Math 151 , Day 33, Friday, April 20, 2007 After Class, Modified 11pm Fri. 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 Monday:

DIST. OF XBAR(S) 
These problems use only the mean and standard deviation.   
  p. 280, 11.7 (Teen cholesterol )
  p. 280, 11.8 (lab measurements)  For (b) they mean "what should n be?'

These problems use  the "sample mean of n independent observations from a normal distribution has a normal distribution." theorem (p. 278) 
ADD, from class:  For a population of temperatures with mean 98.6 degrees and standard deviation .6 degree (.7 is better, I'm told),
we found in class that  the probability that the value of a sample of 1 is above 98.8 = P(Z>.2/.6) = .3707
   Probability that the mean of a sample of 4 is above 98.8 is P(Z> .2/.3) = .2514, since s.d. of Xbar is .6/2 = .3
   Probability that the mean of a sample of 9 is above 98.8 is P(Z> .2/.2) = .1587 , since s.d. of Xbar is .6/3 = .2
   Probability that the mean of a sample of 16 is above 98.8 is P(Z> .2/.15) = .0918 , since s.d. of Xbar is .6/4 = .15
  Find also:  Probability that the mean of a sample of 36 is above 98.8.
                   Probability that the mean of a sample of 100 is above 98.8.
  p. 280, 11.9 NAEP math scores  (n = 1, n = 4)
  p. 290, 11.37 and 11.39 Pollutants in auto exhausts  For 11.39:  You might want to know L so that if you tested your 25 cars and found a high value of x-bar, you would be able to compare it with L; if it was greater than L, you would go back to the manufacturer and say "I  believe you sold me a batch of bad cars, because the chances of getting an average emission level this high if the exhaust system is working properly is only 1 in 100. It is more reasonable to believe the exhaust system is not working, than that we "are" that 1 in 100 possibility."
  p. 289-90 11.36 and 11.38 Glucose testing  If we use this cutoff level L to say that people (with a mean of 4 tests) over L "have diabetes", then the chances of declaring that someone "has diabetes" when they really are OK (with mean 125mg/dl) is .05.  .05 or 5% is the chance of a "false positive" using this protocol, when the real mean is 125.
Note:  11.38 and 11.39 are "backward" Normal distribution problems:  going from proportion/probability to x (here L).

These problems use the Central Limit theorem (p. 281) 
  p. 185, 11.10 What does the CLTh say?
  p. 286 , 11. 12 SAT scores, n = 1 and 70
  p. 286, 11.13, insurance (Hint: find P(Xbar> $275))
  p. 298, 11.41 auto accidents
  p. 298, 11.42 airplane overloads  (Hint: to do the problem you have to assume all the seats are taken.  Maybe not a reasonable assumption, but if there are empty seats, there's likely not a problem with overweight.)

A. (preliminary for Ch. 14) Get 4 slips from the Birkenstock box (Will be outside my door Friday.).  Record them, return them.  HW:  Find their mean xbar. Now xbar is your "point estimate" of the unknown mean of the numbers in the box.
 Calculate  xbar - .841, xbar +.841.  This is your "point estimate" plus or minus a "margin of error" of .841.  
     (xbar - .841, xbar +.841) is your "interval estimate" for the unknown mean of the box.

 + + + + + + +Postpone 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.  Tentative Sample Exam . (Handed out.)  Solutions linked here when ready.
Exam 3 returned:  Get yours if you haven't.  (Discussion at 10:30 today, Mac 321)  Comments   Solutions .
HW questions from Day 30?
(preliminary for Ch. 14) Get 4 slips from the Birkenstock box In class and outside my door.  Record them, return them.  HW:  Find their mean xbar. Now xbar is your "point estimate" of the unknown mean of the numbers in the box.
 Calculate  xbar - .841, xbar +.841.  This is your "point estimate" plus or minus a "margin of error" of .841.  
     (xbar - .841, xbar +.841) is your "interval estimate" for the unknown mean of the box. 
<>~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
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.
  Got through the basics of chapter 11: enough to do the homework, I hope.
Behavior of sample means:  Details
     --  Your x-bars from sample of size 4 (11.6)
     --  10.56: From a population with mean .65  (X = 0 for tail, 1 for head): Many suspicious results.  Mean = .5?   Here's what should have happened. What happened last term: Samples of size n = 20  gave proportions (xbars) from .4 to .85.  Samples of size n = 320 gave proportions from .6 to .68, about a quarter as wide a spread.  320 is 16 times 20.  Square root of 16 = 4. So according to the rule for standard deviations, the s.d. for n = 320 should be 1/4 that of  the s.d. for n = 20.   Looks about right.

"Fuzzy Central Limit Theorem:"
Data whose variation is due to  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 take 4 slips from the Birkenstock box,  write them down, return slips.
  HW:   find 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???

Next time:  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.
      You'll Record them next time on the sheet going around, and draw the interval on the graph transparency going around.
         If xbar = 8.0       7.159|_____________8.0_____________|8.841

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.  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)

    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'll be making 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 Table C. Better TableC 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 box contains numbers from a normally distributed population, with population standard deviation 2.
You will each construct 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.

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