Inference for Distributions
Marc H. Mehlman
[email protected]
University of New Haven
Based on Rare Event Rule: “rare events happen – but not to me”.
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Inference for Distributions
1 / 42
Table of Contents
1
t Distribution
2
CI for µ: σ unknown
3
t-test: Mean (σ unknown)
4
t test: Matched Pairs
5
Two Sample z–Test: Means (σX and σY known)
6
Two Sample t–test: Means (σX and σY unknown)
7
Pooled Two Sample t–test: Means (σX = σY unknown)
8
Two Sample F –test: Variance
9
Chapter #7 R Assignment
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t Distribution
t Distribution
t Distribution.
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t Distribution
If X1 , · · · , Xn is a normal random sample
σ
√
X̄ ∼ N µ,
⇒
n
X̄ − µ
√ ∼ N(0, 1).
σ/ n
If the random sample is not normal, but n ≥ 30, the above is also true
(approximately) by the CLT.
However, typically one does not know what σ equals so one is tempted to
use s instead of σ. This gives
Definition (Student t Distribution)
If a random sample is normal or the sample size is ≥ 30
X̄ − µ
√ ∼ t(n − 1),
S/ n
where t is the Student t Distribution with n − 1 degrees of freedom.
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t Distribution
The t Distributions
When comparing the density curves of the standard Normal distribution
and t distributions, several facts are apparent:
 The density curves of the t distributions are
similar in shape to the standard Normal
curve.
 The spread of the t distributions is a bit
greater than that of the standard Normal
distribution.
 The t distributions have more probability in
the tails and less in the center than does
the standard Normal.
 As the degrees of freedom increase, the t
density curve approaches the standard
Normal curve ever more closely.
We can use Table D in the back of the book to determine critical values t* for t
distributions with different degrees of freedom.
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t Distribution
Robustness
The t procedures are exactly correct when the population is exactly
Normal. This is rare.
The t procedures are robust to small deviations from Normality, but:

The sample must be a random sample from the population.

Outliers and skewness strongly influence the mean and therefore the t
procedures. Their impact diminishes as the sample size gets larger
because of the Central Limit Theorem.
As a guideline:



When n < 15, the data must be close to Normal and without outliers.
When 15 > n > 40, mild skewness is acceptable, but not outliers.
When n > 40, the t statistic will be valid even with strong skewness.
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CI for µ: σ unknown
CI for µ: σ unknown
Confidence intervals for µ when σ is
unknown.
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CI for µ: σ unknown
Confidence intervals
A confidence interval is a range of values that contains the true
population parameter with probability (confidence level) C.
We have a set of data from a population with both µ and σ unknown. We
use x̅ to estimate µ, and s to estimate σ, using a t distribution (df n − 1).

C is the area between −t* and t*.

We find t* in the line of Table.

The margin of error m is:
m=t∗s/ √ n
C
m
−t*
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m
t*
8 / 42
CI for µ: σ unknown
Definition (standard error)
def S
The standard error of the sample mean for σ unknown is SEx̄ = √ .
n
Theorem (CI for µ, σ unknown)
Assume n ≥ 30 or the population is normal. Let
s
def
margin of error = m = t ? (n − 1) ? √ = t ? (n − 1) ? SEx̄ .
n
Then the confidence interval is x̄ ± m.
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CI for µ: σ unknown
Example
The meters of total rainfall for Jupa, Beliana in the first decade of each of the last sixteen centuries is given below:
3.790155
4.116425
3.628361
2.771781
3.989105
6.243354
5.124677
5.040272
4.227491
6.821760
3.183561
6.170435
5.286963
5.439190
3.323666
5.206938
Find a 90% confidence interval for the mean rainfall per first decade of each century.
