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STAT 312


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Chapter 7 - Statistical Intervals Based on a Single Sample
Chapter 8 - Tests of Hypotheses Based on a Single Sample
Chapter 9 - Inferences Based on Two Samples
Chapter 10 - Analysis of Variance
Chapter 11 - Multifactor Analysis of Variance
Chapter 12 - Simple Linear Regression and Correlation
(see section 5.2)
Chapter 13 - Nonlinear and Multiple Regression
Chapter 14 - Goodness-of-Fit Tests and Categorical Data
Analysis
Chapter 15 - Distribution-Free Procedures
Chapter 16 - Quality Control Methods
STAT 312
 Chapter 7 - Statistical Intervals Based on a Single Sample
 Chapter 8 - Tests of Hypotheses Based on a Single Sample
STAT 312
 Chapter 7 - Statistical Intervals Based on a Single Sample
• 7.1 - Basic Properties of Confidence Intervals
• 7.2 - Large-Sample Confidence Intervals for a Population Mean and Proportion
• 7.3 - Intervals Based on a Normal Population Distribution
• 7.4 - Confidence Intervals for the Variance and Standard Deviation of a Normal Pop
 Chapter 8 - Tests of Hypotheses Based on a Single Sample
• 8.1 - Hypotheses and Test Procedures
• 8.2 - Z-Tests for Hypotheses about a Population Mean
• 8.3 - The One-Sample T-Test
• 8.4 - Tests Concerning a Population Proportion
• 8.5 - Further Aspects of Hypothesis Testing
STAT 312
 Chapter 7 - Statistical Intervals Based on a Single Sample
• 7.1 - Basic Properties of Confidence Intervals
• 7.2 - Large-Sample Confidence Intervals for a Population Mean and Proportion
• 7.3 - Intervals Based on a Normal Population Distribution
• 7.4 - Confidence Intervals for the Variance and Standard Deviation of a Normal Pop
 Chapter 8 - Tests of Hypotheses Based on a Single Sample
• 8.1 - Hypotheses and Test Procedures
• 8.2 - Z-Tests for Hypotheses about a Population Mean
• 8.3 - The One-Sample T-Test
• 8.4 - Tests Concerning a Population Proportion
• 8.5 - Further Aspects of Hypothesis Testing
STAT 312
 Chapter 7 - Statistical Intervals Based on a Single Sample
• 7.1 - Basic Properties of Confidence Intervals
• 7.2 - Large-Sample Confidence Intervals for a Population Mean and Proportion
• 7.3 - Intervals Based on a Normal Population Distribution
• 7.4 - Confidence Intervals for the Variance and Standard Deviation of a Normal Pop
 Chapter 8 - Tests of Hypotheses Based on a Single Sample
• 8.1 - Hypotheses and Test Procedures
• 8.2 - Z-Tests for Hypotheses about a Population Mean
• 8.3 - The One-Sample T-Test
• 8.4 - Tests Concerning a Population Proportion
• 8.5 - Further Aspects of Hypothesis Testing
• Right-cick on
image for full .pdf
article
• Links in article
to access datasets
Example: One Mean
“Random Variable”
X = Age (years)
POPULATION
Women in U.S. who
have given birth
Present: Assume that X follows a
“normal
distribution”
in
the
population, with std dev σ = 1.5 yrs,
but unknown mean μ = ? That is, X ~ N(μ, 1.5).
Estimate the parameter value μ.
“Parameter Estimation”
Improve
this
point
estimate of μ to an
“interval estimate” of μ,
via the…
“Sampling Distribution of X ”
standard
deviation
σ = 1.5
This is referred to as
a “point estimate” of
μ from the sample.
mean μ = ???
size n = 400
{x1, x2, x3, x4, … , x400}
FORMULA
mean x = 25.6
Population Distribution of X
Sampling Distribution of X
If X ~ N(μ, σ), then…
 

X ~ N  ,
 , for any sample size n.
n

POPULATION
X = Age of women in U.S.
who have given birth
“standard error”

n
Z
1.5 yrs
 .075 yrs
400
X 
Z
 n
X 

standard deviation
σ = 1.5 yrs
X
μ

X
μ
Sampling Distribution of X
|
X -d μ
d 
X
X +d
X
To obtain an “interval estimate” of μ we
first ask the following general question:
“standard error”
 1.5 yrs

 .075 yrs
n
400
Suppose X is any random sample mean.
Find a “margin of error” (d) so that
there is a 95% probability that the
interval X - d, X + d contains μ.

