Respuesta :
Answer:
The calculated value |t| = | - 2.6932 | = 2.6932 > 1.9803 at 0.05 level of significance
Alternative hypothesis is accepted
The average of machine two is faster than machine one
Step-by-step explanation:
Step(i):-
Given sample size n₁ = n₂ = 60 minutes
The average of first sample 'x⁻₁ = 73.8
The standard deviation of the first sample 'S₁ ' = 5.2 cans per minute
The average of second sample 'x⁻₂ = 76.1
The standard deviation of the second sample 'S₂ ' = 4.1 cans per minute
step(ii):-
Null Hypothesis : H₀: 'x⁻₁ = 'x⁻₂
Alternative Hypothesis : H₁ : 'x⁻₁ > 'x⁻₂
Test statistic
[tex]t = \frac{x^{-} _{1} - x^{-} _{2} }{\sqrt{\frac{S^{2} _{1} }{n_{1} } +\frac{S^{2} _{2} }{n_{2} } } }[/tex]
[tex]t = \frac{73.8 - 76.1 }{\sqrt{\frac{(5.2)^{2} }{60 } +\frac{(4.1)^{2} }{60 } } }[/tex]
t = - 2.6932
|t| = | - 2.6932 | = 2.6932
Step(iii):-
Degrees of freedom
ν = n₁ + n₂ -2 = 60 +60 -2 = 118
t₀.₀₅ = 1.9803
The calculated value |t| = | - 2.6932 | = 2.6932 > 1.9803 at 0.05 level of significance
Final answer:-
Null hypothesis is rejected at 0.05 level of significance
Alternative hypothesis is accepted
The average of machine two is faster than the average of machine one
