Question

Let Y1, Y2, . . . , Y5 be a random sample of size 5 from a normal population with mean 0 and

variance 1. Let Y(bar) = 1/5 sum of 1 to 5 yi, and let Y6 be another independent observation from the same population.
Identify the distributions of the following random variables and explain your reasoning.

V = √5Y6 / √W
X = 2Y6 / √U

Answer 1

The random variable V follows a Student's t-distribution with n-1 degrees of freedom, and the random variable X also follows a Student's t-distribution with n-1 degrees of freedom.

Step-by-step explanation:

The distribution of a random variable V = √5Y6 / √W can be identified by examining the properties of the random variables involved. In this case, Y6 is a single observation from a normal population with mean 0 and variance 1, and W is the sum of squares of the random sample Y1, Y2, . . . , Y5. Since Y6 and W are independent and follow a normal distribution, V follows a Student's t-distribution with n-1 degrees of freedom, where n is the sample size.

Similarly, the distribution of a random variable X = 2Y6 / √U can be determined by considering the properties of Y6 and U. Y6 is a single observation from a normal population with mean 0 and variance 1, and U is the sum of squares of the random sample Y1, Y2, . . . , Y5. Since Y6 and U are independent and follow a normal distribution, X follows a Student's t-distribution with n-1 degrees of freedom, where n is the sample size.