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IRISSAK [1]
3 years ago
6

A researcher plans to conduct a significance test at the 0.01 significance level. She designs her study to have a power of 0.90

at a particular alternative value of the parameter of interest. The probability that the researcher will commit a Type II error for the particular alternative value of the parameter at which she computer the power is'
Mathematics
1 answer:
Rasek [7]3 years ago
6 0

Answer: 0.10

Step-by-step explanation: The type 2 error is committed when the alternative hypothesis is rejected when it should have been accepted causing the researcher to accept the null hypothesis which is false.

Power is the probability of avoiding a type 2 error. That is ;

Power = 1 - P(type 2 error)

Given that power = 0.90 ; P(type 2 error) = probability of committing a type 2 error.

P(type 2 error)' = 1 - P(type 2 error) = Probability of not committing or avoiding a type 2 error

0.90 = 1 - P(type 2 error)

P(type 2 error) = 1 - 0.90

P(type 2 error) = 0.10

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slega [8]

Answer:

5/6 because 1/2 plus 1/3 is 5/6, meaning that B is the correct answer.

Hope this helps!

5 0
2 years ago
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A tank with a capacity of 1000 L is full of a mixture of water and chlorine with a concentration of 0.02 g of chlorine per liter
faltersainse [42]

At the start, the tank contains

(0.02 g/L) * (1000 L) = 20 g

of chlorine. Let <em>c</em> (<em>t</em> ) denote the amount of chlorine (in grams) in the tank at time <em>t </em>.

Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of

(<em>c</em> (<em>t</em> )/(1000 + (10 - 25)<em>t</em> ) g/L) * (25 L/s) = 5<em>c</em> (<em>t</em> ) /(200 - 3<em>t</em> ) g/s

In case it's unclear why this is the case:

The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after <em>t</em> s there will be (1000 + 10<em>t</em> ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time <em>t</em> is (1000 - 15<em>t </em>) L. Then the concentration of chlorine per unit volume is <em>c</em> (<em>t</em> ) divided by this volume.

So the amount of chlorine in the tank changes according to

\dfrac{\mathrm dc(t)}{\mathrm dt}=-\dfrac{5c(t)}{200-3t}

which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5<em>t</em> )^(5/3), then integrate both sides to solve for <em>c</em> (<em>t</em> ):

\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{200-3t}=0

\dfrac1{(200-3t)^{5/3}}\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{(200-3t)^{8/3}}=0

\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{c(t)}{(200-3t)^{5/3}}\right]=0

\dfrac{c(t)}{(200-3t)^{5/3}}=C

c(t)=C(200-3t)^{5/3}

There are 20 g of chlorine at the start, so <em>c</em> (0) = 20. Use this to solve for <em>C</em> :

20=C(200)^{5/3}\implies C=\dfrac1{200\cdot5^{1/3}}

\implies\boxed{c(t)=\dfrac1{200}\sqrt[3]{\dfrac{(200-3t)^5}5}}

7 0
3 years ago
State the gradient of the line perpendicular to the line y=3x-2
kherson [118]

Answer:

-1/3

Step-by-step explanation:

3 is the gradient of the other line

m1×m2=-1 (equation for perpendicular lines)

m1×3= -1

m1= -1/3

Pls confirm if its correct

correct me if wrong

thanks! :)

7 0
2 years ago
If a point at (3,-4) is reflected over the X-axis to form a new point What will be the coordinate of the new point? (-3,-4), (3,
ira [324]

Answer:

(3,4) because only the y would change when it is being reflected over the x axis :)

Step-by-step explanation:

5 0
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The sum of four consecutive numbers is 230. What is the largest of these numbers
vodka [1.7K]

Answer:

n + n+1 + n+2 + n+3 = 230

4n = 224

n = 56 and largest number = 59


Step-by-step explanation:


7 0
3 years ago
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