Horng-Chyi HorngStatistics II 91 Inference on the Variance of a Normal Population (I) H 0 : 2 = 0 H 1 : 2 0 , where 0 is a specified.

Horng-Chyi HorngStatistics II1

Inference on the Variance of a Normal Population (I)

H0: 2 = 0

H1: 2 0 , where 0

is a specified constant. Sampling from a normal distribution with unknown mean

and unknown variance 2, the quantity

has a Chi-square distribution with n-1 degrees of freedom. That is,

2

22 1

Sn

212

22 ~

1

n

Sn


Inference on the Variance of a Normal Population (II)

Let X1, X2, …, Xn be a random sample for a normal distribution with unknown mean and unknown variance 2. To test the hypothesis

H0: 2 = 0

H1: 2 0 , where 0

is a specified constant.

We use the statistic

If H0 is true, then the statistic has a chi-square distribution with n-1 d.f..


k

k

k

xexk

xf xkk

2

addition,In

freedom. of degrees ofnumber theis

0 2/2

1

:ondistributi square-chi of PDF

2

2/12/2/


The Reasoning

For H0 to be true, the value of 02 can not be too large or

too small.

What values of 02 should we reject H0? (based on

value)

What values of 02 should we conclude that there is not

enough evidence to reject H0?



Example 8-11An automatic filling machine is used to fill bottles with liquid detergent. A random sample of 20 bottles results in a sample variance of fill volume of s2 = 0.0153 (fluid ounces)2. If the variance of fill volume exceeds 0.01 (fluid ounces)2, an unacceptable proportion of bottles will be underfilled and overfilled. Is there evidence in the sample data to suggest that the manufacturer has a problem with underfilled and overfilled bottles? Use = 0.05, and assume that fill volume has a normal distribution.



Hypothesis Testing on Variance - Normal Population

H 1 Te s t S ta tis t ic R e je c t H 0 i f

0

2 21,2/1

20

21,2/

20 or nn

> 0

2 21,

20 n

< 0

2

20

220

1

Sn

21,1

20 n


Finding P-Values

Steps:

1. Find the degrees of freedom (k = n-1)in the the 2-table.

2. Compare 02 to the values in that row and find the

closest one.

3. Look the value associated with the one you pick. The p-value of your test is equal to this value.

In example 8-11, 02 = 29.07, k = n-1 = 19, 0.05 < P-Value

< 0.10 because the 2 value associated with (k = 19, = 0.10) is 27.20 while the 2 value associated with (k = 19, = 0.05) is 30.14


P-Values of Hypothesis Testing on Variance

H 1 T e s t S t a t i s t i c P - V a l u e

0

2 20

21

20

21 ,min*2 nn PPP

> 0

2 20

21 nPP

< 0

2

20

220

1

Sn

20

21 nPP


The Operating Characteristic Curves- Chi-square test

Use to performing sample size or type II error calculations. The parameter is defined as:

for various sample sizes n, where denotes the true value of the standard deviation.

Chart VI I,j,k,l are used in chi-square test. (pp. A16-A17)

0


Example 8-12



Construction of the C.I. on the Variance

In general, the distribution of

is chi-square with n-1 d.f.

2

22 1

Sn

Use the properties of t with n-1 d.f.,

111

11

1

21,2/1

22

21,2/

2

21,2/2

22

1,2/1

21,2/

221,2/1

nn

nn

nn

SnSnP

SnP

P


Formula for C.I. on the Variance


Formula for C.I. on the Variance- One-sided


Example 8-13 Reconsider the bottle filling machine problem in Example 8-11. Find a 95% upper-C.I. on the variance?

N = 20, s2 = 0.0153

Therefore, 2 (20-1)0.0153/10.117 = 0.0287

The 95% upper-C.I. on the variance is 0.0287.

In addition, the 95% upper-C.I. on the standard deviation is 0.0287 = 0.17.

117.10219,95.0


Inference on a Population Proportion(I)

H0: p = p0

H1: p p0 , where p0 is a specified constant. P = X/n, in which X is a binomial variable, i.e., the number

of success in n trials. E(X) = np, V(X) = npq = np(1-p)

If H0 is true, then using the normal approximation to the binomial, the quantity

follows the standard normal distribution(Z).

n

pp

pP

pnp

npXZ

00

0

00

00

11


Inference on a Population Proportion (II)

Let x be the number of observations in a random sample of size n that belongs to the class associated with p. To test the hypothesis

H0: p = p0

H1: p p0 , where p0 is a specified constant.

We use the statistic

tion.ple proporis the samn

xpwhere

npp

pp

pnp

npxZ

11 00

0

00

00


The Reasoning

For H0 to be true, the value of Z0 can not be too large or too small.

What values of Z0 should we reject H0? (based on value)

What values of Z0 should we conclude that there is not enough evidence to reject H0?


Example 8-14A semiconductor manufacturing produces controllers used in automobile engine applications. The customer erquires that the process fallout or fraction defective at a critical manufacturing step not exceed 0.05 and that the manufacturing demonstrate process capacity at this level of quality using = 0.05. The semiconductor manufacturer takes a random sample of 200 devices and finds that four of them are defective. Can the manufacturer demonstrate process capacity for the customer?


