posted by Clay .
Let x be a random variable representing dividend yield of Australian bank stocks. We may assume that x has a normal distribution with sigma =2.3%. A random sample of 17 Australian bank stocks has a sample mean of x bar 7.74%. For the entire Australian stock market, the mean dividend yield is Mu=6.7% Do these data indicate that the dividend yield of all Australian bank stocks is higher than 6.7%? Use Alpha-0.05. 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?
Since x has a normal distribution, you can use a z-test for your data.
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
Hypotheses (percentages are converted to decimal form):
Ho: µ = .067 -->null hypothesis
Ha: µ > .067 -->alternate hypothesis
z = (.0774 - .067)/(.023/√17)
I'll let you finish the calculation.
Using a z-table at 0.05 level of significance for a one-tailed test (alternate hypothesis shows a specific direction), find your critical or cutoff value to reject the null.
Does the test statistic calculated above exceed the critical value from the z-table? If it does not, you cannot reject the null hypothesis. If it does, reject the null and accept the alternate hypothesis (the test will be statistically significant).
I hope this will help get you started.
if the correlation coefficient is 0.790, what is the explained variation