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:
<u><em>Step(i):-</em></u>
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
<u><em>Step(ii):-</em></u>
<u><em>Null hypothesis:H₀:</em></u>' μ' = 0.079
<u><em>Alternative Hypothesis</em></u> :μ' > 0.079
Test statistic
![t = \frac{x^{-} -mean}{\frac{S}{\sqrt{n} } }](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bx%5E%7B-%7D%20-mean%7D%7B%5Cfrac%7BS%7D%7B%5Csqrt%7Bn%7D%20%7D%20%7D)
![t = \frac{0.0989-0.079}{\frac{0.023}{\sqrt{11} } }](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B0.0989-0.079%7D%7B%5Cfrac%7B0.023%7D%7B%5Csqrt%7B11%7D%20%7D%20%7D)
t = 2.8840
Degrees of freedom
ν = n-1 = 11-1 =10
<em>Level of significance </em>
![t_{\frac{\alpha }{2} } = t_{\frac{0.05}{2} } = t_{0.025}](https://tex.z-dn.net/?f=t_%7B%5Cfrac%7B%5Calpha%20%7D%7B2%7D%20%7D%20%3D%20t_%7B%5Cfrac%7B0.05%7D%7B2%7D%20%7D%20%3D%20t_%7B0.025%7D)
![t_{0.025} , 10 = 3.5814](https://tex.z-dn.net/?f=t_%7B0.025%7D%20%2C%2010%20%3D%203.5814)
<u><em>Step(iii):-</em></u>
<em>The calculated t - value t = 2.88 > 3.58 at 0.025 level of significance</em>
<em>Therefore null hypothesis is rejected</em>
<em>The data indicate that the dividend yield of all Australian bank stocks is higher than 7.9%</em>