How much data is needed so that it is accurate to do point biserial correlation using SPSS?
Not yet a Doc, still a PhD student but working hard toward the degree. Glad I can help.
Is your question related to the estimation of the point biserial correlation? In this case, you will find useful the following link:
http://ncss.wpengine.netdna-cdn.com/wp-content/themes/ncss/pdf/Procedures/PASS/Confidence_Intervals_for_Point_Biserial_Correlation.pdf
If your interest is in testing hypothesis I can give you another link.
best regards.
Point biserial is just a special case of the Pearson product-moment correlation. In fact, the same data may be plugged into any software or calculator that performs a Pearson correlation and the answer will be the same as that obtained from textbook formulas for pt-biseral. Sample size is important for two reasons: precision of the estimate, and power of a statistical test that population correlation = 0. Bigger N is better on both counts. Minimum N to compute and test a correlation is 3 cases, but nobody would want to rely on such a small N. How much bigger can be ascertained if you're able to quantify the desired level of precision (for a parameter estimate) or the desired level of statistical power (against some specific effect size alternative).
I agree with David. For any statistical tests, the sample size will be the larger the better.
Thank you all colleagues who participated in answering this question. I think it is clear now how to handle point biserial
you need 3 pairs. Three continuous values and three dichotomous values.
Professor Muayyad
It is clear that to calculate a point biserial correlation you need 3 or more data points. To do a point biserial analysis you need more than that. how many? I think this is your question. This is an estimation problem and the solution depends on the variance of the estimator.
The link I of my first answer gives the variance of the sample point biserial correlation. You can isolate $n$ from this formula:
n = (ro^2 + 2 * P * (1 - P) * (2 - 3 * ro^2)) * (1 - ro^2)^2 / (4 * Sg^2 * P * (1 - P))
where ro is the population point biserial correlation, P is the proportion of group coded X=1 and Sg is the standard deviation in the formula.
If we denote the (1 - alpha/2) quantile by Za and r the sample point biserial correlation, the confidence interval (CI) is given by (r - Za * Sg, r + Za * Sg). Thus Sg is the key to calculate the CI. The length of the interval is 2 * Za * Sg.
Give some arbitrary but reasonable values to ro and P. For example, if you find that 40% of the subjects in the population have X = 1, set P = 0.4. More important are the values you assign to ro. Values near zero require very large n. Values near -1 or +1 require moderate or small n. You can work this in Excel.
Example, Assume P = 0.4 and suppose you want a CI length of 0.08 and a 0.95 confidence coefficient.
2 * Za * Sg = 0.08 Then Sg = 0.08 / (2 * Za) = 0.02040854
Give some values to ro and apply in the formula above to calculate n. Next I write the code in R language:
alpha = 0.05
Za = qnorm(1 - alpha/2)
IntervalLength = 0.08
Sg = IntervalLength / (2 * Za)
P = 0.4
ro = 0.8
n = (ro^2 + 2 * P * (1 - P) * (2 - 3 * ro^2)) * (1 - ro^2)^2 / (4 * Sg^2 * P * (1 - P))
n
You obtain n = 219.89 -> 220
change ro to 0.2 and you obtain n = 2172.114 -> 2173
Values of ro near to zero require very large samples.
As you see, reasonable values of ro and Interval length are very important to calculate the sample size needed in your study.
If you are more interested in hypothesis testing, the procedure changes.
Jorge Ortiz
Thank you all colleagues above who provided a helpful answers. I think now I can manage following your suggestions
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