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 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?

Explanation / Answer

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.

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

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