Description Usage Arguments Value Author(s) Examples
View source: R/crawford.test.R
Neuropsychologists often need to compare a single case to a small control group. However, the standard two-sample t-test does not work because the case is only one observation. Crawford and Garthwaite (2012) demonstrate that the Crawford-Howell (1998) t-test is a better approach (in terms of controlling Type I error rate) than other commonly-used alternatives. .
1 | crawford.test.freq(patient, controls)
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patient |
Single value (patient's score). |
controls |
Vector of values (control's scores). |
Returns a data frame containing the t-value, degrees of freedom, and p-value. If significant, the patient is different from the control group.
Dan Mirman, Dominique Makowski
1 2 3 4 | library(psycho)
crawford.test.freq(patient = 10, controls = c(0, -2, 5, 2, 1, 3, -4, -2))
crawford.test.freq(patient = 7, controls = c(0, -2, 5, 2, 1, 3, -4, -2))
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Note: Many functions of the 'psycho' package have been (improved and) moved to other packages of the new 'easystats' collection (https://github.com/easystats). If you don't find where a function is gone, please open an issue at: https://github.com/easystats/easystats/issues
The Crawford-Howell (1998) t-test suggests that the patient's score (10.00) is significantly different from the controls (M = 0.38, SD = 2.97, t(7) = 3.05, p p = 0.019). The patient's score is higher than 98.15% of the control population.
The Crawford-Howell (1998) t-test suggests that the patient's score (7.00) is not significantly different from the controls (M = 0.38, SD = 2.97, t(7) = 2.10, p p = 0.074). However, the null hypothesis of no difference can be rejected at a one-tailed 5% significance level (one-tailed p p = 0.037). The patient's score is higher than 92.62% of the control population.
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