Math 151 , Day 33, Friday, Nov.9, 2007 after class.. hit reload...

HW Day33 Finish Ch. 11 (see day 32, 31, 30, 29).  Start Ch 14; read first to p. 354.  Then reread.  Know (memorize if necessary) the "boxes" pp. 346 and 347 Continue with computational method, how C, z*, n, and margin of error m relate.  Last, p. 355, choosing n for a desired C and m.
Check p. 356; in this order: intro: 14.12, 14.13.  Then calculating:  14.11, 14, 15,  Then relationship 14.18, 19, 20.  Finally sample size 14.17
If you haven't, Memorize the 3 yellow-headed boxes on p. 278, 281 (mean and s.d. of sampling dist. of X-bar, Normality & Central Limit Th.)
Hand in   Nothing new. Read Chapter 14, up to p. 354. Work on reviewing for the exam; sample exam (You can do everything but #6 and #7 now.)
  Postpone all:
Chapter 14, Confidence intervals
p. 348 14.2  margin of error, interval
p. 348 14.3 Applet:  , percent of captures of true mean, C = 80%.
p. 361, 14.38 Applet:  , percent of captures of true mean. C = 90, 95, 99%  Also, Notice the comparative lengths of the intervals!
p. 360 14.34 and 14.35  explaining confidence

Use the ConfidenceInterval.xls Excel spreadsheet to check your computations of confidence intervals below; but do them by hand, as you'll need to for exams.
p. 352, 14.5 analyzing pharmaceuticals
p. 353, 14.6 IQ Test scores.  The sample mean is about 105.84, to check your calculator's result.
p. 359, 14.27 wine stinks
Read, 
to discuss
Optional 
Exam 4 a week from today.  Covers what we cover Monday.   Sample Exam . (Handed out.)  Solutions linked here.

HW questions from Day 32?

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Chapter 11:  See Day 29 for notes
(they're all there...)
The job is to understand the behavior of sample means, and be able to compute with them.
Quiz practice   Quiz answers
Central Limit theorem: 
for large n, sampling distribution of the mean (distribution of Xbars from all possible SRS's) is approximately normal.
   "Large??"  Day 32

Reviewed behavior of sample means.  Got to here Friday.
"Fuzzy Central Limit Theorem:"

Data whose variation is due to adding  many   small    independent   random influences will have an approximately normal distribution.
  Balls and pins, heights of women, etc.  (p. 281, after the yellow box)
- - - - - - - - - - - - - - - - - - - - - - - - - - - -
Chapter 14, beginning:
 SAMPLE from an UNKNOWN population.  Each person took 4 slips from the Birkenstock box,  for
  HW:   found the mean, and your mean + .841.
     Your mean is your best guess at the real mean, based on your sample.  It's not going to be exactly right.  So you build in a fudge factor.
     Your  mean + .841. is your  "Interval Estimate" of the mean of the Birkenstock population.  Does it capture the real mean???

 Your "estimate" of the (unknown) population mean µ of the numbers in the shoebox is your sample mean plus or minus the "fudge factor/margin of error" .841.   It's a "Confidence interval" estimate.
      You Recorded them on the sheet going around, and drew the interval on the graph transparency going around.
         If xbar = 8.0       7.159|_____________8.0_____________|8.841
Remember: xbar is the statistic that estimates the parameter µ

Introduction to Inference: Chapter 14, Confidence intervals
Statistical Inference: drawing conclusions about a population from sample data.
    Requires: Random sample or Randomized experiment.  (Simple Random Sample usually)

"Simple conditions" to develop concepts.
     -- SRS. Most important, now and forever
No "difficulties", no bias   (Population is at least 10 to 20 times as big as sample)
    -- Variable X is perfectly Normal, mean  µ, s.d. sigma.  (We'll extend from this later)
   -- µ is unknown, but sigma is known!  (we'll remove the sigma-known condition later)

First example:  Use sample mean xbar  to "estimate" (unknown) population mean µ
 Mean of 4 grades (HW#11.6) estimates population mean of all 10 ("known" µ = 69.4)  E.g. 69.75,  64.25,  73.5
(Each is a "point estimate")

Interval estimate:  xbar + margin of error (fudge factor)  estimates population mean µ (69.4)
  Won't get a true answer for all samples, but a bigger margin of error gives a better chance at being true

    69.75 + 1:   "µ is between 68.75 and 70.75"  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

Confidence interval estimate of a(n unknown) population parameter: (pp. 346-7)

Confidence level C:  example C = 90%.  A 90% confidence interval is one made by a method that has success rate 90% at capturing the real mean.  For any particular interval, we don't know if it's one of the 90% that contain the real mean or one of the 10% that miss.
Applet:  Confidence intervals.     You each made one from the shoebox.

Next:  What method do we use?
Confidence Interval of the form  estimate + margin-of-error  for the mean with Confidence level C: (pp.349-50) (Table A, or Table C, t dist., z* row) Example:  Sample of size 9 from a Normal population with unknown mean and pop. s.d. sigma = 6,  xbar = 12.
  Find a 90% CI estimate for the unknown mean µ: 
              z* = 1.645  (See TableC  (back flyleaf of text)  Also Normal Distribution. Applet,  2 tailed, less precision, or Table A)
             (sigma)/ sqrt(n) = 6/3=2, so m = 3.290;
                       CI is 12 + 3.290, or  8.710 to 15.290.
    Check your calculations with the ConfidenceInterval.xls Excel spreadsheet

The Birkenstock shoebox 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 (See Table C), so margin of error m is .841 times 1= .841.  (Note.  It should probably be .842--an error copied down for decades?)
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.  
How many people captured the true mean?
(as of Wed.)
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%, 8/16 = 50%, 7/14 = 50%.  Math 251, 5/10 = 50%. Combined, 127/204 = 62%.  This class 11/14=79%. All combined, 138/218 = 63%
Quite variable for small samples, but settling down?) 
Your results, graphed (combined with Math 251):  CI's   Compare with
Applet:  Confidence intervals.


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