Download Chapter 7: Statistical Intervals Based on a Single Sample

Chapter 7: Statistical Intervals Based on a Single
Sample
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March 23, 2014
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Chapter Overview
Basic Properties of Confidence Intervals (CIs)
Large-Sample CI for Population Mean and Proportion
CI for mean µ
CI for proportion p
One-sided intervals
Intervals Based on a Normal Population Distribution
t distribution
One sample t CI for mean µ
CI for the Variance and Standard Deviation of a Normal Population
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Point Estimate vs. Confidence Interval
To estimate a parameter µ of a normal distribution. Given the observed
value x1 , x2 , · · · , xn of a random sample X1 , X2 , · · · , Xn .
Point Estimate
Find an point estimate of µ using the sample mean x̄ =
x1 +x2 +···+xn
n
For different observed values, we may have different estimates for µ.
Which estimate is closer to the true value? No idea :(
Instead, we may provide an interval of values of µ
Include the true value of µ with a certain “level of confidence”
Point estimate ± the error of estimation
Narrow interval → precise estimate
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Random interval centered at X̄
Let X1 , X2 , · · · , Xn be a random sample from a normal distriution with µ and σ.
We know that regardless of sample size n, X̄ is normally distributed with expected
√
value µ and standard deviation σ/ n. [see PROPOSITION on text p224]
Z=
X̄ − µ
X̄ − µX̄
=
√σ
σX̄
n
0.95 = P (−1.96 < Z < 1.96)
−1.96 <
=P
X̄ − µ
√σ
n
!
< 1.96
σ
σ
= P X̄ − 1.96 · √ < µ < X̄ + 1.96 · √
n
n
A random interval X̄ − 1.96 · √σn , X̄ + 1.96 · √σn
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Confidence Interval and Confidence Level
Figure : Fifty different 95% CIs for µ (red ones do not include µ).
CIs allow us to estimate population parameters using a range of values.
A confidence level is a measure of the degree of the relaibility of the interval.
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Definition of 95% Confidence Interval (CI) for µ
After observing X1 = x1 , X2 = x2 , · · · , Xn = xn from a normal distribution
with known σ, we compute the observed sample mean x̄ and a fixed interval. The
95% CI for populaiton mean µ is
σ
σ
x̄ − 1.96 √ , x̄ + 1.96 √
n
n
or
σ
x̄ ± 1.96 √
n
or with 95% confidence
σ
σ
x̄ − 1.96 √ < µ < x̄ + 1.96 √
n
n
Example
A normal population has unknown µ and σ = 2.0, if n = 31 and x̄ = 80.0, what
is the 95% CI?
σ
2.0
x̄ ± 1.96 √ = 80.0 ± 1.96 √ = 80.0 ± .7 = (79.3, 80.7)
n
31
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Definition of 95% Confidence Interval (CI) for µ
After observing X1 = x1 , X2 = x2 , · · · , Xn = xn from a normal distribution
with known σ, we compute the observed sample mean x̄ and a fixed interval. The
95% CI for populaiton mean µ is
σ
σ
x̄ − 1.96 √ , x̄ + 1.96 √
n
n
or
σ
x̄ ± 1.96 √
n
or with 95% confidence
σ
σ
x̄ − 1.96 √ < µ < x̄ + 1.96 √
n
n
Example
A normal population has unknown µ and σ = 2.0, if n = 31 and x̄ = 80.0, what
is the 95% CI?
σ
2.0
x̄ ± 1.96 √ = 80.0 ± 1.96 √ = 80.0 ± .7 = (79.3, 80.7)
n
31
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Interpreting a CI
Given a 95% CI, is it correct to say that µ falls in the CI with
probability 0.95? Why?
No. Look at the probability P X̄ − 1.96 ·
√σ
n
< µ < X̄ + 1.96 ·
√σ
n
,
when substitute X̄ with observed value x̄, no randomness left.
For a particular sample, either µ is in the interval or µ is not. For
instance, it is not correct to say P (79.3 ≤ µ ≤ 80.7) = 0.95 in the
previous exmaple.
This statement under P has no chance element for a given sample.
Therefore, we talk about being confident (in the method).
A precise way to interpret CI is: with 95% confidence, µ falls in the
interval calculated.