Solution: The following normal quartile plot indicates
Normal Q−Q Plot
that the data comes from a normal (or at least almost normal) distribution. Since
σ is unknown and the distribution is close to normal with no outliers, R gives us
●
●
6
●
> mean(mdat)-qt(0.95,15)*sd(mdat)/sqrt(16)
[1] 4.124238
> mean(mdat)+qt(0.95,15)*sd(mdat)/sqrt(16)
[1] 5.171278
5
Sample Quantiles
●
●
●
●
●
●
4
●
●
Thus a 90% confidence interval is (4.124238, 5.171278). Notice
●
●
●
3
●
●
−2
−1
0
1
2
Theoretical Quantiles
Marc Mehlman (University of New Haven)
> t.test(mdat,mu=4.5,conf.level=0.90)
One Sample t-test
data: mdat
t = 0.4948, df = 15, p-value = 0.6279
alternative hypothesis: true mean is not equal to 4.5
90 percent confidence interval:
4.124238 5.171278
sample estimates:
mean of x
4.647758
Inference for Distributions
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CI for µ: σ unknown
Example
The meters of total rainfall for Jupa, Beliana in the first decade of each of the last sixteen centuries is given below:
3.790155
4.116425
3.628361
2.771781
3.989105
6.243354
5.124677
5.040272
4.227491
6.821760
3.183561
6.170435
5.286963
5.439190
3.323666
5.206938
Find a 90% confidence interval for the mean rainfall per first decade of each century.
Solution: The following normal quartile plot indicates
Normal Q−Q Plot
that the data comes from a normal (or at least almost normal) distribution. Since
σ is unknown and the distribution is close to normal with no outliers, R gives us
●
●
6
●
> mean(mdat)-qt(0.95,15)*sd(mdat)/sqrt(16)
[1] 4.124238
> mean(mdat)+qt(0.95,15)*sd(mdat)/sqrt(16)
[1] 5.171278
5
Sample Quantiles
●
●
●
●
●
●
4
●
●
Thus a 90% confidence interval is (4.124238, 5.171278). Notice
●
●
●
3
●
●
−2
−1
0
1
2
Theoretical Quantiles
Marc Mehlman (University of New Haven)
> t.test(mdat,mu=4.5,conf.level=0.90)
One Sample t-test
data: mdat
t = 0.4948, df = 15, p-value = 0.6279
alternative hypothesis: true mean is not equal to 4.5
90 percent confidence interval:
4.124238 5.171278
sample estimates:
mean of x
4.647758
Inference for Distributions
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t-test: Mean (σ unknown)
t-test: Mean (σ unknown)
t-test: Mean (σ unknown)
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t-test: Mean (σ unknown)
Theorem (t–test for the Mean (σ unknown))
Given a random sample, X1 , · · · , Xn , where either
the random sample was sampled from a normal population or
the sample size n ≥ 30,
let the test statistic be
T =
X̄ − µ0
√ .
S/ n
Then T ∼ t(n − 1) under H0 : µ = µ0 . The p–value of a test of H0
1
versus H1 : µX > µ0 is P(T ≥ t).
2
versus H2 : µX < µ0 is P(T ≤ t).
3
versus H3 : µX 6= µ0 is 2P(T ≥ |t|).
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t-test: Mean (σ unknown)
Example
The number of hours of Sesame Street that American 4 year olds watch each year is assumed to
2
distributed normally. Twenty-five 4 year olds are randomly sampled and one finds x̄ = 125 and sX
= 100. What is the p–value
for a test of
H0 : µX = 120
versus
HA : µX > 120?
Solution: The population is normally distributed so the test statistic is
t =
125 − 120
= 2.5
√
10/ 25
comes from t(24) under H0 . Thus the p–value is
> 1-pt(2.5,24)
[1] 0.009827088
It seems unlikely that the average number of Sesame Street watching hours is 120 or less.
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t-test: Mean (σ unknown)
Example
The number of hours of Sesame Street that American 4 year olds watch each year is assumed to
2
distributed normally. Twenty-five 4 year olds are randomly sampled and one finds x̄ = 125 and sX
= 100. What is the p–value
for a test of
H0 : µX = 120
versus
HA : µX > 120?
Solution: The population is normally distributed so the test statistic is
t =
125 − 120
= 2.5
√
10/ 25
comes from t(24) under H0 . Thus the p–value is
> 1-pt(2.5,24)
[1] 0.009827088
It seems unlikely that the average number of Sesame Street watching hours is 120 or less.
