Let X be a random variable representing dividend yield of Au

Let X be a random variable representing dividend yield of Australian bank stocks. We may assume that X has a normal distribution with standard deviation sigma = 2.4 % . A random sample of 19 Australian bank stocks has a sample mean of x(bar) = 8.71%. For the entire Australian stock market, the mean dividend yield is mu = 5.9%. Do these data indicate that the dividend yield of all Australian bank stocks is higher than 5.9%? Use alpha = .05. Are the data statistically significant at the given level of significance? Based on your answers, will you reject or fail to reject the null hypothesis?

Solution

Let mu be the population mean

Ho: mu=5.9 (i.e. null hypothesis)

Ha: mu> 5.9 (i.e. alternative hypothesis)

The test statistic is

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

=(8.71-5.9)/(2.4/sqrt(19))

=5.10

It is a right-tailed test.

Given a=0.05, the critical value is Z(0.05) = 1.645 (from standard normal table)

Since Z=5.10 is larger than 1.645, we reject the null hypothesis.

So we can conclude that the data is statistically significant at the given level of significance