6.5 One and Two sample Inference for Proportions

• np>5; n(1-p)>5• n independent trials;• X=# of successes• p=probability of a success• Estimate:

n

Xp ˆ

Mean and variance of p̂

22

1 1ˆ ( ) *

1 (1 ) (1 )ˆ( ) ( ) ( ) ( )

XEp E EX np p

n n nX np p p p

Var p Var Var Xn n n n

When n is large, approximate probabilities for can be found using the normal distribution with the same mean and standard deviation.

p̂

• An approximate confidence interval for p is

ˆ ˆ(1 )ˆ

p pp z

n

Sample Size• The sample size required to have a certain

probability that our error (plus or minus part of the CI) is no more than size ∆ is

2 21 1(1 ) ( ) * ( )

2 2

z zn p p

ˆ ˆ(1 )p pz

n

If you know p is somewhere …• If thenmaximum p(1-p)=0.3(1-0.3)=0.21

• If thenmaximum p(1-p)=0.4(1-0.4)=0.24

3.0p

2

20.21

zn

6.0p

2

20.24

zn

• Estimate p(1-p) by substitute p with the value closest to 0.5

(0, 0.1), p=0.1(0.3, 0.4), p=0.4(0.6, 1.0), p=0.6

Example

• A state highway dept wants to estimate what proportion of all trucks operating between two cities carry too heavy a load

• 95% probability to assert that the error is no more than 0.04

• Sample size needed if1. p between 0.10 to 0.252. no idea what p is

Solution

1. ∆=0.04, p=0.25

Round up to get n=451

2. ∆=0.04, p(1-p)=1/4

n=601

1.96z

19.450004

96.1)75.0(25.0

2

2

n

1.96z

25.60004.04

96.12

2

n

Tests of Hypotheses

• Null H0: p=p0

• Possible Alternatives: HA: p<p0

HA: p>p0

HA: pp0

Test Statistics

• Under H0, p=p0, and

• Statistic:

is approximately standard normal under H0 .

Reject H0 if z is too far from 0 in either direction.

n

pp

n

ppp

)1()1( 00ˆ

n

pp

ppppz

p )1(

ˆˆ

00

0

ˆ

0

Rejection Regions

Alternative Hypotheses

HA: p>p0 HA: p<p0 HA: pp0

Rejection Regions

z>z z<-z z>z/2 or

z<-z/2

Equivalent Form:

0 0

0 0 0 0

ˆ*

(1 ) (1 )

p p X npnz

np p np p

n

Example

• H0: p=0.75 vs HA: p0.75• =0.05• n=300• x=206• Reject H0 if z<-1.96 or z>1.96

68667.0300

206ˆ

n

Xp

Observed z value

• Conclusion: reject H0 since z<-1.96• P(z<-2.5 or z>2.5)=0.0124< a reject H0.

0.68667 0.752.5

0.75(1 0.75)300

206 2252.5

300(0.75)(1 0.75)

z

or

z

Example

• Toss a coin 100 times and you get 45 heads• Estimate p=probability of getting a headIs the coin balanced one? a=0.05Solution: H0: p=0.50 vs HA: p0.50

45.0100

45ˆ p

Enough Evidence to Reject H0?• Critical value z0.025=1.96• Reject H0 if z>1.96 or z<-1.96

• Conclusion: accept H0

15

5

)50.01)(50.0(100

)50.0(10045

z

Another example

• The following table is for a certain screening test

91010Results Negative

80140Result Positive

Cancer AbsentCancer Present

FNA status

Truth = surgical biopsy

Total

220

920

Total 150 990 1140

True positive 140sensitivity 0.93

True Positives False Negatives 150

• Test to see if the sensitivity of the screening test is less than 97%.

• Hypothesis

• Test statistic

0 0

0

: .97

: .97

H p p

Ha p p

0 0

ˆ 0 0

estimated proportion-prestated proportion

standard error of the estimated proportion

ˆ ˆ 140 150 .972.6325

(1 ) .97 (1 .97)150

p

z

p p p p

SE p p

n

• Check p-value when z=-2.6325, p-value = 0.004

• Conclusion: we can reject the null hypothesis at level 0.05.

What is the conclusion?

One word of caution about sample size:• If we decrease the sample size by a factor of 10,

911Results Negative

814Result Positive

Cancer AbsentCancer Present

FNA status

Truth = surgical biopsy

Total

22

92

Total 15 99 114

True positive 14sensitivity 0.93

True Positives False Negatives 15

And if we try to use the z-test,

0 0

ˆ 0 0

estimated proportion-prestated proportion

standard error of the estimated proportion

ˆ ˆ 14 15 .970.8324

(1 ) .97 (1 .97)15

p

z

p p p p

SE p p

n

P-value is greater than 0.05 for sure (p=0.2026). So we cannot reach the same conclusion.

And this is wrong!

So for test concerning proportions

We want

np>5; n(1-p)>5