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t-test: Mean (σ unknown)
Example
> set.seed(1234)
> ntab=rnorm(99,5,1)
> t.test(ntab,mu=4.6)
One Sample t-test
data: ntab
t = 2.2299, df = 98, p-value = 0.02804
alternative hypothesis: true mean is not equal to 4.6
95 percent confidence interval:
4.624239 5.016220
sample estimates:
mean of x
4.820229
> t.test(ntab,mu=4.6,alternative="greater")
One Sample t-test
data: ntab
t = 2.2299, df = 98, p-value = 0.01402
alternative hypothesis: true mean is greater than 4.6
95 percent confidence interval:
4.65623
Inf
sample estimates:
mean of x
4.820229
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t test: Matched Pairs
t test: Matched Pairs
Matched Pairs
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t test: Matched Pairs
Matched pairs t procedures
Sometimes we want to compare treatments or conditions at the
individual level. The data sets produced this way are not independent.
The individuals in one sample are related to those in the other sample.

Pre-test and post-test studies look at data collected on the same sample
elements before and after some experiment is performed.

Twin studies often try to sort out the influence of genetic factors by
comparing a variable between sets of twins.

Using people matched for age, sex, and education in social studies
allows us to cancel out the effect of these potential lurking variables.
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t test: Matched Pairs
Theorem (Matched Pairs)
Let (X1 , Y1 ), · · · , (Xn , Yn ) be a bivariate random sample and define
def
Dj = Xj − Yj . Assume n ≥ 30, or the Dj ’s are normal (or pretty much
so). Let
H0 : µD = d.
The test statistic is
T =
D̄ − d
√ ∼ t(n − 1)
sD / n
for H0 .
A (1 − α)100% CI for µD is
sD
d¯ ± t ? (n − 1) ? √ .
n
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t test: Matched Pairs
Example (Does lack of caffeine increase depression?)
Randomly selected caffeine-dependent individuals were deprived of all caffeine–rich foods and assigned to receive daily pills. At
one time the pills contained caffeine and, at another time they were a placebo. Depression was assessed quantitatively (higher
scores represent greater depression). Find the p–value of the test
H0 : µdiff = 0
Subject
Caffeine
Placebo
Difference
1
5
16
11
2
5
23
18
3
4
5
1
versus
4
3
7
4
5
8
14
6
HA : µdiff ≥ 0
6
5
24
19
7
0
6
6
8
0
3
3
9
2
15
13
10
11
12
1
11
1
0
-1
Solution:
The normal quartile plot to the left indicates
that the data comes from a normal (or at least almost normal) distribution.
Normal Q−Q Plot
●
●
15
> cdat=c(5,5,4,3,8,5,0,0,2,11,1)
> pdat=c(16,23,5,7,14,24,6,3,15,12,0)
> tstat=(mean(pdat-cdat)-0)/(sd(pdat-cdat)/sqrt(11))
> tstat
[1] 3.530426
> 1-pt(tstat,10)
[1] 0.002721472
●
10
Sample Quantiles
●
●
5
●
●
●
●
0
●
●
−1.5
−1.0
−0.5
0.0
0.5
1.0
1.5
It is very likely that caffeine deprivation causes an increase in rates of depression.
Theoretical Quantiles
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t test: Matched Pairs
Example (Does lack of caffeine increase depression?)
Randomly selected caffeine-dependent individuals were deprived of all caffeine–rich foods and assigned to receive daily pills. At
one time the pills contained caffeine and, at another time they were a placebo. Depression was assessed quantitatively (higher
scores represent greater depression). Find the p–value of the test
H0 : µdiff = 0
Subject
Caffeine
Placebo
Difference
1
5
16
11
2
5
23
18
3
4
5
1
versus
4
3
7
4
5
8
14
6
HA : µdiff ≥ 0
6
5
24
19
7
0
6
6
8
0
3
3
9
2
15
13
10
11
12
1
11
1
0
-1
Solution:
The normal quartile plot to the left indicates
that the data comes from a normal (or at least almost normal) distribution.
