Answer:
Test statistic Z= 1.713
The calculated Z- value = 1.7130 < 2.576 at 0.01 level of significance
Null hypothesis is accepted
There is no difference between the mean annual salary of all lawyers in a city is different from $110,000
Step-by-step explanation:
<u><em>Step(i):-</em></u>
A researcher wants to test if the mean annual salary of all lawyers in a city is
different from $110,000
Mean of the Population μ = $110,000
Sample size 'n' = 53
Mean of the sample x⁻ = $114,000.
standard deviation of the Population = $17,000,
Level of significance = 0.01
Null hypothesis :
There is no difference between the mean annual salary of all lawyers in a city is different from $110,000
H₀: x⁻ = μ
Alternative Hypothesis : x⁻ ≠ μ
<u><em>Step(ii)</em></u>:-
Test statistic
![Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7Bx%5E%7B-%7D-mean%20%7D%7B%5Cfrac%7BS.D%7D%7B%5Csqrt%7Bn%7D%20%7D%20%7D)
![Z = \frac{114000-110000}{\frac{17000}{\sqrt{53} } }](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B114000-110000%7D%7B%5Cfrac%7B17000%7D%7B%5Csqrt%7B53%7D%20%7D%20%7D)
Z = 1.7130
Tabulated value Z = 2.576 at 0.01 level of significance
The calculated Z- value = 1.7130 < 2.576 at 0.01 level of significance
Null hypothesis is accepted
There is no difference between the mean annual salary of all lawyers in a city is different from $110,000