We will consider three approaches: linear interpolation, logarithmic interpolation, and harmonic interpolation.

Linear Interpolation

With linear interpolation, the value we are looking for is calculated by

Linear interpolation

which can also be calculated using the Real Statistics formula

=INTERPOLATE(.025,.02,.05,.522,.447,0)

Here the 0 argument indicates that linear interpolation is being used.

Logarithmic Interpolation

With logarithmic interpolation, the value we are looking for is calculated by

log interpolation

which can also be calculated using the Real Statistics formula

=INTERPOLATE(.025,.02,.05,.522,.447,1)

Here the 1 argument indicates that log interpolation is being used. This is the default value for the INTERPOLATE function.

This is the preferred type of interpolation for the alpha argument. It is occasionally useful for the sample size argument n and the number of variables k.

Harmonic Interpolation

With harmonic interpolation, the value we are looking for is calculated by

Harmonic interpolation

which can also be calculated using the Real Statistics formula

=INTERPOLATE(.025,.02,.05,.522,.447,2)

Here the 2 argument indicates that harmonic interpolation is being used.

This is generally the preferred type of interpolation for the number of variables k, degrees of freedom df, and sample size n.

Table Lookup Functions

The type of interpolation in the calculations of the following functions is specified by the final argument (denoted interp): RhoCRIT, TauCRIT, MCRIT, WCRIT, DCRIT, LCRIT, KSCRIT, KS2CRIT, ADCRIT, QCRIT, DLowerCRIT, DUpperCRIT, and SRankCRIT. interp can take the value FALSE for linear interpolation or TRUE (default) for the recommended type of interpolation.

The recommended type of interpolation is log for alpha and harmonic for df. Generally, harmonic interpolation is used for n and k, although in some situations log interpolation is used instead. Only occasionally is linear interpolation the recommended type of interpolation

RhoCRIT(20,.025,2,TRUE) = .5037                  RhoCRIT(20,.025,2,FALSE) = .5095

References

Zar. J. H. (2010) Biostatistical analysis 5th Ed. Pearson
https://www.pearson.com/us/higher-education/program/Zar-Biostatistical-Analysis-5th-Edition/PGM263783.html

Laurencelle, L. and Dupuis, F-A. (2002) Statistical tables, explained and applied. World Scientific
https://books.google.it/books/about/Statistical_Tables_Explained_and_Applied.html?id=-onhswEACAAJ&redir_esc=y

33 thoughts on “Interpolation”

    • Hello Fajri,
      Since 294 is so close to 300, you can probably use the values in the table for 300.
      If you use the DLowerCRIT and DUpperCRIT functions provided by the Real Statistics software you get the values 1.788 and 1.829 for n = 294 instead of 1.791 and 1.831 for n = 300.
      Charles

      Reply

  7. Hi
    Do you have the general formula for the linear interpolation of the critical value?

    Can you show me the calculation for the critical value of alpha = 0.05 and n = 47 if the critical value is .287563 for alpha = .05 and n = 45 and the critical value is .273243 for alpha = .05 and n = 50?

    Thanks.

    Reply

    • Hi,
      The approach is as described on the webpage, except that you are interpolating n instead of alpha. The formula would be
      .273243+(50-47)/(50-45)*(.287563-.273243) (assuming that I copied the formula properly)
      You can also use the Real Statistics formula =Interpolate(47,45,50,.287563,.273243)
      Charles

      Reply

      • Hi
        am I correct in my computation in this interpolation for the critical value of alpha = 0.05 and n = 44?

        if the critical value is 0.304 for alpha = .05 and n = 40 and the critical value is 0.288 for alpha = .05 and n = 45?

        0.288 + (45-44)/(45-40) * (0.304-0.288) = Critical Value of 0.2912

        Thanks.

        Reply

  8. Hi Charles,

    How do you apply the linear Interpolation to calculate the p-value using the Shaprio-Wilks test in excel? I have downloaded your resource pack but cannot seem to produce the same value in your Shapiro-Wilks example.

    Reply

  10. hi charles,
    would you help me, how to make interpolation for Durbin Watson table, when n=1210, and k =1, with alpha = 5 % ?

    Reply

  12. Am i missing something here? The “Interpolate” function is not an equation within the downloaded tool set. Please let me know if im wrong

    Reply

  13. Hi Charles,
    I am confused on how to work the interpolation.
    My W = 0.9162, n = 4, the W is 0.792 for 0.1 and 0.935 for 0.5.
    How do I find out my p value?

    Reply

    • It depends on the actual W value. If, for example W = .930, then using linear interpolation,
      p-value = .05 + (.1-.05)*(.930-.927)/(.939-.927).
      Charles

      Reply

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