Respuesta :
Answer:
Test statistic
Z = 2.820 > 1.96 at 0.05 level of significance
Null hypothesis is rejected at 0.05 level of significance
The average salary is actually higher than the reported the average of American Medical Association
Step-by-step explanation:
Step(i):-
Given data
sample size 'n' = 64
mean of the Population( μ) = $189,121
Standard deviation of the Population(σ) = $26,975
Mean of the sample (x⁻) = $198,630
Step(ii):-
Null hypothesis : H₀ : there is no significance difference between the means
Alternative Hypothesis :H₁ : The average salary is actually higher than the reported the average of American Medical Association
Test statistic
[tex]Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }[/tex]
[tex]Z = \frac{198630 -189121}{\frac{26,975}{\sqrt{64} } }[/tex]
Z = 2.820
Level of significance
∝ =0.05
[tex]Z_{\alpha } = Z_{0.05} =1.96[/tex]
Final answer:-
Z = 2.820 > 1.96 at 0.05 level of significance
Null hypothesis is rejected at 0.05 level of significance
The average salary is actually higher than the reported the average of American Medical Association
