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Mathematics & Statistics
Statistics
Topic 7
Estimation
Topic Goals
After completing this topic, you should be able
to:
 Distinguish between a point estimate and a
confidence interval estimate
 Construct and interpret a confidence interval
estimate for a single population mean using both
the Z and t distributions
 Form and interpret a confidence interval estimate
for a single population proportion
 Form and interpret a confidence interval estimate
for a single population variance
Estimation
 Last week we looked at the distribution of
sample statistics...
 ...given the value of a population parameter.
 In real-life situations we don’t know the true
value of the population parameter...
 ...so the question is:
 Can we say something about the value of the
population parameter...
 ...given an observed value of a sample statistic?
Confidence Intervals
Content of this topic
 Confidence Intervals for the Population
Mean, μ
 when Population Variance σ2 is Known (Section 8.2)
 when Population Variance σ2 is Unknown (Section 8.3)
 Confidence Intervals for the Population
Proportion, p (large samples) (Section 8.4)
 Confidence Intervals for the Population
Variance, σ2 (Section 9.4)
Definitions
 An estimator of a population parameter is
 a random variable that depends on sample
information . . .
 whose value provides an approximation to this
unknown parameter
 A specific value of that random variable is called
an estimate
Point Estimates
We can estimate a
Population Parameter …
with a Sample
Statistic
(a Point Estimate)
Mean
μ
x
Proportion
p
p̂
Variance
Variance
σ2
s2
Unbiasedness
 A point estimator θ̂ is said to be an
unbiased estimator of the parameter  if the
expected value, or mean, of the sampling
distribution of θ̂ is ,
E(θˆ )  θ
 Examples:
 The sample mean is an unbiased estimator of μ
 The sample variance is an unbiased estimator of σ2
 The sample proportion is an unbiased estimator of p
Unbiasedness
(continued)
 θ̂1 is an unbiased estimator, θ̂ 2 is biased:
θ̂2
θ̂1
θ
θ̂
Most Efficient Estimator
 Suppose there are several unbiased estimators of 
 The most efficient estimator or the minimum variance
unbiased estimator of  is the unbiased estimator with the
smallest variance
 Let θ̂1 and θ̂2 be two unbiased estimators of , based on
the same number of sample observations. Then,
ˆ )  Var(θˆ )
 θ̂1 is said to be more efficient than θ̂ 2 if Var(θ
1
2
 The relative efficiency of θ̂1 with respect to θ̂2 is the ratio
of their variances:
Var(θˆ 2 )
Relative Efficiency 
Var(θˆ )
1
Point and Interval Estimates
 A point estimate is a single number,
 a confidence interval provides additional
information about variability
Lower
Confidence
Limit
Point Estimate
Width of
confidence interval
Upper
Confidence
Limit
Confidence Intervals
 How much uncertainty is associated with a
point estimate of a population parameter?
 An interval estimate provides more
information about a population characteristic
than does a point estimate
 Such interval estimates are called confidence
intervals
Confidence Interval Estimate
 An interval gives a range of values:
 Takes into consideration variation in sample
statistics from sample to sample
 Based on observation from 1 sample
 Gives information about closeness to
unknown population parameters
 Stated in terms of level of confidence
 Can never be 100% confident
Confidence Interval and
Confidence Level
 If P(a <  < b) = 1 -  then the interval from a
to b is called a 100(1 - )% confidence
interval of .
 The quantity (1 - ) is called the confidence
level of the interval ( between 0 and 1)
 In repeated samples of the population, the true value
of the parameter  would be contained in 100(1 )% of intervals calculated this way.
 The confidence interval calculated in this manner is
written as a <  < b with 100(1 - )% confidence
Estimation Process
Random Sample
Population
(mean, μ, is
unknown)
Sample
Mean
X = 50
I am 95%
confident that
μ is between
40 & 60.
Confidence Level, (1-)
(continued)
 Suppose confidence level = 95%
 Also written (1 - ) = 0.95
 A relative frequency interpretation:
 From repeated samples, 95% of all the
confidence intervals that can be constructed will
contain the unknown true parameter
 A specific interval either will contain or will
not contain the true parameter
 The procedure used leads to a correct interval in
95% of the time...
 ...but this does not guarantee anything about a
particular sample.
General Formula
 The general formula for all confidence
intervals is:
Point Estimate  (Reliability Factor)(Standard deviation)
 The value of the reliability factor
depends on the desired level of
confidence
Confidence Intervals
Confidence
Intervals
Population
Mean
σ2
Known
Population
Proportion
σ2 Unknown
Population
Variance
Confidence Interval for μ
(σ2 Known)
 Assumptions
 Population variance σ2 is known
 Population is normally distributed...
 ....or large sample so that CLT can be used.
 Confidence interval estimate:
x  z α/2
σ
σ
 μ  x  z α/2
n
n
(where z/2 is the normal distribution value for a probability of /2 in
each tail)
Example
 A sample of 11 circuits from a large normal
population has a mean resistance of 2.20
ohms. We know from past testing that the
population standard deviation is 0.35 ohms.
 Determine a 95% confidence interval for the
true mean resistance of the population.
Example
(continued)
 A sample of 11 circuits from a large normal
population has a mean resistance of 2.20
ohms. We know from past testing that the
population standard deviation is .35 ohms.
 Solution:
σ
x z
n
 2.20  1.96 (.35/ 11)
 2.20  .2068
1.9932  μ  2.4068
Interpretation
 We are 95% confident that the true mean
resistance is between 1.9932 and 2.4068
ohms
 Although the true mean may or may not be
in this interval, 95% of intervals formed in
this manner will contain the true mean
Margin of Error
 The confidence interval,
x  z α/2
σ
σ
 μ  x  z α/2
n
n
 Can also be written as x  ME
where ME is called the margin of error
ME  z α/2
σ
n
 The interval width, w, is equal to twice the margin of
error
Finding the Reliability Factor, z/2
 Consider a 95% confidence interval:
1   .95
α
 .025
2
Z units:
X units:
α
 .025
2
z = -1.96
Lower
Confidence
Limit
0
Point Estimate
z = 1.96
Upper
Confidence
Limit
 Find z.025 = 1.96 from the standard normal distribution table
Common Levels of Confidence
 Commonly used confidence levels are 90%,
95%, and 99%
Confidence
Level
80%
90%
95%
98%
99%
99.8%
99.9%
Confidence
Coefficient,
Z/2 value
.80
.90
.95
.98
.99
.998
.999
1.28
1.645
1.96
2.33
2.58
3.08
3.27
1 
Intervals and Level of Confidence
Sampling Distribution of the Mean
/2
Intervals
extend from
1 
/2
x
μx  μ
x1
to
100(1-)%
of intervals
constructed
contain μ;
σ
xz
n
100()% do
not.
σ
xz
n
x2
Confidence Intervals
Summary:Finding a confidence
interval for μ (σ known)
 Choose confidence level 1-α (e.g. .95).
 Find an interval [a,b] such that P(a<μ<b)=1-α.
 a and b are determined by
X  z / 2

