Answer:
The calculated t-test = 1.717 < 1.6604 at 0.05 level of significance
Null hypothesis is accepted at 0.05 level of significance
A company has developed a training procedure is significantly improve scores
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given sample mean (x⁻ ) = 517
sample standard deviation (s) =90
Mean of the Population (μ) =500
<em>Null Hypothesis :- H₀ : </em> (μ) =500
Alternative Hypothesis : H₁ : μ ≠ 500
<u><em>Step(ii)</em></u>:-
Test statistic


t = 1.717
<em>Level of significance</em>
∝ = 0.05
t₀.₀₅ = 1.6604
The calculated t-test = 1.717 < 1.6604 at 0.05 level of significance
<u><em>Conclusion</em></u>:-
Null hypothesis is accepted at 0.05 level of significance
A company has developed a training procedure is significantly improve scores