Given that the
desired percentage of sio2 in a certain type of aluminous cement is
5.5. to test whether the true average percentage is 5.5 for a particular
production facility, 16 independently obtained samples are analyzed.
suppose that the percentage of sio2 in a sample is normally distributed
with σ = 0.32 and that [tex]\bar{x}=5.24[/tex].
To investigate whether this indicate conclusively that the true average
percentage differs from 5.5.
Part A:
From the question, it is claimed that the
desired average percentage of sio2 in a certain type of aluminous cement is
5.5 and we want to test whether the information from the random sample indicate conclusively that the true average
percentage differs from 5.5.
Therefore, the null hypothesis and the alternative hypothesis is given by:
[tex]H_0:\mu=5.5 \\ \\ H_a:\mu\neq5.5[/tex]
Part B:
The test statistics is given by:
[tex]z= \frac{\bar{x}-\mu}{\sigma/\sqrt{n}} \\ \\ =\frac{5.25-5.5}{0.32/\sqrt{16}} \\ \\ = \frac{-0.25}{0.32/4} = -\frac{0.25}{0.08} \\ \\ =-3.125[/tex]
Part C:
The p-value is given by
[tex]P(z\ \textless \ -3.125)=1-P(z<3.125)=1-0.99911=0.00089[/tex]
Part D:
Because the p-value is less than the significant level α, we reject the null hypothesis and conclude that "There is sufficient evidence
to conclude that the true average percentage differs from the
desired percentage."
Part E:
If the true average percentage is μ = 5.6 and a level α = 0.01 test based on n =
16 is used, what is the probability of detecting this departure
from H0? (Round your answer to four decimal
places.)
The probability of detecting the departure
from [tex]H_0[/tex] is given by
[tex]1-\phi\left(z_{1-\frac{\alpha}{2}}+ \frac{\mu_0-\mu_1}{\sigma/\sqrt{n}} \right)+\phi\left(-z_{1-\frac{\alpha}{2}}+ \frac{\mu_0-\mu_1}{\sigma/\sqrt{n}} \right) \\ \\ =1-\phi\left(z_{1-\frac{0.01}{2}}+ \frac{5.5-5.6}{0.32/\sqrt{16}} \right)+\phi\left(-z_{1-\frac{0.01}{2}}+ \frac{5.5-5.6}{0.32/\sqrt{16}} \right) \\ \\ =1-\phi\left(z_{1-0.005}+ \frac{-0.1}{0.32/4} \right)+\phi\left(-z_{1-0.005}+ \frac{-0.1}{0.32/4} \right)[/tex]
[tex]=1-\phi\left(z_{0.995}+ \frac{-0.1}{0.08} \right)+\phi\left(-z_{0.995}+ \frac{-0.1}{0.08} \right) \\ \\ =1-\phi(2.576-1.25)+\phi(-2.576-1.25) \\ \\ =1-\phi(1.326)+\phi(-3.826) \\ \\ =1-0.90758+0.00007 \\ \\ =0.0925[/tex]
Part F:
What value of n is required to satisfy
α = 0.01 and β(5.6) = 0.01? (Round your answer up
to the next whole number.)
The value of n is required to satisfy
α = 0.01 and β(5.6) = 0.01 is given by
[tex]n=\left[ \frac{\sigma(z_{0.005}+z_{0.01})}{\mu_0-\mu} \right]^2 \\ \\ = \left[\frac{0.32(-2.576-2.326)}{5.5-5.6} \right]^2 \\ \\ =\left[\frac{0.32(-4.902)}{-0.1} \right]^2=\left[\frac{-1.56864}{-0.1} \right]^2 \\ \\ =(15.6864)^2=247[/tex]
