Question

F(x|theta) = thetax^(theta+1) for 0 < x < 1 and 0 elsewhere where theta > 0. For I = 1,…,n let Yi = -ln(Xi) and let T = ∑ Yi = ∑ [-ln(Xi)].

Suppose the sample size is n=20. Starting with the probability statement: P( < 2thetaT < ) = .95 Derive the formula for a 95% confidence interval for θ. Be sure to replace a and b with the appropriate numbers for your confidence interval.

Answer 1

To derive a 95% confidence interval for θ, calculate the critical t-value (2.093 for 19 degrees of freedom) and use it to scale the mean of T. The formula involves dividing the sum of all -ln(Xi) by the product of the critical t-value and sample size.

Step-by-step explanation:

To construct a 95% confidence interval for θ, given that T is the sum of the transformed sample data Σ1, …, Σn, and the sample size is n=20, we start with t-distribution and use probability inequality.

Given that the degrees of freedom (df) are n - 1 = 19, we obtain the critical t-value (ta) for a two-tailed 95% confidence interval from either a t-table or calculator. In this case, the critical value is ta = 2.093.

To find the confidence interval, we multiply the critical t-value by the standard deviation of the sampling distribution of T and divide the mean of T. The formula would look like this:

95% confidence interval for θ = (ΣXi / (2.093 * n), ΣXi / (2.093 * n))

In the context of the stated problem, ΣXi refers to the sum of all -ln(Xi) for i = 1 … n.

The statement of being '95 percent confident' refers to the likelihood that the procedure used will capture the true parameter value in long run use over many samples.