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Week 2Sampling Distributions & Confidence Intervals
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ObjectivesBy the end of this lesson, you should be able to:
• Explain the important role of the normal distribution as a sampling distribution
• Explain the general concepts of estimating the parameters of a population or a probability distribution
• Understand the central limit theorem
• Construct point and interval estimation of a parameter
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Statistics in Action
• It is helpful to put statistics in the context of a general process of investigation:
1. Identify a question or problem.
2. Collect relevant data on the topic.
3. Analyze the data.
4. Form a conclusion.
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Population & Sample
• We collect a sample of data to better understand the characteristics of a population.
• A variable is a characteristic we measure for each individual or case.
• The overall quantity of interest may be the mean, median, proportion, or some other summary of a population.
• These population values are called parameters.
• We estimate the value of a parameter by taking a sample and computing a numerical summary called a statistic based on that sample.
• Note that the two p's (population, parameter) go together and the two s's ( sample, statistic) go together.
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Fundamental Data Descriptions
Random Sampling:
Definition
A population consists of the totality of the observations with which we are concerned.
Definition
A sample is a subset of a population.
• Each observation in a population is a value of a random variable X having
some probability distribution f(x).
• To eliminate bias in the sampling procedure, we select a random sample in
the sense that the observations are made independently and at random.
• The random sample of size n is: X1, X2, …, Xn . It consists of n observations
selected independently and randomly from the population.
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Some Important Statistics:
Definition:
Any function of the random sample X1, X2, …, Xn is called a statistic.
Location Measure of a Sample:
Definition
If X1, X2, …, Xn represents a random sample of size n, then the sample mean is
defined to be the statistic:
n
X
n
XXXX
n
ii
n
121 (unit)
is a statistic because it is a function of the random sample
X1, X2, …, Xn.
· has same unit of X1, X2, …, Xn.
· measures the central tendency in the sample (location).
X
X
X
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Variability in the Sample:
Definition
If X1, X2, …, Xn represents a random sample of size n, then the sample variance is
defined to be the statistic:
1
)()()(
1
)( 222
211
2
2
n
XXXXXX
n
XX
S n
n
ii (unit)2
Note:
· S2 is a statistic because it is a function of the random sample
X1, X2, …, Xn.
· S2 measures the variability in the sample.
1
)(1
2
2
n
XX
SS
n
ii
(unit)
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Normal Distribution
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Normal DistributionThe normal distribution is one of the most important continuous distributions.
Many measurable characteristics are normally
or approximately normally distributed, such as,
height and weight.
The graph of the probability density function (pdf)
of a normal distribution, called the normal curve,
is a bell-shaped curve.
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2.5% 2.5%
5% region of rejection of null hypothesis
Non directional
Two Tail
body temperature, shoe sizes, diameters of trees,
Wt, height etc…
IQ
68%
95%
13.5%13.5%
Normal Distribution:
half the scores above
mean…half below
(symmetrical)
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The pdf of the normal distribution depends on two parameters: mean = E(X)= and
variance =Var(X) = 2.
If the random variable X has a normal distribution with mean and variance 2, we
write:
X ~ Normal(,) or X ~ N(,)
The pdf of X ~ Normal(,) is given by:
0
;2
1),;()(
2
2
1
x
exnxf
x
The location of the normal
distribution depends on and its
shape depends on .
Suppose we have two normal
distributions:
_______ N(1, 1)
----------- N(2, 2) 1 < 2, 1=2
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1 = 2, 1<2 1 < 2, 1<2
Some properties of the normal curve f(x) of N(,):
1. f(x) is symmetric about the mean .
2. f(x) has two points of inflection at x= .
3. The total area under the curve of f(x) =1.
4. The highest point of the curve of f(x) at the mean .
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Areas Under the Normal Curve of X~N(,)
The probabilities of the normal distribution N(,) depends on and .
a
-
dxf(x))aX(P
b
dxf(x) b)P(X b
a
dxf(x) b)XP(a
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Areas Under the Normal Curve:Definition
The Standard Normal Distribution:
•The normal distribution with mean =0 and variance 2=1 is called the standard normal
distribution and is denoted by Normal(0,1) or N(0,1). If the random variable Z has the
standard normal distribution, we write Z~Normal(0,1) or Z~N(0,1).