Pμ-d

P X -d < μ <
X
μ
< X<

X + d  = 0.95
μ + d  = 0.95
+d 
 -d
P
< Z<
 = 0.95
s.e. 
 s.e.
Z
X 
s.e.
Sampling Distribution of X
d = (z.025)(s.e.) = (1.96)(.075 yrs) = 0.147 yrs

Pμ-d
|
X -d μ
d 
X +d
X

μ + d  = 0.95
X
P X - d < μ < X + d = 0.95
“standard error”
 1.5 yrs

 .075 yrs
n
400
< X<
Z
+d 
 -d
P
< Z<
 = 0.95
s.e. 
 s.e.

Pμ-d

μ +0.95
d  = 0.95
P X -d < μ < X +d
X
μ
< X<
standard normal
= 0.95distribution
N(0, 1)
+d 

P
< Z<
 = 0.95
s.e. 
 s.e.
0.025
-d
-z.025
X 
s.e.
X 
Z  0.025 Z
s.e.
+z.025
IMPORTANT DEF’NS and FACTS
d is called the “95% margin of error”
and is equal to the product of the
“.025 critical value” (i.e., z.025 = 1.96)
times the “standard error” (i.e., n ).
The “confidence level” is 95%.
The “significance level” is 5%.
For any random sample mean X,
the “95% confidence interval” is
( X - margin of error, X + margin of error).
It contains μ with probability 95%.
In this example, the 95% CI is
( X - 0.147, X + 0.147).
For instance, if a particular sample
yields x = 25.6 yrs, the 95% CI is
(25.6 – 0.147, 25.6 + 0.147) =
(25.543, 25.747) yrs.
It contains μ with 95% “confidence.”
d = (z.025)(s.e.) = (1.96)(.075 yrs) = 0.147 yrs

Pμ-d
|
X -d μ
d 
X +d
X

μ + d  = 0.95
X
P X - d < μ < X + d = 0.95
< X<
Z
+d 
 -d
P
< Z<
 = 0.95
s.e. 
 s.e.

Pμ-d

μ +0.95
d  = 0.95
P X -d < μ < X +d
< X<
standard normal
= 0.95distribution
N(0, 1)
+d 

P
< Z<
 = 0.95
s.e. 
 s.e.
0.025
-d
-z.025
X 
s.e.
X 
Z  0.025 Z
s.e.
+z.025
IMPORTANT DEF’NS and FACTS
d = (z.025)(s.e.) d= =(1.96)(.075
(zα/2)(s.e.) yrs) = 0.147 yrs
“100(1
– α)%of
margin
d is called the “95%
margin
error”of error”
and is equal to the product of the
“α/2
“.025 critical value” (i.e., zz.025
α/2)= 1.96)
times the “standard error” (i.e., n ).
The “confidence level” is 95%.
1 – α.
The “significance level” is 5%.
α.
For any random sample mean X,
“100(1
– α)% “confidence
the “95%
confidence
interval” interval”
is
( X - margin of error, X + margin of error).
It contains μ with probability 95%.
1 – α.
In this example, the 95% CI is
( X - 0.147, X + 0.147).
For instance, if a particular sample
yields x = 25.6 yrs, the 95% CI is
(25.6 – 0.147, 25.6 + 0.147) =
(25.543, 25.747) yrs.
It contains μ with 95% “confidence.”

Pμ-d
|
X -d μ
d 
X +d
X

1-α
μ + d  = 0.95
X
P X - d < μ < X + d = 0.95
1–α
< X<
Z
+d 
 -d
P
< Z<
1-α
 = 0.95
s.e. 
 s.e.

