Part A
The probability of making a type ii error is equal to 1 minus the power of a hypothesis testing.
The power of a hypothesis test is given by:
![\beta(\mu')=\phi\left[z_{\alpha/2}+ \frac{\mu-\mu'}{\sigma/\sqrt{n}} \right]-\phi\left[-z_{\alpha/2}+ \frac{\mu-\mu'}{\sigma/\sqrt{n}} \right]](https://tex.z-dn.net/?f=%5Cbeta%28%5Cmu%27%29%3D%5Cphi%5Cleft%5Bz_%7B%5Calpha%2F2%7D%2B%20%5Cfrac%7B%5Cmu-%5Cmu%27%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%20%5Cright%5D-%5Cphi%5Cleft%5B-z_%7B%5Calpha%2F2%7D%2B%20%5Cfrac%7B%5Cmu-%5Cmu%27%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%20%5Cright%5D)
Given that the
machine is overfilling by .5 ounces, then

, also, we are given that the sample size is 30 and the population standard deviation
is = 0.8 and α = 0.05
Thus,
![\beta(16.5)=\phi\left[z_{0.025}+ \frac{-0.5}{0.8/\sqrt{30}} \right]-\phi\left[-z_{0.025}+ \frac{-0.5}{0.8/\sqrt{30}} \right] \\ \\ =\phi\left[1.96+ \frac{-0.5}{0.1461} \right]-\phi\left[-1.96+ \frac{-0.5}{0.1461} \right] \\ \\ =\phi(1.96-3.4233)-\phi(-1.96-3.4233) \\ \\ =\phi(-1.4633)-\phi(-5.3833)=0.07169](https://tex.z-dn.net/?f=%5Cbeta%2816.5%29%3D%5Cphi%5Cleft%5Bz_%7B0.025%7D%2B%20%5Cfrac%7B-0.5%7D%7B0.8%2F%5Csqrt%7B30%7D%7D%20%5Cright%5D-%5Cphi%5Cleft%5B-z_%7B0.025%7D%2B%20%5Cfrac%7B-0.5%7D%7B0.8%2F%5Csqrt%7B30%7D%7D%20%5Cright%5D%20%5C%5C%20%20%5C%5C%20%3D%5Cphi%5Cleft%5B1.96%2B%20%5Cfrac%7B-0.5%7D%7B0.1461%7D%20%5Cright%5D-%5Cphi%5Cleft%5B-1.96%2B%20%5Cfrac%7B-0.5%7D%7B0.1461%7D%20%5Cright%5D%20%5C%5C%20%20%5C%5C%20%3D%5Cphi%281.96-3.4233%29-%5Cphi%28-1.96-3.4233%29%20%5C%5C%20%20%5C%5C%20%3D%5Cphi%28-1.4633%29-%5Cphi%28-5.3833%29%3D0.07169)
Therefore, the probability of making a type II error when the machine is overfilling by .5 ounces is 1 - 0.07169 = 0.9283
Part B:
From part A, the power of the statistical test when the machine is
overfilling by .5 ounces is 0.0717.
Answer:
1) 5 2) 6 3) 1.25 or -1.75
Step-by-step explanation:
1. x^2 - 25 = 0
x^2 - 25 = 0
First, add 25 to both sides!
x^2 = 25
Now square root both sides.
x = 5
2. (x-2)^2 = 16
First, square root both sides.
(x-2) = 4
Add 2 on both sides.
(x) = 6
x = 6
3. Quadratic Formula = -b ±
/ 2a
Substitute in:
-6 ±
/ 2(2)
-6±
/ 4
Both forms:
-6 +
/ 4
5 / 4 = 1.25
Negative:
-6 - 1 / 4
-7 / 4 = -1.75
Answer:
B) A/l
Step-by-step explanation:
Isolate the variable, w. Note the equal sign, what you do to one side, you do to the other.
Divide l from both sides of the equation:
A = l * w
(A)/l = (l)/l * w
w = A/l
B) A/l is your answer.
~
Answer:
less likely
Step-by-step explanation:
Probability is between 0 and 1 so 1 is guaranteed to happen and 0 is guaranteed to not happen
Since
, we can rewrite the integral as

Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:

Both integrals are quite immediate: you only need to use the power rule

to get
![\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E11-3t%5E2%5C%3Bdt%20%3D%20%5Cleft%5Bt-t%5E3%5Cright%5D_0%5E1%2C%5Cquad%20%5Cint_1%5E4%202t%5C%3B%20dt%20%3D%20%5Cleft%5Bt%5E2%5Cright%5D_1%5E4)
Now we only need to evaluate the antiderivatives:
![\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15](https://tex.z-dn.net/?f=%5Cleft%5Bt-t%5E3%5Cright%5D_0%5E1%20%3D%201-1%5E3%3D0%2C%5Cquad%20%5Cleft%5Bt%5E2%5Cright%5D_1%5E4%20%3D%204%5E2-1%5E2%3D15)
So, the final answer is 15.