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Raoul LePage Professor STATISTICS AND PROBABILITY www.stt.msu.edu/~lepage click on STT315_F06. Week 9-25-06 and some preparation for exam 2. suggested exercises solutions given in text 3-33, 3-41, 3-42 (except b, c, h, m, n), 3-43, 3-49, 3-57 (except c, d), 3-59, 3-61, 3-63, 3-65. - PowerPoint PPT Presentation
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1Week 9-25-06 and some preparation for exam 2.
2Week 9-25-06 and some preparation for exam 2.
3
NORMAL DISTRIBUTIONBERNOULLI TRIALS
BINOMIAL DISTRIBUTIONPOISSON DISTRIBUTION
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note thepoint of inflexion
note thebalance point
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SD=15
MEAN = 100
point ofinflexion
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5
50
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6.3
39.7
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~68%
Illustrated for theStandard NormalMean=0, SD=1
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~95%
Illustrated for theStandard normalMean=0, SD=1
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15
100
~68/2=34%
~95/2=47.5%
130 85
11
15
100
~68/2=34%
~95/2=47.5%
130 85
12
15
IQ Z
100
0
1
Standard Normal
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P(Z > 0) = P(Z < 0 ) = 0.5
P(Z > 2.66) = 0.5 - P(0 < Z < 2.66) = 0.5 - 0.4961 = 0.0039
P(Z < 1.92) = 0.5 + P(0 < Z < 1.92) = 0.5 + 0.4726 = 0.9726
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x p(x)
1 p (1 denotes “success”)0 q (0 denotes “failure”) __ 1
0 < p < 1q = 1 - p
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P(success) = P(X = 1) = pP(failure) = P(X = 0) = q
e.g. X = “sample voter is Democrat” Population has 48% Dem.
p = 0.48, q = 0.52P(X = 1) = 0.48
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P(S1 S2 F3 F4 F5 F6 S7) = p3 q4
just write P(SSFFFFS) = p3 q4
“the answer only depends upon howmany of each, not their order.”
e.g. 48% Dem, 5 sampled, with-repl:P(Dem Rep Dem Dem Rep) = 0.483 0.522
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e.g. P(exactly 2 Dems out of sample of 4)= P(DDRR) + P(DRDR) + P(DDRR)+ P(RDDR) + P(RDRD) + P(RRDD)= 6 .482 0.522 ~ 0.374.
There are 6 ways to arrange 2D 2R.
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e.g. P(exactly 3 Dems out of sample of 5)= P(DDDRR) + P(DDRDR) + P(DDRRD)+ P(DRDDR) + P(DRDRD) + P(DRRDD) + P(RDDDR) +P(RDDRD) + P(RDRDD) + P(RRDDD) = 10 .483 0.522 ~ 0.299.
There are 10 ways to arrange 3D 2R.Same as the number of ways to select 3 from 5.
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5! ways to arrange 5 things in a line
Do it thus (1:1 with arrangements): select 3 of the 5 to go first in line, arrange those 3 at the head of line then arrange the remaining 2 after.
5! = (ways to select 3 from 5) 3! 2!So num ways must be 5! /( 3! 2!) = 10.
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Let random variable X denote the number of“S” in n independent Bernoulli p-Trials.By definition, X has a Binomial Distributionand for each of x = 0, 1, 2, …, n P(X = x) = (n!/(x! (n-x)!) ) px qn-x
e.g. P(44 Dems in sample of 100 voters) = (100!/(44! 56!)) 0.4844 0.52100-44 = 0.05812.
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n!/(x! (n-x)!) is the count of how manyarrangements there are of a string of x letters“S” and n-x letters “F.”.
