Transcript 2. Statistical Inference

ECON 251
Research Methods
2. Statistical Inference: Single
Population Mean and Proportion
(Review)
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Descriptive Statistics vs. Inferential Statistics
• Descriptive statistics: calculating summary characteristics of data.
• Inferential statistics: Using sample summary measures to estimate
population characteristics.
Descriptive Statistics
sample
population
Summarize
the data
Summarize the
data
Inferential Statistics
Sample:
Population
Characteristics
are unknown
Find
summarizing
measures
Inference
In descriptive
statistics we
summarize the data
from a population or
a sample of it.
Data on population is
NOT available. We
take a sample and
use its summarizing
measures to estimate
the unknown
population
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characteristics.
Statistical Inference Review
 There are two procedures for making inference
• Hypothesis Testing (HT) and Estimation
 In estimation, we attempt to estimate the value of the parameter in
either of two ways:
―Point Estimator
– A point estimator draws inference about a population by estimating the value of
an unknown parameter using a single value or a point.
―Interval Estimator
– An interval estimator draws inference about a population by estimating the
value of an unknown parameter using an interval.
– We use intervals so we can be precise about our degree of certainty regarding
the sample statistics proximity to the population parameter.
 HT involves testing a specific belief about the value of the parameter
 HT concepts are the foundation for estimation as well, so we begin
there.
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4 Steps For Hypothesis Testing
Step 1
Step 2
Step 3
Set up
alternative &
null hypotheses
Calculate the
test statistic
Find critical values
(Rejection region method)
Step 4
Find the p-value
(P-value method)
Make a decision
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Step One: Set up alternative & null hypotheses
 The purpose of hypothesis testing is to determine whether

there is enough statistical evidence in favor of a certain
belief about a population parameter.
There are two hypotheses (about a population parameter(s))
• H0 - the null hypothesis [for example, H0: m = 5]
• H1 - the alternative hypothesis [for example, H1: m > 5]
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Step One: Set up alternative & null hypotheses
 The alternative hypothesis is most important, it is what you
are trying to prove. Always start by stating the alternative
first.
• The alternative can involve >, < or ≠
• The alternative establishes whether the test is one-tailed or
two-tailed.
• The alternative establishes the location of the rejection
region(s).
• Once you have correctly defined the alternative, the null is
easy to establish.
 We always assume the null is true, therefore H0 MUST
contain =, and may contain ≥, ≤.
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Step Two: Calculating test statistics
 Population Mean w/ Sigma known
z
xm

