Question

et x be a random variable representing dividend yield of bank stocks. We may assume that x has a normal distribution with σ = 2.1%. 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  = 5.38%. Suppose that for the entire stock market, the mean dividend yield is μ = 4.3%. Do these data indicate that the dividend yield of all bank stocks is higher than 4.3%? Use α = 0.01.

(a)

What is the level of significance? (Enter a number.)


State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test?

H₀: μ = 4.3%; H₁: μ ≠ 4.3%; two-tailed

H₀: μ > 4.3%; H₁: μ = 4.3%; right-tailed   

H₀: μ = 4.3%; H₁: μ < 4.3%; left-tailed

H₀: μ = 4.3%; H₁: μ > 4.3%; right-tailed

(b)

What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.

The Student's t, since n is large with unknown σ.

The standard normal, since we assume that x has a normal distribution with unknown σ.    

The standard normal, since we assume that x has a normal distribution with known σ.

The Student's t, since we assume that x has a normal distribution with known σ.

Compute the z value of the sample test statistic. (Enter a number. Round your answer to two decimal places.)

(c)

Find (or estimate) the P-value. (Enter a number. Round your answer to four decimal places.)

Answer 1

a) The level of significance, we are already given in the problem here as 0.01. Therefore 1% is the level of significance here.

b) As we are trying to test here whether the dividend yield of all bank stocks is higher than 4.3%, therefore the null and the alternative hypothesis here are given as:

As we are testing it from the up-side, therefore this is a right tailed test that is a one tailed test here.

b) As we are given the population standard deviation here, we would use the standard normal distribution here. The standard normal, since we assume that x has a normal distribution with known σ.

The test statistic here is computed as:

Therefore 1.63 is the test statistic value here.

c) As this is an upper tailed test, the p-value here is computed from the standard normal tables here as:

p = P( Z > 1.63) = 0.0519

therefore 0.0519 is the required p-value here.