px qn-x is the shared probability of each stringof x letters “S” and n-x letters “F.”(define 0! = 1, p0 = q0 = 1 and the formula goesthrough for every one of x = 0 through n)
is short for the arrangement count =
23Week 9-25-06
24Week 9-25-06
n = 10, p = 0.4mean = n p = 4sd = root(n p q) ~ 1.55
25Week 9-25-06
n = 30, p = 0.4mean = n p = 12sd = root(n p q) ~ 2.683
26Week 9-25-06
n = 100, p = 0.4mean = n p = 40sd = root(n p q) ~ 4.89898
27Week 9-25-06
p(x) = e-mean meanx / x!for x = 0, 1, 2, ..ad infinitum
28Week 9-25-06
e..g. X = number of times ace of spades turns up in 104 tries X~ Poisson with mean 2p(x) = e-mean meanx / x!e.g. p(3) = e-2 23 / 3! ~ 0.18
29Week 9-25-06
e.g. X = number of raisins in MY cookie. Batter has 400 raisins and makes 144 cookies. E X = 400/144 ~ 2.78 per cookie
p(x) = e-mean meanx / x!e.g. p(2) = e-2.78 2.782 / 2! ~ 0.24(around 24% of cookies have 2 raisins)
30Week 9-25-06
THE FIRST BEST THING ABOUT THE POISSON IS THAT THE MEAN ALONE TELLS US THE ENTIRE DISTRIBUTION!note: Poisson sd = root(mean)
31Week 9-25-06
E X = 400/144 ~ 2.78 raisins per cookie
sd = root(mean) = 1.67 (for Poisson)
32Week 9-25-06
THE SECOND BEST THING ABOUT THE POISSON IS THAT FOR A MEAN AS SMALL AS 3 THE NORMAL APPROXIMATION WORKS WELL.
mean 2.78
1.67 = sd = root(mean)Special to Poisson
33Week 9-25-06
E X = 127.8 accidentsIf Poisson then sd = root(127.8) = 11.3049 and the approx dist is:
mean 127.8 accidents
~sd = root(mean) = 11.3Special to Poisson
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Week 9-25-06
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The overwhelming majority of samples of n from a population of N can stand-in for the population.
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The overwhelming majority of samples of n from a population of N can stand-in for the population.
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Sample size n must be “large.” For only a few characteristics at a
time, such as profit, sales, dividend.SPECTACULAR FAILURES MAY OCCUR!
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With-replacement
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With-replacementvs without replacement.
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This sample is obviously “not representative.”
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Rule of thumb: With and without replacement are about the same if
root [(N-n) /(N-1)] ~ 1.
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They would have you believe the population is {8, 9, 12, 42}
and the sample is {42}. A SET is a collection of distinct entities.
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IF THE OVERWHELMING MAJORITY OF SAMPLES ARE “GOOD SAMPLES”
THEN WE CAN OBTAIN A “GOOD” SAMPLE BY
RANDOM SELECTION.
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Digits are made to correspond to letters. a = 00-02 b = 03-05 …. z = 75-77Random digits then give random letters. 1559 9068 … (Table 14, pg. 809)15 59 90 68 etc… (split into pairs) f t * w etc… (take chosen letters)For samples without replacement just pass over any duplicates.
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The Great Trick is far more powerful than we have seen.
A typical sample closely estimates such things as a population mean or the shape
of a population density.But it goes beyond this to reveal how
much variation there is among sample means and sample densities. A typical sample not only A typical sample not only
estimates population quantities. estimates population quantities.
It estimates the sample-to-It estimates the sample-to-sample variations of its own sample variations of its own
estimates.estimates.
46
The average account balance is $421.34 for a random with-replacement sample of 50 accounts.
We estimate from this sample that the average balance is $421.34 for all accounts.
From this sample we also estimate
and display a “margin of error”
$421.34 +/- $65.22 = $421.34 +/- $65.22 = . .
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NOTE: Sample standard deviation smay be calculated in several equivalent ways,
some sensitive to rounding errors, even for n = 2.
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The following margin of error calculation for n = 4 is only an illustration. A sample of four
would not be regarded as large enough.Profits per sale = {12.2, 15.3, 16.2, 12.8}.Mean = 14.125, s = 1.92765, root(4) = 2.Margin of error = +/- 1.96 (1.92765 / 2)
Report: 14.125 +/- 1.8891.A precise interpretation of margin of error will be given later in the course, including the role of 1.96. The interval 14.125 +/- 1.8891 is called a “95%
confidence interval for the population mean.”
We used: (12.2-14.125)2 + (15.3-14.125)2 + (16.2-14.125)2 + (12.8-14.125)2 = 11.1475.
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A random with-replacement sample of 50 stores participated in a test marketing. In 39 of these 50 stores (i.e. 78%) the new package
design outsold the old package design.
We estimate from this sample that 78% of all stores will sell more of new vs old.
We also estimate a “margin of error +/- 11.5%+/- 11.5%
Figured: 1.96 root(pHAT qHAT)/root(n) =1.96 root(.78 .22)/root(50) = 0.114823 in Binomial setup
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A sample of only n = 600 from a population of N = 500 million.
(FINE resolution)
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