Math 151 , Spring 2006, Day 31 Monday, April 17 Hit reload After  class

Do-overs for Exam 2 are due next class!
You found the 68% and 95% CI's for your sample., Please add your results to our list:  (with your initials)
# of 1's, p-hat, SE(p-hat), p-hat + SE, ME for 95% = 1.96SE, p-hat + 1.96SE
Also draw your 68% CI on the graph circulating    |------o------|

Class  was spent on HW from Ch 18, and going slowly through the calculation of a 95% CI for #13, with meaning. 
Will continue (finish?) Ch 18 Wed.
A level C confidence interval for a parameter  is an interval, usually of the form estimate + margin of error,
  found from data, in such a way that
C% of all random samples will yield intervals that capture the true parameter value.

Rule for ME:   ME = z* SE(p-hat), where z* is the "critical value" from the Standard Normal table that has C% of the area in the symmetric central interval between -z* and +z*.
Level C confidence interval for population proportion p:  "One -proportion  z-interval"

(Why it works:  later.)
Example:  You drew a sample of size n =30. p is the (unknown) proportion of 1's in the shoebox. You found the sample  proportion, and you calculated the SE for the sample proportion. Use z* =1.  Then C is about 68%.
Calculate:  if I got 12/30, p-hat = .400.  "q-hat" = 1 - p-hat = 1-.4 = .6.   SD formula: square root of (p ·q/n)= square root of (.4 ·.6/30) = square root of .008 = .089 = SE(p-hat).
68% Confidence Interval:  .400 + .089, or  (.311, .489).
Whose intervals captured the real proportion?  (Expect roughly 68% of you to do so.)

Usually, want higher Confidence Level:  90%, 95%, 99%....
     For 95%:  z* = (approximately 2) = 1.96
           (How?  95% in the middle.  2.5% in each tail.  .0250 is to the left of what?? -1.96.)
           && Shortcut: Table T, p. A-53, bottom two rows.  ("infinity" row is the Standard Normal values) 
               (Compare to your values from the Normal table)
      z*·SE(p-hat) = 1.96·.089 = .174  95% Confidence Interval:  .400 + .174, or  (.226, .574).

Questions on HW?

Note Trade-off:  Higher Confidence ---Wider interval (bigger ME. Less "precision")

Assumptions/conditions:  Assumes Central Limit Theorem for proportions is appropriate.
  Independence:¿¿Data values shouldn't affect each other.   ¿¿ Randomization helps!   ¿¿n < 10% of population.
  Sample Size:  Expect at least 10 successes and 10 failures (rephrase of  np, nq > 10)

  BIAS?  Here's why we studied bias in sampling.  Biases or other bad sampling methods can make our computations worthless! p. 363.

Reprise:  Level C confidence interval for population proportion p: 
 "One -proportion  z-interval"   Chapter 19
 

Note Trade-off:  Higher Confidence ---Wider interval (bigger ME. Less "precision")
Desire:  Small Margin of Error ME + High confidence C.  p. 361-2
But they grow and shrink together: High confidence--Low precision ; High precision (small ME)--low confidence.

Way out:  increase n, the sample size.  (Shrinks SE)  How big a sample size for desired ME and C?  
   Plan ahead:  Decide on desired ME and C (thus z*).  Guesstimate p (p=1/2 requres largest sample size--safest).
Solve equation for n.   (Some results pre-calculated, p. 362)
Notes:  --To cut ME in half, need 4 times the sample size.  Certainty/precision are expensive!
    -- If you're sure your p will be far from 1/2,  you can get a smaller n by using a closer guesstimate for p.

Green shoebox:  To get a 90% CI, ME = .04:  use p = 1/2 = .5.    z* = 1.645.
          n = (1.645²) ( .5· .5) / (.04² )  =  2.706025 · .25/.0016= .67650625/.0016 = 422.8  Round UP! to 423.
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

Why does it work??  Why does the ME calculated this way give intervals that capture the real p C% of the time??
   Think about the Sampling distribution of p-hat.  It's Normal, center at the real (population) p. SD(p-hat) is its standard deviation. SE(p-hat) approximates SD(p-hat)
Now ME = z*SE(p-hat),  where + z* cut off the center C% of the standard normal model.
So, in the Sampling distribution model, Realp+ ME  spans the center C% of this normal curve.
So the probability that p-hat falls in the range Realp+ME  is C%; That is, with many random samples, the proportion of p-hats that fall in the range Real p+ME  is C%.
That is, the proportion of p-hats that are within the distance ME of p---is C%

Now:  If p-hat is within ME of p, then p is within ME of p-hat.  The "arms" (+ ME ) that a p-hat interval sticks out from p-hat will capture p, if and only if p-hat is within ME of p.  But the proportion of p-hats that do that is C%.