Respuesta :
Answer:
The test statistic value t = 1.64 < 2.0086 at 0.05 level of significance
Null hypothesis is accepted
The mean portfolio value for all senior citizen shareholders in this region might not be the same as the mean value reported for their counterparts across the nation
Step-by-step explanation:
Step(i):-
Given mean of the population (μ) = $183,000
Given mean of the sample (x⁻) = $198,000
Given standard deviation of the sample (S) = $65,000.
Mean of the sample size 'n' = 51
level of significance α = 0.05
Step(ii):-
Null hypothesis : H₀ : There is no significance difference between the means
Alternative Hypothesis :H₁: There is significance difference between the means
Test statistic
[tex]t = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }[/tex]
[tex]t = \frac{198,000- 183,000 }{\frac{ 65,000}{\sqrt{51} } }[/tex]
t = 1.64
Step(iii)
Degrees of freedom ν = n-1 = 51-1 =50
t₀.₀₅ = 2.0086
The calculated value t = 1.64 < 2.0086 at 0.05 level of significance
Null hypothesis is accepted
Final answer:-
There is no significance difference between the means
The mean portfolio value for all senior citizen shareholders in this region might not be the same as the mean value reported for their counterparts across the nation