Normal Q−Q Plot
●
●
15
> cdat=c(5,5,4,3,8,5,0,0,2,11,1)
> pdat=c(16,23,5,7,14,24,6,3,15,12,0)
> tstat=(mean(pdat-cdat)-0)/(sd(pdat-cdat)/sqrt(11))
> tstat
[1] 3.530426
> 1-pt(tstat,10)
[1] 0.002721472
●
10
Sample Quantiles
●
●
5
●
●
●
●
0
●
●
−1.5
−1.0
−0.5
0.0
0.5
1.0
1.5
It is very likely that caffeine deprivation causes an increase in rates of depression.
Theoretical Quantiles
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t test: Matched Pairs
Example (cont.)
Using R, one can find the p–value two other ways too.
> t.test(pdat-cdat,mu=0,alternative="greater")
One Sample t-test
data: pdat - cdat
t = 3.5304, df = 10, p-value = 0.002721
alternative hypothesis: true mean is greater than 0
95 percent confidence interval:
3.583269
Inf
sample estimates:
mean of x
7.363636
> t.test(pdat,cdat,paired=TRUE,mu=0,alternative="greater")
Paired t-test
data: pdat and cdat
t = 3.5304, df = 10, p-value = 0.002721
alternative hypothesis: true difference in means is greater than 0
95 percent confidence interval:
3.583269
Inf
sample estimates:
mean of the differences
7.363636
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t test: Matched Pairs
Without any assumptions of population distribution types or sample sizes one has:
Theorem (Sign Test for Matched Pairs)
Ignore pairs with differences of zero and let n be the count of the remaining pairs. The test statistic is the count, X , of pairs
with a positive difference. To find the p–value of a test note that X ∼ BIN(n, 1/2) under the hypotheses H0 the median of
the two matched pairs are the same.
Example
In the previous example concerning caffeine deprivation, ten out of eleven subjects experienced an increase in depression.
Consider the signed test for matched pairs,
H0
:
median depression with caffeine = median depression with placebo
HA
:
median depression with caffeine < median depression with placebo.
versus
Noticing that x = 10, the p–value is
P(X ≥ 10) = P(X = 10) + P(X = 11) =
10 11 1
11
1 1
11
1
1 0
1−
+
1−
10
2
2
11
2
2
or, using R,
> 1-pbinom(9,11,1/2)
[1] 0.005859375
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Two Sample z–Test: Means (σX and σY known)
Two Sample z–Test: Means (σX and σY known)
Two Sample z–Test: Means
(σX and σY known)
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Two Sample z–Test: Means (σX and σY known)
Theorem (Two Sample z–Test: Means (σX and σY known))
Let X1 , · · · , XnX and Y1 , · · · , YnY be independent random samples where
each random sample is either
sampled from a normal population or
has sample size ≥ 30,
Let
H0 : µ X = µ Y
with test statistic
X̄ − Ȳ
Z=q 2
∼ N(0, 1)
σX
σY2
nX + nY
for H0 .
A (1 − α)100% confidence interval for µX − µY is
s
σX2
σ2
x̄ − ȳ ± z ?
+ Y.
nX
nY
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Two Sample z–Test: Means (σX and σY known)
Example
The range of Brand X wireless routers is distributed as N(µX , 17.7) and the
range of Brand Y wireless routers is distributed as N(µY , 15.3). Suppose ten Brand X routers
are sampled and X̄ = 131.2 meters. Further suppose that nine Brand Y routers are sampled and
Ȳ = 141.5 meters. Find the p–value of the test of H0 : µX = µY versus HA : µX < µY .
Solution:
131.2 − 141.5
z= q
= −1.360229.
2
17.72
+ 15.3
10
9
which comes from N(0, 1) under H0 . The p–value for the test is
> zstat=(131.2-141.5)/sqrt((17.7^2/10)+(15.3^2/9))
> zstat
[1] -1.360229
> pnorm(zstat,0,1)
[1] 0.08687867
One must accept H0 at the 0.05 significance level and reject H0 at the 0.10 significance level.
