Question

Let x be a random variable representing dividend yield of bank stocks. We may assume that x has a normal distribution with σ = 3.2%. A random sample of 10 bank stocks gave the following yields (in percents). 5.7 4.8 6.0 4.9 4.0 3.4 6.5 7.1 5.3 6.1 The sample mean is x = 5.38%. Suppose that for the entire stock market, the mean dividend yield is μ = 4.2%. Do these data indicate that the dividend yield of all bank stocks is higher than 4.2%? Use α = 0.01.

(a) What is the level of significance?

Answer 1

To determine if the dividend yield of all bank stocks is higher than 4.2%, we perform a hypothesis test. With a calculated z-score of 12.17 and a critical value of 2.33, we reject the null hypothesis and conclude that the dividend yield is higher than 4.2%.

Step-by-step explanation:

To determine if the dividend yield of all bank stocks is higher than 4.2%, we need to perform a hypothesis test using the given data. The null hypothesis (H0) is that the mean dividend yield is 4.2%, while the alternative hypothesis (Ha) is that the mean dividend yield is greater than 4.2%. The level of significance (α) is given as 0.01.

We can calculate the test statistic, which is the z-score, using the formula: z = (x - μ) / (σ / sqrt(n)).

Substituting the given values, we have: z = (5.38 - 4.2) / (0.032 / sqrt(10)). Calculating this, we get a z-score of 12.17.

Next, we can find the critical value, which corresponds to the α level of significance. Since α = 0.01, the critical value is 2.33 (obtained from the standard normal distribution table).

Since the calculated z-score (12.17) is greater than the critical value (2.33), we reject the null hypothesis. This means that the data indicate that the dividend yield of all bank stocks is indeed higher than 4.2%.