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 known” and “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+1r2r1+3r3

7 9 11 13531Xscale(µ=7,=2)

r3r2r1 0 +1 +2 +3 Zscale(µ=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 = ??????

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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

r

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

r

6.130.255

r

6.2683.241

d

d

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

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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

d

d

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