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Interpreting a CI - Continue
Note that:
It is the confidence interval itself that is random (i.e. it changes for
each sample).
µ is just a fixed number (the population mean). It is non-random, but
usually unknown.
The 95% Confidence Level
The 95% in a 95% C.I. could be interpreted as
P (the next sample will give a C.I. that contains µ) = 0.95.
The 95% means that we used a method that captures the true mean 95%
of the time.
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Example
We want to estimate the mean µ of a normal population with given
σ = 2.0 and a random sample X1 , X2 , · · · , Xn . Let us use µ̂ = X̄.
1
2
What is the distribution of X̄? What is the distribution of
X̄−µ
Find c such that P −c < √σ < c = 0.95?
X̄−µ
√σ
n
?
n
3
Given the observed sample {2,3,1,6,5,7,10,4,9,8}. The estimate of µ
is x̄ = 5.5. What’s the 95% confidence interval for µ?
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Notation zα/2
Let zα/2 denote the value such that P (Z > zα/2 ) = P (Z < −zα/2 ) = α/2, where
Z ∼ N (0, 1).
That is, the area between −zα/2 and zα/2 under the standard normal curve is
1 − α, which means P (−zα/2 < Z < zα/2 ) = 1 − α.
Find the following zα/2 values:
For α = 0.01, zα/2 =
For α = 0.05, zα/2 =
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Other Confidence Level
Definition
A 100(1 − α)% CI for the mean µ of a normal population when the value
of σ is known
σ
σ
x̄ − zα/2 · √ , x̄ + zα/2 · √
n
n
Several most commonly used zα/2 ’s:
Confidence Level
α/2
zα/2
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90%
0.05
1.645
Confidence Intervals
95%
0.025
1.960
99%
0.005
2.576
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From 95% CI to 90% CI
P
−1.96 <
i.e.,
P
−z0.05/2 <
X̄ − µ
√σ
n
X̄ − µ
√σ
n
!
< 1.96
= 0.95
!
< z0.05/2
= 0.95
The 95% CI of the mean µ is
σ
x̄ ± z0.05/2 √
n
Similarly,
P
−z0.10/2 <
X̄ − µ
√σ
n
!
< z0.10/2
= 0.90
The 90% CI of the mean µ is
σ
x̄ ± z0.10/2 √
n
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Confidence Level, Sample Size and Precision
Review A (1 − α)100% CI for the mean µ of a normal population when
the value of σ is known
σ
σ
x̄ − zα/2 · √ , x̄ + zα/2 · √
n
n
Want narrow CI with high confidence level.
Normal population, with unknown mean µ, and standard deviation
σ = 2.0. Sample of size 25 yields x̄ = 1.0.
1
What’s the 100%, 95% and 90% CI for µ?
2
Suppose a sample of size 100 also yields x̄ = 1.0, what’s the 90% CI?
3
Which of the above intervals is narrower?
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Finding the Sample Size
Want to estimate a normal population mean µ with σ = 25. What sample size is
necessary to ensure that the 95% CI has a width of (at most) 10?
Answer
Width of CI=x̄ + zα/2 · √σn − x̄ + zα/2 · √σn = 2zα/2 · √σn
Here 10 = 2zα/2 √25n and α = 1 − 0.95 = 0.05 (z0.025 = 1.96)
√
n = 2 · 1.96
25
10
n = 96.04, the sample size is at least 97.
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Finding the Sample Size
Want to estimate a normal population mean µ with σ = 25. What sample size is
necessary to ensure that the 95% CI has a width of (at most) 10?
Answer
Width of CI=x̄ + zα/2 · √σn − x̄ + zα/2 · √σn = 2zα/2 · √σn
Here 10 = 2zα/2 √25n and α = 1 − 0.95 = 0.05 (z0.025 = 1.96)
√
n = 2 · 1.96
25
10
n = 96.04, the sample size is at least 97.
General Formula
The sample size n necessary to ensure an interval width w for confidence level
(1 − α)100% is
σ 2
n = 2zα/2 ·
w
[hint: we want w = 2zα/2 · √σn ]
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Finding the Sample Size
Want to estimate a normal population mean µ with σ = 25. What sample size is
necessary to ensure that the 95% CI has a width of (at most) 10?
