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 Alpha-0.05. 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?
if the correlation coefficient is 0.790, what is the explained variation
To determine if the dividend yield of all Australian bank stocks is higher than 6.7%, we can perform a one-sample t-test by following these steps:
Step 1: State the hypotheses.
The null hypothesis (H₀): The dividend yield of all Australian bank stocks is equal to 6.7%.
The alternative hypothesis (H₁): The dividend yield of all Australian bank stocks is higher than 6.7%.
Step 2: Set the significance level.
The significance level (α) is given as 0.05.
Step 3: Compute the test statistic.
The test statistic for a one-sample t-test is calculated using the formula:
t = (x̄ - μ) / (s / √n)
Where:
x̄ = sample mean (7.74%)
μ = population mean (6.7%)
s = standard deviation of the sample (2.3%)
n = sample size (17)
Plugging in the values:
t = (7.74 - 6.7) / (2.3 / √17)
Step 4: Determine the critical value.
Since this is a one-tailed test (alternative hypothesis is that the dividend yield is higher), we need to find the critical value corresponding to a one-tailed t-test with 16 degrees of freedom (n - 1) at α = 0.05. Checking a t-table or using software, the critical value is approximately 1.745.
Step 5: Make a decision.
If the calculated t-value is greater than the critical value, we can reject the null hypothesis. Otherwise, fail to reject it.
Step 6: Calculate the test statistic.
Plugging in the values from step 3:
t = (7.74 - 6.7) / (2.3 / √17)
t ≈ 3.712
Step 7: Compare the test statistic with the critical value.
Since the calculated t-value (3.712) is greater than the critical value (1.745), we can reject the null hypothesis.
Step 8: Make a conclusion.
The data indicate that the dividend yield of all Australian bank stocks is indeed higher than 6.7% at the given level of significance (α = 0.05). Therefore, we reject the null hypothesis and conclude that there is statistical evidence to suggest a higher dividend yield for Australian bank stocks compared to the entire Australian stock market.
To determine whether the dividend yield of all Australian bank stocks is higher than 6.7%, we can conduct a hypothesis test using a t-test.
Let's set up the hypotheses:
Null Hypothesis (H₀): The mean dividend yield of all Australian bank stocks is 6.7% (μ = 6.7%).
Alternative Hypothesis (H₁): The mean dividend yield of all Australian bank stocks is higher than 6.7% (μ > 6.7%).
We can use the sample mean and the population standard deviation to calculate the test statistic.
Step 1: Calculate the standard error of the mean (SE):
SE = σ / √n
= 2.3% / √17
Step 2: Calculate the test statistic (t):
t = (x̄ - μ) / SE
= (7.74% - 6.7%) / (2.3% / √17)
Step 3: Determine the critical value:
To determine the critical value at α = 0.05 with a one-tailed test, we need to find the t-value with (n - 1) degrees of freedom. In this case, we have 17 - 1 = 16 degrees of freedom.
Using a t-table or a t-distribution calculator, a t-value of approximately 1.746 can be found for a one-tailed test at α = 0.05 with 16 degrees of freedom.
Step 4: Compare the test statistic with the critical value:
If the absolute value of the test statistic is greater than the critical value, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.
Now, let's calculate the test statistic and compare it to the critical value.
t = (7.74% - 6.7%) / (2.3% / √17)
= 1.04 / (2.3% / √17)
≈ 1.04 / 0.5601
≈ 1.856
Since the obtained test statistic (1.856) is greater than the critical value (1.746), we reject the null hypothesis.
Therefore, the data suggest that the dividend yield of all Australian bank stocks is statistically higher than 6.7% at the given level of significance (α = 0.05).
Since x has a normal distribution, you can use a z-test 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 z-table at 0.05 level of significance for a one-tailed 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 z-table? 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.