Pμ-d

–dα = 0.95
μ 1+0.95
P X -d < μ < X +d
< X<
standard normal
= 0.95distribution
N(0, 1)
+d 

P
< Z<
 = 0.95
s.e. 
 s.e.
0.025
α/2-d
-z-z.025
α/2
X 
s.e.
Xα/2

Z  0.025
s.e. Z
+z.025
+z
α/2
IMPORTANT DEF’NS and FACTS
d = (zα/2)(s.e.)
“100(1
– α)%of
margin
d is called the “95%
margin
error”of error”
and is equal to the product of the
“α/2
“.025 critical value” (i.e., zz.025
α/2)= 1.96)
times the “standard error” (i.e., n ).
The “confidence level” is 95%.
1 – α.
The “significance level” is 5%.
α.
For any random sample mean X,
“100(1
– α)% “confidence
the “95%
confidence
interval” interval”
is
( X - margin of error, X + margin of error).
It contains μ with probability 95%.
1 – α.
What happens if we change α?
Example: α = .01,
.05, 1 – α = .99
.10,
.95
.90
-2.575 -1.96 -1.645

Pμ-d
Why not ask for α = 0, i.e., 1 – α = 1?
Because then the critical values → ± ∞.
X +d
X

μ +d = 1-α
X
P X - d < μ < X + d = 0.95
1–α
< X<
Z
+d 
 -d
P
< Z<
 = 1-α
s.e. 
 s.e.

Pμ-d

–dα = 0.95
μ 1+0.95
P X -d < μ < X +d
+1.645 +1.96 +2.575
|
0
|
X -d μ
d 
< X<
standard normal
= 0.95distribution
N(0, 1)
+d 

P
< Z<
 = 0.95
s.e. 
 s.e.
0.025
α/2-d
-z-z.025
α/2
X 
s.e.
Xα/2

Z  0.025
s.e. Z
+z.025
+z
α/2
IMPORTANT DEF’NS and FACTS
95% margin of error
(z.025)(s.e.) = (1.96)(.075 yrs) = 0.147 yrs
In this example, the 95% CI is
( X - 0.147, X + 0.147).
For instance, if a particular sample
yields x = 25.6 yrs, the 95% CI is
(25.6 – 0.147, 25.6 + 0.147) =
(25.543, 25.747) yrs.
It contains μ with 95% “confidence.”
μ
|
 0.147 

P  μ - 0.147 <

μ + 0.147  = 0.95
P X - 0.147 < μ < X + 0.147 = 0.95
?
X1
X<
0.95
X2
0.025
standard normal
distribution
N(0, 1)
0.025
X3
X4
In principle, over the long run, the probability that
a random interval contains μ will approach 95%.
X5
X6
+1.96
-1.96
… etc…
BUT….
Z
IMPORTANT DEF’NS and FACTS
In this example, the 95% CI is
( X - 0.147, X + 0.147).
For instance, if a particular sample
yields x = 25.6 yrs, the 95% CI is
(25.6 – 0.147, 25.6 + 0.147) =
(25.543, 25.747) yrs.
It contains μ with 95% “confidence.”
μ
|
95% margin of error
(z.025)(s.e.) = (1.96)(.075 yrs) = 0.147 yrs

