MATH 251, P&SI, Fall 2007, Wed. Oct.31, Day 29.After class.

Read 6.1; I'll go through "Choosing the sample size" (391-2) next time, but read Cautions, p. 393-4 carefully. I won't lecture on it or outline it on the webpage.

Hand in:  I think these are all doable.  Bring questions, if you find them not so.
Linear combinations of independent normal r.v.'s (as p.365-6) 
5.51 X+Y couples.
5.49 sum of 4 blocks
5.44 hole and shaft (Y-X)
5.64 screw-on caps break
5.46 difference of means--trustworthiness
5.47 difference of means--general

Sec. 6.1, confidence intervals, p. 396 ff.
6.20, 21 Use Applet:  Confidence Interval

using formula: 
6.5, 6.7 biomarker, C
6.13, 6.14 study habits
6.3, 6.4 changes of m with n and C
6.9  schoolchildren/third grade

Cautions (pp. 393-4) 
6.30 kindergarten, margin of error
6.29ab four intervals (Notice, these questions are about a binomial distribution, B(4, .95))

Read, discuss 
5.45 diff. of means


6.31 Gallup, margin of error

Optional 
(more practice) 
6.6, 6.8 OC
You took 4 Numbers (random sample) from the Birkenstock box:  Found mean xbar.  Found xbar + .841.  This is your interval estimate of the unknown mean of the box's population.  ("margin of error" is .841) (Returned your numbers afterward.)
Add your values to the list, and graph your interval on the transparency circulating.
     If xbar = 8.0       7.159|_____________8.0_____________|8.841

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.)


Homework questions? Day 28

Continuing Central Limit Theorem and friends: Day 27 , Linear combinations of Normal R.V.s Day 28

Ch. 6.1: 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")

Confidence intervals (sec. 6.1) This is one of the two big ideas of inference that we will study.  Chapter 7 will extend this simple idealized situation, so this needs to be firmly in place.

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

Review from here Friday, with HW
A level C  Confidence interval
estimate of a(n unknown) population parameter:

Confidence Interval of the form  estimate + margin-of-error  for the mean µ with Confidence level C: (p.388) (Table A, or Table D (back flyleaf), t dist. bottom 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,  (sigma)/ sqrt(n) = 6/3=2, so m = 3.290;
                       CI is 12 + 3.290, or  8.710 to 15.290.

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.  
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%, 8/16 = 50%, 7/14 = 50%.  Combined, 122/194 = 63% .This class 5/10=50%.  Combined, 127/204 = 62%
Quite variable for small samples, but settling down?)

Extension:  If n is  large, we can use the formula even if population is not normal.
     (Because only the distribution of Xbar is used, and Xbar is normal!  Central Limit Theorem)
Cautions read pp. 393-4

New for Friday:

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.


Sievers home  Math251-Fall07/Day2s29.htm  9pm   10/31/07
This page belongs to Sally Sievers who is solely responsible for its content. Please see our statement of responsibility.