Math 151 , Spring 2006, Day 28 Mon. April 10 Hit
reload After class
Day 28: Reading: Ch. 18, to p. 341 for proportions,
then the rest. ActivStats 18-1 is extremely good for the concept of the
sampling distribution of the proportion..
Hand in
Probability, Ch. 15. p. 299
3 Homes, 5 Amenities (Use Step-by-step p. 290 to
pattern on)
Chapter 18, Sampling
distributions:
proportions, (D&V p.
350ff.)
1, 3 Coin tosses
5 More coin
9 Loans
13 Apples
14 Genetic defect
If you didn't get your 30 slips of paper
from the shoebox; if you didn't simulate 30 flips of the coin (3 times)
(A & B, Day 27), do it now and bring
your results to class to share.
Postpone the
rest: Sampling distributions: means (D&V p. 350ff.)
17, 18 Sampling distribution
19 GPAs
20 home values
A) With your 30 slips of paper from the
green shoebox; you counted how many 1's you have--Calculate the
proportion
p-hat for your sample. 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 Bring these estimates of
p and
SD(p-hat) to class next time. 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.
|
Read,
to
discuss |
Optional |
Ask me for your exam
if you weren't here Fri. Comments.
Add your dots to the dotplots
circulating, and your proportions to the pad of paper:
A) The proportion of 1's from a sample of size 30 from the shoebox
B) The proportion of Heads from a sample of size 30 simulated coinflips
(3 samples each)
Homework questions? Day
27, Day 26
Probability rules
Day 27
Equally likely probabilities: k
possible
outcomes, each equally likely --> each has probability
1/k (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."
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
Day 27
y-bar, the mean of some variable (e.g.
sample
height)
Take n independently sampled values from a population with
population
proportion p:
Sampling distribution of p-hat (read: distribution of p-hats
from all possible samples) is well modeled by
N(p,
)
(if
n < 10% of population,& np and nq >10)
Example: Flip "fair" coin 25 times. p = .5, s.d.
= 0.1.
Probability that my experiment will produce 15 or more heads?
15/25 = .60. P( p-hat > .60)=
P(Z >
1) = 16%, using the 68-95-99.5 rule.
10 or fewer
heads? 10/25 = .40. P(
p-hat < .40)=
P(Z < -1) = 16%, using the 68-95-99.5 rule.
Look at your coinflips. n = 30, so
for SD(p-hat):
pq =
1/4.
pq/n = 1/120=.008333. sqrt(pq/n) = 0.091 = SD(p-hat).
Some issues with "granularity"
-- only certain values are possible. But general shape looks
right.
Start
here Wed.
Sampling distribution of the
mean,
y-bar:
Distribution of all means from all possible random samples of size
n from a population.
Need Random Sample, Independence (in particular,
for sampling without replacement, n < 10% of population.)
Population has mean µ and
standard deviation 
.
Whatever the shape of the population
distribution that we draw the sample from, 
The mean of the y-bars = the
mean of the population
The standard deviation of the
y-bars =
the s. d. of the population divided by the square root of n.
* y-bar "hits" the population mean on
average
(doesn't systematically go too high or too low.)
* Averages are less variable than individual
observations. Averages from large samples are less variable than
averages
from smaller samples (because of dividing by the square root of n)
IF the population is Normal,the
sampling distribution of the y-bars is Normal.
The Central
Limit Theorem (CLT)
In any
case,
for "large" n, the sampling distribution of the y-bars is
Approximately
Normal.
Example: "Normal" body
temperature
98.6 deg. on average. (Assume this is true.)
Assume normal distribution, & s.d.among many
people is 0.6.
Probability that one
individual's
normal temperature is below 98.0 degrees?(P(Z
<-1)
Take SRS
of 9 people. Sampling distribution of the mean? N(98.6,
0.6/3) = N(98.6, 0.2)
Probability that the mean is below 98.0? (P(Z
<-3))
Probability that one
(random)
healthy individual's normal temperature is above 98.8?
Probability that the mean of a sample
of 4 is above 98.8?
Probability that the mean of a sample
of 36 is above 98.8?
Probability that the mean of a sample
of 100 is above 98.8?
Note as n grows, SD shrinks, but only by
square
root of n!
SPSS simulation
Word
: average
of
spinners
which
can land on any number between 0 and 1.
Population--one spinner. distribution flat between
0 and 1, mean .49 s.d. = .29
n = 2, Average of 2 spinners is Xbar. Distributionof
x-bars is triangular between 0 and 1, mean .50, s.d.
.21.
.29/sqrt(2) =.205
n = 4, Average of 4 spinners is Xbar. Distributionof
x-bars is normalish between 0 and 1, mean .50, s.d.
.15.
.29/sqrt(4) =.145
n = 15, Average of 15 spinners is Xbar. Distributionof
x-bars is normal between 0 and 1, mean .50, s.d.
.09.
.29/sqrt(15) =.076
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