Answer:
The calculated value t = 2.038< 2.145 at 0.05 level of significance
Null hypothesis is accepted
There is the average rate is less than μ ≤ 4
Step-by-step explanation:
Step(i):-
The Population of the mean 'μ' =4
sample size 'n' = 16
sample mean 'x⁻' = 4.53
given sample standard deviation 's' = 1.04
level of significance α = 0.05
Step(ii):-
Null hypothesis:H₀ : There is no significance difference between two means
Alternative hypothesis : H₁: There is significance difference between two means
Test statistic
[tex]t = \frac{x^{-} - mean}{\frac{S}{\sqrt{n} } }[/tex]
[tex]t = \frac{4.53-4}{\frac{1.04}{\sqrt{16} } }[/tex]
t = 2.038
Degrees of freedom ν = n-1 = 16-1 =15
t₀.₀₂₅ = 2.145
Conclusion:-
The calculated value t = 2.038< 2.145 at 0.05 level of significance
Null hypothesis is accepted
There is the average rate is less than μ ≤ 4