Let x be a random variable representing the dividend yeal of

Let x be a random variable representing the dividend yeal of Australian bank stocks. We may assume that X has a normal distribution with a standard deviation sigma = 2.8%. Now, suppose we wish to test the null hypothesis that mu = 6.4% against the alternative that mu >6.4% using a level of significance of alpha - .05. To do so, a random sample of 16 Australian bank stocks is observed and has a sample mean of x(bar) = 8.91%. What is the p-value associated with this test?

Solution

It is a right-tailed test.

The test statistic is

Z=(xbar-mu)/(s/vn)

=(8.91-6.4)/(2.8/sqrt(16))

=3.59

So the p-value= P(Z>3.59) =0.0002 (from standard normal table)