P  μ - 0.147 <

μ + 0.147  = 0.95
P X - 0.147 < μ < X + 0.147 = 0.95
X<
 0.147 
X1
In practice, only a single, fixed
interval is generated from a single
random sample, so technically,
“probability” does not apply.
NOW, let us introduce and
test a specific hypothesis…
0.95
0.025
standard normal
distribution
N(0, 1)
0.025
+1.96
-1.96
In principle, over the long run, the probability that
a random interval contains μ will approach 95%.
BUT….
Z
STAT 312
 Chapter 7 - Statistical Intervals Based on a Single Sample
• 7.1 - Basic Properties of Confidence Intervals
• 7.2 - Large-Sample Confidence Intervals for a Population Mean and Proportion
• 7.3 - Intervals Based on a Normal Population Distribution
• 7.4 - Confidence Intervals for the Variance and Standard Deviation of a Normal Pop
 Chapter 8 - Tests of Hypotheses Based on a Single Sample
• 8.1 - Hypotheses and Test Procedures
• 8.2 - Z-Tests for Hypotheses about a Population Mean
• 8.3 - The One-Sample T-Test
• 8.4 - Tests Concerning a Population Proportion
• 8.5 - Further Aspects of Hypothesis Testing
Study Question:
Has “age at first birth”
of women in the U.S.
changed over time?
POPULATION
Women in U.S. who
have given birth
Statistical Inference
and
Hypothesis Testing
“Random Variable”
X = Age at first birth
“Null Hypothesis”
Year 2010: Suppose we know that
X follows a “normal distribution”
(a.k.a. “bell curve”) in the population.
That is, X ~ N(25.4, 1.5).
standard
deviation
μ < 25.4
σ = 1.5
• public education,
awareness programs
• socioeconomic
conditions, etc.
Present: Is H0: μ = 25.4 still true?
Or,
is
the
“alternative
hypothesis” HA: μ ≠ 25.4 true?
i.e., either μ < 25.4 or μ > 25.4 ?
(2-sided)
μ > 25.4
Does
the
sample
statistic x = 25.6 tend to
support H0, or refute H0
in favor of HA?
mean μ = 25.4
{x1, x2, x3, x4, … , x400}
FORMULA
mean x = 25.6
Objective: Hypothesis Testing… via Confidence Interval
We have now seen:
95% CONFIDENCE INTERVAL FOR µ
 = 25.4 25.543
x = 25.6
25.747
“point estimate” for μ
BASED ON OUR SAMPLE DATA, the true value of μ today is
between 25.543 and 25.747, with 95% “confidence.”
FORMAL CONCLUSIONS:
NOTE THAT THE CONFIDENCE INTERVAL ONLY DEPENDS
ON THE SAMPLE, NOT A SPECIFIC NULL HYPOTHESIS!!!
The 95% confidence interval corresponding to our sample mean does not contain the “null
value” of the population mean, μ = 25.4.
Based on our sample data, we may reject the null hypothesis H0: μ = 25.4 in favor of the
two-sided alternative hypothesis HA: μ ≠ 25.4, at the α = .05 significance level.
INTERPRETATION:
According to the results of this study, there exists a
statistically significant difference between the mean ages at first birth in 2010 (25.4 yrs)
and today, at the 5% significance level. Moreover, the evidence from the sample data
suggests that the population mean age today is older than in 2010, rather than younger.
Objective: Hypothesis Testing… via Confidence Interval
What
if…?
95%
CONFIDENCE
INTERVAL FOR µ
We
have
now seen:
25.053
x = 25.2
“point estimate” for μ
95% CONFIDENCE INTERVAL FOR µ
25.347  = 25.4 25.543
x = 25.6
25.747
“point estimate” for μ
BASED ON OUR SAMPLE DATA, the true value of μ today is
25.543 and 25.347,
25.747, with 95% “confidence.”
between 25.053
FORMAL CONCLUSIONS:
NOTE THAT THE CONFIDENCE INTERVAL ONLY DEPENDS
ON THE SAMPLE, NOT A SPECIFIC NULL HYPOTHESIS!!!
The 95% confidence interval corresponding to our sample mean does not contain the “null
value” of the population mean, μ = 25.4.
Based on our sample data, we may reject the null hypothesis H0: μ = 25.4 in favor of the
two-sided alternative hypothesis HA: μ ≠ 25.4, at the α = .05 significance level.
INTERPRETATION:
According to the results of this study, there exists a
statistically significant difference between the mean ages at first birth in 2010 (25.4 yrs)
and today, at the 5% significance level. Moreover, the evidence from the sample data
suggests that the population mean age today is older
thanthan
in 2010,
rather
thanthan
younger.
younger
in 2010,
rather
older.
Confidence Interval
Objective: Hypothesis Testing… via Acceptance
Region
the null hypothesis H0: μ = 25.4
is indeed true, then…
“Null”
Sampling Distribution of X
95% margin of error
(z.025)(s.e.) = (1.96)(.075 yrs) = 0.147 yrs


P  μ25.253
-- 0.147
μ25.4
+ 0.147
0.95
P  25.4
0.147 <<< XXX <<< 25.547
+ 0.147
 = =0.95
 = 0.95
P X - 0.147 < μ < X + 0.147 = 0.95
“standard error”
 1.5 yrs

