Ch9. Inferences Concerning Proportions

Outline

Estimation of Proportions

Hypothesis concerning one Proportion

Hypothesis concerning several proportions

Analysis of r*c tables

Goodness of fit

Estimation

In acceptance sampling we are concerned with the proportion of defectives in a lot, and in life testing we are concerned with the percentage of certain components which will perform satisfactorily during a stated period of time.

It should be clear from these examples that problems concerning proportions, percentages, or probabilities are really equivalent.

Estimation

The point estimator of the population proportion, itself, is usually the sample proportion X/n.

If the n trials satisfy the assumptions underlying the binomial distribution(P105), we know the mean and the standard deviation of the number of success is given by and (1 )np pnp

Estimator

The mean and the standard deviation of the proportion of success (namely, of the sample proportion) are given by

npp

n and (1 ) (1 )np p p p

n n

The first of these results shows that the sample proportion is an unbiased estimator of the binomial parameter p.

Confidence interval

Construction of confidence interval for the binomial parameter p (estimator).

We first define 0x and 1x

0

0

( ; , ) / 2x

k

b k n p

and

such that

1

( ; , ) / 2n

k x

b k n p

Thus, we assert with a probability of approximate , and at least , that the inequality

0 1( ) ( )x p x x p

1 1

EX

Suppose we want to find approximate 95% confidence interval for p for samples of size n=20.

0( ; , ) 0.025B x n p 0x and 1

x can be determined by

11 ( 1; , ) 0.025B x n p

p 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

- 0 1 3 5 7 9 11 14

6 9 11 13 15 17 19 20 -

0x

1x

Confidence interval

For n is large

We construct approximate confidence intervals for the binomial parameter p by using the normal approximation to the binomial distribution. With the probability 1

/ 2 / 2(1 )

X npz z

np p

/ 2 / 2

(1 ) (1 )x x x x

x xn n n nz p zn n n n

This yields

EX.

If x=36 of n=100 persons interviewed are familiar with the tax incentives for installing certain energy-saving devices, construct a 95% confidence interval for the corresponding true proportion.

Solution: x/n =36/100=0.36

hence 0.36(1 0.36) 0.36(1 0.36)0.36 1.96 0.36 1.96

100 100p

0.266 0.454p

Maximum error

The error when we use X/n as estimator of p is given by |X/n -p|

Again using the normal distribution, we can assert with probability that the inequality

1

/ 2

(1 )| |X p p

p zn n

Maximum error of estimate

/ 2

(1 )p pE z

n

EX

In a sample survey conducted in a large city, 136 of 400 persons answered yes to the question of whether their city’s public transportation is adequate. With 99% confidence, what can we say about the maximum error if x/n=0.36 is used as an estimate of the corresponding true proportion?

/ 2

(1 ) 0.34 0.662.575 0.061

400

p pE z

n

Sample size determine

2/ 2(1 )( )z

n p pE

2/ 21( )4

zn

E

If p is unknown

If p is known

9.2 Hypothesis

The test of null hypothesis that a proportion equals some specified constant is widely used in sampling inspection, quality control, and reliability verification.

Statistic for large sample test concerning p

0

0 0(1 )

X npZ

np p

Null hypothesis 0p p

Criterion Region for testing (Large sample)

Alternative hypothesis

Reject null hypothesis if

0p p

0p p

0p p

0p p

Z z

Z z

/ 2 / 2Z z or Z z

EX

In a study designed to investigate whether certain detonator used with explosives in coal mining meet the requirement that at least 90% will ignite the explosive when charged, it is found that 174 of 200 detonators function properly. Test the null hypothesis p=0.9 again the alternative p<0.9 at the 0.05 level of significance.

Solution

1. Null hypothesis:

Alternative hypothesis

2. Level of significance: 0.05

3. Criterion: Reject the null hypothesis if Z<-1.645

4. Calculation:

5. The null hypothesis cannot be rejected. 6. P-value: 0.079 > level of significance 0.05

0.9p 0.9p

0

0 0

174 200(0.9)1.41

(1 ) 200(0.9)(0.1)

X npZ

np p

Hypothesis concerning several proportions

We compare the consumer response to two different products, when we decide whether the proportion of defectives of a given process remains constant from day to day.

Testing

1 2 kp p p p

Large-sample test

We require independent random samples of size if the corresponding number of successes are

the test we should use is based on the fact that

1) Large samples the sampling distribution of is approximately the standard normal distribution

2) The square of random variable having the standard normal distribution with 1 degree of freedom

3) The sum of k independent random variables having chi-square distribution with 1 degree of freedom is a random variable having the chi-square distribution with k degrees of freedom. (Proves are not required)

1 2, , , kn n n

1 2, , , kX X X

(1 )i i

i

i i

X npZ

np p

Cont.2

2

1

( )

(1 )

ki i i

i i i i

x n p

n p p

Is a value of random variable having approximately the chi-square distribution with k degrees of freedom.

In practice, we substitute for the pi, which under the null hypothesis are all equal, the pooled estimate

1 2

1 2

ˆ k

k

x x xp

n n n

The null hypothesis should be rejected if the difference between the and are large, the critical region is

where the number of degrees of freedom is k-1.

ix ˆin p2 2

Another approach

Sample1 ... Sample k Total

successes ...

Failures ...

Total ..

1x kx

1 1n x k kn x

1n kn

n xx

n

: 1, 2, 1, ,ijo i j k Define the observed cell frequency

The expected number of successes and failures for the j-th sample are estimated by

1 ˆj je n p and 2 ˆ(1 )j je n p

thus22

2

1 1

( )kij ij

i j ij

o e

e

Samples of three kind of materials, subjected to extreme temperature changes, produced the results should in the following table

A B C Total

successes 41 27 22 90

Failures 79 53 78 210

Total 120 80 100 300

Use the 0.05 level of significance to test whether, the probability of crumbling is the same of the three kinds of materials.

Solution

1. Null hypothesis:

Alternative hypothesis: are not all equal

2. Level of significance: 0.05

3. Criterion: Reject the null hypothesis if , degree 2

4. Calculation:

5. The null hypothesis cannot be rejected.

1 2 3p p p

2 5.991

11

90 12036

300e

12

90 8024

300e

90 3ˆ300 10

p 575.42

3013

e

Statistics for test concerning difference between two proportions

1 2

1 2

1 2

1 1ˆ ˆ(1 )( )

X Xn n

Z

p pn n

For large samples, is a random variable having approximately the standard normal distribution.

1 1 2 2

1 2 1 1 2 2/ 2

1 2 1 2

(1 ) (1 )x x x x

x x n n n nz

n n n n

Confidence interval for the values of 1 2p p

9.4 Analysis of r*c tables

The key random variable

22

1 1

( )r cij ij

i j ij

o e

e

is chi-square distribution with (r-1)(c-1) degrees of freedom

9.5 Goodness of fit

Goodness of fit: try to compare an observed frequency distribution with the corresponding values of an expected, or theoretical, distribution.

22

1

( )ki i

i i

o e

e

is a random variable has the chi-square distribution with k-m degrees of freedom, where k is the number of terms in the formula and m is the number of quantities, obtained from the observed data that are needed to calculate the expected frequencies.

EX

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