1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
AleksAgata [21]
3 years ago
10

Kitchen salespersons average $10,000 per week in sales. Manager proposes a compensation plan with new selling incentives. Manage

r hopes that the results of a trial selling period will enable him to conclude that the compensation plan increases the average sales per salesperson. Required:a. Develop the appropriate null and alternative hypotheses. b. What is the type I error in this situation? What are the consequences of making this error? c. What is the type II error in this situation? What are the consequences of making this error?
Mathematics
1 answer:
Oxana [17]3 years ago
8 0

Answer:

a) The alternative hypothesis has to reflect the claim that the research want to test. In this case, we want to test if there is enough evidence that the incentives increase the average sales per salesperson.

That is, if the average sales are significantly higher than $10,000 per salesperson.

The null hypothesis will state that the average sales per salesperson is not significantly higher than $10,000, representing the base case.

Then, we can write the hypothesis as:

H_0: \mu=10,000\\\\H_a: \mu>10,000

b) A Type I error happens when a true null hypothesis is rejected.

In this case, it would represent concluding that the incentive program is effective in increasing sales when it is not really the case.

The significative difference would be only due to chance and not to a real difference.

c) A Type II error happens when a false null hypothesis failed to be rejected.

In this case, it would represent that the sample statistic gives no enough evidence to support the claim that the incentive program is effective, although it is, in fact, effective. The conclusion will be, erroneously, that the program of incentives is not effective.

Step-by-step explanation:

You might be interested in
A box is known to hold exactly 125 in3. Which side dimensions produce the most accurate measurement of the box?
guapka [62]

Answer: 25 and 5 sowwy if i'm wrong.

Step-by-step explanation:

So we know that the box is 125 in3. What ___ times ____ is equal to = 125

So we should name the multiples of 125 which is : 1,5,25,125

25 times 5 is = 125 so there's your answer.

5 0
4 years ago
Read 2 more answers
Which expression correctly shows how to use The binomial theorem to determine the fourth term in the expansion of (2xy^3+y)^7?
monitta
3 plus 7 packs for you and
8 0
3 years ago
Read 2 more answers
Solve using elimination<br> x+y-2z=8<br> 5x-3y+z=-6<br> -2x-y+4z=-13
Free_Kalibri [48]
So here is your answer with LaTeX issued format interpretation. Full process elucidated briefly, below:

\begin{alignedat}{3}x + y - 2z = 8 \\ 5x - 3y + 2 = - 6 \\ - 2x - y + 4z = - 13 \end{alignedat}

For this equation to get obtained under the impression of those variables we have to eliminate them individually for moving further and simplifying the linear equation with three variables along the axis.

Multiply the equation of x + y - 2z = 8 by a number with a value of 5; Here this becomes; 5x + 5y - 10z = 40; So:

\begin{alignedat}{3}5x + 5y - 10z = 40 \\ 5x - 3y + z = - 6 \\ - 2x - y + 4z = - 13 \end{alignedat}

Pair up the equations in a way to eliminate the provided variable on our side, that is; "x":

5x - 3y + z = - 6

-

5x + 5y - 10z = 40
______________

- 8y + 11z = - 46

Therefore, we are getting.

\begin{alignedat}{3}5x + 5y - 10z = 40 \\ - 8y + 11z = - 46 \\ - 2x - y + 4z = - 13 \end{alignedat}

Multiply the equation of 5x + 5y - 10z = - 40 by a number with a value of 2; Here this becomes; 10x + 10y - 20z = 80.

Multiply the equation of - 2x - y + 4z = - 13 by a number with a value of 5; Here this becomes; - 10x - 5y + 20z = - 65; So:

\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 8y + 11z = - 46 \\ - 10x - 5y + 20z = - 65 \end{alignedat}

Pair up the equations in a way to eliminate the provided variables on our side, that is; "x" and "z":

- 10x - 5y + 20z = - 65

+
10x + 10y - 20z = 80
__________________

5y = 15

\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 8y + 11z = - 46 \\ 5y = 15 \end{alignedat}

Multiply the equation of - 8y + 11z = - 46 by a number with a value of 5; Here this becomes; - 40y + 55z = - 230.

