Math151 Day34

Math 151 , Day 34, Monday, April 23, 2007 After class. hit reload...

HW Day34 . Ch 14; read first to p. 354. Then reread. Know (memorize if necessary) the "boxes" pp. 346 and 347. Continue with computational method pp 349-50, how C, z*, n, and margin of error m relate.
Check p. 356; in this order: intro: 14.12, 14.13. Then calculating: 14.11, 14, 15, Then relationship 14.18, 19, 20. READ also Ch. 16, pp. 387-391, remembering that all our knowledge about sampling still applies. (ignore "significance test" parts.)Check, pp.406-716.19, 21, 22, 23, 25, 26 (Test to here.)

Last, p. 355, choosing n for a desired C and m. Check, Finally sample size 14.17

Moore Ch. 14, Day 34 Hand in Wednesday (Yes, all)
p. 348 14.2 margin of error, interval
p. 348 14.3 Applet: Confidence Interval , percent of captures of true mean, C = 80%.
p. 361, 14.38 Applet: Confidence Interval , 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; 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
p. 354, 14.7 n and margin of error
p. 354, 14.8 C and margin of error
p. 358, 14. 21, 22, & 23 Hotel managers' personalities
p. 360, 14.30 & 32 Study times
p. 361, 14.36 Crime, Margins of error

Added: p. 406, 16.29 (Hotel managers again)

Read,
to discuss

p. 361, 14.37 newspaper poll

Optional

Exam 4 this Friday. Covers Ch. 10 p. 250 on (Discrete and Continuous models and R.V.'s), Ch. 11 up to p. 286 only, Ch. 14 to p.354, Ch. 16 to p. 391. (i.e. thru tonight's HW).   Sample Exam is good as written. (Handed out Fri. Outside my door..) Solutions .
   Sign up Wed. for 10:30 start: Confirm with me any other time to take it.

Homework questions? Sampling distribution of the (sample) mean, Central Limit Theorem. HW Day 33
Note: 11.38 and 11.39 are "backward" Normal distribution problems: going from proportion/probability to x (here L). I didn't discuss these in class. We'll postpone discussing these till Chapter 15 (Probably Monday.)
I forgot to say this in class: How big does n have to be before the sampling distribution of the x-bars is "normal enough"? The more symmetrical and unimodal the population, the smaller n needs to be. However, unless a distribution is quite skewed or bumpy, by n = 25 it's roughly normal. See Day 29, bottom, play with applet.

Behavior of sample means: Your pooled class data:
     -- 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???

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.
Record them 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 (population distribution) 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")

"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."

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).

Fall 2001: 65% (15 of 23) xbars recorded were within 4 of µ. (between 65.4 and 73.4).

78% (18 of 23) xbars recorded were within 5 of µ. (between 64.4 and 74.4).

Spring 2002: 83% (35 of 42) xbars recorded were within 4 of µ. (between 65.4 and 73.4).

90% (38 of 42) xbars recorded were within 5 of µ. (between 64.4 and 74.4).

(I excluded impossible xbars)

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 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        If you don't knowµ, you don't know if it's true or 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.
Applet: Confidence intervals. You made one from the shoebox.

Confidence Interval CI 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 (Moore, back flyleaf.)

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 Table C. Better Table C. 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 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.
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%. Combined, 115/180 = 64% This class:7/14 =50%. Combined, 122/194 = 63%...Quite variable for small samples, but settling down?)

Got to here: Homework will demonstrate the concepts below:
Relation of m (margin of error, half width), C (confidence level), and n (sample size), (and sigma)
    C and z* get bigger and smaller together (bigger C means bigger z*, and vice versa) (standard normal sketch)

,    m = z* (sigma)/ sqrt(n)
    Want bigger C? Must accept bigger m. Trade off confidence vs. accuracy.
    But bigger n will make smaller m. This makes sense: bigger sample size, more info-->more accurate estimate.
            (square root makes it Expensive: have to quadruple n to make m half as big)
    So smaller m, margin of error, can be achieved only by
        » accepting lower confidence level (smaller C, smaller z*),
        » or by increasing sample size (bigger n).
        » Sigma: We can't change it, it comes with the population. But smaller sigma (more population variability) will give smaller m (narrower CI), i.e. more accuracy in prediction (for the same C and n).

Science projects directed by Prof. Wahl: Experiments on chickens bred to be "identical"--very low variability from one to the other. Therefore very small samples suffice.

More time? "Backward" problems from Ch. 11 (see HW)

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