Answer
Width of CI=x̄ + zα/2 · √σn − x̄ + zα/2 · √σn = 2zα/2 · √σn
Here 10 = 2zα/2 √25n and α = 1 − 0.95 = 0.05 (z0.025 = 1.96)
√
n = 2 · 1.96
25
10
n = 96.04, the sample size is at least 97.
General Formula
The sample size n necessary to ensure an interval width w for confidence level
(1 − α)100% is
σ 2
n = 2zα/2 ·
w
[hint: we want w = 2zα/2 · √σn ]
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Review: Confidence Level, Sample Size and Precision
Recall that
A (1 − α)100% CI for the mean µ of a normal population when σ is known
σ
σ
x̄ − zα/2 · √ , x̄ + zα/2 · √
n
n
The width of the above CI is w = 2zα/2 ·
√σ .
n
Larger sample size n results in narrower CI, lower confidence level
results in narrower CI.
Confidence level (a measure of the degree of the relaibility of the
interval) cannot be too high. 100% will result in (−∞, ∞).
One strategy: specify desired confidence level and interval width then
determine sample size.
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Large-Sample CIs
Large-Sample CI for Population Mean and Proportion
CI for mean µ
CI for proportion p
One-sided intervals
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Large-Sample CI for Population Mean
Let X1 , X2 , · · · , Xn be a random sample from an unknown population having
unknown mean µ and unknown standard deviation σ. How to find (1 − α)100% CI?
By CLT, X̄ is approximately normal distributed with mean µ, std dev
X̄ − µ
Z=
√σ
n
√σ .
n
∼ N (0, 1)
We find CI by:
P
−zα/2 <
X̄ − µ
√σ
n
!
< zα/2
≈1−α
Then (1 − α)100% CI for µ is:
σ
σ
X̄ − zα/2 √ , X̄ + zα/2 √
n
n
In the expression σ is still unknown. For large sample sizes, we may replace it with
sample standard deviation S:
S
S
X̄ − zα/2 √ , X̄ + zα/2 √
n
n
For a given sample x1 , x2 , · · · , xn , plug in X̄ = x̄ and S = s in the above expression.
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Example
Given a sample from an unknown distribution with mean µ and standard
deviation σ, the sample size is 196, sample mean x̄ = 2.0, and sample
standard deviation s = 3.0. Find the 95% CI of the population mean.
Answer
s
x̄ ± zα/2 √
n
3.0
2.0 ± 1.96 √
196
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Example
Given a sample from an unknown distribution with mean µ and standard
deviation σ, the sample size is 196, sample mean x̄ = 2.0, and sample
standard deviation s = 3.0. Find the 95% CI of the population mean.
Answer
s
x̄ ± zα/2 √
n
3.0
2.0 ± 1.96 √
196
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Large Sample Confidence Interval for a Population Proportion p
Suppose X ∼ Bin(n, p) (Binomial), the sampling distribution of the
sample proportion p̂ = X/n is
!
r
p(1 − p)
approx
p̂ ∼ N µp̂ = p, σp̂ =
n
for large sample size n (np ≥ 10 and n(1 − p) ≥ 10).
p̂ − p
q
∼ N (0, 1)
p(1−p)
n
To find (1 − α)100% CI, we get
P
−zα/2 < p
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p̂ − p
p(1 − p)/n
!
< zα/2
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≈1−α
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Finding the CI for a Population Proportion p
Solving the inequality, we get the (1 − α)100% CI of p:
r
p̂ +
lower confidence limit=
2
zα/2
2n
1+
p̂ +
upper confidence limit=
2
zα/2
2n
p̂(1−p̂)
n
− zα/2
+
1+
4n2
2
zα/2
n
r
+ zα/2
2
zα/2
p̂(1−p̂)
n
+
2
zα/2
4n2
2
zα/2
n
When n is quite large, the above limits can be approximated by:
r
p̂(1 − p̂)
lower confidence limit=p̂ − zα/2
n
r
p̂(1 − p̂)
upper confidence limit=p̂ + zα/2
n
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The 100(1 − α)% CI for Population Proportion p
The 100(1 − α)% CI for the population proportion p for a sample of
size n, with sample proportion p̂, is given by:
r
p̂ +
2
zα/2
2n
± zα/2
1+
p̂(1−p̂)
n
+
2
zα/2
4n2
2
zα/2
n
Although this formula was based on a large-sample distribution of p̂, it
works quite well even when the sample size n is reasonably small.