Multiply the equation of 5y = 15 by a number with a value of 8; Here this becomes; 40y = 120; So:

\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 40y + 55z = - 690 \\ 40y = 120 \end{alignedat}

Pair up the equations in a way to eliminate the provided variables on our side, that is; "y":

40y = 120

+

- 40y + 55z = - 230
_________________

55z = - 110

\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 40y + 55z = - 230 \\ 55z = - 110 \end{alignedat}

Solving for the variable of 'z':

\mathsf{55z = - 110}

\bf{\dfrac{55z}{55} = \dfrac{-110}{55}}

Cancel out the common factor acquired on the numerator and denominator, that is, "55":

z = - \dfrac{\overbrace{\sout{110}}^{2}}{\underbrace{\sout{55}}_{1}}

\boxed{\mathbf{z = - 2}}

Solving for variable "y":

\mathbf{\therefore \quad - 40y - 55 \big(- 2 \big) = - 230}

\mathbf{- 40y - 55 \times 2 = - 230}

\mathbf{- 40y - 110 = - 230}

\mathbf{- 40y - 110 + 110 = - 230 + 110}

Adding the numbered value as 110 into this equation (in previous step).

\mathbf{- 40y = - 120}

Divide by - 40.

\mathbf{\dfrac{- 40y}{- 40} = \dfrac{- 120}{- 40}}

\mathbf{y = \dfrac{- 120}{- 40}}

\boxed{\mathbf{y = 3}}

Solve for variable "x":

\mathbf{10x + 10y - 20z = 80}

\mathbf{Since, \: z = - 2; \quad y = 3}

\mathbf{10x + 10 \times 3 - 20 \times (- 2) = 80}

\mathbf{10x + 10 \times 3 + 20 \times 2 = 80}

\mathbf{10x + 30 + 20 \times 2 = 80}

\mathbf{10x + 30 + 40 = 80}

\mathbf{10x + 70 = 80}

\mathbf{10x + 70 - 70 = 80 - 70}

\mathbf{10x = 10}

Divide by this numbered value \mathbf{10} to get the final value for the variable "x".

\mathbf{\dfrac{10x}{10} = \dfrac{10}{10}}

The numbered values in the numerator and the denominator are the same, on both the sides. This will mean the "x" variable will be left on the left hand side and numbered values "10" will give a product of "1" after the division is done. On the right hand side the numbered values get divided to obtain the final solution for final system of equation for variable "x" as "1".

\boxed{\mathbf{x = 1}}

Final solutions for the respective variables in the form of " (x, y, z) " is:

\boxed{\mathbf{\underline{\Bigg(1, \: \: 3, \: \: - 2 \Bigg)}}}

Hope it helps.
8 0
3 years ago
Read 2 more answers
Most everyday situations involving chance and likelihood are examples of ______.
aleksklad [387]
I believe the answer is probability 
4 0
3 years ago
Read 2 more answers
Plz help me well mark brainliest if you are correct!.....
Alex17521 [72]

Answer:

The first one is the answer to the question

4 0
3 years ago
Other questions:
  • Write each number as a product using the GCF as a​ factor, and apply the Distributive Property.
    14·1 answer
  • Solve 6( x + 20) = 168
    5·2 answers
  • The soccer club has planned a trip to a tournament. The cost of the van is $210. When 3 students who are not members of the club
    10·1 answer
  • <img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B3x%20%2B%2013%7D%20%20%3D%20%20%5Csqrt%7B6x%20%2B%201%7D%20" id="TexFormula1" tit
    5·2 answers
  • Write a unit rate that compares the quantities describe 50 hockey players on 5 teams
    10·1 answer
  • HELLLOOOO HELLLLLLPPPPPPPPPPPPPP
    7·2 answers
  • Place them in ascending order please ​
    12·1 answer
  • Anthony borrowed $15 from his mom to pay for food at the school dance. He bought one snow cone for $3, and then used the rest of
    11·2 answers
  • The table shows some values of f(x) and g(x) for different values of x:
    6·1 answer
  • SSS geometry definition
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!