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(All D&V) Finish these--conditional probability and independence (p. 299): 8 Pets Make a 2-way table using the counts given. Use it to find the conditional probabilities. 9 Health, 10 Death penalty Mixture of conditional/not conditional problems. Read carefully! 23 Health, independence? 22 Pets, independence 29 a Absenteeism IF you didn't do it already, do A, Day 26. (Simulate 25 coinflips; repeat 10 times.) Make the dotplot by hand; remember, this is not SPSS's "Dotplot" graph. See D&V p. 39. READ; will be assigned next time:
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Read,
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Homework questions? Day
26
Continue Conditional probability and independence Day
26
A and B are independent <--> P(B|A) = P(B). Knowing
A is true/not has no effect on chances of B.
Sampling DistributionsCh.
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"
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 = .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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