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Two Sample z–Test: Means (σX and σY known)
Example
The range of Brand X wireless routers is distributed as N(µX , 17.7) and the
range of Brand Y wireless routers is distributed as N(µY , 15.3). Suppose ten Brand X routers
are sampled and X̄ = 131.2 meters. Further suppose that nine Brand Y routers are sampled and
Ȳ = 141.5 meters. Find the p–value of the test of H0 : µX = µY versus HA : µX < µY .
Solution:
131.2 − 141.5
z= q
= −1.360229.
2
17.72
+ 15.3
10
9
which comes from N(0, 1) under H0 . The p–value for the test is
> zstat=(131.2-141.5)/sqrt((17.7^2/10)+(15.3^2/9))
> zstat
[1] -1.360229
> pnorm(zstat,0,1)
[1] 0.08687867
One must accept H0 at the 0.05 significance level and reject H0 at the 0.10 significance level.
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Two Sample t–test: Means (σX and σY unknown)
Two Sample t–test: Means (σX and σY unknown)
Two Sample t–test: Means
(σX and σY unknown)
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Two Sample t–test: Means (σX and σY unknown)
Theorem (Conservative Two Sample t-test: Means (σX and σY unknown))
Let X1 , · · · , XnX and Y1 , · · · , YnY be independent random samples. Assume either
the random samples were sampled from a normal population or
the sum of the sample sizes, nX + nY , are ≥ 30.
Let
H 0 : µX = µY
The (conservative) test statistic is
X̄ − Ȳ
T = q 2
∼ t(min(nX − 1, nY − 1))
SX
S2
+ nYY
nX
for H0 .
A (1 − α)100% confidence interval for µX − µY is
s
?
x̄ − ȳ ± t (min(nX − 1, nY − 1))
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sX2
s2
+ Y.
nX
nY
25 / 42
Two Sample t–test: Means (σX and σY unknown)
Theorem (Welch’s t Test)
Let X1 , · · · , XnX and Y1 , · · · , YnY be independent random samples. Assume either
the random samples were sampled from a normal population or
the each the sample sizes are ≥ 5.
Let
H0 : µX = µY
The test statistic is
T = s
X̄ − Ȳ
S2
X
nX
∼ t (df)
2
S
+ nY
Y
where
2
sX
nX
df =
1
nX −1
s2
X
nX
2
s
+ nY
!2
Y
!2
+ n 1−1
Y
s2
Y
nY
!2
for H0 . When using a table, let the degrees of freedom be largest integer that is less than or equal to df for a slightly more
conservative test (statistical software can handle non–integer degree of freedoms).
A (1 − α)100% confidence interval for µX − µY is
v
u 2
us
s2
x̄ − ȳ ± t (df)t X + Y .
nX
nY
?
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Two Sample t–test: Means (σX and σY unknown)
Robustness:
Two sample t–tests tend to be more robust than One sample t–test. This
is especially true when either the
two sample sizes are equal or
the two populations have the similarly shaped distributions.
When either (or both) of these conditions hold, the guidelines for using the
Conservative Two Sample t–test or Welch’s t–test can be relaxed.
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Two Sample t–test: Means (σX and σY unknown)
Example
Does smoking damage the lungs of children exposed to
parental smoking?
Forced Vital Capacity (FVC) is the volume (in milliliters) of air
that an individual can exhale in 6 seconds.
FVC was obtained for a sample of children not exposed to parental smoking and
for a sample of children exposed to parental smoking.
Parental smoking
Mean FVC
s
n
Yes
75.5
9.3
30
No
88.2
15.1
30
Is the mean FVC lower in the population of children exposed to parental
smoking? We test:
H0: µsmoke = µno <=> (µsmoke − µno) = 0
Ha: µsmoke < µno <=> (µsmoke − µno) < 0 (one-sided)
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Two Sample t–test: Means (σX and σY unknown)
Example (cont.)
