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Statistics 300:Elementary Statistics
Sections 7-2, 7-3, 7-4, 7-5
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Parameter Estimation
• Point Estimate–Best single value to use
• Question–What is the probability this
estimate is the correct value?
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Parameter Estimation
• Question–What is the probability this
estimate is the correct value?
• Answer–zero : assuming “x” is a
continuous random variable
–Example for Uniform Distribution
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If X ~ U[100,500] then
• P(x = 300) = (300-300)/(500-100)
• = 0
100 300 400 500
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Parameter Estimation
• Pop. mean – Sample mean
• Pop. proportion– Sample proportion
• Pop. standard deviation– Sample standard deviation
x
s
p
p
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Problem with Point Estimates
• The unknown parameter (, p, etc.) is not exactly equal to our sample-based point estimate.
• So, how far away might it be?
• An interval estimate answers this question.
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Confidence Interval
• A range of values that contains the true value of the population parameter with a ...
• Specified “level of confidence”.
• [L(ower limit),U(pper limit)]
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Terminology
• Confidence Level (a.k.a. Degree of Confidence)
– expressed as a percent (%)
• Critical Values (a.k.a. Confidence Coefficients)
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Terminology
• “alpha” “” = 1-Confidence–more about in Chapter 7
• Critical values–express the confidence level
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Confidence Interval for lf is known (this is a rare situation)
nzE
Ex
2
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Confidence Interval for lf is known (this is a rare
situation)if x ~N(?,)
n
zx
2
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Why does the Confidence Interval for
look like this ?
n
zx
2
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score.-z a into
valuean make
),(~
xn
Nx
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)(
is expression
score-z general The
x
z
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n
unchanged
x
x
x
is
and
: is
,for
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n
xz
x
is on based
score-z a so
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%952)(
2
:statementy probabilit a Make
n
xP
Using the Empirical Rule
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Normal Distribution
0 . 0 0
0 . 0 5
0 . 1 0
0 . 1 5
0 . 2 0
0 . 2 5
0 . 3 0
0 . 3 5
0 . 4 0
0 . 4 5
-3 -2 -1 0 1 2 3
Value of Observation
Re
lati
ve
lik
elih
oo
d
2
2
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Check out the “Confidence z-scores”
on the WEB page.
(In pdf format.)
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Use basic rules of algebra to rearrange the parts of
this z-score.
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95.022
:statementy probabilit theManipulate
nx
nP
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95.022
:statementy probabilit theManipulate
nx
nxP
Confidence = 95% = 1 - 95% = 5%
= 2.5% = 0.025
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95.022
terms theoforder
thechange and (-1)by hrough multiply t
:statementy probabilit theManipulate
nx
nxP
Confidence = 95% = 1 - 95% = 5%
= 2.5% = 0.025
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Confidence Interval for lf is not known (usual situation)
n
stx
2
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Sample Size Neededto Estimate within E,with Confidence = 1-
2
2ˆ
E
Zn
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Components of Sample SizeFormula when Estimating
• Z/2 reflects confidence level
– standard normal distribution
• is an estimate of , the standard deviation of the pop.
• E is the acceptable “margin of error” when estimating
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Confidence Interval for p
• The Binomial Distribution gives us a starting point for determining the distribution of the sample proportion : p
trials
successes
n
xp ˆ
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For Binomial “x”
npq
np
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For the Sample Proportion
x is a random variablen is a constant
xnn
xp
1ˆ
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Time Out for a Principle:
If is the mean of X and “a” is a constant, what is the mean of aX?
Answer:
a
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Apply that Principle!
• Let “a” be equal to “1/n”
• so
• and
n
XX
naXp
1ˆ
pnpn
npaa xp
1
)(ˆ
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Time Out for another Principle:
If is the variance of X and “a” is a constant, what is the variance
of aX?
Answer:
2x
222xaX a
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Apply that Principle!
• Let x be the binomial “x”
• Its variance is npq = np(1-p), which is the square of is standard deviation
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Apply that Principle!
• Let “a” be equal to “1/n”
• so
• and
n
XX
naXp
1ˆ
)(/1 2222ˆ npqna Xp
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Apply that Principle!
n
pq
and
n
pqnpq
n
p
p
ˆ
2ˆ
21
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When n is Large,
n
pqpNp ,~ˆ
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What is a Large “n”in this situation?
• Large enough so np > 5
• Large enough so n(1-p) > 5
• Examples:– (100)(0.04) = 4 (too small)
–(1000)(0.01) = 10 (big enough)
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Now make a z-score
npq
ppz
ˆ
And rearrange for a CI(p)
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Make a probability statement:
ˆ1.96 1.96 95%
p pP
pqn
Using the Empirical Rule
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Normal Distribution
0 . 0 0
0 . 0 5
0 . 1 0
0 . 1 5
0 . 2 0
0 . 2 5
0 . 3 0
0 . 3 5
0 . 4 0
0 . 4 5
-3 -2 -1 0 1 2 3
Value of Observation
Re
lati
ve
lik
elih
oo
d
2
2
![Page 41: Statistics 300: Elementary Statistics Sections 7-2, 7-3, 7-4, 7-5](https://reader036.vdocument.in/reader036/viewer/2022062519/5697bfa41a28abf838c973cd/html5/thumbnails/41.jpg)
Use basic rules of algebra to rearrange the parts of
this z-score.
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Manipulate the probability statement:
Step 1: Multiply through by :
ˆ1.96 1.96 0.95
pq
n
pq pqP p p
n n
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Manipulate the probability statement:
ˆStep 2: Subract from all parts of the expression:
ˆ ˆ1.96 1.96 0.95
p
pq pqP p p p
n n
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Manipulate the probability statement:
Step 3: Multiply through by -1:
(remember to switch the directions of < >)
ˆ ˆ1.96 1.96 0.95pq pq
P p p pn n
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Manipulate the probability statement:
Step 4: Swap the left and right sides to
put in conventional < < form:
ˆ ˆ1.96 1.96 0.95
p
pq pqP p p p
n n
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Confidence Interval for p(but the unknown p is in the formula. What can we do?)
n
pqzp
2ˆ
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Confidence Interval for p(substitute sample statistic for p)
n
qpzp
ˆˆˆ
2
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Sample Size Neededto Estimate “p” within E,
with Confid.=1-
qpE
Zn ˆˆ
2
2
2
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Components of Sample SizeFormula when Estimating “p”
• Z/2 is based on using the standard normal distribution
• p and q are estimates of the population proportions of “successes” and “failures”
• E is the acceptable “margin of error” when estimating
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Components of Sample SizeFormula when Estimating “p”
• p and q are estimates of the population proportions of “successes” and “failures”
• Use relevant information to estimate p and q if available
• Otherwise, use p = q = 0.5, so the product pq = 0.25
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Confidence Interval for starts with this fact
then
square) (chi ~
1 22
2
sn
),(~ if Nx
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What have we studied already that connects with
Chi-square random values?
square) (chi ~
1 22
2
sn
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2
2
2 2
2
2
11 1
xn
n s n
x
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22
2
2 a sum of squared
standard normal values
x x
z
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Confidence Interval for
2
2
2
2
1
1
L
R
snUB
snLB