Math 151 , Fall 2005, Day 28 Mon. Oct. 31 Hit
reload After class (some
additions)
Day 28(Mon. Oct. 31): 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..
No new homework to hand
in Wed: But please read Chapter 18!
Hang on to your count of 1's from 30 slips
of paper from the shoebox. If you didn't get your sample of 30, stop
by my office (Mac 102) and get one, or do it before class Wed.
Hand in Friday?(All
D&V) p. 350
Chapter 18, Sampling distributions:
proportions
1, 3 Coin tosses
5 More coin
9 Loans
13 Apples
14 Genetic defect
Sampling distributions: means
17, 18 Sampling distribution
19 GPAs
20 home values
A) If you 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. 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 |
Exam 2 returned: See comments,
and a way to earn back some points
If you have time, take 30 slips of
paper at random from the green shoebox. Count how many 1's you have.
Record and keep this. Return slips to shoebox.
Homework questions? Day
27
Start here Wednesday!
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 flips. Some issues with "granularity"
-- only certain values are possible. But general shape looks right.
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 µand
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: 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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