Skip to main content
Cross Validated

Return to Answer

added 95 characters in body
Source Link
EdM
EdM
  • 101.5k
  • 11
  • 102
  • 303

The simplest way to deal with this is to recognize that the correlation coefficient of 0.50 estimated from 49 observations (as described in comments) is quite compatible with the value of 0.39 that you estimate from the imputed data.

The Fisher transformation is a standard way to estimate confidence intervals for correlation coefficients. The hyperbolic arctangent of the correlation coefficient $\rho$ is distributed approximately normally with standard error (SE) of $1/\sqrt{(n-3)}$, where $n$ is the sample size.

For a correlation coefficient estimate of 0.5 based on 49 observations, the 95% confidence interval (at $\pm 1.96$ SE) after back-transforming to the correlation scale is:

SE <- 1/sqrt(49-3)
tanh(atanh(0.5) + 1.96*SE)
# [1] 0.6849035
tanh(atanh(0.5) - 1.96*SE)
# [1] 0.2545947

The value of 0.39 estimated from the imputations is well within the 95% confidence interval (0.25, 0.68) of the estimate of 0.50 from the 49 cases. It's less than 1 SE away.

(atanh(0.5)-atanh(0.39))/SE
# [1] 0.9326118

The match seems quite good. Why complicate things further?

The simplest way to deal with this is to recognize that the correlation coefficient of 0.50 estimated from 49 observations (as described in comments) is quite compatible with the value of 0.39 that you estimate from the imputed data.

The Fisher transformation is a standard way to estimate confidence intervals for correlation coefficients. The hyperbolic arctangent of the correlation coefficient $\rho$ is distributed approximately normally with standard error (SE) of $1/\sqrt{(n-3)}$, where $n$ is the sample size.

For a correlation coefficient estimate of 0.5 based on 49 observations, the 95% confidence interval (at $\pm 1.96$ SE) after back-transforming to the correlation scale is:

SE <- 1/sqrt(49-3)
tanh(atanh(0.5) + 1.96*SE)
# [1] 0.6849035
tanh(atanh(0.5) - 1.96*SE)
# [1] 0.2545947

The value of 0.39 estimated from the imputations is well within the 95% confidence interval (0.25, 0.68) of the estimate from the 49 cases. The match seems quite good. Why complicate things further?

The simplest way to deal with this is to recognize that the correlation coefficient of 0.50 estimated from 49 observations (as described in comments) is quite compatible with the value of 0.39 that you estimate from the imputed data.

The Fisher transformation is a standard way to estimate confidence intervals for correlation coefficients. The hyperbolic arctangent of the correlation coefficient $\rho$ is distributed approximately normally with standard error (SE) of $1/\sqrt{(n-3)}$, where $n$ is the sample size.

For a correlation coefficient estimate of 0.5 based on 49 observations, the 95% confidence interval (at $\pm 1.96$ SE) after back-transforming to the correlation scale is:

SE <- 1/sqrt(49-3)
tanh(atanh(0.5) + 1.96*SE)
# [1] 0.6849035
tanh(atanh(0.5) - 1.96*SE)
# [1] 0.2545947

The value of 0.39 estimated from the imputations is well within the 95% confidence interval (0.25, 0.68) of the estimate of 0.50 from the 49 cases. It's less than 1 SE away.

(atanh(0.5)-atanh(0.39))/SE
# [1] 0.9326118

The match seems quite good. Why complicate things further?

Source Link
EdM
EdM
  • 101.5k
  • 11
  • 102
  • 303

The simplest way to deal with this is to recognize that the correlation coefficient of 0.50 estimated from 49 observations (as described in comments) is quite compatible with the value of 0.39 that you estimate from the imputed data.

The Fisher transformation is a standard way to estimate confidence intervals for correlation coefficients. The hyperbolic arctangent of the correlation coefficient $\rho$ is distributed approximately normally with standard error (SE) of $1/\sqrt{(n-3)}$, where $n$ is the sample size.

For a correlation coefficient estimate of 0.5 based on 49 observations, the 95% confidence interval (at $\pm 1.96$ SE) after back-transforming to the correlation scale is:

SE <- 1/sqrt(49-3)
tanh(atanh(0.5) + 1.96*SE)
# [1] 0.6849035
tanh(atanh(0.5) - 1.96*SE)
# [1] 0.2545947

The value of 0.39 estimated from the imputations is well within the 95% confidence interval (0.25, 0.68) of the estimate from the 49 cases. The match seems quite good. Why complicate things further?