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

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 =  2.8840

Degrees of freedom

ν = n-1 = 11-1 =10

Level of significance

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%

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