•The pdf of Z~N(0,1) is given
by:
2
2
1
2
1)1,0;()(
z
eznzf
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•The standard normal distribution, Z~N(0,1), is very important
because probabilities of any normal distribution can be
calculated from the probabilities of the standard normal
distribution.
•Probabilities of the standard normal distribution Z~N(0,1) of
the form P(Za) are tabulated.
P(Za) =
a
dzf(z)
a
-
z2
1
dze2π
1 2
= from the table
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Probabilities of Z~N(0,1):
Suppose Z ~ N(0,1).
P(Za) =From
Table (A.3)P(Zb) = 1P(Zb) P(aZb) =
P(Zb)P(Za)
Note: P(Z=a)=0 for every a .
· We can transfer any normal distribution X~N(,) to the
standard normal distribution, Z~N(0,1) by using the following
result.
Result: If X~N(,), then N(0,1)~X
Z
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Example:
Suppose Z~N(0,1).
(1)P(Z1.50)=0.9332
Z 0.00 0.01 …
:
1.5 0.9332
:
(2) P(Z0.98)=1P(Z0.98)=1 0.8365= 0.1635
Z 0.00 … 0.08
: : :
: … …
0.9 0.8365
(3)P(1.33 Z2.42)= P(Z2.42) P(Z1.33)= 0.9922 (1-0.9082)= 0.9004
Z … 0.02 0.03
: :
1.3 0.9082
:
2.4 0.9922(4) P(Z0)=P(Z 0)=0.5
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Example:
Suppose Z~N(0,1). Find the value of k suchthatP(Zk)= 0.0207.Solution:
Probability is less than 0.5 K is negativeFind Z with Prob.=1-0.0207=0.9793 k = 2.04
Z … 0.04
: :
2.0 0.9793
:
Probabilities of X~N(,):
Result: X ~N(,)
~
XZ
aZ
aXaX
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aZPaXP)1
aZP1aXP1aXP)2
aZP
bZPaXPbXPbXaP)3
4) P(X=a)=0 for every a.
5) P(X) = P(X)=0.5
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Example:
Suppose that the hemoglobin level for healthy adults males has a normal distribution
with mean =16 and variance 2=0.81 (standard deviation =0.9).
(a) Find the probability that a randomly chosen healthy adult male has hemoglobin
level less than 14.
(b) What is the percentage of healthy adult males who have hemoglobin level less than
14?
Solution:
Let X = the hemoglobin level for a healthy adult male
X ~ N(,)= N(16, 0.9).
9.0
1614ZP
14ZP)14 P(X
= P(Z 2.22)=1-0.9868=0.0132
(a)
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(b) The percentage of healthy adult males who
have hemoglobin level less than 14 is
P(X 14) 100% = 0.01320 100% =1.32%
Therefore, 1.32% of healthy adult males have
hemoglobin level less than 14.
Example:
Suppose that the birth weight of babies has a normal distribution with mean =3.4 and
standard deviation =0.35.
(a) Find the probability that a randomly chosen baby has a birth weight between 3.0 and
4.0 kg.
(b) What is the percentage of babies who have a birth weight between 3.0 and 4.0 kg?
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Solution:
X = birth weight of a baby
= 3.4 = 0.35 (2 = 0.352 = 0.1225)
X ~ N(3.4,0.35 )
(a) P(3.0<X<4.0)=P(X<4.0)P(X<3.0)
0.3ZP
0.4ZP
35.0
4.30.3ZP
35.0
4.30.4ZP
= P(Z1.71) P(Z 1.14)= 0.9564 0.1271= 0.8293
(b) The percentage of babies who have a birth weight between 3.0 and 4.0 kg is
P(3.0<X<4.0) 100%= 0.8293 100%= 82.93%
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Notation:
P(ZZA) = A
Result:
ZA = Z1A
Example:
Z ~ N(0,1)
P(ZZ0.025) = 0.025
P(ZZ0.95) = 0.95
P(ZZ0.90) = 0.90
Example:
Z ~ N(0,1)
Z0.025 = 1.96
Z0.95 = 1.645
Z0.90 = 1.285
Z … 0.06
: :
1.9 0.975
P(ZZ0.025) = 0.025
Z0.025 = 1.96
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Example
In an industrial process, the diameter of a ball bearing is an important component part.
