A major home improvement store conducted its biggest brand recognition campaign in the company's history. A series of new television advertisements featuring well-known entertainers and sports figures were launched. A key metric for the success of television advertisements is the proportion of viewers who "like the ads a lot". A study of 1,189 adults who viewed the ads reported that 230 indicated that they "like the ads a lot." The percentage of a typical television advertisement receiving the "like the ads a lot" score is believed to be 22%. Company officials wanted to know if there is evidence that the series of television advertisements are less successful than the typical ad (i. e.. if there is evidence that the population proportion of "like the ads a lot" for the company's ads is less than 0.22) at a 0.01 level of significance. Referring to the above, the null hypothesis will be rejected if the test statistic is:

Step-by-step explanation:

We would set up the hypothesis test.

For the null hypothesis,

P = 0.22

For the alternative hypothesis,

P < 0.22

This is a left tailed test

Considering the population proportion, probability of success, p = 0.22

q = probability of failure = 1 - p

q = 1 - 0.22 = 0.78

Considering the sample,

Sample proportion, p = x/n

Where

x = number of success = 230

n = number of samples = 1189

p = 230/1189 = 0.19

We would determine the test statistic which is the z score

z = (p - P)/√pq/n

z = (0.19 - 0.22)/√(0.22 × 0.78)/1189 = - 2.5

Recall, population proportion, p = 0.22

We want the area to the left of 0.22 since the alternative hypothesis is lesser than 0.22. Therefore, from the normal distribution table, the probability of getting a proportion < 0.22 is 0.00621

So p value = 0.00621

Since alpha, 0.01 > than the p value, 0.00621, then we would reject the null hypothesis.

Referring to the above, the null hypothesis will be rejected if the test statistic is - 2.5

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.