Question

Let x be a random variable representing dividend yield of Australian bank stocks. We may assume that x has a normal distribution with $ppg__cognero__Chapter8TestStats__media__$ %. A random sample of 23 Australian bank stocks has a sample mean of $ppg__cognero__Chapter8TestStats__media__$ %. For the entire Australian stock market, the mean dividend yield is $ppg__cognero__Chapter8TestStats__media__$ %. Do these data indicate that the dividend yield of all Australian bank stocks is higher than 8.9%? What is the value of the test statistic? Round your answer to three decimal places.

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Select one:

a. 7.793

b. –0.339

c. –1.625

d. 0.339

e. –7.793

Answer 1

Solution:

The provided sample mean is $\bar X$ = 5% and the known population standard deviation is σ = 2.4%,

and the sample size is n = 23

We have to test claim that, mean dividend yield $\mu$ > 8.9%

The following null and alternative hypotheses need to be tested:

H0 :

= 8.9

Ha:

ul> 8.9

This corresponds to a right-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Test statistic:

The Z-statistic is computed as follows: X – uo 5 – 8.9 z= = = -7.793 Ovn 2.4/ 23

Hence option e is correct

Done