Let x be a random variable representing dividend yield of Australian bank stocks. We may assume that x has a normal distribution with s 5 2.

Question

Let x be a random variable representing dividend yield of Australian bank stocks. We may assume that x has a normal distribution with s 5 2.4%. A random sample of 10 Australian bank stocks gave the following yields. 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 5.38%. For the entire Australian stock market, the mean dividend yield is m 5 4.7% (Reference: Forbes). Do these data indicate that the dividend yield of all Australian bank stocks is higher than 4.7%

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Athena 1 week 2021-09-15T21:09:07+00:00 1 Answer 0

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    2021-09-15T21:10:34+00:00

    Answer:

    No. There is not enough evidence to support the claim that the dividend yield of all Australian bank stocks is higher than 4.7%.  

    The P-value is 0.185.

    Step-by-step explanation:

    This is a hypothesis test for the population mean.

    The claim is that the dividend yield of all Australian bank stocks is higher than 4.7%.

    Then, the null and alternative hypothesis are:

    H_0: \mu=4.7\\\\H_a:\mu> 4.7

    The significance level is assumed to be 0.05.

    The sample has a size n=10.

    The sample mean is M=5.38.

    The standard deviation of the population is known and has a value of σ=2.4.

    We can calculate the standard error as:

    \sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{2.4}{\sqrt{10}}=0.759

    Then, we can calculate the z-statistic as:

    z=\dfrac{M-\mu}{\sigma_M}=\dfrac{5.38-4.7}{0.759}=\dfrac{0.68}{0.759}=0.896

    This test is a right-tailed test, so the P-value for this test is calculated as:

    P-value=P(z>0.896)=0.185

    As the P-value (0.185) is bigger than the significance level (0.05), the effect is not significant.

    The null hypothesis failed to be rejected.

    There is not enough evidence to support the claim that the dividend yield of all Australian bank stocks is higher than 4.7%.

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