| Hand in:
Do: 7.8 ADD
(CI by hand)
Do: 7.5 SPSS
luck CI
Notice the interesting distribution of the sample. What does the
CI not tell you?
Postpone: 7.31, 7.32 SPSS vit. c, test and CI's; matched pairs +. For 32b, you have to re-express the 5 "after" numbers as percents (e.g. Sample 1: 20/98 = 20.4%...) and then find a new CI of this data set. Postpone: 7.39 SPSS C: Factory to Haiti, matched pairs The answers weirdly assume you'll do Haiti - Factory, when it seems more natural to do Factory - Haiti; and the SPSS file is set up to do Factory - Haiti. Do it the natural way. Do: 7.50 motor insulation (logarithm) |
Read,
discuss
|
Optional
(more practice) |
Will show Friday:
SPSS: Transform/Compute
(first
handout)
"tables": CDF functions take value x, give Prob of being
less than or equal to x (like our book's Normal table)
IDF functions take probability p, give value x such that the
probability
of being less than or equal to x is p. The help in SPSS on
these
is sloppy, leaves off the "or equal to." Irrelevant for continuous
distributions,
crucial for discrete ones.
One-sample t procedures by hand: Day
34 Milk bacteria, see bottom 5370,
4890, 5100, 4500, 5260, 5150, 4900, 4760, 4700, 4870
SPSS: Analyze>Compare
Means>One-Sample
T Test: Test value = 4800.
P-value
is labeled "Sig (2-tailed)"--divide by 2 for 1 tail (if observed is in
correct direction)
Analyze>Descriptive Statistics>Explore.
Statistics
button, set Confidence level.
Will
show Friday: also SPSS
for it. MATCHED PAIRS t
procedures:
(get for free!)
before--after, left
hand--right
hand, Drug A vs. Drug B on the same person or on a matched pair
(Sec.
3.1,pp. 207-8)
For each pair, find the difference
in the observed values. Then treat these differences as if
they
are "the" data set, from a normal population, and do One-sample t
procedures.
Usually the null hypothesis will be "µ
= 0", there is "no difference" between the treatments.
Example: wax paper sandwich
bags:
Is the wax layer the same inside and out?
25 bags: measure (wax outside - wax inside)
for each. (pounds per square foot). n = 25
Differences: xbar
= .093, s =
.723
SEM = .723/5 = .1446
H0 : µ
= 0 (mean
difference
is
0)
t = (.093 - 0)/SEM
= .093/.1446
= .643.
Ha : µ
Not = 0 (there is a
difference)
t is less than .685 (d.f. = 24) which is right-tail t* for probability
.25
Because test is 2-sided, double the tail: .50. P value is >
.50.
No evidence for difference.
Done: Robustness
of t-procedures: A confidence interval or
significance
test is called robust if the confidence level or P-value
doesn't
change very much when the assumptions of the procedure are
violated.
pp. 462-465.
t-procedures are quite robust against
nonnormality.
But
sensitive
to outliers. Look at data. Need SRS!
Details: n <15
t ok unless data clearly not normal, or if there are outliers.
n > 15 t ok unless there is strong skewness, or
outliers.
n > 40 or so: t ok even if there is skewness.
(Outliers?
I suggest trying with and without them, see what changes).
Matched-pairs data (differences) are often more normal in
shape than the separate variables ("oddness" is often the same for both
items in a pair, and disappears in subtraction. Another reason
why
this is a nice experimental design. ) (Not true in the
full-moon
example.)
What if t's not suitable?
Skewness: Try log or other transformation, work
on transformed data. (Sadly, CI's can't be transformed back.
Because
µlog(X) is not equal to log(µX) )
Outliers or other nonnormality: Distribution-free/
nonparametric procedures. Usually less power than
distribution-based.
(Uses less information, duh!) Often based on binomial or similar
models.
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