Let X be a random variable representing dividend yield of Australian bank stocks

Question

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?

Explanation / Answer

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

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.