Question

Let x be a random variable representing dividend yield of Australian bank stocks. We may assume that x has a normal o=2.1% distribution with A random sample of 24 Australian bank stocks has a mean x=6.19% For the entire Australian stock market, the mean dividend yield is u=5.8% Do these data indicate that the dividend yield of all Australian bank stocks is higher than 5.8%? Use a=0.01 What is the level of significance?

Answer 1

Solution:- The level of significance is 0.01.

From the above test we have sufficient evidence in the favor of the claim that the dividend yield of all Australian bank stocks is higher than 5.8%.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u < 5.8
Alternative hypothesis: u > 5.8

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 0.4287
z = (x - u) / SE

z = 0.91

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a z statistic test statistic of 0.91.

Thus the P-value in this analysis is 0.1814.

Interpret results. Since the P-value (0.1814) is greater than the significance level (0.01), we cannot reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that the dividend yield of all Australian bank stocks is higher than 5.8%.