Download 7.3,_8.3_-_One_Mean

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
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 (e.g., histogram, Q-Q plot) and
in fact formally testable from the sample data.
If violated (e.g., skewed) or inconclusive (e.g., small
sample size), then a transformation (e.g. logarithm)
or “distribution-free” 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!
Power and Sample Size
• IF the null hypothesis H : μ = μ
0
•
•
•
0
is true,
“Null Distribution”
then we should expect a random sample
mean to lie in its “acceptance region” with
probability 1 – α, the “confidence level.”
That is,
P(Accept H0 | H0 is true) = 1 – α.
Therefore, we should expect a random
sample mean to lie in its “rejection region”
with probability α, the “significance level.”
That is,
P(Reject H0 | H0 is true) = α.
1
/2
/2
H0:  = 0
“Type 1 Error”
Rejection
Region
Acceptance Region
for H0
Rejection
Region
μ0 + zα/2 (σ / n)
Power and Sample Size
“Null Distribution”
“Alternative Distribution”
• IF the null hypothesis H : μ = μ is false,
0
0
then the “power” to correctly reject it in
favor of a particular alternative HA: μ = μ1 is
1–
1
/2
/2

H0:  = 0
Rejection
Region
Acceptance Region
for H0
P(Reject H0 | H0 is false) = 1 – .
Thus,
P(Accept H0 | H0 is false) = .
“Type 2 Error”
HA: μ = μ1
Rejection
Region
μ0 + zα/2 (σ / n)
μ1 – z (σ / n)
Set them equal to each other, and solve for n…
 z /2  z 
| 1  0 |
n
, where  


 

2
Given:
• X ~ N(μ , σ ) Normally-distributed population random variable,
with unknown mean, but known standard deviation
• H0: μ = μ0
Null Hypothesis value
• HA: μ = μ1
Alternative Hypothesis specific value
• 
significance level (or equivalently, confidence level 1 – )
• 1–
power (or equivalently, Type 2 error rate  )
Then the minimum required sample size is:
N(0, 1)
 z /2  z 
| 1  0 |
n
,
where







2
1

z
Example: σ = 1.5 yrs, μ0 = 25.4 yrs,  = .05  z.025 = 1.96
Suppose it is suspected that currently, μ1 = 26 yrs.
Want 90% power of correctly rejecting H0 in favor of HA, if it is false
 1 –  = .90   = .10  z.10 = 1.28
qnorm(.9)
 = |26 – 25.4| / 1.5 = 0.4
2
1.96 + 1.28 
n

So… minimum sample size required is

  65.61
0.4
n  66
Given:
• X ~ N(μ , σ ) Normally-distributed population random variable,
with unknown mean, but known standard deviation
• H0: μ = μ0
Null Hypothesis value
• HA: μ = μ1
Alternative Hypothesis specific value
• 
significance level (or equivalently, confidence level 1 – )
• 1–
power (or equivalently, Type 2 error rate  )
Then the minimum required sample size is:
N(0, 1)
 z /2  z 
| 1  0 |
n
,
where







2
1

z
Example: σ = 1.5 yrs, μ0 = 25.4 yrs,  = .05  z.025 = 1.96
Suppose it is suspected that currently, μ1 = 26 yrs.
Want 95%
90% power of correctly rejecting H0 in favor of HA, if it is false
1.28
.90   = .05
.10  z.05
 1 –  = .95
 = |26 – 25.4| / 1.5 = 0.4
.10 = 1.645
qnorm(.975)
qnorm(.9)
2 2
1.645 
1.96 + 1.28
65.61
81.225
So… minimum sample size required is n  
0.4
0.4  

n  66
82
Given:
• X ~ N(μ , σ ) Normally-distributed population random variable,
with unknown mean, but known standard deviation
• H0: μ = μ0
Null Hypothesis value
• HA: μ = μ1
Alternative Hypothesis specific value
• 
significance level (or equivalently, confidence level 1 – )
• 1–
power (or equivalently, Type 2 error rate  )
Then the minimum required sample size is:
N(0, 1)
 z /2  z 
| 1  0 |
n
,
where







2
1

Example: σ = 1.5 yrs, μ0 = 25.4 yrs,  = .05  z.025 = 1.96
z
Suppose it is suspected that currently, μ1 = 25.7
26 yrs.
yrs.
Want 95% power of correctly rejecting H0 in favor of HA, if it is false
 = |25.7
|26 – –
25.4|
25.4|
/ 1.5
/ 1.5
==
0.4
0.2
 1 –  = .95   = .05  z.05 = 1.645
qnorm(.975)
2
1.96 +
+ 1.645
1.645  2
1.96

 81.225
324.9
So… minimum sample size required is n  



0.2
0.4


n  325
82
Given:
• X ~ N(μ , σ ) Normally-distributed population random variable,
with unknown mean, but known standard deviation
• H0: μ = μ0
Null Hypothesis value
• HA: μ = μ1
Alternative Hypothesis specific value
• 
significance level (or equivalently, confidence level 1 – )
• 1–
power (or equivalently, Type 2 error rate  )
Then the minimum required sample size is:
N(0, 1)
 z /2  z 
| 1  0 |
n
,
where







2
Example: σ = 1.5 yrs, μ0 = 25.4 yrs,  = .05  z.025 = 1.96
1

z
Suppose it is suspected that currently, μ1 = 25.7 yrs.
With n = 400, how much power exists to correctly reject H0 in favor of HA, if it is false?

