In 1995, Derek Jeter's batting average was 0.250; in 1996, his batting average was 0.314. David Justice's batting average in 1995 was 0.253 and was 0.321 in 1996. The team statistician claims Jeter's overall average was better than Justice's average. What may make this claim possible?

Sampling error

Cause-and-effect relationship

Convenience error

Simpson's Paradox

Confounding

The correct option is;

Simpson Paradox

Step-by-step explanation:

The phenomenon whereby particular trends are prevalent in small data portions but are not evident or an inverse trend is observe when the portions are joined together is known as Simpson's paradox.

Whereby the data for calculating the bating averages as found online are given as follows;

Season,                       Derek Jeter                           David Justice

1995,                           12/48 = 0.250                         104/411 ≈ 0.253

1996,                           183/582 ≈ 0.314                      45/140 ≈ 0.321

The overall hits to the overall bats ratio are;

,         (183 + 12)/(582 + 48) ≈0.310   (104+45)/(411+140) = 0.27

Which shows that  Derek Jeter's overall average was better than Justice's average  

step-by-step explanation:

compare

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.