Hand in Monday Please read Ch. 15--lots
of jargon and weird stuff! |
Read, to discuss |
Optional (more practice)
|
Your shoebox
results so far: Write your samples, and their xbars (one on each
pad--yellow or
white), say whether you think the mean of that box is > 20 (Y/N).
Exam 4 this Friday, next class
. Covers Ch. 10 , Ch. 11 up to p. 286 only, Ch. 14 all, Ch. 16
thru
p. 391. (i.e. thru Day 34 HW).
Sample Exam (Handed out
Fri.
Outside my door..) Solutions
.
Sign up Today for early (>9:30)
start:
Confirm with
me any other time to take it.
Homework questions? sample size: Day 34
Why do CI's work? CI's Day 34
Exam Review (big ideas only!)
Probability: Prob. of event = proportion of times it
would happen in a long sequence of repetitions of he experiment.
Discrete and Continuous Random Variables
Probability that a single individual chosen at random has
some characteristic = proportion in population with that characteristic.
Take a Sample from
a Population (hope
sample is like population)
Statistic
estimates Parameter
(summary numbers calculated from above)
xbar
estimates
µ
Sample means Xbar: all possible xbars from all possible
SRS's (of size n) from the population.
Population has mean µ, s.d. sigma. X, sample of size 1 ,
has distribution of population.
Behavior: Let n get larger. Xbar will get
closer and closer to µ. -->LLN (Law of large
numbers)
Fixed n. Xbars distribution:
Mean = µ , S.d. is sigma ÷ (square root of n)
If X is Normal, so is Xbar.
In any case, if n is "large", Xbar is approx. Normal -->Central
Limit Theorem
Estimating unknown µ using an xbar:
(estimating a parameter using a statistic)
CI (confidence interval) level C, margin
of error m.
No particular CI is
guaranteed to successfully capture µ; But the Method will produce
a "true" result C% of the time.
For µ: (SRS,
pop. Normal, sigma known)
xbar +
m, m = z* (sigma ÷ (square root of n)),
+z* cut off central C% in standard normal dist.
Plan ahead: for
desired C and m, known sigma, find n needed to achieve those.
In practice: Need SRS or reasonable facsimile!!! Large n
allow substituting s from data for unknown parameter sigma. Large
n allows using CI formula even if pop. not very normal (CLTh).
Links:
Normal and Xbar, compared
Sampling dist; Central limit
th. Rice
U. Applet
TableA(Normal),
TableC(t distribution)
ConfidenceInterval
Questions? Sample exam Solutions .
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = ="Statistics means
never having to say you're
certain."
Confidence interval Estimation made our best guess at an
unknown population mean.
Testing will investigate a claim made that the
unknown
mean is actually a particular value.
~~~~~~~~~~~~~~~~
Ch. 15: "Significance tests use
an elaborate
vocabulary, but the basic idea is simple: an outcome that would
"rarely" happen if a claim were true--is good evidence that the claim
is
NOT true." (p.363 top)
Suppose someone claims that the average height of Wells women over the
years is 70" (5'10"). I take samples (151 classes) every
year. This year my sample has mean 65.67" (n = 20ish). Standard
deviation for heights of women in population is supposed to be about
2.5" , so s.d. for means from samples of 20 is about 2.5/4.48= 0.56. IF
the real mean is 70", my sample is astonishingly unusual
(65.67-70)/0.56= -4.33 /0.56 = -7.73, 7.73 s.d's below the mean.
Conclude the
claim is Not true.
- - - - - - - - - - - - - - - - - - - - -
- - -
Extended Standard Normal Table--"Normal Tails"
(also from Weblinks page, )
z
P(Z <
z)
P(Z > z) = same in scientific notation: E-03 = 10-3
3.00
.9986501019683700
.0013498980316301 1.35E-03
4.00
.9999683287581670
.0000316712418331 3.17E-05
5.00
.9999997133484280
.0000002866515718 2.87E-07
6.00
.9999999990134120
.0000000009865877 9.87E-10
7.00
.9999999999987200
.0000000000012799 1.28E-12
8.00
.9999999999999990
.0000000000000007 6.66E-16 Below this, machine
can't compute.
If your assumptions lead you to a(n almost)
impossible
z value, question your assumptions!
(The basis of significance/hypothesis testing)
-
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
- -
Need machinery to analyze less "obvious" results--build in effect
of
standard deviation (if s.d. were 10" would my sample still be
inconsistent with the claim?) and sample size (if n were only 4 would
that change my result?) .
Do 15.2 p. 365 Are older students like traditionals, or
higher (on average) on this measure?: Normal, s.d. =
30. Claim:
pop. mean = 115. n = 25. IF mean is really
115, Xbars are N(115, 6). Sketch!
xbar = 118.6 .
118.6-115=3.6. This is 3.6/6 = 0.6 s.d.'s above the mean, a
pretty typical kind of value.
xbar =
125.8 125.8-115=10.8. This is 10.8/6 = 1.8
s.d's
above the mean, high enough to be pretty unusual (how unusual?) if the
mean is
really 115.
xbar = 139
139-115=24. This
is 24/6 = 4 s.d.'s above the mean, unreasonably high if the
mean is really 115.