n
 Population Mean w/ Sigma unknown
 Population Proportion
xm
t
s
n
z
pˆ  p
p(1  p)
n
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Step Two: Calculating test statistics
 The “standardization” formulas provide the test statistic.
They convert our sample statistic from the sampling
distribution to the “standardized” distribution (t or z in this
case).
• There are millions of sampling distributions. Rather than
knowing everything about every one of those distributions, we
“standardize” our statistic thereby moving it from the sampling
distribution and placing it on the “standardized” distribution.
• We know everything there is to know about the “standardized”
distribution. Because the test statistic is on the “standardized”
distribution, we can compare the test statistic to a critical
value, or the area associated with the test statistic (p-value) to
alpha.
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Step Three: Find critical value or p-value
 You need to decide which method you are going to use to
make your decision.
• If you are doing the calculations by hand, you will frequently
use the rejection region (critical value) method.
• The critical value will either be given to you (exams, in class
examples) or you would find it in excel (NORMSINV, TINV).
• P-value method will frequently be used when you are using
software to do your calculations, as most programs provide
these values. You can also find them in excel (NORMSDIST,
TDIST).
• In the latter case, be sure you can identify the p-value
graphically as well.
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Decision Rule: rejection region (critical value) method
Reject H0 if the test statistic is more extreme than the critical value
Given the significance level (probability of type I error) = a
Critical values
Two sided alternative
 za
Rejection region
Critical value
za
2
2
Rejection region
One sided (upper tail) alternative
za
Critical value
Rejection region
One sided (lower tail) alternative
z
Rejection region a
In case of t distribution we will have t a 2 & t a respectively.
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Decision Rule: p-value method
P-value is "the amount of evidence in favor of the alternative hypothesis.” The
smaller the p-value, the more evidence in favor of the alternative (and the more
likely you will reject H0 ). P-value is most commonly compared to a of 5% for
Reject/DNR decision:
Reject H0 if the p-value is smaller than the significance level
Two sided alternative
Each p-value/2
p-value = the area to the
right of tm
-|tm|
|tm|
One sided (upper tail) alternative
tm
p-value = the area to the
left of tm
One sided (lower tail) alternative
tm
(tm=test statistic; Same holds true for |Zm| & Zm )
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Using the p-value is the most common method of making your decision as
most computer software provides this value. However, you must graph your
distribution before making a final determination.
Three steps to finding the p-value from a graph.
1. Find the test statistic
2. Draw an arrow from the test statistic to the extreme end of nearest
rejection region
3. If a two-tailed test, do this on the opposite side of the distribution as
well.
The area of the graph which has an arrow through it, is the p-value.
Try showing the p-value graphically in these 4 examples. In each case, assume
that the critical value is 3.2:
H0: m = 5;
H1 : m > 5
Test stat = 7
H0: m = 5;
H1: m > 5;
Test stat = 3
H0: m = 5;
H1 : m < 5
Test stat = 3
H0: m = 5;
H1 : m ≠ 5
Test stat = 7
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Step Four: Make your decision
 Make one of the following two conclusions based on the test:
• Reject the null hypothesis in favor of the alternative
hypothesis.
 There ___ enough evidence to infer that the alternative is true
• Do not reject the null hypothesis in favor of the alternative
hypothesis.
 There _______ enough evidence to infer that the alternative
is true
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Errors
 Two types of errors are possible when making a
decision:
• Type I error - reject H0 when H0 is true.
• Type II error - do not reject H0 when H0 is false.
States of Nature
DNR H0
Reject H0
H0 is true
H0 is false
1- a
Type II error, 
Type I error, a,
Significance level
1- ,
power of test
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Analogy: Hypothesis testing is similar to a jury trial
 Assume innocent until proven guilty
• Assume H0 is true until proven otherwise
 Court either finds defendant guilty (Reject H0) or not guilty (DNR H0)
• Courts do not prove a person innocent (Accept H0); rather if “not guilty”
just not enough evidence to prove guilty; similarly, if we DNR H0, we are
not saying H0 is true, only that there is not enough evidence for us to
believe otherwise.
 Level of proof required to establish guilty verdict? What if you convict
an innocent person?
• Identical to establishing significance level of test.
• Type I error (a) is equivalent to convicting an innocent person. We focus
•
on a, rather than worry about a Type II error () – releasing a guilty
person.
“Beyond a reasonable doubt” is court of law norm
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Errors
 It would be desirable to reduce both types of errors at the


same time. But this is NOT possible.
There is a trade off between a and . As we try to decrease
a,  will increase and vice versa.
Because the consequences of a Type I error are in most
circumstances considered to be of greater concern than a
Type II error (sending innocent person to jail is worse than
letting a guilty person go), we focus on controlling the size of
the Type I error.
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Errors
 Standard in statistics varies depending upon the issue at
stake:
•
•
•
•
______________ evidence = 1% significance level
__________ evidence = 1.001-5% significance level
__________ evidence = 5.001-10% significance level
__________ evidence = 10.001% or higher significance level
 Unless stated specifically to the contrary, assume we are
using a = .05 in all problems.
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Examples—Hypothesis Testing
#1 A Nielsen survey estimated in the year 2000 that the mean
number of hours of television viewing per household was
7.25 hours per day. The survey involved 250 households. The
sample data had a standard deviation of 2.5 hours per day. In
1990, it was determined that the population mean of viewing
per household was 6.70 hours per day. Has TV viewing
increased since 1990?
(t249,0.005=2.596, t249,0.01=2.34, t249,0.025=1.9695, t249,0.05=1.651);
(z0.005=2.58, z0.01=2.33, z0.025=1.96, z0.05=1.645)
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Hypothesis Testing – 4 Step Solution
 Identify the alternative and null hypotheses.
H0: m
H1: m
 Calculate the test statistic
Z
xm