 .075 yrs
n
400
0.95
… and out
here…
… we would expect a
random sample mean
X to lie in here, with
95% probability…
0.025
…with 5%
probability.
0.025
95% ACCEPTANCE REGION FOR H0
X
|
X
μ
25.4
25.253
μ = 25.4
25.547
Objective: Hypothesis Testing… via Acceptance Region
We have now seen:
95% ACCEPTANCE REGION FOR H0
25.253
 = 25.4
25.547 x = 25.6
IF H0 is true, then we would expect a random sample mean x
to lie between 25.253 and 25.547, with 95% probability.
FORMAL CONCLUSIONS:
NOTE THAT THE ACCEPTANCE REGION ONLY DEPENDS
ON THE NULL HYPOTHESIS, NOT ON THE SAMPLE!!!
The 95% acceptance region for the null hypothesis does not contain the sample mean of
x = 25.6.
Based on our sample data, we may reject the null hypothesis H0: μ = 25.4 in favor of the
two-sided alternative hypothesis HA: μ ≠ 25.4, at the α = .05 significance level.
INTERPRETATION:
According to the results of this study, there exists a
statistically significant difference between the mean ages at first birth in 2010 (25.4 yrs)
and today, at the 5% significance level. Moreover, the evidence from the sample data
suggests that the population mean age today is older than in 2010, rather than younger.
Approximately 95% of the sample
mean values are contained between
and
  2 n
  2 n
Sample 1
Sample 2
Sample 3
Sample 4
Sample 5
etc…


n is called the 95% margin of error
X
x1


2
x2
x3

x4

x5
Approximately 95% of the sample
mean values are contained between
and
  2 n
  2 n
Sample 1
Sample 2
Sample 3
Sample 4
Sample 5


n is called the 95% margin of error
X
x1


2
x2
x3

x4

x5
Approximately
Approximately95%
95%of
ofthe
theintervals
sample
mean
contained
x  2values
 n arefrom
x  2between
 n
to
and
  2 n
  2 n
contain , and approx 5% do not.
Sample 1
Sample 2
Sample 3
Sample 4
Sample 5


n is called the 95% margin of error
X
x1


2
x2
x3

x4

x5
Objective:
Objective: Hypothesis
Hypothesis Testing…
Testing… via
via Acceptance
“p-value” - measures
Region the strength of the rejection
the null hypothesis H0: μ = 25.4
is indeed true, then…
what is the probability of obtaining
a random sample mean that is
as, or more, extreme than the one
actually obtained?
i.e., 0.2 yrs OR MORE away from
μ = 25.4, ON EITHER SIDE (since the
alternative hypothesis is 2-sided)?
25.6
X  25.4
ZZ
.075
s.e.
= 2(0.00383)
= 0.0077
> 1-pnorm(2.6667)
0.95
> pnorm(-2.6667)
> pnorm(25.2, 25.4, .075)
[1] 0.003830001
0.025
0.00383


= P  X  25.2   P  X  25.6 
= 2 P  X  25.6  = 2 P  Z  2.667
= P X  25.2 U X  25.6
0.025
95% ACCEPTANCE REGION FOR H0
0.00383
|
25.2
25.253
μ = 25.4
25.547
x = 25.6
- measures
Objective:
Objective 2:Hypothesis
Hypothesis
Testing…
Testing…
via
via
“p-value”
Acceptance
Region the strength of the rejection
the null hypothesis H0: μ = 25.4
is indeed true, then…
what is the probability of obtaining
a random sample mean that is
as, or more, extreme than the one
actually obtained?
i.e., 0.2 yrs OR MORE away from
μ = 25.4, ON EITHER SIDE (since the
alternative hypothesis is 2-sided)?


= P  X  25.2   P  X  25.6 
= 2 P  X  25.6  = 2 P  Z  2.667
= P X  25.2 U X  25.6
25.6
X  25.4
ZZ
.075
s.e.
= 2(0.00383)
= 0.0077
10.95
–α
α/2
0.025
0.00383
α0.025
/2
100(1 – α)% ACCEPTANCE REGION FOR H0
95% ACCEPTANCE REGION FOR H0
0.00383
|
25.2
-zα/2
25.253
μ = 25.4
+zα/2
25.547
x = 25.6
- measures
Objective:
Objective 2:Hypothesis
Hypothesis
Testing…
Testing…
via
via
“p-value”
Acceptance
Region the strength of the rejection
the null hypothesis H0: μ = 25.4
is indeed true, then…
what is the probability of obtaining
a random sample mean that is
as, or more, extreme than the one
actually obtained?
i.e., 0.2 yrs OR MORE away from
μ = 25.4, ON EITHER SIDE (since the
alternative hypothesis is 2-sided)?


