The Easiest solution isn

1
The Easiest solution isnt always the best
solution, even in Math Should we always believe
what we are taught in the classroom?
2
Purpose

• Statisticians use a well selected sample to
  estimate an unknown value of a population.
• The unknown value may be the mean income, or the
  proportion of defective products, or proportion
  of yes responses.
• Estimating an unknown population proportion is
  the topic of interest.

3
Background/symbols

• p Population proportion (unknown)
• n of subjects/ objects randomly selected.
• X of subjects/ objects in the sample with Yes
  responses.
• p sample proportion x/n
• Traditionally p is used as an estimate of p.
• Is there a better alternative?

4

• We often provide an interval estimate of p,
• p error of estimation Confidence interval
• A well known interval is 95 confidence.
• To determine error , we need to understand how p
  value varies from sample to sample.

5
About p

• p fixed value of a population, while
• p varies from sample to sample, and thus it
  has a distribution. What we know is
• Under certain conditions,
• p is normally distributed with a mean value of
  p, and standard deviation of vp(1-p)/n
• A normally distributed value can be changed to a
  standard normal score ,called a z score.
• A well known result is that about 95 of the z
  scores fall between -2 and 2.

6

• Lets standardize p of yes /n,
• ( p- mean)/ std dev z ( standard normal)
• ( p- p)/ vp(1-p)/n z
• ( p- p)/ vp(1-p)/n 2 ( p is unknown).
• Note We need to solve the above equation for p
• Easy approach for non math majors solve for p
  in numerator
• p p- 2 vp(1-p)/n,
• p p 2 vp(1-p)/n,
• (p- 2 vp(1-p)/n , p 2 v p(1-p)/ n)
• Makes an approximate 95 confidence interval of
  p. ( Mathematically incorrect)

7
Lets try again

• ( p- p)/ vp(1-p)/n 2
• Solve it the right way by squaring both sides,
  and solving the quadratic equation for p.
• We get two solutions of p, ( mathematically
  tedious)
• Those solutions make the 95 interval of p.This
  interval is very tedious, and lacks logical
  explanation.
• we take the average of those solutions, we get
• p ( of yes 2)/ (n4) our new estimate p

8
A new interval of p

• Recall old interval of p
• (p- 2 vp(1-p)/n , p 2 v p(1-p)/ n)
• An alternative interval of p
• (p -2 v p(1- p)/n , p 2 v p(1- p)/n
• Recall p ( of yes 2)/ n4
• p ( of yes)/n

9
How good is the new interval

• Simulation results coming soon.