91Ó°ÊÓ

Please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. Compute the \(z\) value of the sample test statistic. (c) Find (or estimate) the \(P\) -value. Sketch the sampling distribution and show the area corresponding to the \(P\) -value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \(\alpha ?\) (e) Interpret your conclusion in the context of the application. Let \(x\) be a random variable representing dividend yield of Australian bank stocks. We may assume that \(x\) has a normal distribution with \(\sigma=2.4 \% .\) A random sample of 10 Australian bank stocks gave the following yields. The sample mean is \(\bar{x}=5.38 \% .\) For the entire Australian stock market, the mean dividend yield is \(\mu=4.7 \%\) (Reference: Forbes). Do these data indicate that the dividend yield of all Australian bank stocks is higher than \(4.7 \% ?\) Use \(\alpha=0.01\)

Step by step solution

01

Determine Hypotheses and Type of Test

The level of significance is given as \(\alpha = 0.01\). The null hypothesis \(H_0\) is that the mean dividend yield of Australian bank stocks is \(\mu = 4.7\%\). The alternate hypothesis \(H_1\) is that the mean dividend yield is higher than \(4.7\%\), i.e., \(\mu > 4.7\%\). This is a right-tailed test.

02

Choose the Sampling Distribution

Since the population standard deviation \(\sigma\) is known (\(2.4\%\)), the normal distribution (specifically the \(z\)-distribution) is appropriate for the test. Also, the sample size is \(n = 10\), which is reasonable for assuming normality when the population is normal.

03

Calculate the Test Statistic

The test statistic \(z\) is calculated using the formula: \[z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}\]. Given \(\bar{x} = 5.38\%\), \(\mu = 4.7\%\), \(\sigma = 2.4\%\), and \(n = 10\), we have: \[z = \frac{5.38 - 4.7}{2.4 / \sqrt{10}} \approx \frac{0.68}{0.758}\approx 0.897\].

04

Determine the P-value

Consult the standard normal distribution table (z-table) for the calculated \(z\)-value of \(0.897\). This provides a \(P\)-value of approximately \(0.1841\) for a right-tailed test.

05

Decision - Reject or Fail to Reject?

Since the \(P\)-value (\(0.1841\)) is greater than \(\alpha = 0.01\), we fail to reject the null hypothesis. The data are not statistically significant at the \(0.01\) level.

06

Interpret the Conclusion

The data do not provide sufficient evidence to conclude that the mean dividend yield of all Australian bank stocks is higher than \(4.7\%\) at the \(\alpha = 0.01\) significance level.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Significance Level

The significance level, often denoted by the Greek letter \( \alpha \), is a threshold that determines the probability of rejecting the null hypothesis when it is actually true. In hypothesis testing, this level is set before any data is analyzed. It quantifies the maximum acceptable probability of making a Type I error, which is rejecting a true null hypothesis.

For example, in the given problem, the significance level is set at \( \alpha = 0.01 \), which means there is a 1% risk of incorrectly rejecting the null hypothesis. This low significance level suggests a stringent criterion to conclude that the data is statistically significant. In practice, researchers choose \( \alpha \) depending on how much risk they are willing to take. Common values are 0.05, 0.01, and 0.10. By clearly understanding and setting this level, you ensure that any conclusions drawn from the experiment are consistent with the desired level of confidence.

Null and Alternate Hypotheses

The null hypothesis \( H_0 \) and the alternate hypothesis \( H_1 \) are essential in hypothesis testing as they form the basis of the test of significance. These are the competing statements about the population that we are testing.

• The **null hypothesis** \( H_0 \) represents the status quo or the default position. It is the hypothesis that there is no effect or no difference. In the example, \( H_0 \) states that the mean dividend yield of Australian bank stocks is \( \mu = 4.7\% \).
• The **alternate hypothesis** \( H_1 \) is the claim that we are testing against the null hypothesis. It is what we believe might be true instead. Here, \( H_1 \) claims that the mean dividend yield is higher than \( 4.7\% \), so \( \mu > 4.7\% \).

Our task is to use sample data to determine which hypothesis is more likely to be true. We perform a right-tailed test because we are interested in whether the mean dividend yield is greater than a specific value.

Sampling Distribution

A sampling distribution is a probability distribution of a statistic obtained through a large number of samples drawn from a specific population. It is essential in hypothesis testing because it allows us to determine how likely it is to observe a test statistic that is at least as extreme as the one computed from the sample data.

In the given problem, since the population standard deviation \( \sigma \) is known to be \( 2.4\% \), and the sample size \( n = 10 \) does not violate the assumptions of normality (given that \( x \) is normally distributed), we use the normal distribution or specifically the \( z \)-distribution for hypothesis testing. This choice is appropriate because it efficiently estimates the probability of falling within a particular range of the sampling distribution.

• The **test statistic** is calculated using the formula: \[ z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \]

Plugging in the values: \( \bar{x} = 5.38\% \), \( \mu = 4.7\% \), \( \sigma = 2.4\% \), and \( n = 10 \), the \( z \) value comes out to approximately \( 0.897 \). This statistic helps us determine where our sample mean falls relative to the hypothesis being tested.

P-value Calculation

The P-value is a measure used in hypothesis testing to help determine the strength of the evidence against the null hypothesis. It represents the probability of obtaining a result at least as extreme as the one that was actually observed, assuming that the null hypothesis is true.

To calculate the P-value, you consult a standard normal distribution table (z-table) using the calculated \( z \)-value from the test statistic. For a right-tailed test like the one in our example, the calculated \( z \)-value was approximately \( 0.897 \). Checking the z-table, this corresponds to a P-value of around \( 0.1841 \).

Because the P-value (\( 0.1841 \)) is greater than our significance level \( \alpha = 0.01 \), we do not reject the null hypothesis. This implies that the data does not provide sufficient evidence to support the claim that the mean dividend yield is more than \( 4.7 \% \). Thus, the result is not statistically significant, and any observed differences could well be due to random chance.