Someone please help me, I do not understand how to solve this problem:
Let x be a random variable representing dividend yield of bank stocks. We may assume that x has a normal distribution with σ = 2.0%. 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 = 4.9%. Do these data indicate that the dividend yield of all bank stocks is higher than 4.9%? Use α = 0.01.
(a) What is the level of significance?
(b) What is the value of the sample test statistic? (Round your answer to two decimal places.)
(c) Find (or estimate) the P-value. (Round your answer to four decimal places.)
(a) The level of significance, denoted by α, is the probability of making a Type I error, which is rejecting a true null hypothesis. In this case, α = 0.01, which means we are willing to accept a 1% chance of wrongly concluding that the dividend yield of bank stocks is higher than 4.9% when it is not.
(b) To find the sample test statistic, we need to calculate the test statistic called the t-score. The formula for the t-score is:
t = (x - μ) / (σ / √n)
where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
In this case, x = 5.38%, μ = 4.9%, σ = 2.0%, and n = 10.
Plugging in these values, we get:
t = (5.38% - 4.9%) / (2.0% / √10) ≈ 0.219
So, the value of the sample test statistic is approximately 0.219.
(c) To find the P-value, we need to compare the test statistic (t-score) to the critical value(s) based on the given significance level (α = 0.01) and the degrees of freedom (df = n - 1).
In this case, the degrees of freedom is df = 10 - 1 = 9.
Using a standard normal distribution table, we can find that the critical value corresponding to a significance level of 0.01 for a two-tailed test is approximately ±2.821.
Since our test statistic (t = 0.219) is within the range of -2.821 to +2.821, it means it falls within the acceptance region, and the P-value is greater than 0.01.
To estimate the P-value, we can use statistical software or a calculator that provides the option to find the P-value directly from the t-distribution. Alternatively, we can use a t-table.
Using a t-table with 9 degrees of freedom, we find that the t-value of 0.219 corresponds to a P-value of approximately 0.8308.
Therefore, the P-value is approximately 0.8308.