When n is quite large, the CI reduces to
r
p̂(1 − p̂)
p̂ ± zα/2
n
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Example
Among 10000 cats in Indiana state, 20% are found to be long-hairs. What
is the 99% CI for the proportion p of long-hairs in Indiana?
Hint
r
p̂ − zα/2
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p̂(1 − p̂)
, p̂ + zα/2
n
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r
p̂(1 − p̂)
n
!
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Example
Among 10000 cats in Indiana state, 20% are found to be long-hairs. What
is the 99% CI for the proportion p of long-hairs in Indiana?
Hint
r
p̂ − zα/2
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p̂(1 − p̂)
, p̂ + zα/2
n
Confidence Intervals
r
p̂(1 − p̂)
n
!
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One-sided Confidence Intervals (Confidence Bounds)
The 100(1 − α)% upper confidence bound for µ is
s
µ < x̄ + zα √
n
The 100(1 − α)% lower confidence bound for µ is
s
µ > x̄ − zα √
n
Confidence Level
α
zα
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90%
0.10
1.28
Confidence Intervals
95%
0.05
1.645
99%
0.01
2.33
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CI for Normal Population with Unknown µ and Unknown σ
t-distribution and properties
One sample t CI for normal mean µ with unknown µ
and unknown σ
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CI For Normal Mean µ with Unknown µ and Unknown σ
Let X1 , X2 , · · · , Xn be a random sample from a normal distriution with
unknown µ and σ. To find a CI for µ, We need to introduce a new
distribution.
Theorem
When X̄ is the mean of a random sample of size n, and S is the sample
standard deviation from a normal distribution with mean µ, then the rv:
T =
X̄ − µ
√S
n
has a t distribution with degree of freedom(df) n − 1, denoted by tn−1 .
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Properties of t Distribution
The density curve of a t with df ν:
Bell-shaped
More spread out than the standard normal (z) curve − heavier tails
t curve becomes less spread out when ν increases
t distribution becomes standard normal distribution when df ν → ∞
(because as n increases, s → σ)
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Properties of t Distribution
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tα,ν Notation and t Table
The notation tα,ν is the value on the measurement axis for which the
area under the t curve with df=ν to the right of tα,ν is α; tα,ν is
called t critical value.
The tα,ν values are tabulated.
Exercises
1 Find t
0.1,10 , t0.001,30 , t0.05,120 .
2 Determine t critical value that will capture the desired t curve when
Central area= 0.95, df=16
Lower-tail area= 0.1, df=16
3
For a rv T which follows a t distribution
with df=n − 1, what is
P −tα/2,n−1 < T < tα/2,n−1 ?
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The One-sample t Confidence Interval
2
Given a random sample X1 , X2 , ..., Xn from
an unknown normal distri. N (µ,
σ ), we
have T = X̄−µ
∼ tn−1 . For a given α, P −tα/2,n−1 < X̄−µ
< −tα/2,n−1 = 1 − α.
S
S
√
√
n
n
Proposition
Let x̄ and s be the sample mean and sample standard deviation computed
from a random sample from a normal population. The (1 − α)100% CI for
µ is
s
s
x̄ − tα/2,n−1 · √ , x̄ + tα/2,n−1 · √
n
n
If the sample size is large, the critical value can be taken from the
standard normal table.
The t CIs are robust to small or even moderate deviations from
normality unless n is quite small.
Compare to the (1 − α)100%
CI for normal mean µ when σ is known,
x̄ − zα/2 √σn , x̄ + zα/2 √σn .
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Example
A random sample of size 10 from a normal population yields:
2,3,1,6,5,7,10,4,9,8. Find the 95% CI for the normal mean µ.
Answer
s
s
x̄ − tα/2,n−1 · √ , x̄ + tα/2,n−1 · √
n
n
α = 0.05, t0.05/2,10−1 = t0.025,9 = 2.262,
x̄ =
s
s=
2 + 3 + ··· + 8
= 5.5
10
22 + 32 + · · · + 82 −
10 − 1
(2+3+···+8)2
10
= 3.0277
so the 95% CI is:
3.0277
3.0277
(5.5 − 2.262 · √
, 5.5 + 2.262 · √
) = (3.3343, 7.6657)
10
10
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Example
A random sample of size 10 from a normal population yields:
2,3,1,6,5,7,10,4,9,8. Find the 95% CI for the normal mean µ.
