Posted by Clay on .
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 Alpha0.05. 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?

statistics 
MathGuru,
Since x has a normal distribution, you can use a ztest for your data.
Formula:
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
Calculating:
z = (.0774  .067)/(.023/√17)
I'll let you finish the calculation.
Using a ztable at 0.05 level of significance for a onetailed 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 ztable? 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. 
statistics 
Anonymous,
if the correlation coefficient is 0.790, what is the explained variation