2. Let X be a random variable representing the dividend yield of Australian bank stocks. We may assume that X has a normal distribution with Now, suppose we wish to test the null hypothesis that against the alternative that using a level of significance of Ą = .05. To do so, a random sample of 16 Australian bank stocks is observed and has a sample mean of What is the value of the test statistic?
To find the value of the test statistic, we need to compute the test statistic formula for a hypothesis test involving the mean of a normal distribution.
The formula for the test statistic when testing a null hypothesis about the mean of a population with a known standard deviation is given by:
Z = (X - µ) / (σ / √n)
Where:
Z stands for the test statistic,
X is the sample mean,
µ represents the population mean under the null hypothesis,
σ is the known population standard deviation,
and √n is the square root of the sample size.
In this case, the null hypothesis states that the population mean dividend yield is 6.8%, and the alternative hypothesis is that the population mean is greater than 6.8%. The level of significance is 0.05, denoted as α = 0.05.
Given:
Sample mean, X = 7.3%
Population mean under the null hypothesis, µ = 6.8%
Population standard deviation, σ = 0.9% (mentioned in the question)
Sample size, n = 16
Calculating the test statistic:
Z = (X - µ) / (σ / √n)
Z = (7.3 - 6.8) / (0.9 / √16)
Z = (0.5) / (0.9 / 4)
Z = (0.5) / (0.225)
Z = 2.22
Therefore, the value of the test statistic (Z) is 2.22.