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 = 2.3%. A random sample of 11 Australian bank stocks has a mean x = 9.89%. For the entire Australian stock market, the mean dividend yield is μ = 7.9%. Do these data indicate that the dividend yield of all Australian bank stocks is higher than 7.9%? Use a = 0.05. What is the level of significance?
a. 0.025
b. 0.050
c. 0.100
d. 0.900
e. 0.975

Answer 1

Answer:

a) 0.025 level of significance

The calculated t - value t = 2.88 > 3.58 at 0.025 level of significance

Therefore null hypothesis is rejected

The data indicate that the dividend yield of all Australian bank stocks is higher than 7.9%

Step-by-step explanation:

Step(i):-

Let 'x' has a normal distribution

Given sample size 'n' =11

Mean of the sample (x⁻)  = 9.89% = 0.0989

Standard deviation of the sample (s) = 2.3% = 0.023

Mean of the Population ' μ' = 7.9% = 0.079

Step(ii):-

Null hypothesis:H₀:' μ'  = 0.079

Alternative Hypothesis :μ'  > 0.079

Test statistic  

         $t = \frac{x^{-} -mean}{\frac{S}{\sqrt{n} } }$

        $t = \frac{0.0989-0.079}{\frac{0.023}{\sqrt{11} } }$

       t =  2.8840

Degrees of freedom

ν = n-1 = 11-1 =10

Level of significance

$t_{\frac{\alpha }{2} } = t_{\frac{0.05}{2} } = t_{0.025}$

$t_{0.025} , 10 = 3.5814$

Step(iii):-

The calculated t - value t = 2.88 > 3.58 at 0.025 level of significance

Therefore null hypothesis is rejected

The data indicate that the dividend yield of all Australian bank stocks is higher than 7.9%