Power = 1 –  = P Z   z /2   n


 P Z   1.96  0.2 400

 P  Z  2.04 = 0.9793, i.e., 98%
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
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
… is actually a family of distributions, indexed
by the degrees of freedom, labeled tdf.
Z ~ N(0, 1)
t10
tt3
2
t1
William S. Gossett
(1876 - 1937)
As the sample size n gets large, tdf converges to the standard normal
distribution Z ~ N(0, 1). So the T-test is especially useful when n < 30.
… is actually a family of distributions, indexed
by the degrees of freedom, labeled tdf.
Z ~ N(0, 1)
t4
.025
William S. Gossett
(1876 - 1937)
1.96
As the sample size n gets large, tdf converges to the standard normal
distribution Z ~ N(0, 1). So the T-test is especially useful when n < 30.
Lecture Notes Appendix…
or…
qt(.975, 4)
qt(.025, 4, lower.tail = F)
[1] 2.776445
… is actually a family of distributions, indexed
by the degrees of freedom, labeled tdf.
Z ~ N(0, 1)
t4
.025
William S. Gossett
(1876 - 1937)
.025
1.96
2.776
Because any t-distribution has heavier tails than the Z-distribution,
it follows that for the same right-tailed area value, t-score > z-score.
If n is small,
T-score > 2.
… the “T-score" increases (from ≈ 2 to a
max of 12.706 for a 95% confidence level)
as n decreases  larger margin of error
 less power to reject, even if a genuine
statistically significant difference exists!
If n is large,
T-score ≈ 2.
Given:
X = Age at first birth ~ N(μ , σ )
• H0: μ = 25.4 yrs Null Hypothesis
• HA: μ ≠ 25.4 yrs Alternative Hypothesis
Previously…
σ = 1.5 yrs, n = 400, x  25.6 yrs
statistically significant at  = .05
Now suppose that σ is unknown, and n < 30.
Example: n = 16, x  25.9 yrs, s = 1.22 yrs
• standard error (estimate) =
• .025 critical value = t15, .025
s
1.22 yrs

 0.305 yrs
16
n
Lecture Notes Appendix…
Given:
X = Age at first birth ~ N(μ , σ )
• H0: μ = 25.4 yrs Null Hypothesis
• HA: μ ≠ 25.4 yrs Alternative Hypothesis
Previously…
σ = 1.5 yrs, n = 400, x  25.6 yrs
statistically significant at  = .05
Now suppose that σ is unknown, and n < 30.
Example: n = 16, x  25.9 yrs, s = 1.22 yrs
• standard error (estimate) =
s
1.22 yrs

 0.305 yrs
16
n
• .025 critical value = t15, .025 = 2.131
95% Confidence Interval =
(25.9 – 0.65, 25.9 + 0.65) =
(25.25, 26.55) yrs
p-value = 2 P ( X  25.9)
 2P(T15  1.639)
 Test Statistic: T15 
25.9 - 25.4
0.305
95% margin of error
= (2.131)(0.305 yrs)
= 0.65 yrs
Lecture Notes Appendix…
Given:
X = Age at first birth ~ N(μ , σ )
• H0: μ = 25.4 yrs Null Hypothesis
• HA: μ ≠ 25.4 yrs Alternative Hypothesis
Previously…
σ = 1.5 yrs, n = 400, x  25.6 yrs
statistically significant at  = .05
Now suppose that σ is unknown, and n < 30.
Example: n = 16, x  25.9 yrs, s = 1.22 yrs
• standard error (estimate) =
s
1.22 yrs

 0.305 yrs
16
n
• .025 critical value = t15, .025 = 2.131
95% Confidence Interval =
(25.9 – 0.65, 25.9 + 0.65) =
(25.25, 26.55) yrs
p-value = 2 P ( X  25.9)
 2 P(T15  1.639)
= 2 (between .05 and .10)
= between .10 and .20.
95% margin of error
= (2.131)(0.305 yrs)
= 0.65 yrs
The 95% CI does contain the null value μ = 25.4.
The p-value is between .10 and .20, i.e., > .05.
(Note: The R command
2 * pt(1.639, 15, lower.tail = F)
gives the exact p-value as .122.)
Not statistically significant; small n gives low power!
Edited R code:
y = rnorm(16, 0, 1)
z = (y - mean(y)) / sd(y)
x = 25.9 + 1.22*z
Generates a normally-distributed random
sample of 16 age values…
sort(round(x, 1))
24.0 24.2 24.6 24.7 25.1 25.6 25.6 25.7 25.9 26.0 26.4 27.0 27.0 27.2 27.6 28.0
c(mean(x), sd(x))
[1] 25.90
1.22
… with sample mean = 25.9 and sample sd = 1.22.
t.test(x, mu = 25.4)
One Sample t-test
data: x
t = 1.6393, df = 15, p-value = 0.1219
alternative hypothesis: true mean is not equal to 25.4
95 percent confidence interval:
25.24991 26.55009
sample estimates:
mean of x
25.9
See…
http://pages.stat.wisc.edu/~ifischer/Intro_Stat/Lecture_Notes/6__Statistical_Inference/HYPOTHESIS_TESTING_SUMMARY.pdf