MATH 251, Probability and Statistics I, Fall 2005, Wed. Nov. 2, Day 29After classcorrected

Quiz Monday: Knowing and using: Binomial distribution formula (p. 349 bottom)
mean and st. dev. for Binomial:  X (count), p-hat (proportion)   (Summary p.350)
mean and st. dev. for X-bar from SRS of size n   (Summary p. 368)
Normal?  Central limit theorem: says yes for all of the above distributions, approximately, for n large.
If population(s) normal to start with, linear combinations stay normal (including X-bar), mean and s.d. follow algebra rules (as last quiz.)
Mean and variance quizzes back:  Main problems:
     Mean and s.d. of a distribution (population) are actually completely parallel to our computations for data, viewed correctly:
          X    1   2   4
          p   .2  .2  .6
# out of 10    2   2   6 Consider a population of 10, with these proportions.
 Mean:  (1+1+2+2+4+4+4+4+4+4)/10 = (1·2 + 2·2 + 4·6)/10 = 1·.2 + 2·.2 + 4·.6 = 3
 Population variance (divide by n)     [Sample variance (divide by n-1)--the one we learned in Chapter 1]
 = [(1-3)²+(1-3)²+ (2-3)²+(2-3)² +(4-3)²+(4-3)²+(4-3)²+(4-3)²+(4-3)²+(4-3)²+(4-3)²]/10
= [(1-3)²·2 + (2-3)²·2 + (4-3)²·6]/10 = [(1-3)²·.2 + (2-3)²·.2 + (4-3)²·.6]

Variance of X - Y = Variance of X +(-1)Y = Variance of X + Y, because (-1)² = 1.
Pesky added constant.  Another way to remember what  the variance of a +bX is.
    Let Y = a, with probability 1 (no matter what happens, a is the result).
  y |  a          Mean of Y = a.   Variance of Y = 0
  p |  1          Then sigma²_{(a + bX)} = sigma²_{(Y + bX)} = sigma²_(Y) + sigma²_(bX) = 0 +  sigma²_(bX) = 0+ b²sigma²_(X)

Introduction to Statistical Inference:
  Requires: Random sample or Randomized experiment.  (Our theory: Simple Random Sample usually)
First example:  Use sample mean xbar  to "estimate" (unknown) population mean µ

 Mean of 4 grades (HW#3.75, sec. 3.4) estimates population mean of all 10 ("known"= 69.4)
       E.g. 69.75,  64.25,  73.5    (Each is a "point estimate")

• "xbar IS µ"   Never true exactly
• "xbar is close to µ"  True for most xbars, depending on "close", and sample size n.
• "xbar is probably close to µ" ["probably"?? It is or it isn't. We have "confidence" ours is a close one]
• "xbars are usually close to µ" True.
• If we have only one sample -- one xbar, we have NO IDEA if ours is one of the "close" ones.
• Quantify "usually" and "close."

Fall 2002: 33% (16 of 48) xbars  recorded were within 1 of µ. (between 68.4 and 70.4).
         83% (40 of 48) xbars  recorded were within 4 of µ. (between 65.4 and 73.4).
         94% (45 of 48) xbars  recorded were within 5 of µ. (between 64.4 and 74.4).
Note tradeoff between "close"(accuracy) and "usually" (confidence)

Interval estimate:  xbar + margin of error (fudge factor)  estimates population mean µ (69.4)

    69.75 + 1:   "µ is between 65.75 and 73.75"  True
    69.75 + 4:   "µ is between 65.75 and 73.75"  True
     73.5 + 4:    "µ is between 69.5 and 77.5"  False
     73.5 + 5:    "µ is between 68.5 and 78.5"  True
      64.25 + 4:   "µ is between 60.25 and 68.25"  False
      64.25 + 5:   "µ is between 59.25 and 69.25"  False

A level C  Confidence interval estimate of a(n unknown) population parameter:

• a  Interval computed from the sample data,
• by a method  that has probability C of producing an interval that will capture the true, unknown, parameter.

(Many repeated samples would capture the unknown paramater  C - proportion of the time. Unfortunately, we only get one sample, as a rule.)

• the estimate  is xbar
• margin of error m is :  z* times Standard deviation of sample mean
z* from Normal table.  Probability C is between -z* and +z*.

Standard deviation of sample mean:  Sigma /sqrt(n)
Must know standard deviation of population!
or, If sample size is large, use s (standard deviation calculated from sample)

The Birkenstock box contains numbers from a normally distributed population, with population standard deviation 2.
You each constructed a 60% confidence interval for the unknown mean:
    n = 4.
    Standard deviation of sample mean = 2/sqrt(4) = 2/2 = 1
    z* for C = 60% is .841, so margin of error m is .841 times 1= .841.
To get the z* for C = 60% from the normal table, note that this is the middle 60%, which leaves 40% to be split between the 2 tails.  So 20% above z*,  and 80% below.  Go into the body of table A, find 80%= .8000 is between values .7995 and .8023, closer to .7995.  The z value with .7995 below it is .84.  Table D gives it more precisely as  .841.
Start here Friday:  How many people captured the true mean?
( previous classes,11/20 = 55% ,  22/29= 76%.   9/18 = 50% , 11/20 = 55%,  15/22= 68%,  16/24 = 67%
16/18 = 88%  7/13 = 54%.  Combined, 107/164 = 65%
Quite variable for small samples, but settling down?)

Applet:  Confidence Interval shows it's not the individual interval that C describes, but the Method.

Why does the formula work? (pp. 387-8, briefer there)
1. A particular xbar is within m of the population mean, if and only if the interval xbar + m contains the population mean.
     P(µ - m < Xbar < µ + m) =  P(Xbar -m < µ  < Xbar + m ) = Prob. that CI contains µ.

2. We choose z* (and from it m) so that the probability that Xbar is within m of the population mean   is C.

       P(µ - m < Xbar < µ + m) = C
How? Probability C is between -z* and +z* in the standard normal table,
between -z* ·(s.d. of Xbar) and +z* ·(s.d. of Xbar) around µ,  in the normal distribution of Xbar.
 m = z* ·(s.d. of Xbar)
(Recall: standardized z to raw x:  x = µ + z (sigma),
raw = mean + z (st. dev.) .  xbar is the raw.)
 C = P(µ - m < Xbar < µ + m) =  P(Xbar -m < µ ,  and µ  < Xbar +m ) = Prob. that CI contains µ.