n
 Find the critical value or p-value.
• Z0.05 =
 Make the decision
• _______ H0 in favor of the alternative. There is ___________
proof that TV viewing has increased since 1990.
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#2 The owners of Subway claim that their stores average
$875,000 in annual sales. You used this information in
deciding to open a store in Delaware. Your store, however, has
not come even close to these annual sales figures. You want
to prove that you were misled, and that the average figure for
all stores is actually less than 875,000. You collect annual
sales figures from 70 randomly selected stores. The average
in your sample turns out to be $856,000, with a standard
deviation of $24,000. You also know from a friend who is in
management of a similar franchise, that you can count on the
standard deviation of sales being $28,000. Can you prove your
claim?
(t69,0.005=2.649, t69,0.01=2.382, t69,0.025=1.995, t69,0.05=1.667);
(t70,0.005=2.648, t70,0.01=2.381, t70,0.025=1.994, t70,0.05=1.667);
(z0.005=2.58, z0.01=2.33, z0.025=1.96, z0.05=1.645)
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#3 Your company is considering opening a retail store in
Fairbanks Alaska, but will only do so if average daily
spending per capita is higher there than in the rest of the
country. According to recent data, the average US
household spends $90 per day. A sample was taken in
Fairbanks. From a sample of 49, the average daily
expenditure was $84.50, and the standard deviation was
$14.50. Should you open a store in Fairbanks? You have a
lot riding on this decision, you need to be sure of your
conclusion.
(t48,0.005=2.68, t48,0.01=2.41, t48,0.025=2.01, t48,0.05=1.68);
(z0.005=2.58, z0.01=2.33, z0.025=1.96, z0.05=1.645)
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#4 Microsoft Outlook is believed to be the most widely used email manager. A Microsoft executive claims that Microsoft
Outlook is used by more than 75% of Internet users. A Merrill
Lynch study involving 300 respondents, reported that 72%
use Microsoft Outlook. Is there enough evidence here to
disprove the executive’s claim?
(t299,0.005=2.592, t299,0.01=2.339, t299,0.025=1.968, t299,0.05=1.65);
(z0.005=2.58, z0.01=2.33, z0.025=1.96, z0.05=1.645)
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#5 A fast-food restaurant plans a special offer that will enable
customers to purchase specially designed drink glasses
featuring well-known cartoon characters. If more than 15%
of the customers will purchase the glasses, the special offer
will be implemented. A preliminary test has been set up at
several locations, and 88 of 500 customers purchased the
glasses. Should the special offer be introduced?
(t87,0.005=2.634, t87,0.01=2.37, t87,0.025=1.988, t87,0.05=1.663);
(z0.005=2.58, z0.01=2.33, z0.025=1.96, z0.05=1.645)
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#6 For a new newspaper to be financially viable, it has to
capture more than 12% of the Toronto market. In a survey
conducted among 400 randomly selected prospective
readers, 58 participants indicated they would subscribe to
the newspaper. Can the publisher conclude that the
proposed newspaper will be financially viable at the 10%
significance level?
(t57,0.005=2.665, t57,0.01=2.39, t57,0.025=2.00, t57,0.05=1.67);
(z0.005=2.58, z0.01=2.33, z0.025=1.96, z0.05=1.645, z0.1=1.282)
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Confidence Interval Estimation – 4 Steps
 Confidence interval estimation relies on the same concepts
and relationships as does hypothesis testing. A simple four
step approach to these problems can also be helpful.
1. We begin by calculating the point estimate from our sample
data.
2. To establish the appropriate interval width, find the upper
and lower limits on the standardized distribution associated
with your confidence level.
3. Use the confidence interval formulas to place them on the
sampling distribution.
4. Place the sample statistic at the center of the interval and
the confidence interval is complete.
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Confidence Interval Formulas
 Population Mean w/ Sigma known
 Population Mean w/ Sigma unknown
 Population Proportion
x  za / 2
x  ta / 2
pˆ  za / 2

n
s
n
pˆ (1  pˆ )
n
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



1 – a is the confidence level associated with the interval
Sample statistic is used as the center of the interval
W: width of the interval; 2 x W: total length of the interval
UCL (Upper Confidence Limit) and LCL (Lower Confidence Limit)
are found using the critical value associated with a/2
 Confidence interval width for mean is a function of:
•
•
•
•
use of t distribution or z distribution
level of confidence chosen (positively related)
 of the sampling distribution (positively related)
sample size (negatively related)
 Population parameter can lie outside of interval – in fact, we
know it will a % of the time
 If interested in establishing a confidence interval of a specific width
and level of confidence, calculate the least number that is required
to be in your sample to achieve your objective ahead of time.
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Example – Confidence Interval Estimation
#7 As a new Subway franchisee, you are estimating your expected annual
sales. You have annual sales figures from 70 randomly selected stores.
The average in your sample turns out to be $856,000, with a standard
deviation of $24,000. The population standard deviation is 28,000. You
want a 90% and 95% confidence interval around your estimate.
(t69,0.005=2.649, t69,0.01=2.382, t69,0.025=1.995, t69,0.05=1.667);
(t70,0.005=2.648, t70,0.01=2.381, t70,0.025=1.994, t70,0.05=1.667);
(z0.005=2.58, z0.01=2.33, z0.025=1.96, z0.05=1.645)
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Confidence Interval Estimation – 4 Steps
1. Calculate the point estimate.
x
2. Find the upper and lower limits on the standardized
distribution associated with your confidence level.
• For 1–a = 90%; Z0.05 =
3. Use the confidence interval formulas to place the upper
and lower limits on the sampling distribution.
x  za / 2

