By NEAS - 8/5/2018 9:59:28 PM
MS Module 19 Correlation Fisher transformation practice exam questions
(The attached PDF file has better formatting.)
X and Y are a bivariate normal distribution from which a sample of 20 observations is taken. The sample correlation between X and Y is 0.82.
We test the null hypothesis H0: ρ = 0.71. The alternative hypothesis is Ha: ρo > 0.71.
Question 19.1: Fisher transform of observed correlation ρ
What is the Fisher transform of the observed correlation ρ?
Answer 19.1: 0.5 × ln( (1 + 0.82) / (1 – 0.82) ) = 1.1568
(Fisher transform = ½ × ln( (1+ρ)/(1–ρ) ) )
Question 19.2: Fisher transform of correlation ρ0 assumed in null hypothesis
What is the Fisher transform of the correlation ρ0 assumed in the null hypothesis?
Answer 19.2: 0.5 × ln( (1 + 0.71) / (1 – 0.71) ) = 0.8872
(Fisher transform = ½ × ln( (1+ρ)/(1–ρ) ) )
Question 19.3: Variance of Fisher transform
What is the variance of the Fisher transform?
Answer 19.3: 1 / (20 – 3) = 0.058824
(variance of the Fisher transform = 1/(number of observations – 3) )
Question 19.4: Standard deviation of Fisher transform
What is the standard deviation of the Fisher transform?
Answer 19.4: 0.0588240.5 = 0.2425
(standard deviation = square root of variance)
Question 19.5: z value
What is the z value to test the null hypothesis?
Answer 19.5: (1.1568 – 0.8872) / 0.2425 = 1.1118
(z value = (Fisher transform of sample ρ – Fisher transform of ρo) / standard deviation of Fisher transform)
Question 19.6: Confidence interval of the Fisher transform
What is the 90% confidence interval for the true value of the Fisher transform of the correlation?
Answer 19.6: confidence interval = Fisher transform ± critical z value × standard deviation of Fisher transform
● lower bound: 1.1568 – 1.645 × 0.2425 = 0.758
● upper bound: 1.1568 + 1.645 × 0.2425 = 1.556
Question 19.7: Confidence interval of the correlation
What is the 95% confidence interval for the true value of the correlation?
Answer 19.7: The inverse of the Fisher transform is ( e2x – 1) / ( e2x + 1).
● lower bound: (exp(2 × 0.758) – 1) / (exp(2 × 0.758) + 1) = 0.640
● upper bound: (exp(2 × 1.556) – 1) / (exp(2 × 1.556) + 1) = 0.915
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