Answer
s
s
x̄ − tα/2,n−1 · √ , x̄ + tα/2,n−1 · √
n
n
α = 0.05, t0.05/2,10−1 = t0.025,9 = 2.262,
x̄ =
s
s=
2 + 3 + ··· + 8
= 5.5
10
22 + 32 + · · · + 82 −
10 − 1
(2+3+···+8)2
10
= 3.0277
so the 95% CI is:
3.0277
3.0277
(5.5 − 2.262 · √
, 5.5 + 2.262 · √
) = (3.3343, 7.6657)
10
10
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More Intervals on Normal Populations
Prediction interval (PI) for a single observation
CI for variance σ 2 and standard deviation σ
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Prediction Interval for a Single Future Value
ColorredIn many applications, we wish to predict a single value of a
variable using a given sample. Suppose we have a random sample from a
normal population X1 , · · · , Xn . Now we wish to predict the the value of
Xn+1 , a single future observation. We need a new rv:
T =
X̄ − Xn+1
q
S 1 + n1
It can be shown that: T ∼ tn−1 .
By solving the below inequality for Xn+1


X̄
−
X
n+1
P −tα/2,n−1 < q
< tα/2,n−1  = 1 − α
S 1 + n1
we can find the (1 − α)100% PI. The prediction level is (1 − α).
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Prediction Interval for a Single Obs
Proposition
A PI for a single observation to be selected from a normal population
distribution is
r
1
x̄ ± tα/2,n−1 · s 1 +
n
The prediction level is (1 − α)100%.
Interpretation
Similar to the CI: we are (1 − α)100% sure that a future obs will fall in the
PI Or, when the PI is calculated sample after sample, in the long run, 95%
will include the true value of the future obs.
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Example
A random sample of size 10 from a normal population yields: x̄ = 21.90,
s = 4.134, given t0.025,9 = 2.262, find:
95% CI for the normal mean µ, how do you interpret the CI?
4.134
4.134
21.90 − 2.262 √ , 21.90 + 2.262 √
10
10
⇒ (18.94, 24.86)
95% PI for a single future value, how do you interpret the PI?
!
r
r
1
1
21.90 − 2.262 × 4.134 1 + , 21.90 + 2.262 × 4.134 1 +
10
10
⇒ (12.09, 31.71)
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CIs for Variance and Standard Deviation of a Normal
Given a sample from a normal population, want to estimate the CI of the
variance σ 2 or std dev σ. Need a new distribution: χ2 .
Theorem
Let X1 , X2 , · · · , Xn be a random sample from a normal distribution
with parameters µ and σ 2 . Then the rv:
P
(n − 1)S 2
(X̄ − Xi )2
=
2
σ
σ2
has a χ2 distribution with (n − 1) df. Denoted by χ2n−1 .
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χ2 Distribution
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χ2 Distributions and χ2α,ν Notation
χ2 curves
A group of positive distributions: x ≥ 0. Not symmetric.
Each has a positive skew with long upper tail.
different df ν has different density shape, and density curve becomes
more symmetric when ν increases
χ2α,ν Notation
χ2α,ν is called a χ2 critical value, denote the number on the
measurement axis such that α of the area under the χ2 curve with ν
degrees of freedom lies to the right of χ2α,ν .
χ2α,ν values are tabulated.
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Examples
1
Find χ20.1,12 , χ20.05,15 , χ20.95,15 .
2
Determine the following
2
The 95% percentile of the
χ dist2with df2 ν = 15.
2
P 10.52 < χ < 40.65 , where χ is a χ rv with df ν = 25.
3
For a random sample of size n from a normal population with
variance σ 2 . What is:
(n − 1)S 2
2
2
P χ1−α/2,n−1 <
< χα/2,n−1 ?