= P  X  25.2   P  X  25.6 
= 2 P  X  25.6  = 2 P  Z  2.667
= P X  25.2 U X  25.6
25.6
X  25.4
ZZ
.075
s.e.
= 2(0.00383)
= 0.0077
10.95
–α
Very informally, the p-value of a sample is the probability (hence a
α / 2number between 0 and 1) that it “agrees” with the null hypothesis.α0.025
/2
0.025
Hence a very small p-value indicates strong evidence against the
0.00383 null hypothesis.
0.00383
The smaller
the p-value,REGION
the stronger
the evidence,
100(1 – α)%
ACCEPTANCE
FOR
H0
95%
ACCEPTANCE
REGION
and the more
“statistically
significant”
the findingFOR
(e.g., pH<0 .0001).
|
25.2
-zα/2
25.253
μ = 25.4
+zα/2
25.547
x = 25.6
If p-value < ,
then reject H0;
significance!
 = .05
... But interpret it
correctly!
~ Summary of Hypothesis Testing for One Mean ~
Assume the population random variable is normally distributed, i.e., X  N(μ, σ).
NULL HYPOTHESIS H0: μ = μ0 (“null value”)
ALTERNATIVE HYPOTHESIS HA: μ  μ0
i.e., either μ < μ0 or μ > μ0 (“two-sided”)
 CONFIDENCE INTERVAL
Test null hypothesis at
significance level α.
zα/2
σ
n
Compute the sample mean x.
Compute the 100(1 – α)% “margin of error” = (critical value)(standard error)
Then the 100(1 – α)% CI =
(x - margin of error, x + margin of error).
Formal Conclusion:
Reject null hypothesis at level α, if CI does not contain μ0. Statistical significance!
Otherwise, retain it.
~ Summary of Hypothesis Testing for One Mean ~
Assume the population random variable is normally distributed, i.e., X  N(μ, σ).
NULL HYPOTHESIS H0: μ = μ0 (“null value”)
ALTERNATIVE HYPOTHESIS HA: μ  μ0
i.e., either μ < μ0 or μ > μ0 (“two-sided”)
 ACCEPTANCE REGION
Test null hypothesis at
significance level α.
zα/2
σ
n
Compute the sample mean x.
Compute the 100(1 – α)% “margin of error” = (critical value)(standard error)
Then the 100(1 – α)% AR =
(μ0 - margin of error, μ0 + margin of error).
Formal Conclusion:
Reject null hypothesis at level α, if AR does not contain x. Statistical significance!
Otherwise, retain it.
~ Summary of Hypothesis Testing for One Mean ~
Assume the population random variable is normally distributed, i.e., X  N(μ, σ).
NULL HYPOTHESIS H0: μ = μ0 (“null value”)
ALTERNATIVE HYPOTHESIS HA: μ  μ0
i.e., either μ < μ0 or μ > μ0 (“two-sided”)
 p-value
Compute the sample mean x.
x - μ0  (shown)
Compute the z-score

σ n or 
Test null hypothesis at
significance level α.
Z ~ N(0, 1)
If +, then the p-value = 2 P(Z ≥ z-score ).
If –, then the p-value = 2 P(Z ≤ z-score ).
z-score
Formal Conclusion:
Reject null hypothesis if p < α. Statistical significance! Otherwise, retain it.
Remember: “The smaller the p-value, the stronger the rejection, and the more
statistically significant the result.”
Objective: Hypothesis Testing… 1-sided tests
2-sided test
H0: μ = 25.4
HA: μ  25.4


