| Hand in
Monday (Day 25 HW if you didn't complete it for today; especially Normal problems) <>A) You've got your 30 slips of paper from the green shoebox and counted how many 1's you have--Calculate the proportion p-hat for your sample = (# of 1's ÷ 30). Bring to class. If you didn't get the slips in class, the shoebox is outside my door. Come and pick 30 at random; count number of 1's; return slips to shoebox; proceed as above. B) (added after class) Go to http://bcs.whfreeman.com/bps3e, Statistical Applets, Probability Applet. Set the Toss at 30 times, leave the probability of heads at .5. Toss. Record the proportion of heads. Reset, and repeat twice more. (I just did it and got 12/30=.40, 15/30 = .50, 19/30 = .63). Bring your 3 results to class to share. Postpone the rest: (With your slips from the shoebox) Also plug in p-hat (instead of p) to the formula for SD(p-hat): so if you got 12/30, p-hat = .4. "q-hat" = 1 - p-hat = 1-.4 = .6. SD formula: square root of (p ·q/n)= square root of (.4 ·.6/30) = square root of .008 = .089 Save these estimates of p and SD(p-hat) for a later class.. Probability, Ch. 15. p. 299 Chapter 18, Sampling distributions:
proportions
p. 350 |
Read,
to discuss |
Optional |
Try to do in class today: take 30 slips of paper at random from the green shoebox. Count how many 1's you have. Record and keep this for HW. Return slips to shoebox.
Homework questions? Day 26 P(a < x < b) = the probability
that the outcome x is between a and b
is the area under the model's density curve, between a and b.
is the proportion of x's which would come up between a and b
if we did the phenomenon a zillion times.
We declare P(a) = 0 (In a continuous model, getting precisely
a
is utterly unlikely; can't even measure that well),
so P(a < x
< b) = P(a < x < b)
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- - - - - -
Our most important probability model: NORMAL Model
family.
Same techniques as before, only we ask "probability that one chosen at
random..." instead of "proportion of all..." Review Normal
techniques:
See Day 26 for links
= = = = = = = = = = = = = = = = = =
Start here Monday
Back to Chapter 16, Probability rules
Equally likely probabilities: If there are k
possible
outcomes, and each is equally likely, each has probability 1/k (Dice,
coin,
etc. Plenty of things are not equally likely.)
"General addition rule": (Remember:) If A and B
are disjoint events, then P( A or B) = P(A) + P(B)
If A and B can happen together, P(A and B) is not 0.
Then P(A or B) = P(A) + P(B) - P(A and B).
"General addition rule."
Use Venn diagram to see. Example
A thousand people are interviewed by the census bureau, and the results
tabulated in this two way table.
Working Status vs. Sex.
|
Restated as proportions of the
whole:
|
Now to Chapter 18, probability applied to
simple
random samples!
Sampling Distributions Ch.
18
Take a Sample from a population. SRS!.
Imagine (simulate) what would happen if you
took "all possible" SRS's. For each sample, calculate
a
statistic. We'll look at 2 important ones:
p-hat, the proportion of some
characteristic
(e.g. proportion who are sophomores, proportion of thumbtacks point-up)
x-bar, the mean of some variable (e.g.
sample
height).
We expect/hope these will be "close" to the corresponding parameters
in the population. Quantify "expect", "close":
Need from the sample:
Each individual's chance of being chosen is independent
of each other's chance. (Random sample)
Need to not "use up" too much
of the population. Sample size < 10% of Population size is
good
enough.
Sample proportion(s) first:
Suppose true proportion is p (prob. of "up", for
thumbtack).
Let 1-p = q (prob. of "down"). Take SRS of size n.
The mean of all p-hats from all possible SRS's
is p.
The standard deviation of all such p-hats is 
.
If n is big enough (need bigger if p or q closer to 0), then
the shape of the distribution of all p-hats is (approximately) Normal!
Summary: Take n independently sampled values
from
a population with population proportion p:
Sampling distribution of p-hat (read: p-hats from all possible
samples) is well modeled by N(p, )
if 1) n < 10% of population
2) np >10 and nq > 10
"success/failure condition"--n "big enough"
Use the "Normal approximation to binomial" Applet at http://bcs.whfreeman.com/scc to get a sense of the shape of the sampling distribution of p-hat. The number of "successes" is graphed here, and we're looking at the proportion--just think of the x-axis as going from 0 to 1, instead of from 0 to n. Notice how the square root of n works in the denominator; if we make n 4 times as big (from n=10 to n=40), the "spread" of the distribution of p-hat is half as wide. Also notice that bigger n, or more middling p, makes the distribution more normal.
Example: Flip "fair" coin 25 times. Probability
of
Heads = p = 1/2. q = 1-p = 1/2. n =
25
SD(p-hat): pq =
1/4.
pq/n = 1/100. sqrt(pq/n) = 1/10 = 0.1 = SD(p-hat)
Probability that my experiment will produce 15 or more heads?
15/25 = .60. P( p-hat > .60)?
Can we use the normal model? 1)
Population
is infinite. OK. 2) np = nq = 12.5, which are >
10.
OK.
Standardize .60: z = [(value -
mean)/s.d.] = [(.60 - .50)/.1] = .1/.1 = 1. .60 is one s.d. above
the mean.
P(Z >
1) = 16%, using the 68-95-99.5
rule.
See p. 340 for another example.
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