A car company claimed that it improved the gas mileage of a particular model of car between 2011 and 2012. A consumer advocacy group obtains a random sample of 16 cars from 2011 and finds the average number of miles driven on exactly 25 gallons of gas in laboratory conditions is 574.4 miles, with a standard deviation of 43.1 miles. The consumer advocacy group also obtains a random sample of 24 cars from 2012 and finds the average number of miles driven on exactly 25 gallons of gas in laboratory conditions is 604.6 miles, with a standard deviation of 27.8 miles. The 99% confidence interval for the difference in mean miles driven on 25 gallons of gas (2012 model – 2011 model) is (–3.95, 64.35).

Does this confidence interval support the company’s assertion that their gas mileage has increased?

a)Yes, we are 99% confident that the true mean difference in miles driven per 25 gallons is 30.2 miles.

b)Yes, the fact that most of the values in the confidence interval are positive means that the true mean number of miles driven on 25 gallons is higher in 2012.

c)Yes, the 2012 sample mean of 604.4 miles per 25 gallons is 30.2 more miles than the 2011 sample mean of 574.4, which is an increase of 1.208 miles per gallon.

d)No, the confidence interval contains 0, so we are 99% confident that the true mean increase in number of miles driven on 25 gallons is 0.

e)No, the confidence interval contains 0, so 0 is a plausible value for the true mean increase in number of miles driven on 25 gallons. An increase of 0 would indicate the gas mileage has not increased.

16

step-by-step explanation:

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.