|
Hand in: I think these are all doable. Bring
questions, if you find them not so. Sec. 6.1, confidence intervals, p. 396 ff. using formula: Cautions (pp. 393-4) |
Read,
discuss
5.45 diff. of means
|
Optional
(more practice) 6.6, 6.8 OC |
Quiz Monday(?): Knowing and using: Binomial
distribution formula (p. 349 bottom)
mean and st. dev. for Binomial: X (count),
p-hat (proportion) (Summary p.350)
mean and st. dev. for X-bar from SRS of size
n (Summary p. 368)
Normal? Central limit theorem: says
yes for all of the above distributions, approximately, for n large.
If population(s) normal to start with, linear
combinations stay normal (including X-bar), mean and s.d. follow
algebra rules (as last quiz.)
Homework questions? Day 28
Continuing Central Limit Theorem and friends: Day 27 , Linear combinations of Normal R.V.s Day 28
Ch. 6.1: Introduction to Statistical
Inference:
Requires: Random sample or Randomized
experiment.
(Our theory: Simple Random Sample usually)
First example: Use sample mean
xbar
to "estimate" (unknown) population
mean µ
Mean of 4 grades (HW#3.75, sec.
3.4)
estimates population mean of all 10 ("known"= 69.4)
E.g. 69.75,
64.25, 73.5 (Each
is a "point estimate")
Interval estimate: xbar + margin of error (fudge factor) estimates population mean µ (69.4)
69.75 + 1: "µ is
between
65.75 and 73.75" True
69.75 + 4: "µ is
between
65.75 and 73.75" True
73.5 + 4:
"µ
is between 69.5 and 77.5" False
73.5 + 5:
"µ
is between 68.5 and 78.5" True
64.25 + 4:
"µ
is between 60.25 and 68.25" False
64.25 + 5:
"µ
is between 59.25 and 69.25" False
Review from here
Friday, with HW
A level
C
Confidence interval
estimate of a(n unknown) population parameter:
(Table A, or Table D
(back flyleaf), t dist. bottom row)
The Birkenstock box contains numbers from a
normally distributed population, with population standard deviation 2.
You each constructed a 60% confidence interval for the unknown mean:
n = 4.
Standard deviation of sample mean = 2/sqrt(4) = 2/2
= 1
z* for C = 60% is .841, so margin of error m is .841
times 1= .841.
To get the z* for C = 60% from the normal table, note
that this is the middle 60%, which leaves 40% to be split
between the 2 tails. So 20% above z*, and 80%
below. Go into the body of table A, find 80%= .8000 is between
values .7995 and .8023, closer to .7995. The z value with .7995
below it is .84. Table D gives it more precisely as .841.
How many people captured the true mean?
Previous
classes,11/20 = 55% , 22/29= 76%. 9/18 = 50% , 11/20
= 55%, 15/22= 68%, 16/24 = 67%
16/18 = 88%, 7/13 = 54%, 8/16 = 50%,
7/14 = 50%. Combined, 122/194 = 63% .This class 5/10=50%.
Combined, 127/204 = 62%
Quite variable for small samples, but settling
down?)
Extension: If n is large, we can
use the formula even if population is not normal.
(Because only the distribution of Xbar is
used, and Xbar is normal! Central Limit Theorem)
Cautions read pp. 393-4
New for Friday:
Applet:
Confidence Interval shows it's not the individual interval that C
describes, but the Method.
Why does the formula work? (pp. 387-8, briefer there)
1. A particular xbar is within m of the population mean,
if and only if the interval xbar + m contains the
population mean.
P(µ - m < Xbar < µ + m)
= P(Xbar -m < µ < Xbar + m ) = Prob. that CI
contains µ.
2. We choose z* (and from it m) so that the probability that Xbar is within m of the population mean is C.
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