Respuesta :
Answer:
The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.
The null hypothesis is rejected.
There is significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units
Step-by-step explanation:
Step:-(1)
Given data the samples of material 1 gave an average (coded) wear of 85 units with a sample standard deviation of 4
Mean of the first sample x₁⁻ =85
standard deviation of the first sample S₁ = 4
Given data the samples of material 2 gave an average of 81 and a sample standard deviation of 5.
Mean of the first sample x₂⁻ =81
standard deviation of the first sample S₂ = 5
Step :-2
Null hypothesis: H₀: there is no significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units
Alternative hypothesis :H₁: there is significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units
Assume the populations to be approximately normal with equal variances.σ₁² =σ₂²
The test statistic
[tex]Z= \frac{x_{1} -x_{2} }{\sqrt{\frac{S^2_{1} }{n_{1} } +\frac{S^2_{2} }{n_{2} } } }[/tex]
Given n₁=n₂=60.
[tex]Z= \frac{85-81 }{\sqrt{\frac{(4)^2 }{60 } +\frac{5^2 }{60} } }[/tex]
On calculation, we get
Z = [tex]\frac{4}{\sqrt{0.6833} }[/tex]
z = 4.8389
The tabulated value Z =1.96 at 0.05 level of significance.
The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.
The null hypothesis is rejected.
Conclusion:-
there is significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units.
