Let x be a random variable representing dividend yield of bank stocks. We may as
ID: 3243824 • Letter: L
Question
Let x be a random variable representing dividend yield of bank stocks. We may assume that x has a normal distribution with sigma = 2.4%. A random sample of 10 bank stocks gave the following yields (in percents). 5.7 4.8 6.0 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 mu = 4.9%. Do these data indicate that the dividend yield of all bank stocks is higher than 4.9%? Use alpha = 0.01. (a) What is the level of significance? 01 State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? H_0: mu > 4.9%: H_1: mu = 4.9%: right-tailed H_0: mu = 4.9%: H_1: mu > 4.9%: right-tailed H_0: mu = 4.9%: H_1: muExplanation / Answer
Solution:-
xbar = 5.38
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: < 4.9%
Alternative hypothesis: > 4.9%
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.
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 = 0.759
z = (x - ) / SE
z = 0.6325
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a z statistic test statistic of 0.6325. We use the z Distribution Calculator to find P(z > 0.6325) = 0.2643
Thus the P-value in this analysis is 0.2643
Interpret results. Since the P-value (0.2643) is greater than the significance level (0.01), we cannot reject the null hypothesis.
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