= P  X  25.2   P  X  25.6 
= 2 P  X  25.6  = 2 P  Z  2.667
p-value = P X  25.2 U X  25.6
In this case,  = .05 is split evenly
between the two tails, left and right.
= 2(.00383) = .0077
1-sided tests
“Right-tailed”
H0: μ  25.4
HA: μ > 25.4
Here, all of  = .05
is in the right tail.
0.95
0.025
.00383
0.025
95% ACCEPTANCE REGION FOR H0
.00383
|
25.2
25.253
μ = 25.4
25.547
x = 25.6
Objective: Hypothesis Testing… 1-sided tests
2-sided test
H0: μ = 25.4
HA: μ  25.4


= P  X  25.2   P  X  25.6 
= 2 P  X  25.6  = 2 P  Z  2.667
p-value = P X  25.2 U X  25.6
In this case,  = .05 is split evenly
between the two tails, left and right.
= 2(.00383) = .00383
.0077
1-sided tests
“Right-tailed”
H0: μ  25.4
HA: μ > 25.4
Here, all of  = .05
is in the right tail.
0.95
0.05
0.025
0.025
.00383
.00383
95% ACCEPTANCE
95% ACCEPTANCE
REGION
REGION
FOR HFOR
H0
0
|
25.2
25.253
μ = 25.4
25.547
?
x = 25.6
Objective: Hypothesis Testing… 1-sided tests
2-sided test
H0: μ = 25.4
HA: μ  25.4


25.2 
= P  X  25.2   P  X  25.6
25.2  = P
2P
= 2 P  X  25.6
 Z Z 2.667
p-value = P X  25.2 U X  25.6
25.2
In this case,  = .05 is split evenly
between the two tails, left and right.
1-sided tests
“Right-tailed”
H0: μ  25.4
HA: μ > 25.4
“Left-tailed”
H0: μ  25.4
HA: μ < 25.4
Here, all of  = .05
is in the right tail.
Here, all of  = .05
is in the left tail.
= 2(.00383)
(.99617) = .99617
.00383
0.95
0.05
95% ACCEPTANCE REGION FOR H0
|
x = 25.2
μ = 25.4
?
x = 25.6
STATBOT 312
Subject: basic calculation of p-values for ztest
CALCULATE… from H
00
Test Statistic
x 
“z-score” = x  00

HA: μ < μ0
nn
HA: μ > μ0
HA: μ ≠ μ0?
1 – table entry
table entry
sign of
z-score?
–
2 × table entry
+
22 ×× (1
(1 –– table
table entry)
entry)
Loose Ends…
• Normality check?
•  unknown?
• n=?
Given:
• X ~ N(μ , σ ) Normally-distributed population random variable,
with unknown mean, but known standard deviation
• H0: μ = μ0
Null Hypothesis
• HA: μ ≠ μ0
Alternative Hypothesis (2-sided)
• 
significance level (or equivalently, confidence level 1 – )
• n
sample size
From this, we obtain…
s
n
x1, x2,…, xn
“standard error” s.e.
(estimate)
x
s
sample mean
sample standard deviation
…with which to test the null hypothesis (via CI, AR, p-value).
In practice however, it is far more common that the
true population standard deviation σ is unknown.
So we must estimate it from the sample!
Recall that
s2 

2
(
x

x
)
 i
n 1
SS
df
See Ch. 4.6 (311)
Given:
• X ~ N(μ , σ ) Normally-distributed population random variable,
with unknown mean, but known standard deviation
• H0: μ = μ0
Null Hypothesis
• HA: μ ≠ μ0
Alternative Hypothesis (2-sided)
• 
significance level (or equivalently, confidence level 1 – )
• n
sample size
This introduces additional
variability from one sample to
another… PROBLEM???
From this, we obtain…
s
n
x1, x2,…, xn
Not if n is “large”…say,  30.
“standard error” s.e.
(estimate)
x
s
But what if n < 30? T-test!
sample mean
sample standard deviation
…with which to test the null hypothesis (via CI, AR, p-value).
In practice however, it is far more common that the
true population standard deviation σ is unknown.
So we must estimate it from the sample!
Recall that
s2 

2
(
x

x
)
 i
n 1
SS
df
X ~ N ( ,  )
How do we check that this assumption is
reasonable, when all we have is a sample?
{ x1 , x2 , x3 ,…, x24 }
And what do we do if it’s not, or we can’t tell?
Z ~ N(0, 1)
IF our data approximates a bell
curve, then its quantiles should
“line up” with those of N(0, 1).
Z
X 

