Mathematics, 20.07.2019 08:40 MrSnuffleBuns4277

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 x = 5.24.

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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 $\bar{x}=5.24$ .

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:

$H_0:\mu=5.5 \\ \\ H_a:\mu\neq5.5$

Part B:

The test statistics is given by:

$z= \frac{\bar{x}-\mu}{\sigma/\sqrt{n}} \\ \\ =\frac{5.25-5.5}{0.32/\sqrt{16}} \\ \\ = \frac{-0.25}{0.32/4} = -\frac{0.25}{0.08} \\ \\ =-3.125$

Part C:

The p-value is given by

$P(z\ \textless \ -3.125)=1-P(z$

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 H_0

is given by

$1-\phi\left(z_{1-\frac{\alpha}{2}}+ \frac{\mu_0-\mu_1}{\sigma/\sqrt{n}} \right)+\phi\left(-z_{1-\frac{\alpha}{2}}+ \frac{\mu_0-\mu_1}{\sigma/\sqrt{n}} \right) \\ \\ =1-\phi\left(z_{1-\frac{0.01}{2}}+ \frac{5.5-5.6}{0.32/\sqrt{16}} \right)+\phi\left(-z_{1-\frac{0.01}{2}}+ \frac{5.5-5.6}{0.32/\sqrt{16}} \right) \\ \\ =1-\phi\left(z_{1-0.005}+ \frac{-0.1}{0.32/4} \right)+\phi\left(-z_{1-0.005}+ \frac{-0.1}{0.32/4} \right)$

$=1-\phi\left(z_{0.995}+ \frac{-0.1}{0.08} \right)+\phi\left(-z_{0.995}+ \frac{-0.1}{0.08} \right) \\ \\ =1-\phi(2.576-1.25)+\phi(-2.576-1.25) \\ \\ =1-\phi(1.326)+\phi(-3.826) \\ \\ =1-0.90758+0.00007 \\ \\ =0.0925$

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

$n=\left[ \frac{\sigma(z_{0.005}+z_{0.01})}{\mu_0-\mu} \right]^2 \\ \\ = \left[\frac{0.32(-2.576-2.326)}{5.5-5.6} \right]^2 \\ \\ =\left[\frac{0.32(-4.902)}{-0.1} \right]^2=\left[\frac{-1.56864}{-0.1} \right]^2 \\ \\ =(15.6864)^2=247$

Answer from: Quest

probability for j is 1/4

while

a is 1/6

Answer from: Quest

the answer is; biodiversity resulting from few ancestors

adaptive radiation occurs when a population is subjected to a large natural environment with spatial ecological variations. the parts of the population in the different ecologies begin to adapt to their local environments. due to non-random mating across the large populations, this causes the different populations from the large population to speciate within their local environments. they, therefore, diverge to different species sharing a common ancestor.

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