The buyer sets specifications on the diameter to be 3.00±0.01 cm. The implication is
that no part falling outside these specifications will be accepted. It is known that, in the
process, the diameter of a ball bearing has a normal distribution with mean 3.00 cm
and standard deviation 0.005 cm. On the average, how many manufactured ball
bearings will be scrapped?
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Solution:
=3.00
=0.005
X=diameter
X~N(3.00, 0.005)
The specification limits are:
3.00±0.01
x1=Lower limit=3.000.01=2.99
x2=Upper limit=3.00+0.01=3.01
P(x1<X< x2)=P(2.99<X<3.01)=P(X<3.01)P(X<2.99)
99.2ZP
01.3ZP
005.0
00.399.2ZP
005.0
00.301.3ZP
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= P(Z2.00) P(Z 2.00)
= 0.9772 0.0228
= 0.9544
Therefore, on the average, 95.44% of manufactured ball bearings will be accepted and
4.56% will be scrapped.
Example
Gauges are used to reject all components where a certain dimension is not within the
specifications 1.50±d. It is known that this measurement is normally distributed with
mean 1.50 and standard deviation 0.20. Determine the value d such that the
specifications cover 95% of the measurements.
Solution:
=1.5
=0.20
X= measurement
X~N(1.5, 0.20)
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The specification limits are:
1.5±d
x1=Lower limit=1.5d
x2=Upper limit=1.5+d
P(X> 1.5+d)= 0.025 P(X< 1.5+d)= 0.975
P(X< 1.5d)= 0.025
0.025)d5.1(X
P
025.0)d5.1(
ZP
025.020.0
5.1)d5.1(ZP
025.020.0
dZP
Z … 0.06
: :
-1.9 0.025
20.0
d:Note
96.120.0
d
025.0)20.0
dP(Z
Z0.025
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392.0d
)96.1)(20.0(d
96.120.0
d
The specification limits are:
x1=Lower limit=1.5d = 1.5 0.392 = 1.108
x2=Upper limit=1.5+d=1.5+0.392= 1.892
Therefore, 95% of the measurements fall within the specifications
(1.108, 1.892).
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Sampling Distributions
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Sampling distribution:
Definition
The probability distribution of a statistic is called a sampling
distribution.
· Example: If X1, X2, …, Xn represents a random sample of
size n, then the probability distribution of is called the
sampling distribution of the sample mean .
X
X
Sampling Distributions of Means:
If X1, X2, …, Xn is a random sample of size n taken from a normal distribution with mean and variance
2, i.e. N(,), then the sample mean has a normal distribution with meanX
X
)X(E
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and variance
nXVar X
22)(
· If X1, X2, …, Xn is a random sample of size n from N(,), then ~N(
, ) or ~N(, ).X
X
X
n
N(0,1)~n/
XZ)
n ,N( ~ X·
X
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Central Limit Theorem
If X1, X2, …, Xn is a random sample of size n from any distribution (population) with
mean and finite variance 2, then, if the sample size n is large, the random variable
n
XZ
/
is approximately standard normal random variable, i.e.,
approximately. N(0,1)~n/
XZ
)n
,N( ~X N(0,1)~n/
XZ
We consider n large when n30.
For large sample size n, has approximately a normal
distribution with mean and variance , i.e.,
X
n
2
)n
,N( ~X
approximately.
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Altman, D. G et al. BMJ 1995;310:298
Central Limit Theorem: the larger the sample size, the closer a distribution will approximate the normal distribution or
A distribution of scores taken at random from any distribution will tend to form a normal curve
jagged
smooth
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The sampling distribution of is used for inferences about the
population mean .
The standard deviation of the sampling distribution is called the
standard error and is equal to𝜎
𝑛
X
Example
An electric firm manufactures light bulbs that have a length of life that is approximately
normally distributed with mean equal to 800 hours and a standard deviation of 40
hours. Find the probability that a random sample of 16 bulbs will have an average life
of less than 775 hours.
