Example:
A study is planned to estimate correlation coefficient (r) between marks obtained in final examination and attendance of students. Previous studies have showed approximate estimate of r = 0.60. How many students must be included in the study, if you want to estimate the r within 10% of its true value with 95% confidence level?
Solution:
Here
r = 0.6, ε = 10% = 0.1, Confidence level = 95%
After putting these values, we get required sample size = 488.
What will happen after selecting this sample size, considering that we get the correlation coefficient r = 0.6, as anticipated, in the study?
Let’s calculate 95% confidence interval of r.
For this we need to follow following steps.
1. Transform the correlation coefficient with the Fisher's transformation (R’).
R' = 0.5 * Log ((1+r) / (1 – r))
R’ = 0.5 * Log ((1+0.6) / (1 – 0.6))
R’ = 0.69
2. Calculate Standard error of R’
SE= 1/ Sqrt (N-3) = 1/ SQRT(485) = 0.045
3. Calculate 95% Confidence intervals of R’ as follows.
R’ ± 1.96 * SE
95% CI of R’ = 0.69 ± 1.96 * 0.045 = 0.60 – 0.78
4.Convert back these lower bounds and upper bounds of R’ into original scale, i.e. LB and UB of r as follows
LB of r= (Exp (2*LB of R' )- 1)/(Exp (2*LB of R' )+1)=(Exp(2*0.6)+1)/(Exp(2*0.6)-1)=0.54
UB of r= (Exp (2*UB of R' )- 1)/(Exp (2*UB of R' )+1)=(Exp(2*0.78)+1)/(Exp(2*0.78)-1)=0.66
95 % CI of r is 0.54 – 0.66 (10% towards both sides of r). Hence, as intended, with this sample size, we can estimate r with 10% precision at 95% confidence level.
@ Sachin Mumbare