n
 How to find zα/2?
 Look in table for value such that P(Z> zα/2)=α
 e.g. if 1-α=.95, then zα/2 = 1.96.
Finding the Reliability Factor, z/2
 Consider a 95% confidence interval:
1   .95
α
 .025
2
Z units:
X units:
α
 .025
2
z = -1.96
Lower
Confidence
Limit
0
Point Estimate
z = 1.96
Upper
Confidence
Limit
 Find z.025 = 1.96 from the standard normal distribution table
Example
 Assume that the calorie contents per 100 ml
of Guiness is normally distributed.
 A sample of 11 pints has a mean calorie
content per 100 ml of 35.1. We know from
past testing that the population standard
deviation is 2.35.
 Determine a 95% confidence interval for the
true mean calorie content per 100 ml
Guiness.
Example
(continued)
 A sample of 11 pints from a large normal
population has a mean calorie content per
100 ml of 35.1. We know from past testing
that the population standard deviation is
2.35 calories per 100 ml.
 Solution:
x  z.05/2
σ
n
 35.1  1.96 (2.35/ 11)
 35.1  1.3888
33.7112  μ  36.4888
Interpretation
 We are 95% confident that the true calorie
content per 100 ml is between 33.7112
and 36.4888.
 Although the true mean may or may not be
in this interval, 95% of intervals formed in
this manner will contain the true mean
Example
 Suppose a second sample of 11 pints has a mean
calorie content per 100 ml of 35.9.
 A 95% confidence interval for this sample is:
x  z .05/2
σ
n
 35.9  1.96 (2.35/ 11)
 35.9  1.3888
34.5112  μ  37.2888
Reducing the Margin of Error
ME  z α/2
σ
n
The margin of error can be reduced if
 the population standard deviation can be reduced (σ↓)
 The sample size is increased (n↑)
 The confidence level is decreased, (1 – ) ↓
How is the formula obtained?
 Recall the formula for the confidence interval:
x  z α/2
σ
σ
 μ  x  z α/2
n
n
 How is it obtained?
 A 100(1-α)% confidence interval is an interval [a,b] such
that P(a<μ<b) = 1-α.
 We use the fact that
X 
Z
~ N (0,1)
/ n
Derivation
P( z / 2  Z  z / 2 )  1  


X 
 P  z / 2 
 z / 2   1  
/ n



 

