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 2.4% . A random sample of 19 Australian bank stocks has a sample mean of = 8.71%. For the entire Australian stock market, the mean dividend yield is = 5.9%. Do these data indicate that the dividend yield of all Australian bank stocks is higher than 5.9%? Use = .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

The question is missing the value of standard deviation, there\'s a typo there. I assume it\'s 4%.

Ho: mu = 5.9%
Ha: mu > 5.9%

z = (mu - x) / (sigma / sqrt(n)) = (8.71-5.9)/(4/sqrt(19)) = 3.06

The p-value from Z table corresponding to z = 3.06 is 0.0011

The p-value is less than the level of significance and so the data are statistically significant. Thus, we reject the null hypothesis.