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Lecture

Lecture 2

16 Pages
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Department
Quantitative Methods
Course Code
QMS 202
Professor
Clare Chua

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1/17/2011
1
Chapter 9
Confidence Interval Estimation
Topics covered:
Confidence Interval Estimation for the Mean (1 known)
Confidence Interval Estimation for the Mean (1 unknown)
Confidence Interval Estimation for the Proportion
Determining Sample Size
Objective
Learn how to constructconstruct and
interpretinterpret confidence interval
estimates
estimates
For the mean
For the proportion
Confidence Interval
Estimation
Population Mean PPopulation Proportion E
X Normal and/or nt30?
If no stop.
np t5 and n(1-p) t5?
If no stop.
Vknown? Vunknown?
Limits: Limits:
df=n-1
n
1
zx r
Limits:
p=x/n
n
s
txr
n
)p1(p
zp
r
WEEK 1WEEK 1 WEEK 2WEEK 2
Frequently asked questions
1. “1is knownand 1is unknown
1is given and 1is not given
2. What is Z scores/values?
Go to this link to understand "z-score" :
http://www.youtube.com/watch?v=1xhCL5m4nI0&feature=
BF&list=QL&index=1
http://www.youtube.com/watch?v=s0lLBcARxL4&feature=
mfu_in_order&list=UL
Cut and paste the link on the URL box.
3. What are you estimating? Parameters Por S
4. How do you interpret CI?
Go to this link
http://www.youtube.com/watch?v=hP6flJdoIxc&feature=f
vw
=2
Review
+2+1r2r1+3r3
7 9 11 13531Xscale=7,=2)
r3r2r1 0 +1 +2 +3 Zscale=0,=1)
Convert any normal random variable X to standardized normal
random variable Z.
If X is normally distributed N(, 1), the standardized variable Z has
a standard normal distribution with mean=0 (or =0) and
standard deviation 1 (or 1=1), denoted N(0,1)
V
P
X
Z
% Confidence Interval when 1is given : n
*x z2/
V
rD
We needed “z value” for a % confidence Interval.
90% confidence levelConfidence coefficient of 0.90 Z= ??????
95% confidence levelConfidence coefficient of 0.95 Z= ??????
97% confidence levelConfidence coefficient of 0.97 Z= ??????
99% confidence levelConfidence coefficient of 0.99 Z= ??????
%C fidItlh
iti
s
%
C
on
fid
ence
I
n
t
erva
l
w
h
en 1
i
s no
t
g
i
ven :
n
s
tx1n,
2
D
r
We needed t value” for a % confidence Interval.
Suppose we needed to find t, when sample size (denoted as n) is 25.
90% confidence levelConfidence coefficient of 0.90 t = ??????
95% confidence levelConfidence coefficient of 0.95 t = ??????
97% confidence levelConfidence coefficient of 0.97 t = ??????
99% confidence levelConfidence coefficient of 0.99 t = ??????
www.notesolution.com
1/17/2011
2
TEMPLATE : Confidence Interval for population mean
when 1is known
:
Define : Estimate µ
95% CI :
n
1
Zx
2
.
r
8
1
Note:
Therearethreewaystofindthe
50
8
.
1
9566.10.20 r
5.00.20 r
5.205.19 dd
There
are
three
ways
to
find
the
Confidence interval
1. Use the formula, find z value
using the INVN function
from CASIO calculator
2. Use the formula, find z value
using SPSS function:
IDF.Normal (prob, mean, stdev)
3. Use the INTR function of CASIO
Calculator to find Confidence
Interval directly
TEMPLATE : Confidence Interval for population mean
when 1is unknown
:
Define : Estimate µ
95% CI :
n
s
tx1n,
2
.
r
33
0640
2
0
255
Note:
There are two ways to find the
Confidence interval
1. Using the table to find t value an
d
th flt fidth CI
25
0640
.
2
0
.
255
6.130.255
r
6.2683.241
use
th
e
f
ormu
l
a
t
o
fi
n
d
th
e
CI
.
2. Using the formula, find t value
using SPSS function:
IDF.T (./2, degree of freedom)
and calculator CI using SPSS
compute function.
3. Using the INTR function of CASIO
Calculator to find Confidence
Interval directly
Confidence Interval
Estimation
Population Mean PPopulation Proportion E
X Normal and/or nt30?
If no stop.
np t5 and n(1-p) t5?
If no stop.
Vknown? Vunknown?