Solution:
X= the length of life
=800 , =40
X~N(800, 40)
n=16
800X
1016
40
nX
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)10,800(N)n
,N( ~X
N(0,1)~10
800XZ
n/
XZ
10
800775
10
800XP
10
800775ZP
0062.0
50.2ZP
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Estimation & Confidence Interval
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Estimation Problems
· Suppose we have a population with some unknown
parameter(s).
Example: Normal(,)
and are parameters.
· We need to draw conclusions (make inferences) about the
unknown parameters.
· We select samples, compute some statistics, and make
inferences about the unknown parameters based on the
sampling distributions of the statistics.
Statistical Inference
(1) Estimation of the parameters
Point Estimation
Interval Estimation (Confidence Interval)
(2) Tests of hypotheses about the parameters
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Classical Methods of Estimation:
Point Estimation:
A point estimate of some population parameter is a single value of a statistic .
For example, the value of the statistic computed from a sample of size n is a point
estimate of the population mean .
x X
Interval Estimation (Confidence Interval = C.I.):
An interval estimate of some population parameter is an interval of the form ( , ),
i.e, << . This interval contains the true value of "with probability 1", that is P( << )=1UL LU
L U
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Example of Point Estimation
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Interval Estimation (Confidence Interval) of the Mean ():
An interval estimate of some population parameter is an interval of the form ( , ),
i.e, << . This interval contains the true value of "with probability 1", that is P( << )=1
L U
UL L U
( , ) is called a (1)100% confidence interval (C.I.) for .
1 is called the confidence coefficient
= lower confidence limit
= upper confidence limit
=0.1, 0.05, 0.025, 0.01 (0<<1)
UL
L
U
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Interval Estimation (Confidence Interval) of the Mean ():
If is the sample mean of a random sample of size n
from a population (distribution) with mean and known variance2, then a (1)100% confidence interval for can be calculatedas follows depending on whether the population variance 2 isknown or not.
n/XXn
1i
i
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),(
22n
ZXn
ZX
nZX
2
nZX
nZX
22
where is the Z-value leaving an area
of /2 to the right; i.e., P(Z> )=/2, or
equivalently, P(Z< )=1/2.
2
Z
2
Z2
Z
Note:
We are (1)*100% confident that ),(
22n
ZXn
ZX
(i) First Case: 2 is known:
The Z value is called the Z-score and the test is called the Z-test
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Example
The average zinc concentration recorded from a sample of zinc measurements in 36
different locations is found to be 2.6 gram/milliliter. Find a 95% and 99% confidence
interval (C.I.) for the mean zinc concentration in the river. Assume that the population
standard deviation is 0.3.
Solution:
= the mean zinc concentration in the river.
Population Sample
=?? n=36
=0.3 =2.6
First, a point estimate for is =2.6.
(a) We want to find 95% C.I. for .
= ??
95% = (1)100%
0. 95 = (1)
=0.05
/2 = 0.025
XX
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= Z0.025
= 1.96A 95% C.I. for is
2
Z
nZX
2
nZX
nZX
22
36
3.0)96.1(6.2
36
3.0)96.1(6.2
2.6 0.098 < < 2.6 + 0.098 2.502 < < 2.698 ( 2.502 , 2.698)We are 95% confident that ( 2.502 , 2.698).
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(b) Similarly, we can find that a 99% C.I. for is2.471 < < 2.729
( 2.471 , 2.729)We are 99% confident that ( 2.471 , 2.729)Notice that a 99% C.I. is wider than a 95% C.I. This is a tradeoff betweenaccuracy and precision
Theorem
If is used as an estimate of , we can then be
(1)100% confident that the error (in estimation) will
not exceed
X
nZ
2
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Example:
In previous example, we are 95% confident that the sample mean
differs from the true mean by an amount less than 6.2X
098.036
3.0)96.1(
2
nZ
Note:
Let e be the maximum amount of the error, that is ,
then: nZe
2
nZe
2
e
Zn
2
2
2
eZn
Theorem :
If is used as an estimate of , we can then be (1)100% confident that
the error (in estimation) will not exceed a specified
amount e when the sample size is
X
2
2
eZn
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Solution:
We have = 0.3 , e=0.05. Then by Theorem,
Therefore, we can be 95% confident that a random sample of size n=139 will provide
an estimate differing from by an amount less than e=0.05.