 P 
z / 2  X   
z / 2   1  
n
n






 P  X 
z / 2      X 
z / 2   1  
n
n






 P X 
z / 2    X 
z / 2   1  
n
n


Large Samples
 If the population is not normal....
 ....and the variance is not known....
 The same confidence interval can still be
used...
 ...if the sample is large.
 For in that case the Central Limit Theorem tells
us that the sample mean is approximately
normal with mean μ and standard deviation
σ/√n,...
 ...and s2 ≈ σ2.
Example
 For a sample of 200 tea boxes you observe that
the average weight is 101.0 grams with a
standard deviation of 2.78 grams.
 Determine a 99% confidence interval for the
population mean.
 Solution. Note that the sample is large, so:
 σ2 ≈ s2
 the Central Limit Theorem says that the sample
mean is approximately normal with mean μ and
standard deviation σ/√n≈2.78/√200=.197
Example
 So, we can proceed as if the sample mean is normal
with known variance and apply
x  z α/2
σ
σ
 μ  x  z α/2
n
n
 Therefore:
x  z .01/2
σ
n
 101.0  2.58 (2.78/ 200 )
 101.0  .507
100.49  μ  101.51
Confidence Intervals
Confidence
Intervals
Population
Mean
σ2 Known
Population
Proportion
σ2 Unknown
Population
Variance
Confidence Interval for μ
(σ2 Unknown)
 If the population standard deviation σ is
unknown, we can substitute the sample
standard deviation, s
 This introduces extra uncertainty, since
s is variable from sample to sample
 Therefore we use the t distribution
instead of the normal distribution
Student’s t Distribution
 Consider a random sample of n observations
 with mean x and standard deviation s
 from a normally distributed population with mean μ
 Then the variable
x μ
t
s/ n
follows the Student’s t distribution with (n - 1) degrees
of freedom
Student’s t Distribution
 The t is a family of distributions
 The t-value depends on degrees of
freedom (d.f.)
 Number of observations that are free to vary after
sample mean has been calculated
d.f. = n - 1
Student’s t Distribution
Note: t
Z as n increases
Standard
Normal
(t with df = ∞)
t (df = 13)
t-distributions are bellshaped and symmetric, but
have ‘fatter’ tails than the
normal
t (df = 5)
0
t
Confidence Interval for μ
(σ Unknown)
(continued)
 Assumptions
 Population standard deviation is unknown
 Population is normally distributed
 Use Student’s t Distribution
 Confidence Interval Estimate:
x  t n -1,α/2
s
s
 μ  x  t n -1,α/2
n
n
where tn-1,α/2 is the critical value of the t distribution with n-1 d.f.
and an area of α/2 in each tail:
P(t  t
)  α/2
n 1,α/2
Student’s t Table
Upper Tail Area
df
.10
.05
.025
1 3.078 6.314 12.706
Let: n = 3
df = n - 1 = 2
 = .10
/2 =.05
2 1.886 2.920 4.303
/2 = .05
3 1.638 2.353 3.182
The body of the table
contains t values, not
probabilities
0
2.920 t
t distribution values
With comparison to the Z value
Confidence
t
Level
(10 d.f.)
t
(20 d.f.)
t
(30 d.f.)
Z
____
.80
1.372
1.325
1.310
1.282
.90
1.812
1.725
1.697
1.645
.95
2.228
2.086
2.042
1.960
.99
3.169
2.845
2.750
2.576
Note: t
Z as n increases
Example
 A sample of 11 pints has a mean calorie
content per 100 ml of 35.1, with a standard
deviation of 2.35.
 Determine a 95% confidence interval for the
true mean calorie content per 100 ml
Guiness if the calorie content is normal.
Example
(continued)
 A sample of 11 pints from a large normal
population has a mean calorie content per
100 ml of 35.1, with a standard deviation of
2.35 calories per 100 ml.
 Solution:
x  t10,.05/2
s
n
 35.1  2.228 (2.35/ 11)
 35.1  1.5787
33.5213  μ  36.6787
Interpretation
 We are 95% confident that the true calorie content per
100 ml is between 33.5213 and 36.6787.
 Compare with the previous example (variance known)...
 ...where we were 95% confident that the true calorie
content per 100 ml is between 33.7112 and 36.4888.
 The interval is wider because there is additional
uncertainty over σ.
About Student
 Student is the pseudonym of William Sealy Gosset...
 ...who worked for the Guiness brewery in Dublin...
 ...and needed procedures for quality control of the
brewing process.
 He derived the distribution of
X 
t
s/ n
which is still named after him
Confidence Intervals
Confidence
Intervals
Population
Mean
σ2 Known
Population
Proportion
σ2 Unknown
Population
Variance
Confidence Intervals for the
Population Proportion, p
 An interval estimate for the population
proportion ( p ) can be calculated by
adding an allowance for uncertainty to