Limits: Limits:
df=n-1
n
1
zx r
Limits:
p=x/n
n
s
txr
n
)p1(p
zp
r
WEEK 1WEEK 1 WEEK 2WEEK 2
Interpreting the CI
Read page 369, ….with 95%
confidence, you conclude that the
mean amount of all the sales
ii ibt
LI (l
i
nvo
i
ces
i
s
b
e
t
ween
LI
(l
ower
interval) and UI (upper interval).
Referring to population”
Confidence Interval Estimation
for the
p
ro
p
ortion
pp
We extends the concept of CI to categorical data
Here you are concerned with estimating the proportion of
items in a population having a certain characteristic of
interest (e.g. people who have an iphone)
Confidence Interval Estimation for
the proportion
The unknown population proportion is
represented by S
The point estimate for Sis the sample
proportion
denoted as
p
where p=X/n
proportion
,
denoted
as
p
,
where
p=X/n
Where n = sample size (number of observed items/subjects in a sample)
And X=number of items in the sample having the characteristic of interest
www.notesolution.com
1/17/2011
3
95% CI :
)
p
1
(
p
z
p
r
Note:
There are three ways to find the
Confidence interval
Check the conditions: np > 5 and n(1-p) > 5
State the n and p values
If conditions are satisfied, proceed to find CI
TEMPLATE : Confidence Interval for population
proportion, E
Define : est E= population proportion of _________________
_
95%
CI
:
n
)
p
(
p
z
p
r
50
)24.01(24.0
575.224.0
r
16.024.0r
40.008.0dd
1. Use the formula, find z value
using the INVN function
from CASIO calculator
2. Use the formula, find z value
using SPSS function:
IDF.Normal (prob, mean,
stdev)
3. Use the INTR function of
CASIO
Calculator to find Confidence
Interval directly
Note
n
)p1(p
zp
r
The equation contains a Z statistic since you can
use the normal distribution to approximate the
Binomial distribution when the sample size is
sufficiently large.
That is why we check the condition before we
proceed to calculate the Confidence Interval for S
Example
In a study reported in the Wall Street Journal on
April 4, 1999. The Tupperware Corporation
surveyed 1007 U.S. workers. Of the people
surveyed, 665 indicated that they take their
lunch to work with them. Of these 665 taking
lunch, 200 reported that they take them in
brownbags
brown
bags
.
a) Set up a 95% confidence interval estimate of
the population proportion of U.S. workers who
take their lunch to work with them.
Check the conditions: np > 5 and n(1-p) > 5
State the n and p values
Solution to Question 5-36
E= population proportion of U.S. workers who take their lunch to work with them
a) Set up a 95% confidence interval estimate of the
population proportion of U.S. workers who take their lunch to work with them.
n=1007
p=x/n = 665/1007 = 0.6604
np 1007(06604) 665 0228
Are the conditions satisfied?
np
=
1007(0
.
6604)
=
665
.
0228
n(1-p) = 1007 (1-0.6604) =341.9772
If yes, proceed to find CI
95% CI :
n
)p1(p
zp
r
Note:
There are three ways to find the
Confidence interval
Solution to Question 5-36
est.E= population proportion of U.S. workers who take their lunch to work with them
a) Set up a 95% confidence interval estimate of the
population proportion of U.S. workers who take their lunch to work with them.
Confidence
interval
1. Use the formula, find z value
using the INVN function
from CASIO calculator
2. Use the formula, find z value
using SPSS function:
IDF.Normal (prob, mean, stdev)
3. Use the INTR function of CASIO
Calculator to find Confidence
Interval directl
y
95% CI :
n
)p1(p
zp
r
Solution to Question 5-36
est. E= population proportion of U.S. workers who take their lunch to work with them
a) Set up a 95% confidence interval estimate of the
population proportion of U.S. workers who take their lunch to work with them.
