Transcript Introduction to Inference - Beedie School of Business

Introduction to Inference
Chapter 6
Statistical Inference
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Want to draw conclusions based on
sample data
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i.e.-- say something about an entire
population based on information in
a sample.
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Conclusions are subject to sampling
error
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We want to quantify the margin of
error we are likely to encounter.
Examples
1. I’m interested in estimating the
mean income of loggers in the
interior of BC.
2. Gallup poll.
Confidence intervals
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Our goal is to obtain an estimate
of some parameter of a
population.
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We want our estimate to be of the
general form
Best guess +/- error of
estimation
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To get our estimate, we take a
random sample from the
population and proceed on the
basis of the information we obtain
from the sample.
What is our “best guess”??
How do we obtain our “error of
estimation” ?
Confidence intervals for
population mean
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Best Guess
 Our “best guess” is xbar, the
sample mean.
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How to determine the
estimation error?
Determining the estimation error
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To determine estimation error, we’ll
use our knowledge of sampling
distribution of xbar.
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We know that xbar has mean µ,
with standard deviation, /sqrt(n),
where n is our sample size.
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If our original population is normal,
we also know that xbar is normally
distributed.
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Even if the original population isn’t
normal, xbar approximately follows a
normal distribution if the sample
size, n , is large (by CLT).
Determining the error of
estimation -- continued
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For example, we are about
95% sure that xbar lies in the
range
µ +/- 2 /sqrt(n).
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(*)
E.g., if  = 45 and n = 100, we
are 95% sure that xbar lies in
the range
µ +/- 9. (verify).
How to use (*) to get our
confidence interval??
A confidence interval for
population mean
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It turns out that our 95% c.i.
for µ is just
xbar +/- 2 /sqrt(n).
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Why does this work?
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From our picture ( to be
added in class), we see that
our interval will “trap” µ
approximately 95% percent
of the time.
A numerical
example
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I collect SRS of n = 100
loggers. I find average income
in sample is $17,000.
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Obtain a 95% c.i. for the mean
income of all loggers.
Assume that std. dev. of
loggers incomes is known to
be  = 2500.
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(In Ch 7 we will drop the assumption that 
is known --  will be estimated from the
sample data)
Level C confidence intervals
Tradeoffs
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For a given sample size,
higher level of confidence
leads to wider confidence
interval.
Designing a confidence
interval
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Suppose I want a level C
confidence interval of the form
Xbar +/- m,
where m is a desired margin of
error that I supply in advance. I
also specify C in advance.
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Required sample size is
n = (z*/m)2
where z* is the z value required
for level C confidence
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Where does this formula come
from?
Example--Designing a
confidence interval
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Recall the logger example. I
have  = $2500. Suppose I
want a 99% c.i. of the form
xbar +/- 300. How big a
sample do I need?

What if I want a 99%
confidence interval of form
xbar +/- 150?
Example 6.12 (Text)

