MATH 251, Probability and Statistics I, Fall 2001, Wed. Nov. 7, Day 29
4 Numbers from the Birkenstock box: Find mean xbar.
Find xbar + .841. This is your interval estimate
of the unknown mean of the box's population.
Add your values to the list, and graph your interval on
the transparency circulating.
Quiz Friday: Knowing and using: Binomial
distribution formula (p. 388 bottom)
mean and st. dev. for Binomial: X (count),
p-hat (proportion)
mean and st. dev. for X-bar from SRS of size
n
Normal? Central limit theorem: 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.)
Confidence intervals (sec. 6.1) This is one of the two big
ideas of inference that we will study. Chapter 7 will extend this
simple idealized situation, so this needs to be firmly in place.
Confidence interval estimate of a(n unknown) population parameter:
-
Interval constructed from the data +
-
Confidence level C:
probability that intervals constructed by this method will capture
the true, unknown, parameter.
(Many repeated samples would
capture the unknown paramater C proportion of time.
Unfortunately,
we only get one sample, as a rule.)
Confidence Interval of the form estimate
+
margin-of-error for the mean, Confidence level C: (p.306)
-
the estimate is xbar
-
margin of error m is : z* times Standard deviation
of sample mean
z* from normal table. Probability C is between -z*
and +z*. Draw sketch!
(Table D p. T-11, bottom rows. The row labeled "infinity" is the row
for z* . )
Standard deviation of sample mean: Sigma /sqrt(n)
Must know standard deviation of population!
or, If sample size is large, use s (standard
deviation calculated from sample)
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.
How many people captured the true mean?
(
previous classes,
9/18 = 50% , 11/20 = 55% , 22/29= 76%. Combined, 42/67= 63%
)
Why does the formula work?
-
If a particular xbar is within m of the population mean, then the
interval xbar + m contains the population mean.
& If a particular xbar is farther than m from the population
mean, then the interval xbar + m doesn't contain the
population mean.
-
We choose z* (and from it m) so that the probability
that Xbar is within m of the population mean is
C.
How? Probability C is between -z* and +z* in the standard normal
table,
between -z* ·(s.d. of Xbar) and +z*
·(s.d. of Xbar) around µ, in the normal distribution
of Xbar.
-
Table D, bottom row, is a restating of table A, normal table,
but with probabilities (areas) on the edge, and z values in the body.
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 .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.
--For fixed C, m, (and sigma), find n, the needed sample size.
--Assumptions and warnings
Read 6.1
Hand in:
p. 447, using formula
6.8 ARMSA
6.2 corn variety
6.5 corn again; change sample size.
6.6 Julie's potassium
Sample size, given margin of error m.
--Solve m = z* sigma/ sqrt(n) for n, to derive the formula on p. 443.
p. 450, 6.12 Julie's potassium
6.13 corn again
Cautions (pp. 444-5)
p. 451, 6.20 Rangel election
6.24 radio call-in
6.25 extra runner You didn't do 6.1, so look in the back of
the book for the CI from there. Make a stem and leaf of the
data. Look in the back of the book for the answer to part a.
Answer part b. |
Read, discuss
p. 447 6.1, 6.2 (there are 2.2 pounds to a kg.) What would you
do to get the answers for 6.2 from those for 6.1? |
Optional
(more practice)
6.11 reading ability: CI's and sample size |