Aside:
To find Z value
Z (1-0.95)/2= Z0.05/2
Z0.025InvN(0.025,1,0)=r1.9599
Round the value of Z to 4 decimals
1007
)6604.01(6604.0
9599.16604.0
r
0292.06604.0
r
6896.06311.0
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Description
1/17/2011 Objective Learn how to coonnssttrrucctt and Chapter 9 intteerrprreett confidence interval Confidence Interval Estimation esttmamatess Confidence Interval Estimation for the Mean ( known) For the mean Confidence Interval Estimation for the Mean ( unknown) For the proportion Confidence Interval Estimation for the Proportion Determining Sample Size Frequently asked questions Confidence Interval Estimation 1. is known and is unknown is given and is not given Population Mean 2 Population Proportion - 2. What is Z scores/values? X Normal and/or nJ30? np J 5 and n(1-p) J 5? Go to this link to understand "z-score" : If no stop. If no stop. http://www.youtube.com/watch?v=1xhCL5m4nI0&feature= BF&list=QL&index=1 8 known? 8 unknown? http://www.youtube.com/watch?v=s0lLBcARxL4&feature= Limits: p(1 p) mfu_in_order&list=UL p H z n Cut and paste the link on the URL box. p=x/n x H z 3. What are you estimating? Parameters 2 or 5 Limits: n Limitsx H t 4. How do you interpret CI? df=n-1 n Go to this link http://www.youtube.com/watch?v=hP6flJdoIxc&feature=f WE EEEK 11 WEEEEKK 2 vw Review 8 % Confidence Interval when is xivHnz:, / 2 =2 n We needed z value for a % confidence Interval. 90% confidence leveConfidence coefficient of 0.90??? 95% confidence leveConfidence coefficient oZ= ?????? 97% confidence leveConfidence coefficient oZ= ?????? 99% confidence leveConfidence coefficient oZ= ?????? % Confdence IIntterhen iis nott giiven : s x H t , ,n 1 We needed t value for a % confidence Interval.n H3 H2 H1 +1 +2 +3 Suppose we needed to find t, when sample size (denoted as n) is 25. 1 3 5 7 9 11 13 Xscale(=7, =2) H3 H2 H1 0 +1 +2 +3 Zscale(=0, =1) 90% confidence levConfidence coefficient ot = ?????? Convert any normal random variable X to standardized normal 95% confidence levConfidence coefficient ot = ?????? random varX 2 Z. 97% confidence levConfidence coefficient ot = ?????? Z 8 99% confidence level If X is normally distributed N(, ), the standardized variable Z has Confidence coefficient ot = ?????? standard deviation 1 (or =1), denoted N(0,1)r =0) and 1 www.notesolution.com 1/17/2011 TEMPLATE : : Confidence Interval for population mean TEMPLATE :: Confidence Interval for population mean when is known when is unknown Define : Estimate Define : Estimate 95% CI : s 95% CI : x H t. ,n 1 x H Z . 2 n Note: 2 n Confidence intervalto find the Note: 1. Using the table to find t value and 1.8 Theearehre wayyo ndheth 33 use he fformuld heCII.fiind 20 .0 H 1.9566 Confidence interval 22555 .0 H 22.006440 50 1. Use the formula, find z value 25 2. Using the formula, find t value ufrom CASIO calculatorn using SPSS function: 2. Use the formula, find z value 255 .0 H 13 .6 IDF.T (./2, degree of freedom) 20 .0 H 0.5 using SPSS function: compute function. using SPSS IDF.Normal (prob, mean, stdev) 3. Using the INTR function of CASIO 3. Use the INTR function of CASIO 241 .3 @ @ 268 .6 Calculator to find Confidence Calculator to find Confidence Interval directly 19 .5 @ @ 20 .5 Interval directly Confidence Interval Estimation Interpreting the CI Population Mean 2 Population Proportion - Read page 369, .with 95% X Normal and/or nJ30? np J 5 and n(1-p) J 5? If no stop. If no stop. confidence, you conclude that the mean amount of all the sales 8 known? 8 unknown? iinvoiices ibei ttweenLII(llower Limits: interval) and UI (upper interval). p H z(1 p) p=x/n n Limits:H z n Limitx H t df=n-1 n Referring to population WEEEEK 11 WEEEEKK 2 Confidence Interval Estimation for the proportion The unknown population proportion is represented by 5 Confidence Interval Estimation The point estimate for 5 is the sample for the p propportion prropoortion, deenotedd as p, whhere p==X/nn Where n = sample size (number of observed items/subjects in a sample) We extends the concept of CI to categorical data And X=number of items in the sample having the characteristic of interest Here you are concerned with estimating the proportion of items in a population having a certain characteristic of interest (e.g. people who have an iphone) 2 www.notesolution.com 1/17/2011 TEMPLATE : Confidence Interval for population proportion, - Note Define : est - = population proportion of __________________ State the n and p values p(1 p) p H z Check the conditions: np > 5 and n(1-p) > 5 n If conditions are satisfied, proceed to find CI
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