A study of career paths of hotel general
managers sent questionnaires to a SRS
of 160 hotels belonging to major US hotel
chains. There were 114 responses. The
average time these 114 general
managers had spent with their current
company was 11.78 years. Give a 99%
c.i. for the mean number of years general
managers of major-chain hotels have
spent with their current company. (Take
it as known that the std dev of time with
the company for general managers is 3.2
years).
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A margin of error of +/- 1 year is
considered acceptable. What is the
minimum sample size that would be
required to achieve this level of accuracy
with a confidence level of 99%
Confidence interval
summary
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Assume I have a random sample
and that the population std dev, ,
is known.
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A level C c.i. for the population
mean is
xbar +/- z*/sqrt(n).
(value of z* will depend on C)
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The c.i, is exact when the
underlying population is normal.
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If the pop. is not normal, the c.i. is
approximately correct for large
samples (by CLT).
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We can find the sample size, n,
required to obtain a c.i. with a
specified margin error, m, by
using the formula
n = (z*/m)2
Tests of Significance
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Confidence interval
 Goal is to estimate some
parameter of a population.
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Test of Significance
 Goal is to assess the
evidence provided by the
data in favor of some claim
about the population.
Motivational Example
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Four randomly selected students do a
20 hour SAT prep course at a special
school. After the course, they write the
SAT. Their scores turn out to be 560,
600, 590, 490. We find xbar = 560.
Scores of students who take the
course are known to be normally
distributed with a standard deviation of
50.
National SAT test scores are normally
distibuted with a mean of 500 and a
standard deviation of 50.
Does the sample data provide strong
support for the school’s claim that the
prep course is effective in increasing
SAT scores?
Motivational Example
(cont.)
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Let’s use what we know about
the behavior of xbar to assess
this claim.
In particular, we ask
 What is the probability of
observing a sample mean
of 560 or larger if the
population mean score for
those who took the course
is 500? (i.e., course doesn’t
help on average)
Motivational Example
(cont.)
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What if the sample mean had
been xbar = 700?
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What if the sample mean were
xbar = 520?
Formalization of our
example:
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First, state hypotheses:
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H0: µ = 500
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(course has no effect).
Ha: µ > 500
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(course increases mean score).
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Terminology
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H0 is the null hypothesis.
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Ha is the alternative hypothesis.
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Usually, Ha is the claim we hope
to establish. (Equivalently, H0 is
the claim we want to falsify).
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In our example, the more the
sample mean exceeds 500, the
more evidence we have in favor
of Ha ( i.e. against H0 ).
Formalization of our example: (cont.)
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Second: We compute the test statistic
z = (xbar – mu0)/(sigma/sqrt(n))
= (560 -500)/(50/sqrt(4)) = 2.4
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Third: Assess the strength of the evidence
against the null hypothesis by computing a pvalue.
p-value = Prob (z >=2.4) = 0.0082
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(The p-value is obtained from the z value
that results from the specific form of our
hypotheses.)
Fourth: state a conclusion.
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“strong” evidence in favor of company’s
claim.
Other possible forms of
H0 and Ha in our example
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Suppose that I suspect that the
school has a “reverse” effect and
actually decreases average SAT
performance.
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How would I set up hypotheses
in such a way that a low average
test score will support my
suspicions?
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What if I suspect that the course has
some effect on SAT scores, but I’m
not sure whether it increases or
decreases the average score. How
would I set up appropriate “two sided”
hypotheses for this situation?
 Either a low or a high average
test score will support my
suspicions.
General form of the z test
Example
SSHA is a psychological test that
measures motivation, attitude
towards school, and study habits
of students. Scores range from 0
to 200. The mean score for US
college students is about 115 with
a std. dev of about 30. A teacher
who suspects that older students
have better attitudes towards
school gives the SSHA to 20
students who are at least 30 years
of age. Their mean score is 135.2
(a)
(b)
(c)
State appropriate null and
alternative hypotheses.
Report the p-value of your test and
state your conclusion clearly.
Your test required 2 important
assumptions in addition to the
assumption that sigma = 30. What
are they? Which is more
important?
Another Example
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A study of the pay of corporate
CEOs examined the increase
in cash compensation for the
CEOs of 104 companies,
adjusted for inflation, in a
recent year. The average
inflation adjusted increase in
the sample was xbar = 6.9%
with a sample standard
deviation of s = 55%. Is this
good evidence that the mean
real compensation increased
in the past year?
Because the sample size is
large, s is close to the
population sigma, so it is
reasonable to assume that σ
=55%.
P-values and statistical
significance
Example –statistical significance
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The mean yield of corn in the US is
about 120 bushels per acre. A
survey of 40 farmers this year gives
a sample mean of xbar = 123.8
bushels per acre. We want to know
if this provides good evidence that
the national mean this year is not
120 bushels per acre. Assume that
the farmers surveyed constitute a
SRS from the population of all
commercial corn growers and that
the population has a std dev of
sigma = 10 bushels per acre.
(a) Set up the appropriate
hypotheses. Give the p-value for the
test. Is the result significant at the
5% level?
(b) Are you convinced that the
population mean is not 120 bushels
per acre?
(c) Is your conclusion correct if the
distribution of corn yields is
somewhat non-normal? Why?
Another Example
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A computer has random
number generator that
generates random numbers
uniformly distributed between
0 and 1. If this is true, the
numbers generated come from
a population with µ = 0.5 and 
= .2887. A command to
generate 100 random numbers
gives an average of 0.4365. Is
the generator working
properly?
Another Example
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A union leader claims that
the average school teacher
makes less than $40,000
per year. A random sample
of 400 school teachers
finds a sample mean of
xbar = $39,650. The
standard deviation in school
teachers incomes is known
to be $5,000. Assess the
union leader’s claim.
Relationship between
confidence intervals and two
sided tests
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A level  2-sided significance
test rejects the hypothesis
H0: µ = µ0
and accepts the alternative
Ha: µ ≠ µ0
precisely when the value µ0 lies
outside a level 1-  confidence
interval for µ.
Example (confidence interval
and 2 sided hypothesis test)
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Diameters of a certain machine
part are normally distributed
with a standard deviation of 0.1
mm. A random sample of 25
parts yields an average
diameter of 11.9 mm.
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Find a 99% c.i. for the true
mean diameter.
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Based on your sample, can
you conclude, at the 1% level
of significance, that the true
mean diameter differs from 12
mm?
Comments about hypothesis
testing
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P-values tell us more than
setting, in advance, a fixed
level of significance.
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Statistical significance is not
necessarily the same as
practical significance.
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Statistical testing is not always
valid –e.g. faulty data; bias in
questionnaires etc.
Two types of error
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Example – If there are more
than 32,000 trees on a plot of
land it will be economical to log
the plot. I sample the plot and
get the following 95% c.i. for
the total number of trees
32,500 +/- 3,500
Should I log the lot?
Type I and Type II
errors