Question

Let x be a random variable representing dividend yield of Australian bank stocks. We may assume that x has a normal distribution with σ = 2.0%. A random sample of 23 Australian bank stocks has a sample mean of x = 7.67%. For the entire Australian stock market, the mean dividend yield is μ = 7.6%. Do these data indicate that the dividend yield of all Australian bank stocks is higher than 7.6%? Use α = 0.05. What is the value of the test statistic?

Answer 1

The calculated z-test statistic is approximately 0.1683. To determine its significance, we compare it to the critical value for a one-tailed test with α = 0.05. Additional information, such as the critical value, is needed for a conclusive decision.

To determine if the dividend yield of all Australian bank stocks is higher than 7.6%, we perform a one-sample z-test. The null hypothesis
`(\(H_0\))` is that the population mean dividend yield is 7.6%, and the alternative hypothesis
`(\(H_1\))` is that it is higher.

The formula for the z-test statistic is:

`\[ Z = \frac{\bar{X} - \mu}{(\sigma)/(√(n))} \]`

Given a sample mean
`\(\bar{X} = 7.67\)`, population mean
`\(\mu = 7.6\)`, population standard deviation
`\(\sigma = 2.0\)`, and sample size n = 23, we can calculate:

`\[ Z = (7.67 - 7.6)/((2.0)/(√(23))) \]`

`\[ Z = (0.07)/((2.0)/(√(23))) \]`

`\[ Z = (0.07)/((2.0)/(√(23))) * (√(23))/(√(23)) \]`

`\[ Z = (0.07 * √(23))/(2.0) \]`

`\[ Z \approx (0.07 * 4.795)/(2.0) \]`

`\[ Z \approx (0.33665)/(2.0) \]`

`\[ Z \approx 0.1683 \]`

The calculated z-test statistic is approximately 0.1683. To determine if this value is greater than the critical value for a one-tailed test with a significance level α = 0.05, additional information, such as the critical value, is required.