96.1
2
Z 1393.13805.0
3.096.1
22
2
eZn
Example
How large a sample is required in previous example if we want to be 95% confident
that our estimate of is off by less than 0.05?
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T-Distribution:
· Recall that, if X1, X2, …, Xn is a random sample of size n
from a normal distribution with mean and variance 2, i.e.
N(,), then
N(0,1)~n/
XZ
· We can apply this result only when 2 is known and number
of samples is 30 or more!
If 2 is unknown (or n<30), we replace the population variance
2 with the
sample variance · to have the following
statistic
1
)(1
2
2
n
XX
S
n
ii
nS
XT
/
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Result:
If X1, X2, …, Xn is a random sample of size n from a normal distribution with mean
and variance 2, i.e. N(,), then the statistic
nS
XT
/
has a t-distribution with =n1degrees of freedom (df), and we write T~ t().
Note:
t-distribution is a continuous distribution.
The shape of t-distribution is similar to the shape of
the standard normal distribution.
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Z and T Distributions
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t = The t-value above which we find an area equal to that
is P(T> t ) =
Since the curve of the pdf of T~ t() is symmetric about 0, we
have
t1 = t Values of t are tabulated in Tables.
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Example:
Find the t-value with =14 (df) that leaves an area
of:
(a) 0.95 to the left.
(b) 0.95 to the right.
Solution:
= 14 (df); T~ t(14)
(a) The t-value that leaves an area of 0.95 to the left is
t0.05 = 1.761
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(b) The t-value that leaves an area of 0.95 to the right is
t0.95 = t 1 0.95 = t 0.05 = 1.761
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Example:
For = 10 degrees of freedom (df), find t0.10 and t 0.85 .
Solution:
t0.10 = 1.372
t0.85 = t10.85 = t 0.15 = 1.093 (t 0.15 = 1.093)
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If and are the sample mean
and the sample standard deviation of a random sample of size n from a normal
population (distribution) with unknown variance 2, then a (1)100% confidence
interval for is :
nXXn
ii /
1
n
ii nXXS
1
2 )1/()(
Result:
),(
22n
StX
n
StX
n
StX
2
n
StX
n
StX
22
Interval Estimation (Confidence Interval) of the Mean ():
(ii) Second Case: 2 is unknown (or n is small):Recall:
)1 t(n~n/S
XT
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where is the t-value with =n1 degrees of freedom leaving
an area of /2 to the right; i.e., P(T> )=/2, or equivalently, P(T< )=1/2.
2
t
2
t
2
t
Example
The contents of 7 similar containers of sulfuric acid are 9.8, 10.2, 10.4, 9.8, 10.0, 10.2,
and 9.6 liters. Find a 95% C.I. for the mean of all such containers, assuming an
approximate normal distribution.
Solution:
.n=70.10/
1
nXXn
ii 283.0)1/()(
1
2
n
ii nXXS
First, a point estimate for is 0.10/1
nXXn
ii
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Now, we need to find a confidence interval for . = ??95%=(1)100% 0. 95=(1) =0.05 /2=0.025
= t0.025 =2.447 (with =n1=6 degrees of freedom)
A 95% C.I. for is2
t
n
StX
2
n
StX
n
StX
22
7
283.0)447.2(0.10
7
283.0)447.2(0.10
10.0 0.262< < 10.0 + 0.262 9.74 < < 10.26( 9.74 , 10.26)We are 95% confident that ( 9.74 , 10.26).
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To summarize: Estimation of the Mean ():
Recall:
XXE )(
nXVar X
22)(
n,N~X
N(0,1)~n/
XZ
(2 is known and
n>=30)
)1 t(n~n/S
XT
(2 is unknown or n is
smaller than 30)
We use the sampling distribution of to make
inferences about .
X