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2 years ago

What is the relationship between the difference in the means and the MADs?

Mathematics

1 answer:

lyudmila [28]2 years ago

4 0

Answer:

Its the second one in the photo

Step-by-step explanation:

it's the second because when I clicked it I got it right.

You might be interested in

What is the degree of EMD and what is the degree for DMB

elena55 [62]

EMD = 90 - DMC
EMD = 90 - 20
EMD = 70

Now, DMB = 180 - FMD
DMB = 180 - 90
DMB = 90

Hope this helps!

8 0

3 years ago

The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100,

Gre4nikov [31]

Answer:

A.the type 1 error probability is $\mathbf{\alpha = 0.0244 }$

B. β = 0.0122

C. β = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

$\mathtt{H_o: \mu = 100}$

$\mathtt{H_1: \mu \neq 100}$

A. If the acceptance region is defined as $98.5 < \overline x > 101.5$ , find the type I error probability $\alpha$ .

Assuming the critical region lies within $\overline x < 98.5$ or $\overline x > 101.5$ , for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is $\mu = 100$

∴

$\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}$

$\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}$

when $\mu = 100$

$\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }$

$\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }$

$\mathtt{\alpha = P ( Z 2.25) }$

$\mathtt{\alpha = P ( Z$

From the standard normal distribution tables

$\mathtt{\alpha = 0.0122+( 1- 0.9878) })$

$\mathtt{\alpha = 0.0122+( 0.0122) })$

$\mathbf{\alpha = 0.0244 }$

Thus, the type 1 error probability is $\mathbf{\alpha = 0.0244 }$

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis $\mathtt{H_o}$

Thus;

β = P( type II error) - P( fail to reject $\mathtt{H_o}$ )

∴

$\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }$

Given that $\mu = 103$

$\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }$

$\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }$

$\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }$

$\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}$

From standard normal distribution table

β = 0.0122 - 0.0000

β = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

$\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }$

Given that $\mu = 105$

$\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }$

$\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }$

$\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }$

$\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}$

From standard normal distribution table

β = 0.0000 - 0.0000

β = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

8 0

3 years ago

How do you do this?Can you help me with the steps? I don't really get this a lot

polet [3.4K]

Before you do the steps, change the fractions given to improper fractions:
2 2/3 = 8/3
3 1/6 = 19/6

Now, I think the first step they want you to write is:
8/3 (divided by) 19/6

When you divide two fractions, it's the same as multiplying the first fraction by the reciprocal of the second fraction. (A reciprocal is a fancy way of saying you flip the fraction)
So the next step will look like this:
8/3 x 6/19

With this setup, you can cross simplify the fractions so that they look like this:
8/1 x 2/19
When you multiply this, you get 16/19
Let me know if you have any questions!

4 0

3 years ago

Read 2 more answers

URHENT I NEED HELP WITH THESE I NEED THE FIRST QUESTION.

Goshia [24]

Answer:

259824773358.58 cubic miles and 1 cone would fit into the circle

Step-by-step explanation:

Can i have brainliest? Please?

5 0

2 years ago

Find the quation of the quadratic function with roots 0 and 4 and a vertex at (2,-2)

Nookie1986 [14]

Answer:

y = (1/2)x² -2x

Step-by-step explanation:

recall that the general equation of a quadratic function is:

y = Ax² + Bx + C

given that 0 and 4 are roots, that means when x = 0 and x = 4, then y = 0

From this we can get 2 points on the curve, namely, (0,0) and (4,0)

substituting these points one at a time into the equation above,

for the first point (0,0),

0 = A (0)² + b(0) + C

C = 0

hence the equation becomes: y = Ax² + Bx

for the 2nd known point (4,0)

0 = A(4)² + B(4)

0 = 16A + 4B (divide both sides by 4)

0 = 4A + B

B = -4A ------------(eq 1)

we are given a 3rd point, vertex at (2,-2)

for (2,-2)

-2 = A(2²) + B(2)

-2 = 4A + 2B (divide both sides by 2)

-1 = 2A + B (subtract 2A from both sides)

B = -1 -2A --------(eq 2)

solving the system of equations using the method of your choice in eq1 and eq 2 gives:

A = 1/2 and B = -2

hence the equation is

y = (1/2)x² -2x

7 0

3 years ago