the sample proportion ( p̂ )
 Sample has to be large because we will use
the normal approximation to the binomial
Confidence Intervals for the
Population Proportion, p
(continued)
 Recall that the distribution of the sample
proportion is approximately normal if the
sample size is large, with standard deviation
p(1  p)
σp 
n
 We will estimate this with sample data:
p̂(1  p̂)
s pˆ 
n
Confidence Interval Endpoints
 Upper and lower confidence limits for the
population proportion are calculated with the
formula
p̂  z α/2
p̂(1  p̂)
p̂(1  p̂)
 p  p̂  z α/2
n
n
 where
 z/2 is the standard normal value for the level of confidence desired
 p̂ is the sample proportion
 n is the sample size
Example
 A random sample of 100 people
shows that 25 are left-handed.
 Form a 95% confidence interval for
the true proportion of left-handers
Example
(continued)
 A random sample of 100 people shows
that 25 are left-handed. Form a 95%
confidence interval for the true proportion
of left-handers.
p̂  z α/2
p̂(1  p̂)
p̂(1  p̂)
 p  p̂  z α/2
n
n
25
.25(.75)
25
.25(.75)
 1.96
 p 
 1.96
100
100
100
100
0.1651  p  0.3349
Interpretation
 We are 95% confident that the true
percentage of left-handers in the population
is between
16.51% and 33.49%.
 Although the interval from 0.1651 to 0.3349
may or may not contain the true proportion,
95% of intervals formed from samples of
size 100 in this manner will contain the true
proportion.
General Example
 Trinity news, Issue 9, vol. 54 reports the results
of a poll with 430 respondents on the SU
president election. It is reported that: “Moore
[has] the slightest lead of 31.7% to [...]
Donohue’s 29.2% [...] and Reilly’s 27.2%. Given
the margin of error the narrow gap [...] means
any one [can win]”.
 Compute a 95% confidence interval for the
proportion of votes for Moore.
 What do you think about the paper’s claim
referring to the margin of error?
First question
 We have a large sample: n = 430
 We are asked to compute a 95% confidence level:
1-α=.95 and – thus – α=.05.
 The quantity of interest is a population proportion.
 Let p denote the proportion of votes for Moore.
 So, the appropriate confidence interval is
p̂  z α/2
p̂(1  p̂)
 p  p̂  z α/2
n
p̂(1  p̂)
n
Obtain necessary info and apply
p̂  z α/2
p̂(1  p̂)
 p  p̂  z α/2
n
p̂(1  p̂)
n
Sample proportion: p̂  .317
Use Table 1 or 8 to find z / 2  z.025  1.96
So:
p̂  z α/2
p̂(1  p̂)
p̂(1  p̂)
 p  p̂  z α/2
n
n
.317(1  .317)
.317(1  .317)
 p  .317  1.96
430
430
.317  .0440  p  .317  .0440
.273  p  .361
.317  1.96
We are 95% confident that the true poportion of votes
for Moore is between 27.3% and 36.1%
Second question
We are 95% confident that the true poportion of votes
for Moore is between 27.3% and 36.1%
The score for Donohue is 29.2% and lies inside this
95% confidence interval, Reilly’s 27.2%, however,
does not. So, given the chosen level of confidence
and the resulting margin of error, the race could be
deemed “too close to call” between Moore and
Donohue.
NB The article does not specify the confidence level.
So, we don’t know what “margin of error” it refers to.
Bad statistical practise.
Probability Question
 In a particular city, 20% of mobile phones are
owned by people younger than 15. In addition,
52% of people with a mobile phone have a “Pay
As You Talk” deal. Among those younger than
15, 12% have a “Pay As You Talk” deal.
 What is the probability that a randomly chosen
mobile phone is “Pay As You Talk” and belongs
to someone younger than 15?
Set up the analysis
 Give the events of interest a name:
 Let A be the event that the phone belongs to a person
younger than 15.
 Let B be the event that the phone is “Pay As You Talk”
 What probabilties are given?
 P(A) = .2
 P(B) = .52
 P(B|A) = .12
 What are we asked?
 Joint probability of A and B: P(A∩B)
Find Solution
 We know that:
P(A  B)
P( B | A ) 
P( A)
 Therefore,
P(A  B)  P(B | A)P(A)
 .12 * .2
 .024
Exam Tips
 Move on to next question if stuck
 Method more important than outcome
 don’t worry too much about rounding (not too many
digits)
 computation errors have minor penalty
 Write down all the steps you take
 avoids mistakes
 I can’t give marks if it’s not clear what you’re doing
 Before you answer a question think about your
“plan of attack” before you start writing.
 Be relaxed when you walk in (no cramming).