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Karolina [17]
2 years ago
8

Consider the following analogy: You are a hiring manager for a large company. For every job applicant, you must decide whether t

o hire the applicant based on your assessment of whether he or she will be an asset to the company. Suppose your null hypothesis is that the applicant will not be an asset to the company. As in hypothesis testing, there are four possible outcomes of your decision: (1) You do not hire the applicant when the applicant will not be an asset to the company, (2) you hire the applicant when the applicant will not be an asset to the company, (3) you do not hire the applicant when the applicant will be an asset to the company, and (4) you hire the applicant when the applicant will be an asset to the company.
1. Which of the following outcomes corresponds to a Type I error?
A. You hire the applicant when the applicant will not be an asset to the company.
B. You do not hire the applicant when the applicant will be an asset to the company.
C. You do not hire the applicant when the applicant will not be an asset to the company.
D. You hire the applicant when the applicant will be an asset to the company.
2. Which of the following outcomes corresponds to a Type II error?
A. You hire the applicant when the applicant will not be an asset to the company.
B. You hire the applicant when the applicant will be an asset to the company.
C. You do not hire the applicant when the applicant will be an asset to the company.
D. You do not hire the applicant when the applicant will not be an asset to the company.
As a hiring manager, the worst error you can make is to hire the applicant when the applicant will not be an asset to the company. The probability that you make this error, in our hypothesis testing analogy, is described by:________.
Mathematics
2 answers:
sveticcg [70]2 years ago
8 0

Answer:

1. Option A

2. Option C

Step-by-step explanation:

The null hypothesis is that the applicant will not be an asset to the company, thus you do not hire such applicant

The alternative hypothesis is that the applicant will be an asset to the company and you then hire such applicant.

A type I error occurs when the researcher rejects the null hypothesis when true.

A type II error occurs when the researcher fails to reject the null hypothesis when it is not true.

1. Type I error:

You hire the applicant when the applicant will not be an asset to the company

2. Type II error:

You do not hire the applicant when the applicant will be an asset to the company.

3. Type I error because you rejected the null hypothesis to not hire when the applicant will not be an asset to the company.

rosijanka [135]2 years ago
6 0

Answer:

1. A. You hire the applicant when the applicant will not be an asset to the company.

2. C. You do not hire the applicant when the applicant will be an asset to the company.

Step-by-step explanation:

1. The type I error happens when the null hypothesis is rejected when it is true, in this way we know that the null hypothesis is that the new employee will not be active for the company, so option B is rejected, because it refers that the Applicant if he will be active or for the company, option C is rejected because the inactive employee is rejected, accepting the null hypothesis, option D is rejected because the contracted applicant if active, so the correct answer is A, in which the inactive applicant is hired.

2.

we know that the type II error occurs when the null hypothesis is accepted, being this false, we know that the null hypothesis is to hire an inactive applicant for the company, so option A is not correct, in which the null hypothesis is accepted taking it as true, option B is rejected, in which the contract is made to an active applicant, so the null hypothesis is false and option D is rejected, in which the null hypothesis is rejected, therefore the correct answer It is the C in which the active applicant is not hired.

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LORAN is a long range hyperbolic navigation system. Suppose two LORAN transmitters are located at the coordinates (-100,0) and (
Gre4nikov [31]

Answer:

\frac {(x)^{2}}{8100}+\frac {(y)^{2}}{1900}=1

Step-by-step explanation:

center of the hyperbola  is  (0,0)  = (h, k)

c = the distance form the center to either focal point  = 100

c^{2} =100^{2}=10000

The differences  from the receiver to the transmitters  =  2a

2a =  180  miles

a   =  180/2=90 miles

a^2 =90^{2}= 8100

b^{2}= c^{2} - a^{2}

b^{2}= 100^{2}-90^{2}=1900

b^{2}=1900

The standard form is

\frac {(x-h)^{2}}{a^{2}}+\frac {(y-k)^{2}}{b^{2}}=1

\frac {(x-0)^{2}}{a^{2}}+\frac {(y-0)^{2}}{b^{2}}=1

\frac {(x)^{2}}{8100}+\frac {(y)^{2}}{1900}=1

3 0
2 years ago
New Clarendon Park is undergoing renovations to its gardens. One garden that was originally a square is being adjusted so that o
frosja888 [35]
The answer is 80 square meters.

The square area is expressed as:
A = a²,
where A is the area of the square, and a is the side of the square.

The rectangle area is expressed as:
A₁ = a₁ · b₁,
where A₁ is the area of the rectangle, and a₁ and b₁ are the sides of the rectangle.

After renovations, square garden becomes rectangular.

One side is doubled in length:
a₁ = 2a

The other side is decreased by three meters.
b₁ = a - 3

The new area is 25% than the original square garden:
A₁ = A + 25%A =
     = A + 25/100·A
     = A + 1/25·A
     = a² + 1/25·a²
     = <span>a² + 0.25·a²
</span>     = 1.25·a²

If the starting equation is:
A₁ = a₁ · b₁

Thus, the equation is:
1.25a² = 2a·(<span>a - 3)
</span>1.25a² = 2a · a - 2a · 3
1.25a² = 2a² - 6a

<span>Therefore, the equation that could be used to determine the length of a side of the original square garden is:
</span><u>2a² - 6a = </u><span><u>1.25a²</u></span>


Now, we will solve the equation:
2a² - 6a = 1.25a²
2a² - 1.25a² - 6a = 0
0.75a² - 6a = 0
⇒ a(0.75a - 6) = 0

From here, one of the multiplier must be zero - either a or (0.75a - 6). Since a could not be zero, (0.75a - 6) is:
0.75a - 6 = 0
0.75a = 6
a = 6 ÷ 0.75
a = 8

If the side of the square is 8, then the area of the rectangle is
A₁ = 1.25 · a²
A₁ = 1.25 ·8²
A₁ = 1.25 · 64
A₁ = 80

Therefore, the area of the new rectangle garden is 80 square meters.
4 0
3 years ago
In a basketball tournament, team A scored 6 more points than 3 times as many points as team B scored. Team C scored 45 more poin
Sophie [7]
A + B + C = 476
A = 3B + 6
C = B + 45

now its just a matter of subbing..
A + B + C = 476
(3B + 6) + B + (B + 45) = 476...combine like terms
5B + 51 = 476
5B = 476 - 51
5B = 425
B = 425/5
B = 85 <== team B scored 85

A = 3B + 6
A = 3(85) + 6
A = 255 + 6
A = 261 <=== team A scored 261

C = B + 45
C = 85 + 45
C = 130 <=== team C scored 130
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2 years ago
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Matt has a block. He uses one of the flat surfaces of the block to trace a triangle. What type of solid figure is Matt's block?
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Triangular prism OR  Pyramid
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2 years ago
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tatiyna
If Ian uses 1/20 of the ketchup 4 times, this is equal to 4/20, or \frac{4}{20}.

\frac{4}{20} can be simplified to \frac{1}{5}, which means we need to work out \frac{1}{5} of 12.2

\frac{1}{5} is equivalent to 20%, which as a decimal is 0.2

0.2 x 12.2 = 2.44 - however, this is how much he has used, and we need to figure out how much is left.

12.2 - 2.44 = 9.76

So, Ian has 9.76 oz remaining :)

4 0
2 years ago
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