So 125.8 or (more so!) 139 would be evidence that the mean
for this group (older students) is NOT 115, is in fact higher.
Went over
the above in class: Shoeboxes are outlined for HW above. Continue Monday here:
Shoeboxes
(white and yellow slips): Take a sample of size 4 from
each, record, return numbers.
I claim the mean value
for both shoeboxes is µ = 20. Am
I telling you the truth? I can't remember for sure. I do know that
the distribution in the box is normal, standard deviation is 4.
I do remember that if µ is not 20, then it is greater than 20. µ
> 20.
Take a sample of size 4, find xbar.
Once for each shoebox! (should have found xbar already)
How far from 20 is it? far
enough that I believe the mean is not 20??
Take data. Calculate test statistic,
usually based on one that estimates the parameter in the
hypotheses. For µ, test statistic is the z-score of xbar,
so a big z-score number means that xbar is far from µ.
Is it an unlikely
result if H0 is true? Then that is
evidence
against
H0.
Measuring the strength of the evidence against H0 (a
common measuring stick for all distributions and parameters):
P-value of
a test: The probability, computed assuming
that H0 is true, that the observed outcome would
take a value as extreme or more extreme than what we actually
observed
(if
we could repeat taking-data again). p. 368.
The smaller the P-value, the stronger the data's
evidence against H0 ( for Ha).
For a test of µ , using xbar (sigma
known),
the P-value is
--the area of the tail beyond the observed xbar, in the
direction of Ha (one tail)
(--or twice that area (two-tail).)
We usually calculate it by standardizing the observed xbar (assuming
H0 true) and looking in the normal table. (p. 369 on)
H0: µ =20 Ha:
µ > 20 How far from 20 is your xbar?
Find
z for xbar.
For xbar = 24, z = 2
Is this a far-out value of z?
What is the probability of being farther out, i.e. being in the tail
beyond this z? That's the P-value. P = .0228 Table A
<>Applet: P-value of a
test of significance automates this. (Uses "raw" scale of xbars, rather
than z-scores). Use as check, guide.
How to: At
top, put in H0 value, choose direction of Ha,
put in sample size n, and s.d. of the population sigma.
Do Update (Reset sends back to 'opening" values). The graph
and scale axis show distribution of x-bars assuming H0 is true.
Under the graph, put your "observed" x-bar value in the "I have data..." box
and do Show P. The P-value is the size of the tail, shown in gold.
For HW draw the picture and label the axes both in "raw" and in z-
values. Show direction(s) of the alternative. Mark xbar, z, and
shade the area which is P-value. (And do the calculations of course.)
Example (one sided): H0: µ =1000 hrs. (Average lightbulb life.) Competing bulb:
Show it's better.
Ha:
µ > 1000 hrs. (one-sided)
Sample of size n = 25. Population
sigma = 150 hrs. S.d. of xbars = 150/5
= 30.
Get xbar = 1075
hrs. Are these bulbs better than the "standard?"
z = (1075-1000) ÷ (150/5) = -75/30 = 2.5;
P(Z > 2.5) = .0062 = P-value. More than
6 in a thousand and less than 7 in a thousand. More crudely, Less than 1% chance
of getting a result this high if we did it again--if the real mean is 1000.
Example (one sided): H0:
µ =1000 hrs. (Average lightbulb
life.) Suspect company's cheating:
Show it's worse.
Ha: µ < 1000 hrs.
Sample of size n = 25.
Population sigma = 150 hrs. S.d. of xbars = 150/5 = 30.
Get xbar = 940 hrs. Are these bulbs worse than claimed?
z = (940-1000)
÷ (150/5) = -60/30 = -2.
P(Z <
- 2) = .0228 = P-value
More than 2% and less than 3% chance of getting a result this low (below1000)
if we did it again--if the real mean is 1000.
Will do two-sided
in more detail Wed.
Example (two
sided): H0:
µ =1000 hrs. (Average lightbulb
life.)(Quality control on assembly line--find if it is "off" either
way.)
Ha:
µ Not = 1000 hrs. (two-sided)
Ha: "Alternative hypothesis" A claim
or statement about the population we are trying to find evidence FOR.
A value either much bigger than or much smaller than the H0
value is evidence against H0 & for Ha.
Sample of size n
= 25. Population sigma = 150 hrs. S.d. of xbars = 150/5 = 30.
Get xbar = 940
hrs. Is the quality control "off?"
z = (940-1000) ÷ (150/5) = -60/30 = - 2; P(Z <
- 2) = .0228
P-value (two sided): We measure the probability
of seeing something (again) as extreme as the observed value (or more
so).
So you need to measure the P-value symmetrically
both directions from the observed value--so the P value is double
what it would be for a one-sided test. P-value is approximately 5%; more precisely, 2·.0228
= .0456
So for a test of a mean, the P-value for one-sided is half
that for two sided, IF the result is in the direction of evidence for the alternative.
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