A team of designers was given the task of reducing the defect rate in the manufacture of a certain printed circuit board. The team decided to reconfigure the cooling system. A total of 973 boards were produced the week before the reconfiguration was implemented, and 254 of these were defective. A total of 847 boards were produced the week after reconfiguration, and 95 of these were defective

a. Construct a 90% confidence interval for the decrease in the defective rate after the reconfiguration
b. A quality control engineer claims that the reconfiguration has decreased the proportion of defective parts by more than 0.15. Does the confidence interval contradict this claim?

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.

we have been given a piecewise function which has three parts

we can graph each part one by one.

first part is x+1 which is just like equation y=x+1

restriction with first part is x< 0 so that means we can use any number for x which is less than 0 for example x=-1 and x=-2

now plug x=-1 into y=x+1 to get y

y=x+1=-1+1=0

so the point will be (-1,0)

similarly we get another point (-2,-1)

now we graph a line joining both points but keep in mind that line must be on left side of x=0 because of restriction x< 0

same way we can graph 2nd and third parts

if we have < or > then we use open circle.

if we have < = or > = then we use close circle.

so using above information, final graph can be drawn as: