324 StatLecture Notes

(7 One- and Two-Sample Estimation Problem )

( Book*: Chapter 8 ,pg265)

Probability& Statistics for Engineers & ScientistsBy Walpole, Myers, Myers, Ye

Estimation

1

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ˆ

2

n

XXS

n

XP

n

XX

i

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• Point estimate:

• Is a single numerical value to estimate parameter

• Example :

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ˆ

2

n

XXS

n

XP

n

XX

i

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• Interval estimate

• Is two numerical values to estimate parameter

• It means to be in an interval s.a

• Lower bound upper bound

•(L) (U)

)( UL

Interval estimation

• An interval estimate of a population parameter is an interval of the form where and depend on the value of the statistic for a particular sample and also on the sampling distribution of .Since different samples will generally yield different values of and therefore different values of and . From the sampling distribution of we shall be able to determine and such that the is equal to any positive fractional value we care to specify. If for instance we find and such that:

then we have a probability of of selecting a random sample that will produce an interval containing .

UL ˆˆ ˆL

ˆU

ˆL

ˆU

ˆL ˆ

U

)ˆˆ( ULP

ˆL

ˆU

101)ˆˆ( forP UL

(1 )

• The interval computed from the selected sample, is then called a confidence interval.

• The fraction is called confidence coefficient or the degree of confidence.

• The end points and are called the lower and upper confidence limits.

• For Example:

when we have a confidence interval and so on, that we are confident that is between ,

UL ˆˆ

(1 )100%

(1 )

ˆL

ˆU

05.0 %95

%95 ˆL ˆ

U

Single Sample: Estimating the Mean:4 .9

1-Confidence Interval on µ ( σ² Known):

If is the mean of a random sample of size n from a population with known variance σ² , a confidence interval for µ is given by:

where is the -value leaving an area of to the right

X

%100)1(

1 / 2 1 / 2 (1)X Z X Zn n

1 / 2Z 2

1 / 2 1 / 2ˆ ˆ, (2)L UX Z X Z

n n

):1EX (The mean of the quality point averages of

a random sample of 36 college seniors is calculated to be 2.6.

Find the confidence intervals for the mean of the entire senior class. Assume that the population standard deviation is 0.3.

%95

Solution:

confidence interval for the mean µ :

------------------------------------------------

Thus , we have 95% confident that µ lies between 2.502 and 2.698

36 , 2.6 , 0.3n X

%95

12

1 0.95 0.05 0.025 1.962

Z

12

X Zn

698.2502.2

098.06.2)36

3.0(96.16.2

See Ex 9.2 pg 271

Theorem 9.1:

If is used as an estimate of µ , we can be confident that the error will not be exceed .

For example (1): e= (1.96) (0.3/6)=0.098 or

Theorem (2):

If is used as an estimate of µ , we can be confident that the error will not exceed a specified amount, when the sample size is:

The fraction of n is rounded up to next whole number.

X %100)1(

12

Zn

X %100)1(

e2

12 (3)

Z

ne

):2EX (

How large a sample is required in Ex. (1) if we want to be confident that our estimate of µ is off by less than 0.05?

Solution

n is rounded up to whole number.

%95

12

2

0.05 0.025 1.96,2

0.3 , 0.05

(1.96)(0.3)138.2976 138

0.05

Z

e

n

2- Confidence Interval of when σ² is Unknown n<30 :

If and s are the mean and standard deviation of a random sample , a ²σunknown variance with a normal population from

confidence interval for µ is given by:

where is the t-value with n-1 degrees of freedom leaving an area of to the right.

X

%100)1(

1 , 1 1 , 12 2

(4)n n

S SX t X t

n n

12

t

2

Ex 9.5 pg 275

The contents of 7 similar containers of sulfuric acid are 9.8, 10.2, 10.4, 9.8, 10, 10.2, 9.6 liters. Find a confidence interval for the mean of all such containers assuming an approximate normal distribution.

%95

Solution:

confidence interval for the mean :------------------------------------------------

0.025,6 0.975 , 61 , 1

2

1 , 12

7 , 10 , 0.283 ,

:1 0.95 0.05 0.025 2.4472

0.283( ) 10 (2.447)( ) 10 0.262

7

9.738 10.262 (9.738 10.262) 0.95

n

n

n X S

at t t t

SX t

n

P

=2.447

Two Samples: Estimating the 4 :9Difference between Two Means:

1- Confidence Interval for when Known:

If are the means of independent random samples of size n1 and n2 from populations with known variances respectively, a confidence interval for is given by:

where is the -value leaving an area of to the right .

21 2

2

2

1 and

21 XandX

2

2

2

1 and

%100)1( 21

2 2

1 21 2

11 22

( ) (5)X X Zn n

12

Z 2

EX (4):

A standardized chemistry test was given to 50 girls and 75 boys. The girls made an average grade of 76, while the boys made an average grade of 82. Find a

confidence interval for the difference where is the mean score of all boys and is the mean score of all girls who might take this test. Assume that the population standard deviations are 6 and 8 for girls and boys respectively.

%96 21

1 2

12

2 2

1 21 2

11 22

1 2 1 2

1 0.94 0.04 0.02 2.052

( )

36 64(82 76) (2.05) 6 2.571

50 75

3.429 8.571 (3.429 8.571) 0.96

Z

X X Zn n

P

girls Boys

Solution:

confidence interval for the mean :

-----------------------------------------------------

1 50n

1 76X

1 6

2 75n

2 82X

2 8

96% 1 2

See Ex 9.10 pg286

are the means of independent random samples of size n1 and n2 respectively from approximate normal populations with unknown but equal variances, a confidence interval for is given by:

, where

is the pooled estimate of the population standard deviation and is the t – value with degrees of

freedom leaving an area of to the right.

21 and If XX

%100)1( 21

1 21 ,

1 22

1 1( ) (6)P

vX X t S

n n

2 2

1 1 2 2

1 2

( 1) ( 1)(7)

2P

n S n SS

n n

1 ,2

vt

221 nnv

2

2-Confidence Interval for when Unknown but equal variances:

21 2

2

2

1 and

):288pg-11.9EX (

The independent sampling stations were chosen for this study, one located down stream from the acid mine discharge point and the other located upstream. For 12 monthly samples collected at the down stream station the species diversity index had a mean value and a standard deviation while 10 monthly samples had a mean index value

and a standard deviation . Find a 90% confidence interval for the difference between the population means for the two locations, assuming that the populations are approximately normally distributed with equal variances.

11.31 X 771.01 S

448.02 S04.22 X

Solution:

Station 1 Station 2

1 12n 2 10n

1 3.11X 2 2.04X

1 0.771S 2 0.448S

90% confidence interval for the mean :1 2

1 2

2 2

0.95 , 201 , 2

2

1 2 1 2

11(0.771) 9(0.448)0.646

12 10 2

1 0.90 0.1 0.05 1.7252

1 1(3.11 2.04) (1.725)(0.646) 1.07 0.477

12 10

0.593 1.547 (0.593 1.547) 0.90

P

n n

S

at t t

P

0.05,20

Single Sample Estimating a 10 :9Proportion:

Sample Confidence Interval for P:–Large

If is the proportion of successes in a random sample of size n and an approximate confidence interval for the

binomial parameter is given by:

Where is the Z- value leaving an area of to the right.

%100)1(

p

pq ˆ1ˆ 1

2

ˆ ˆˆ (8)

pqp Z

n

12

Z

2

):6EX(

A new rocket – launching system is being considered for deployment of small, short – rang rockets. The existing system has p=0.8 as the probability of a successful launch. A sample of 40 experimental launches is made with the new system and 34 are successful. Construct a 95% confidence interval for p

Solution:

a 95% confidence interval for p .

34ˆ ˆ40 , 0.85 , 0.15

40n p q

12

12

1 0.95 0.05 0.025 1.962

ˆ ˆ (0.85)(0.15)ˆ 0.85 (1.96) 0.85 (0.111)

40

0.739 0.961 (0.739 0.961) 0.95

at Z

pqp Z

n

p P p

See Ex 9.14 pg 297

:3Theorem

If is used as an estimate of p we can be

confident that the error will not exceed .

p %100)1(

12

ˆ ˆpqe Z

n

:7EX

In Ex. 7, find the error of p .

Solution:

The error will not exceed the following value:

12

ˆ ˆ (0.85)(0.15)(1.96) 0.111

40

pqe Z

n

Theorem :

If is used as an estimate of p we can be

confident that the error will be less than a specified amount e when the sample size is approximately:

Then the fraction of n is rounded up.

p %100)1(

2

1 2

2

ˆ ˆ(9)

Z pqn

e

):8EX(

How large a sample is required in Ex. 7 if we want to be

95% confident that our estimate of p is within 0.02?

Solution:

12

2

2

ˆ ˆ0.02 , 1.96 , 0.85 , 0.15

(1.96) (0.85)(0.15)1224.51 1225

(0.02)

e Z p q

n

Two Samples: Estimating the difference between two proportions

%100)1(

12

Z

• Ex 9.17 pg 301

A certain change n the process for manufacturing component parts is being considered. Samples are taken under both the existing and the new process results in an improvement.

If 75 of 1500 items from the existing process are found to be defective, and 80 of 2000 items from the new process found to be defective, find a 90% confidence interval for the rue difference in the proportions of defectives for the existing and new process respectively.

12

Z