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STatiana [176]
4 years ago
7

Every time you conduct a hypothesis test, there are four possible outcomes of your decision to reject or not reject the null hyp

othesis: (1) You don’t reject the null hypothesis when it is true, (2) you reject the null hypothesis when it is true, (3) you don’t reject the null hypothesis when it is false, and (4) you reject the null hypothesis when it is false.
Consider the following analogy: You are an airport security screener. For every passenger who passes through your security checkpoint, you must decide whether to select the passenger for further screening based on your assessment of whether he or she is carrying a weapon. Suppose your null hypothesis is that the passenger has a weapon. As in hypothesis testing, there are four possible outcomes of your decision: (1) You select the passenger for further inspection when the passenger has a weapon, (2) you allow the passenger to board her flight when the passenger has a weapon, (3) you select the passenger for further inspection when the passenger has no weapon, and (4) you allow the passenger to board her flight when the passenger has no weapon.
1. Which of the following outcomes corresponds to a Type I error?A. You allow the passenger to board his flight when the passenger has a weapon.B. You select the passenger for further inspection when the passenger has no weapon.C. You allow the passenger to board his flight when the passenger has no weapon.D. You select the passenger for further inspection when the passenger has a weapon.2. Which of the following outcomes corresponds to a Type II error?A. You allow the passenger to board his flight when the passenger has no weapon.B. You select the passenger for further inspection when the passenger has no weapon.C. You select the passenger for further inspection when the passenger has a weapon.D. You allow the passenger to board his flight when the passenger has a weapon.As a security screener, the worst error you can make is to allow the passenger to board his flight when the passenger has a weapon. The probability that you make this error, in our hypothesis testing analogy, is described by _____.
Mathematics
1 answer:
almond37 [142]4 years ago
6 0

<u>Answer:</u>

<u>1. A. You allow the passenger to board his flight when the passenger has a weapon.</u>

<u>2. B. You select the passenger for further inspection when the passenger has no weapon.</u>

<u>Explanation:</u>

1. Remember, a Type I error in simple words means that the assumption "the passenger has a weapon" (null hypothesis) is <em>actually true,</em> but the airport security screener <em>incorrectly concluded it is false. </em>In other words, he assumed the passenger had no weapon and allowed the passenger to board his flight <u>when he actually did have one.</u>

<em>2. While, </em><em>a </em><em>Type II error </em><em>means that </em>the assumption "the passenger has a weapon" (null hypothesis) is <em>actually false, </em>but the airport security screener <em>incorrectly concluded it is true. </em>In other words, he assumed the passenger had a weapon and selected the passenger for further inspection <u>when he actually didn't have one.</u>

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The vertex of a parabola is at (-4,-3). If one x-intercept is at -11, what is the other x intercept?
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The equation of the parabola could be written as y-k = a(x-h)^2, where (h,k) is the vertex.  Thus, y-(-3) = a(x+4)^2, or y+3 = a(x+4)^2.

The coordinates of one x-intercept are (-11,0).  Thus, y+3 = a(x+4)^2 becomes 
0+3 = a(-11+4)^2, so that 3 = a(-7)^2, or 3 = 49a.  Therefore, a = 3/49, and the equation of the parabola becomes

y+3 = (3/49)(x+4)^2.

To find the other x-intercept, let y = 0 and solve the resulting equation for x:

0+3 = (3/49)(x+4)^2, or (49/3)*2 = (x+4)^2

Taking the sqrt of both sides, plus or minus 49/3 = x+4.

plus 49/3 = x+4 results in 37/3 = x, whereas

minus 49/3 = x+4 results in x = -61/3.  Unfortunatelyi, this disagrees with what we are told:  that one x-intercept is x= -11, or (-11,0).


Trying again, using the quadratic equation y=ax^2 + bx + c,
we substitute the coordinates of the points (-4,-3) and (-11,0) and solve for {a, b, c}:

-3 = a(-4)^2 + b(-4) + c, or -3 = 16a - 4b + c

 0 = a(-11)^2 - 11b + c, or 0 = 121a - 11b + c

If the vertex is at (-4,-3), then, because x= -b/(2a) also represents the x-coordinate of the vertex,                       -4 = b / (2a), or  -8a = b, or 
0 = 8a + b

Now we have 3 equations in 3 unknowns:

0  =  8a +  1b
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0 = 121a - 11b + c

This system of 3 linear equations can be solved in various ways.  I've used matrices, finding that a, b and c are all zero.  This is wrong.


So, let's try again.  Recall that x = -b / (2a) is the axis of symmetry, which in this case is x = -4.  If one zero is at -11, this point is 7 units to the left of x = -4.  The other zero is 7 units to the right of x = -4, that is, at x = 3.

Now we have 3 points on the parabola:  (-11,0), (-4,-3) and (3,0).

This is sufficient info for us to determine {a,b,c} in y=ax^2+bx+c.
One by one we take these 3 points and subst. their coordinates into 
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0=a(-11)^2 + b(-11) + 1c   =>  0 = 121a - 11b + 1c
-3 = a(-4)^2 +b(-4)  + 1c   =>  -3 = 16a   - 4b  +  1c
0 = a(3)^2   +b(3)    + c     =>   0 = 9a     +3b   + 1c

Solving this system using matrices, I obtained a= 3/49, b= 24/49 and c= -99/49.

Then the equation of this parabola, based upon y = ax^2 + bx + c, is

y = (1/49)(3x^2 + 24x - 99)               (answer)

Check:  If x = -11, does y = 0?

(1/49)(3(-11)^2 + 24(-11) - 99 = (1/49)(3(121) - 11(24) - 99
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y = (1/49)(3x^2 + 24x - 99)               (answer)
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