The test statistic is
> (75.5-88.2)/sqrt(9.3^2/30 + 15.1^2/30)
[1] -3.922419
The conservative test gives 29 for the degrees of freedom and a p–value of
> tstat=(75.5-88.2)/sqrt(9.3^2/30 + 15.1^2/30)
> tstat
[1] -3.922419
> pt(tstat,29)
[1] 0.0002468389
The degrees of freedom for Welch’s Test is
2
2
9.32
+ 15.1
30
30
2 2
1
9.3
1
+ 30−1
30−1
30
15.12
30
2 = 48.23342.
(Use 48 if you looking up the p–value in a table). Using R, Welch’s test gives a p–value of
> pt(-3.922419, 48.23342)
[1] 0.0001385454
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Two Sample t–test: Means (σX and σY unknown)
Example
> dat1=rnorm(25,0.9,2.5)
> dat2=rnorm(43,0.6,1.3)
> t.test(dat1,dat2,mu=0.0)
Welch Two Sample t-test
data: dat1 and dat2
t = 1.2977, df = 29.947, p-value = 0.2043
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.4611103 2.0679355
sample estimates:
mean of x mean of y
1.2426692 0.4392566
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Pooled Two Sample t–test: Means (σX = σY unknown)
Pooled Two Sample t–test: Means (σX = σY unknown)
Pooled Two Sample t–test: Means
(σX = σY unknown)
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Pooled Two Sample t–test: Means (σX = σY unknown)
Theorem (Pooled Two Sample t–test: Means (σ = σX = σY unknown))
Let X1 , · · · , XnX and Y1 , · · · , YnY be independent random samples. Assume σ = σX = σY and that either
the random samples were sampled from a normal population or
the sum of the sample sizes, nX + nY , are ≥ 30.
Define the pooled estimator of σ 2 ,
2
2
(nX − 1)SX
+ (nY − 1)SY
2 def
Sp =
nX + nY − 2
.
Let
H0 : µX = µY
The test statistic is
T =
Sp
X̄ − Ȳ
q
1 + 1
n
n
X
∼ t(nX + nY − 2)
Y
for H0 .
A (1 − α)100% confidence interval for µX − µY is
s
?
x̄ − ȳ ± t (nX + nY − 2)sp
Marc Mehlman (University of New Haven)
1
nX
Inference for Distributions
+
1
nY
.
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Pooled Two Sample t–test: Means (σX = σY unknown)
Example
> atab=rnorm(45,3,2)
> btab=rnorm(50,3.5,2)
> t.test(atab,btab,var.equal=TRUE)
Two Sample t-test
data: atab and btab
t = -1.7116, df = 93, p-value = 0.0903
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-1.3836440 0.1026042
sample estimates:
mean of x mean of y
3.356853 3.997373
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Inference for Distributions
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Two Sample F –test: Variance
Two Sample F –test
Two Sample F –Test: Variance
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Two Sample F –test: Variance
Definition
Let X1 , · · · , Xn be a random sample from N(µX , σ) and let Y1 , · · · , Ym be
an independent random sample N(µY , σ). Then
SX2
∼ F(n − 1, m − 1)
SY2
is the F distribution with n − 1 degrees of freedom for the numerator and
m − 1 degrees of freedom for the denominator.
The F distribution is named after Sir Ronald A. Fisher.
“Even scientists need their heroes, and R. A. Fisher was certainly
the hero of 20th century statistics. His ideas dominated and
transformed our field to an extent a Caesar or an Alexander
might have envied.” – Bradley Efron
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Inference for Distributions
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Two Sample F –test: Variance
Theorem (Two–Sample F Test)
Let X1 , · · · , Xn be a random sample from N(µX , σX ) and let Y1 , · · · , Ym
be an independent random sample from N(µY , σY ). Let H0 : σX = σY .
Then
2
def S
under H0 .
F = X2 ∼ F(n − 1, m − 1)
SY
Furthermore the p–value of H0 versus
1
HA : σX > σY is P(f ≥ F (n − 1, m − 1)).
2
HA : σX < σY is P(f ≤ F (n − 1, m − 1)).
3
HA : σX 6= σY is twice the smaller value from the two alternative
hypothesis above.
Warning
The F–test is not robust with respect normal distribution assumptions.
