For the following hypothesis test, determine the null and alternative hypotheses. Also, classify the hypothesis test as two tailed, left tailed, or right tailed. The mean local monthly bill for cell phone users in this country was $47.79 in 2001. A hypothesis test is to be performed to determine whether last year's mean local monthly bill for cell phone users has decreased from the 2001 mean of $47.79.

The null and alternative hypothesis are:

This is a left tailed test.

Step-by-step explanation:

A hypothesis test is to be performed to determine whether last year's mean local monthly bill for cell phone users has decreased from the 2001 mean of $47.79.

Then, the claim that will be expressed in the alternative hypothesis is that the monthly bill is lower than $47.79.

The null hypothesis will state that the monthly bill does not differ significantly from $47.79.

Then, the null and alternative hypothesis are:

As we only want to test if its lower, we are interested only in the left tail of the distribution. We want to know if the test statistic is below the critical value to conclude if we have evidence for our claim. This is then a left tailed test.

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.