σ2
Dr Sharabati (Purdue University)
Confidence Intervals
Spring 2014
38 / 42
CI for σ 2 and σ
(n − 1)S 2
2
χ21−α/2,n−1 <
<
χ
=1−α
α/2,n−1
σ2
Solve the inequality for σ 2 , we get:
P
(n − 1)S 2
(n − 1)S 2
< σ2 < 2
2
χα/2,n−1
χ1−α/2,n−1
Proposition
A (1 − α)100% CI for variance σ 2 of a normal population is:
!
(n − 1)s2 (n − 1)s2
,
χ2α/2,n−1 χ21−α/2,n−1
A (1 − α)100% CI for std dev σ is then:
!
s
s
(n − 1)s2
(n − 1)s2
,
χ2α/2,n−1
χ21−α/2,n−1
Dr Sharabati (Purdue University)
Confidence Intervals
Spring 2014
39 / 42
Example
A random sample from a normal population of size 10 yields
2,3,1,6,5,7,10,4,9,8. Find the 95% CI for the normal variance σ 2 and std
dev σ.
Answer
(n − 1)s2 (n − 1)s2
,
χ2α/2,n−1 χ21−α/2,n−1
!
Because 95% CI ⇒ α = 0.05, sample size n = 10 ⇒df= 9, we have
χ20.975,9 = 2.700, χ20.025,9 = 19.022. Also from the sample, we can compute
s2 =
(2+3+···+8)2
10
22 +32 +···+82 −
10−1
= 9.167 95% CI for σ 2 :
9 × 9.167 9 × 9.167
,
19.022
2.700
⇒ (4.337, 30.557)
95% CI for σ is then:
√
Dr Sharabati (Purdue University)
√
4.337, 30.557 ⇒ (2.083, 5.528)
Confidence Intervals
Spring 2014
40 / 42
Example
A random sample from a normal population of size 10 yields
2,3,1,6,5,7,10,4,9,8. Find the 95% CI for the normal variance σ 2 and std
dev σ.
Answer
(n − 1)s2 (n − 1)s2
,
χ2α/2,n−1 χ21−α/2,n−1
!
Because 95% CI ⇒ α = 0.05, sample size n = 10 ⇒df= 9, we have
χ20.975,9 = 2.700, χ20.025,9 = 19.022. Also from the sample, we can compute
s2 =
(2+3+···+8)2
10
22 +32 +···+82 −
10−1
= 9.167 95% CI for σ 2 :
9 × 9.167 9 × 9.167
,
19.022
2.700
⇒ (4.337, 30.557)
95% CI for σ is then:
√
Dr Sharabati (Purdue University)
√
4.337, 30.557 ⇒ (2.083, 5.528)
Confidence Intervals
Spring 2014
40 / 42
Summary
CI for Normal Population
A (1 − α)100% CI for the mean µ
when σ is known, x̄ ± zα/2 · √σn
when σ is unknown, x̄ ± tα/2,n−1 ·
√s
n
A (1 − α)100% CI for the variance
2
(n−1)s2
, (n−1)s
χ2α/2,n−1 χ21−α/2,n−1
Large-Sample CI for Population Mean and Proportion
A (1 − α)100% CI for mean µ: x̄ ± zα/2 √sn
A (1 − α)100% CI for proportion p: approx. p̂ ± zα/2
q
p̂(1−p̂)
n
Notation: Let x̄ and s be the sample mean and sample standard deviation
computed from a random sample. Confidence level is (1 − α).
Dr Sharabati (Purdue University)
Confidence Intervals
Spring 2014
41 / 42
Summary
CI for Normal Population
A (1 − α)100% CI for the mean µ
when σ is known, x̄ ± zα/2 · √σn
when σ is unknown, x̄ ± tα/2,n−1 ·
√s
n
A (1 − α)100% CI for the variance
2
(n−1)s2
, (n−1)s
χ2α/2,n−1 χ21−α/2,n−1
Large-Sample CI for Population Mean and Proportion
A (1 − α)100% CI for mean µ: x̄ ± zα/2 √sn
A (1 − α)100% CI for proportion p: approx. p̂ ± zα/2
q
p̂(1−p̂)
n
Notation: Let x̄ and s be the sample mean and sample standard deviation
computed from a random sample. Confidence level is (1 − α).
Dr Sharabati (Purdue University)
Confidence Intervals
Spring 2014
42 / 42