 X    Z
X is a linear function of Z
X ~ N ( ,  )
How do we check that this assumption is
reasonable, when all we have is a sample?
{ x1 , x2 , x3 ,…, x24 }
And what do we do if it’s not, or we can’t tell?
Sample quantiles
Z ~ N(0, 1)
IF our data approximates a bell
curve, then its quantiles should
“line up” with those of N(0, 1).
Z
X 

 X    Z
X is a linear function of Z
• Q-Q plot
• Normal scores plot
• Normal probability plot
X ~ N ( ,  )
How do we check that this assumption is
reasonable, when all we have is a sample?
And what do we do if it’s not, or we can’t tell?
IF our data approximates a bell
curve, then its quantiles should
“line up” with those of N(0, 1).
Z
X 

 X    Z
X is a linear function of Z
• Q-Q plot
• Normal scores plot
• Normal probability plot
qqnorm(mysample)
qqline(mysample)
(R uses a slight variation
to generate quantiles…)
X ~ N ( ,  )
How do we check that this assumption is
reasonable, when all we have is a sample?
And what do we do if it’s not, or we can’t tell?
IF our data approximates a bell
curve, then its quantiles should
“line up” with those of N(0, 1).
Z
X 

 X    Z
X is a linear function of Z
• Q-Q plot
• Normal scores plot
• Normal probability plot
qqnorm(mysample)
qqline(mysample)
(R uses a slight variation
to generate quantiles…)
Formal statistical tests exist; see notes.
Method can be extended to other models
X ~ N ( ,  )
How do we check that this assumption is
reasonable, when all we have is a sample?
And what do we do if it’s not, or we can’t tell?
 Use a mathematical “transformation” of the data (e.g., log, square root,…).
x = rchisq(1000, 15)
hist(x)
y = log(x)
hist(y)
X is said to be “log-normal.”
How do we check that this assumption is
reasonable, when all we have is a sample?
X ~ N ( ,  )
And what do we do if it’s not, or we can’t tell?
 Use a mathematical “transformation” of the data (e.g., log, square root,…).
qqnorm(x, pch = 19, cex = .5)
qqline(x)
qqnorm(y, pch = 19, cex = .5)
qqline(y)
How do we check that this assumption is
reasonable, when all we have is a sample?
X ~ N ( ,  )
And what do we do if it’s not, or we can’t tell?
 Use a mathematical “transformation” of the data (e.g., log, square root,…).
“Cauchy distribution”
1
1
f ( x) 
 1  x2
X ~ N ( ,  )
How do we check that this assumption is
reasonable, when all we have is a sample?
And what do we do if it’s not, or we can’t tell?
 Use a mathematical “transformation” of the data (e.g., log, square root,…).
So then what????
POPULATION
Study Question:
Has “Mean (i.e., average) Age at
First Birth” of women in the U.S.
changed since 2010 (25.4 yrs old)?
“Statistical Inference”
via… “Hypothesis Testing”
Present Day: Assume “Mean Age at
First Birth” follows a normal distribution
(i.e., “bell curve”) in the population.
Population
Distribution

X
H0: pop mean age  = 25.4
(i.e., no change since 2010)
x4
x1
x2
x3
x5
… etc…
x400
The reasonableness of the normality
assumption is empirically verifiable, and in
fact formally testable from the sample data.
If violated (e.g., skewed) or inconclusive
(e.g., small sample size), then “distributionfree” nonparametric tests should be used
instead…
Examples: Sign Test, Wilcoxon Signed Rank
Test (= Mann-Whitney U Test)
POPULATION
Study Question:
Has “Mean (i.e., average) Age at
First Birth” of women in the U.S.
changed since 2010 (25.4 yrs old)?
“Statistical Inference”
via… “Hypothesis Testing”
Present Day: Assume “Mean Age at
First Birth” follows a normal distribution
(i.e., “bell curve”) in the population.
Population
Distribution

X
H0: pop mean age  = 25.4
(i.e., no change since 2010)
x4
x1
x2
x3
x5
… etc…
x400
Sample size n partially depends on the
power of the test, i.e., the desired probability
of correctly rejecting a false null hypothesis
(80% or more). Coming up next!