Marc Mehlman (University of New Haven)
Inference for Distributions
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Two Sample F –test: Variance
Example
The # of tail wags per minute of 11 randomly chosen Labrador dogs at feeding time was tallied
and sample variance of 6.73 was calculated. The # of tail wags per minute of 9 randomly chosen
terriers at feeding time was tallied and sample variance of 25.91 was calculated. Assuming that
the number of wags for both types of dogs are normal, find the p–value of a test of
H0 : σL = σT
def SL2
2
ST
Solution: F =
versus
HA : σL < σT .
∼ F(11 − 1, 9 − 1) under H0 and f =
6.73
25.91
= 0.26. The p–value of the
left–tailed test is
> pf(6.73/25.91,10,8)
[1] 0.02510348
There is reason to reject the hypothesis that variance of tail wagging is the same between
terriers and Labradors in favor of more tail wagging variance for terriers.
Marc Mehlman (University of New Haven)
Inference for Distributions
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Two Sample F –test: Variance
Example
The # of tail wags per minute of 11 randomly chosen Labrador dogs at feeding time was tallied
and sample variance of 6.73 was calculated. The # of tail wags per minute of 9 randomly chosen
terriers at feeding time was tallied and sample variance of 25.91 was calculated. Assuming that
the number of wags for both types of dogs are normal, find the p–value of a test of
H0 : σL = σT
def SL2
2
ST
Solution: F =
versus
HA : σL < σT .
∼ F(11 − 1, 9 − 1) under H0 and f =
6.73
25.91
= 0.26. The p–value of the
left–tailed test is
> pf(6.73/25.91,10,8)
[1] 0.02510348
There is reason to reject the hypothesis that variance of tail wagging is the same between
terriers and Labradors in favor of more tail wagging variance for terriers.
Marc Mehlman (University of New Haven)
Inference for Distributions
37 / 42
Two Sample F –test: Variance
Example
> dat1=rnorm(25,0.9,2.5)
> dat2=rnorm(43,0.6,1.5)
> var.test(dat1,dat2)
F test to compare two variances
data: dat1 and dat2
F = 2.4715, num df = 24, denom df = 42, p-value = 0.009983
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
1.242724 5.280972
sample estimates:
ratio of variances
2.471486
Marc Mehlman (University of New Haven)
Inference for Distributions
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Two Sample F –test: Variance
To use a F table:
1
To use a F table, chose X to be the random variable with the larger
variance (thus the F statistic ≥ 1).
2
If the test is two–sided, use α/2 for the table.
3
When a degree of freedom can not be found in the table, use the
closest value that is less than the desired degree of freedom for a
conservative test.
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Inference for Distributions
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Chapter #7 R Assignment
Chapter #7 R Assignment
Chapter #7 R Assignment
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Inference for Distributions
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Chapter #7 R Assignment
First create the normal datasets:
>
>
>
>
set.seed(4321)
ndat1=rnorm(73,3.1,1.9)
ndat2=rnorm(57,2.9,2.3)
ndat3=rnorm(83,3.5,2.3)
and then do the problems:
1
For dataset ndat1, not knowing the mean is 3.1 or the standard deviation is 1.9,
find the p–value of the test
H0 : µ = 3.7
2
versus
HA : µ < 3.7.
The dataset “immer” (found on my website) contains the barley yield in years
1931 and 1932 where the same field are recorded. The matched pair yield data are
presented in by the columns Y1 and Y2. Find a 97.5% confidence interval for the
difference of the means, µY1 − µY2 , and use a matched pair t–test to find the
p–value of
H0 : µY1 = µY2
versus
HA : µY1 6= µY2 .
Marc Mehlman (University of New Haven)
Inference for Distributions
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Chapter #7 R Assignment
3
Assuming the means and standard deviations of ndat1 and ndat2 are
unknown, use Welch’s t test to test if the means of the two data sets
are equal.
4
Assuming the means and standard deviations of ndat2 and ndat3 are
unknown, but we know the population variances are equal, use pooled
two sample t test to test if the means of the two data sets are equal.
5
Test if the populations that ndat1 and ndat3 came from have the
same variance.
Marc Mehlman (University of New Haven)
Inference for Distributions
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