Answer:
The statistical result of analysis of the parameters strongly suggest that compressive stress of the presented type of concrete is normally distributed
Step-by-step explanation:
Here we have
112.3 97.0 92.7 86.0 102.0 99.2 95.8 103.5 89.0 86.7
Mean = 96.42
Standard deviation = 8.26
Population mean = 100 MPa
Confidence level, α = 0.05
Therefore we have
The hypothesis H₀: μ= 100 alternative
Hₐ: μ ≠ 100
The test statistic is then
[tex]t=\frac{\bar{x}-\mu }{\frac{\sigma }{\sqrt{n}}}[/tex]
Which gives
[tex]t=\frac{96.42-100 }{\frac{8.26 }{\sqrt{10}}}[/tex] = -1.37
Determining the rejection region at α = 0.05
[tex]t<t_{0.05,9} = -1.833[/tex]
Therefore, since [tex]t>t_{0.05,9}[/tex] and with the tcdf (-10⁹⁹, -1.37, 9) = 0.102 > α > 0.05
We fail to reject the null hypothesis, and the test is NOT significant
Therefore, by comparison to a normal distributed of distribution, the compressive stress of this type of concrete is normally distributed.