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snow_tiger [21]
1 year ago
8

The mean annual income of certified welders is normally distributed with a mean of $50,000 and a population standard deviation o

f $2,000. The ship building association wishes to find out whether their welders earn more or less than $50,000 annually. The alternate hypothesis is that the mean is not $50,000. If the level of significance is 0.10, what is the decision rule
Mathematics
1 answer:
Bogdan [553]1 year ago
5 0

Using the p-value method, the decision rule is:

  • |z| < 1.645: do not reject the null hypothesis.
  • |z| > 1.645: reject the null hypothesis.

<h3>What is the relation between the p-value and the test hypothesis?</h3>

Depends on if the p-value is less or more than the significance level:

  • If it is more, the null hypothesis is not rejected.
  • If it is less, it is rejected.

In this problem, we have a two-tailed test, as we are testing if the mean is different of a value. For a significance level of 0.1, the critical value of z(when a p-value of 0.1 is obtained) is of |z| = 1.645, hence the decision rule is:

  • |z| < 1.645: do not reject the null hypothesis.
  • |z| > 1.645: reject the null hypothesis.

More can be learned about p-values at brainly.com/question/13873630

#SPJ1

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A heron is perched in a tree 50 feet above sea level Directly below the heron, a pelican is flying 17 feet above sea level Direc
IceJOKER [234]

Answer:

c my friend easy maths asfdf

Step-by-step explanation:

7 0
2 years ago
Abby filled her goodie bags with 4 cookies and 3 candy bars and spent a total of $10.25 per bag. Marissa filled her goodie bags
Paladinen [302]
Let's start by writing a system of linear equations:
c -> cookies 
cb -> candy bars
(You can use any abbreviations to your preference)

Abby:
4 cookies 
3 candy bars
$10.25 per bag
The equation would be:
4c+ 3cb = $10.25

Marissa:
2 cookies
7 candy bars
$14.75 per bag
The equation would be:
2c + 7cb = $14.75

So our linear equation system would be:
<span>4c+ 3cb = $10.25
</span><span>2c + 7cb = $14.75

I would try to get rid of one variable so I can solve for the other variable. In this case, it is easier to get rid of c since I can multiply the second equations by 2. Then it would subtract the two equations.

(2c + 7cb = $14.75) 2 = 4c + 14 cb = $29.50

      4c + 3cb = $10.25
   -  4c+14 cb = $29.50                    (4c would get canceled.)
---------------------------------
             -11 cb = - $19.25                (Divide by -11 to solve for cb)
</span>             ---------   -------------
              -11           -11
            
             cb = $1.75

Now we know cb (candy bar) cost, we would substitute this value into cb into one of the equations. It doesn't matter which equation you put it in. I will substitute it in the first equations. 

  4c + 3 (1.75) = $10.25
  4c + 5.25 = $10.25                       (Multiply 3 by 1.75)
       -5.25       -5.25                         (Subtract 5.25 on both sides)
             4c = 5                                 (Divide by 4 on both sides to get c)
             ----   ---
              4      4
         
         c= 1.25

Check the work:
4(1.25) + 3(1.75) 
  = $10.25

2(1.25) + 7(1.75)
   = $14.75

Total cost:
cookies = $1.25
candy bars = $ 1.75

Hope this helps! :)

6 0
3 years ago
WILL MARK BRANLIEST IF GOTTEN RIGHT
n200080 [17]

Answer:

110 degrees

have a nice day :D

i suck at math

6 0
2 years ago
Read 2 more answers
If the second number is subtracted from the sum of the first number and 2 times the third number, the result is 1. The thrid num
weeeeeb [17]

Answer:

<h2>x = 0, y = 5, z = 3</h2>

Step-by-step explanation:

x,\ y,\ z-\text{three numbers}\\\\\left\{\begin{array}{ccc}(x+2z)-y=1&(1)\\z+2x=3&(2)\\x+3y+z=18&(3)\end{array}\right\\\\(2)\\z+2x=3\qquad\text{subtract}\ 2x\ \text{from both sides}\\z=3-2x\qquad(*)\\\\\text{Substitute}\ (*)\ \text{to (1) and (3)}\\\\\left\{\begin{array}{ccc}x+2(3-2x)-y=1&\text{use the distributive property}\\x+3y+(3-2x)=18\end{array}\right

\left\{\begin{array}{ccc}x+(2)(3)+(2)(-2x)-y=1\\x+3y+3-2x=18&\text{subtract 3 from both sides}\end{array}\right\\\left\{\begin{array}{ccc}x+6-4x-y=1&\text{subtract 6 from both sides}\\(x-2x)+3y=15\end{array}\right\\\left\{\begin{array}{ccc}(x-4x)-y=-5\\-x+3y=15\end{array}\right\\\left\{\begin{array}{ccc}-3x-y=-5&\text{multiply both sides by 3}\\-x+3y=15\end{array}\right

\underline{+\left\{\begin{array}{ccc}-9x-3y=-15\\-x+3y=15\end{array}\right}\qquad\text{add all sides of the equations}\\.\qquad-10x=0\qquad\text{divide both sides by (-10)}\\.\qquad\boxed{x=0}\\\\\text{Put it to the second equation:}\\-0+3y=15\\3y=15\qquad\text{divide both sides by 3}\\\boxed{y=5}\\\\\text{Put the value of}\ x\ \text{to}\ (*):\\\\z=3-2(0)\\\boxed{z=3}

7 0
3 years ago
39+5h+4g-2h i have been trying 2 work this out 4 ages so plz help
SIZIF [17.4K]
Unless you're given the value of the variables, you can't really solve for anything. The only thing you can do is simplify. 

<span>39+5h+4g-2h
The only like terms are 5h and -2h.

39 + 3h + 4g
This would be the simplified version.</span>
7 0
2 years ago
Read 2 more answers
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