The parameter of interest is the process fraction defective. H0: p = 0.05

H1: p < 0.05

This formulation of the problem will allow the manufacturer to make a strong claim about process capacity if the null hypothesis H0: p = 0.05 is rejected.

= 0.05, x = 4, n = 200, and p0 = 0.05.

To reject H0: p = 0.05, the test statistic Z0 must be less than -z0.05 = -1.645

Conclusion: Since Z0 = -1.95 < -z0.05 = -1.645, we reject H0 and conclude that the process fraction defective p is less than 0.05. The P-value for this value of the test statistic Z0 is P = 0.0256, which is less than = 0.05. We conclude that the process is capable.

95.1

200)95.0(05.0

05.0200/4

1

95.1)95.0)(05.0(200

)05.0(2004

1

00

00

00

00

npp

ppZ

pnp

npxZ


Hypothesis Testing on a Population Proportion

H1 Test Statistic Reject H0 if

p p0 Z0 > z or Z0 < - z

p > p0 Z0 > z

p < p0

npp

ppZ

pnp

npXZ

00

00

00

00

1

or 1

Z0 < -z


P-Values of Hypothesis Testing on a Population Proportion

H1 Test Statistic P-Value

p p0 P = 2*P(Z > |Z0|)

p > p0 P = P(Z > Z0)

p < p0

npp

ppZ

pnp

npXZ

00

00

00

00

1

or 1

P = P(Z < Z0)


How to calculate Type II Error? (I)(H0: p = p0 Vs. H1: p p0)

Under the circumstance of type II error, H0 is false. Supposed that the true value of the population proportion is p. The distribution of Z0 is:

npp

npp

npp

ppN

npp

ppZTherefore

N

npp

ppTrue

/1

/1,

/1~

1 ,

)1,0(~1

0000

0

00

00


How to calculate Type II Error? (II) - refer to section &4.3 (&8.1)

Type II error occurred when (fail to reject H0 while H0 is false)

npp

npp

npp

ppNZZZZ

/)1(

/)1( ,

/)1(~ where

0000

00

20

2

npp

nppZpp

npp

nppZpp

npp

nppZppZ

npp

nppZppP

nppnpp

npp

ppZ

nppnpp

npp

ppZ

nppnpp

npp

ppZ

P

ZZZP

/)1(

/)1(

/)1(

/)1(

/)1(

/)1(

/)1(

/)1(

/)1(/)1(

/)1(

/)1(/)1(

/)1(

/)1(/)1(

/)1(

002/0002/0

002/0002/0

00

00

02/

00

00

00

00

00

02/

2/02/

Therefore,


Formula for Type II Error

Two-sided alternative H1: p p0

One-sided alternative H1: p < p0

One-sided alternative H1: p > p0

npp

nppZpp

npp

nppZpp

/)1(

/)1(

/)1(

/)1( 002/0002/0

npp

nppZpp

/)1(

/)1(1 000

npp

nppZpp

/)1(

/)1( 000


The Sample Size (I)

Given values of and p, find the required sample size n to achieve a particular level of .

2

0

002/

002/0

0002/0

002/0002/0

)1()1(

/)1(

/)1( Then,

Let

0 when /)1(

/)1(

/)1(

/)1(

/)1(

/)1( Since

pp

ppZppZn

npp

nppZppZ

Z

ppnpp

nppZpp

npp

nppZpp

npp

nppZpp


The Sample Size (II)

Two-sided Hypothesis Testing

One-sided Hypothesis Testing


Example 8-15

(1) Consider the semiconductor manufacturer from Example 8-14. Suppose that his process fallout is really p = 0.03. What is the -error for his test of process capacity, which uses n = 200 and = 0.05?

(2) Suppose that the semiconductor manuafcturer was willing to accept a -error as large as 0.10 if the true value of the process fraction defective was p = 0.03. If the manufacturer continue to use = 0.05, what sample size would be required?


(1) Since H1: p < 0.05, therefore

(2) The sample size required for this one-sided alternative is

67.0

)44.0(1

200/)97.0(03.0

200/)95.0(05.0)645.1(03.005.01

/)1(

/)1(1 000

npp

nppZpp

832

05.003.0

)97.0(03.028.1)95.0(05.0645.1

)1()1(

2

2

0

00

pp

ppZppZn


Formula for C.I. on the Population Proportion

Z

npp

pp~

)1(


Formula for One-Sided C.I. on the Population Proportion


Example 8-16

In a random sample of 85 automobile engine crankshaft bearings, 10 have a surface finish that is rougher than the specifications allow. Construct a 95% C.I. on the population proportion p?

Sol:

19.005.0 85

)88.0(12.096.112.0

85

)88.0(12.096.112.0

)1()1(

95

12.085/10/ ,85

025.0025.0

p

p

n

ppzpp

n

ppzp

p is% C.I. on

nxpn


Choice of Sample Size for C.I. on a Population Proportion

where E is the half-width of the C.I.

Since the max value for p(1-p) is 0.25 for 0 p 1, we can use the following formula instead.

||

ppE


Horng-Chyi HorngStatistics II 91 Inference on the Variance of a Normal Population (I) H 0 : 2 = 0 H 1 : 2 0 , where 0 is a specified.

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