Large-Sample Tests for Population Proportions, Study notes of Data Analysis & Statistical Methods

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Lecture 16: Tests about a Population Proportion

Devore: Section 8.

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Large-Sample Tests

Let

p

denote the proportion of individuals or objects in a

either possesses a desired property (S) or it doesn’t (F).population who possess a specified property; thus, each object

Consider a simple random sample

X

1 ,... , X

n

. If the sample

size

n

is small relative to the population size, the number of

successes in the sample

X

has an approximately binomial

distribution. If

n

itself is also large, both

X

and the sample

proportion

ˆp

X/n

are approximately normally distributed

Large-sample tests concerning

p

are a special case of the more

general large-sample procedures for an arbitrary parameter

θ

.

We considered such a large-sample test before for the mean

μ

of an arbitrary distribution.

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hypothesisLet us consider first an upper-tailed test. It means having a null

H 0 : p = p 0

vs. an alternative

H

a

p > p

0 .

Under the null hypothesis, we have

E

p

) =

p

0

and

σ

ˆp

p

0 (

p

0 ) /n

; therefore, for large

n

the test statistic

Z

ˆp

p

0

p

0 (

p

0 ) /n

has approximately standard normal distribution

The rejection region is, clearly,

z

z

α

for a test of

approximately level

α

.

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The lower-tailed test has a rejection region

z

z

α

The two-tailed test has a rejection region

z

| ≥

z

α/

2

. The last

expression is a concise way of saying that

z

z

α/

2

or

z

z

α/

2 .

of the binomial distribution is reasonable:These tests are applicable whenever the normal approximation

np

0

,

n

p

0 )

.

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Type II Error and sample size determination

H Type II Error probability can be computed exactly as before. If

0

is not true, the true proportion

p

p

′

6

p

0

. Under

H a : p = p ′

we have

Z

is still approximately normal; however,

E

Z

p

′

−

p

0

p

0 (

p

0 ) /n

and

V

Z

p

′ (

p

′ ) /n

p

0 (

p

0 ) /n

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formula (before for the mean test. We only give the upper-tailed testThe formulas for the type II error are very similar to what we saw

H

a

p > p

0 )

β

p

′ ) = Φ

[ p 0 − p ′ + z α √ p 0

p

0 ) /n

p

′ (^

p

′ )^ /n

]

and the lower-tailed test formula (

H

a

p < p

0 )

β

p

′ ) = 1

− Φ [ p 0 − p ′ − z α √ p 0

p

0 ) /n

p

′ (^

p

′ )^ /n

]

two-tailed case, the formula is approximate as beforeSample size formulas can also be easily derived. In the

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Small Sample tests

rather than the normal approximation.n is small. They are based directly on the binomial distributionThese are test procedures for proportions when the sample size

Consider the alternative hypothesis

H

a

p > p

0

and let

X

be

yet again the number of successes in the sample size

n

. For a

test level

α

, we find the rejection region from

P

X

c

when

X

Bin

n

;

p

0 ))

− P ( X ≤ c − 1

when

X

Bin

n

p

0 ))

B

c

n, p

0 )

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It is usually not possible to find an exact value of

c

in this case;

the usual way out is to use the

largest rejection region of the

form

c, c

,... , n

satisfying the bound on the Type I error

To compute the Type II error for an alternative

p

′

> p

0 , we first

note that

X

Bin

n, p

′ )

if the alternative is true. Then,

β

p

′ ) =

P

X < c

when

X

Bin

n, p

′ )) =

B

c

−

n, p

′ )

calculation Note that this is a result of a straightforward binomial probability

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We have

x

and

np

0

. Thus, we must

find

c

such that

P

X

c

) = 1

B

c

for

X

Bin

. It is easy to check that the rejection

region will be

.

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