n

4. Place the point estimate at the center of the interval
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Using CI to decide hypothesis tests:
 If you have calculated a confidence interval, and then decide
you also want to test a hypothesis with this information, you
can do so directly provided:
―The hypothesis being tested is two-tailed
―The a from the hypothesis test, and 1–a from the confidence
interval total 1.0
 If these two conditions hold, then determine whether the
hypothesized value in the null hypothesis for the parameter
falls in the interval created. If it does, DNR H0. If it does not
Reject H0.
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Example—Using CI to decide a Hypothesis Test
#8 The owners of Subway claim that their stores average $875,000
in annual sales. You used this information in deciding to open a
store in Delaware. Your store, however, has not come even close
to these annual sales figures. You want to prove that you were
misled, and that the average figure for all stores is actually NOT
875,000. You collect annual sales figures from 70 randomly
selected stores. The average in your sample turns out to be
$856,000, with a standard deviation of $24,000. You also know
from a friend who is in management of a similar franchise, that
you can count on the standard deviation of sales being $28,000.
Can you prove your claim?
(t69,0.005=2.649, t69,0.01=2.382, t69,0.025=1.995, t69,0.05=1.667);
(t70,0.005=2.648, t70,0.01=2.381, t70,0.025=1.994, t70,0.05=1.667);
(z0.005=2.58, z0.01=2.33, z0.025=1.96, z0.05=1.645)
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Estimating n for Confidence Intervals
 Sample sizes required to construct intervals of a certain
degree of confidence and width can be determined by using
one of the following formulas below:
 Sample Size for Means
 Sample Size for Proportions
• a priori idea of
pˆ
• no a priori idea of
pˆ
n
za / 2 2 2
w2
 za / 2 pˆ (1  pˆ ) 

n
W


za / 2 2 (.5) 2
n
w2
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2
Estimating n for Confidence Intervals
 When involving mean:
• Use sample standard deviation from previous study as 
• Use a pilot study to obtain a standard deviation ()
• Use judgment, or best guess
 When involving proportion:
• Use best estimate if confident of a reasonable value for
ˆ
p
― You have an a priori value for sample proportion
• Use 0.5 as
ˆ
p
― You have no a priori value for sample proportion
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Example – Estimating n for Confidence Intervals
#9 The interval you have created for your Subway is a good
start, but you would be more comfortable with a tighter
range for your estimate of sales. You decide that the
maximum you can tolerate is +/- 2,500. What sample size
would you need to collect to obtain a 90% confidence
interval for annual sales with a width of 2,500?
(t69,0.005=2.649, t69,0.01=2.382, t69,0.025=1.995, t69,0.05=1.667);
(t70,0.005=2.648, t70,0.01=2.381, t70,0.025=1.994, t70,0.05=1.667);
(z0.005=2.58, z0.01=2.33, z0.025=1.96, z0.05=1.645)
za / 2 
n

2
w
2
2
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Examples—Confidence Intervals
#10 The Environmental Protection Agency (EPA) has agreed to
give tax rebates to manufacturers of vehicles that get a
combined city and highway gas mileage of at least 32
mpg. A 49 car sample of a new Ford vehicle reveals a
mean of 32.6 mpg. It is believed that the highway gas
mileage for Ford vehicles has a standard deviation of 0.78
mpg.
(t48,0.005=2.68, t48,0.01=2.41, t48,0.025=2.01, t48,0.05=1.68);
(z0.005=2.58, z0.01=2.33, z0.025=1.96, z0.05=1.645)
Construct a 95% confidence interval. Then a 99%
confidence interval
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#11 Redo example #10, but this time, the standard deviation of
the mpg for the 49 cars is 0.83, and there is no credible
information regarding the population standard deviation of
mpg for these vehicles.
Construct a 95% confidence interval. Then a 99%
confidence interval.
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#12 Suppose we have made an interval estimation for the
mean of the population such as: [126.56, 192.41]. If we
realize that the true population mean is 195.7, what should
we conclude?
• The procedure for interval estimation must have been done
•
•
•
•
incorrectly.
We should first standardize the LCL and UCL and then see
if they capture the mean.
The procedure can still be valid, since we allow for a certain
amount of error.
We must use a t distribution instead of a z distribution.
We could never get this result.
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#13 A major news source conducted a poll asking 814 adults to
respond to a series of questions about their feelings toward
the state of affairs within the United States. A total of 562
adults responded “yes” to the question: Do you feel things
are going well in the United States these days?
• A) What is the point estimate of the proportion of the adult
population that feel things are going well in the United
States?
• B) What is the 90% confidence interval for the proportion of
the adult population that feels things are going well in the
United States?
• C) If one wanted to be 95% certain, and have an interval no
wider than 3%, what sample size would be required?
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