Cindy is using division to write a fraction equivalent to 30/100. She tried to divide the numerator and denominator by 3. She go
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Given that t<span>he
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 
.
</span>
<span>To investigate whether this indicate conclusively that the true average percentage differs from 5.5.
Part A:
From the question, it is claimed that </span><span>t<span>he desired average percentage of sio2 in a certain type of aluminous cement is 5.5</span></span> and we want to test whether the information from the random sample <span>indicate conclusively that the true average percentage differs from 5.5.
Therefore, the null hypothesis and the alternative hypothesis is given by:
Part B:
The test statistics is given by:
Part C:
The p-value is given by
Part D:
Because the p-value is less than the significant level α, we reject the null hypothesis and conclude that "</span><span>There is sufficient evidence to conclude that the true average percentage differs from the desired percentage."
Part E:
</span>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
is given by
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](https://tex.z-dn.net/?f=n%3D%5Cleft%5B%20%5Cfrac%7B%5Csigma%28z_%7B0.005%7D%2Bz_%7B0.01%7D%29%7D%7B%5Cmu_0-%5Cmu%7D%20%5Cright%5D%5E2%20%5C%5C%20%20%5C%5C%20%3D%20%5Cleft%5B%5Cfrac%7B0.32%28-2.576-2.326%29%7D%7B5.5-5.6%7D%20%5Cright%5D%5E2%20%5C%5C%20%20%5C%5C%20%3D%5Cleft%5B%5Cfrac%7B0.32%28-4.902%29%7D%7B-0.1%7D%20%5Cright%5D%5E2%3D%5Cleft%5B%5Cfrac%7B-1.56864%7D%7B-0.1%7D%20%5Cright%5D%5E2%20%5C%5C%20%20%5C%5C%20%3D%2815.6864%29%5E2%3D247)
</span>
<span>To investigate whether this indicate conclusively that the true average percentage differs from 5.5.
Part A:
From the question, it is claimed that </span><span>t<span>he desired average percentage of sio2 in a certain type of aluminous cement is 5.5</span></span> and we want to test whether the information from the random sample <span>indicate conclusively that the true average percentage differs from 5.5.
Therefore, the null hypothesis and the alternative hypothesis is given by:
Part B:
The test statistics is given by:
Part C:
The p-value is given by
Part D:
Because the p-value is less than the significant level α, we reject the null hypothesis and conclude that "</span><span>There is sufficient evidence to conclude that the true average percentage differs from the desired percentage."
Part E:
</span>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
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
Answer:
(a) The sample mean startup cost is 107.3 thousand dollars.
The sample mean startup cost is 107.3 thousand dollars.
(b) The 90% confidence interval for the population average startup costs for candy store franchises is ($89.4, $125.2) thousand.
Step-by-step explanation:
The questions asked related to the data are:
(a) Use a calculator with mean and sample standard deviation keys to find the sample mean startup cost x and sample standard deviation s. (Round your answers to one decimal place.)
x = thousand dollars
s = thousand dollars
(b) Find a 90% confidence interval for the population average startup costs μ for candy store franchises. (Round your answers to one decimal place.)
lower limit thousand dollars
upper limit thousand dollars
Solution:
The data provided is:
X = {100, 170, 133, 93, 75, 94, 116, 100, 85}
(a)
Compute the sample mean as follows:
The sample mean startup cost is 107.3 thousand dollars.
Compute the sample standard deviation as follows:
The sample standard deviation of startup cost is 28.9 thousand dollars.
(b)
As the population standard deviation is not known we will use a <em>t</em>-interval.
The critical value of <em>t</em> for 90% confidence level and (n - 1) = 8 degrees of freedom is:
*Use a <em>t</em>-table for the value.
Compute the 90% confidence interval for population mean as follows:
Thus, the 90% confidence interval for the population average startup costs for candy store franchises is ($89.4, $125.2) thousand.