Page 586 - Probability and Statistical Inference

12. Large-Sample Inference 563

and comes up with an approximate level a test against an appropriate alterna-
tive hypothesis. The details are left out for brevity.

Example 12.4.6 (Examples 11.4.1-11.4.2 Continued) Consider the vari-
ables (X , X ), from Example 11.4.1 for employees on their job performance
1 2
scores before and after the training. Assuming a bivariate normal distribution
for (X , X ), we want to test H : ρ = 0.7 versus H : ρ > 0.7 at an approximate
0
1
1
2
5% level. From a set of observed data for 85 employees, we found r = .757.
n
Thus, we have tanh (r) = .98915, tanh (ρ ) = .8673, and hence from (12.4.20),
-1
-1
0
we obtain:
Now, with α = ..05, one has z = 1.645. Since z does not exceed z , we
.05
.05
calc
do not reject the null hypothesis at an approximate 5% level. We conclude that
the correlation coefficient ρ between the job performance scores does not
appear to be significantly larger than 0.7.!
Example 12.4.7 A psychologist wanted to study the relationship between
the age (X , in years) and the average television viewing time per day (X , in
2
1
minutes) in a large group of healthy children between the ages 4 and 12 years.
From the study group, 42 children were randomly picked. For each child, the
television viewing time was recorded every day during the two preceding
weeks and X is the average of the 14 recorded numbers. From this set of 42
2
observations on the pair of random variables (X , X ), we found r = .338.
1
n
2
Assuming a bivariate normal distribution for (X , X ), we wish to obtain an
1
2
approximate 95% confidence interval for ρ, the population correlation coeffi-
cient. From (12.4.19), we claim that tanh (r ) ± is an approximate 95%
-1
n
confidence interval for tanh (ρ). We have tanh (r ) = .35183 and so the inter-
-1
-1
n
-1
val (0.03798, 0.66568) for tanh (ρ) has an approximate 95% confidence. In
other words, the interval (.03796,. 58213) for ρ has an approximate 95%
confidence. !
12.5 Exercises and Complements
12.2.1 (Example 12.2.1 Continued) Suppose that a random variable X has
the Cauchy pdf f(x; θ) = π {1 + (x − θ) } I(x ∈ ℜ) where θ(∈ ℜ) is the
2 -1
-1
unknown parameter. Show that the Fisher information, I (θ) = 2. {Hint: With
y = x−θ, first write I (θ) = π -1 Evaluate the integral by
substituting y = tan(z).}