Math151 Day32

Math 151 , Day 32, Wed. Apr. 16, 2008 hit reload....after class, shoebox results added.

HW Day32 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


*Hand in* Note: 11.38 and 11.39 were "backward" Normal distribution problems: going from proportion/probability to x (here L). We'll discuss them later in the term, probably with Ch 15. 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 *Postpone the rest:* 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

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.)   (Chapter 14)
Add your values to the list, and graph your interval on the graph circulating.   Call out if you fill up the graph: I have another.
     If xbar = 8.0       7.159|_____________8.0_____________|8.841
Quiz after recap.

<>~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Re-RECAP: What is the distribution of the random variable Xbar, when the experiment is to take a simple random sample of size n? This is the distribution of means of all possible SRS's of size n.
We'll call it the "sampling distribution of the (sample) mean" (p. 275-7, then details 278-86)

Whatever the population distribution of X, that we draw the sample from, (see p. 278)
(as long as the population is large compared to the sample (at least 10 to 20 times sample)
The mean of the x-bars = the mean of the population
The standard deviation of the x-bars =
the s. d. of the population divided by the square root of n.
IF the population is Normal, the sampling distribution of Xbar is Normal.
"The Central Limit Theorem"
In any case, for "large" n, the sampling distribution of Xbar is Approximately Normal.

Closed book Quiz now.
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HW questions

Central Limit Theorem...
How large is "large"? How approximate is "approximate"?
    If the population was close to normal, n doesn't need to be very large.
    If the population is not badly skewed or bimodal, n=25 already gives a pretty good approximation to normal.
Author's website applet, Central Limit theorem for a highly skewed dist.
Pictures on overhead.    Rice U. Applet

What if the population is not 10 to 20 times the sample size? The real s.d. of the x-bars will be narrower than the rule above. You may not "get to" normal as a shape. Sample of 4 grades from a population of 10: this semester   last semester All possible samples.

"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 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) (Each mean 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, with our definition of probability. 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."

Previous year: 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)
Won't get a true answer for all samples, but a bigger margin of error gives a better chance at being true
Suppose yours were 69.75, 73.5, 64.25

    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)

an Interval constructed from the data, (usually estimate + margin of error) +
a Confidence level C: where
C = probability that intervals constructed by this method will capture the true, unknown, parameter.
(C is "success rate" for the method -- if you use the method repeatedly)

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.
got about here Wed.
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)

the estimate is xbar
margin of error m is : z* times Standard deviation of sample mean

Probability C is between -z* and +z*

(Table A, or Table C, t dist., z* row)

Standard deviation of sample mean: Sigma /sqrt(n)

know

m = z* (sigma)/ sqrt(n),

CI is xbar + z* (sigma)/ sqrt(n)
.

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? before & after Wed. class
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%, 5/10 = 50%, 11/14=79%. All combined, 138/218 = 63% This class, 8/17 = 47%, Combined 146/235 = 62%
Quite variable for small samples, but settling down?)
Fall results, graphed (combined with Math 251): CI's   This class, CI's. Compare with Applet: Confidence intervals.


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