Respuesta :
Answer:
D) + 3.82
Z- test statistic value = 3.819≅ 3.82
Step-by-step explanation:
Explanation:-
Given data the sample size 'n' = 350
Population mean 'μ' = $8,500
Sample mean 'x⁻' = $8,500
The population standard deviation is 'σ' = $1,200
The level of significance ∝ =0.05
The tabulated value Z₀.₉₅ = 1.96
Null hypothesis: There is no significant difference between the small private liberal arts college and the financial administrator.
x⁻= μ
Alternative hypothesis: x⁻> μ
Test statistic
[tex]Z = \frac{x^{-} -mean }{\frac{S.D}{\sqrt{n} } }[/tex]
[tex]Z = \frac{8745-8500}{\frac{1200}{\sqrt{350} } }[/tex]
On calculation , we get
[tex]Z = \frac{245}{64.14} = 3.819[/tex]
Z- test statistic value = 3.819≅ 3.82
Conclusion:-
The calculated value = 3.82 > 1.96 at 0.05 level of significance.
The null hypothesis is rejected
Alternative hypothesis is accepted
The financial administrator believes that the average cost is higher than the small private liberal arts college
