Download Inferences for proportions • sample proportion ( ˆ p ) fraction of

Chapter 8
1
Inferences for proportions
• sample proportion ( pˆ )
fraction of measurements in the sample that are
“successes”:
€
pˆ = (# successes in sample) / n
• population proportion ( p )
€
fraction of the measurements in the population that
are “successes”
Sampling distribution of pˆ
From an SRS of size n selected from a large population
€
(sample should be no more than 10% of the population)
with population proportion p, we compute the sample
proportion pˆ .
• mean of the sampling distribution is p.
€
• standard deviation of the sampling distribution,
also called the standard error for pˆ , is
p(1− p)
€
n
• distribution becomes more nearly normal as the
€
sample size increases
Chapter 8
2
Procedures for population proportions
From an SRS selected from a large population (and
at most 10% of the size of the population),
determine the sample statistic pˆ .
An approximate level C confidence interval for
€
p is
pˆ(1− pˆ )
pˆ ± z*
n
€
where z* is the upper
€
€
1−C
normal critical value.
2
[TI-83: STAT TESTS 1-PropZInt… ]
Chapter 8
3
One-sample z test for pˆ
Assumptions: SRS selected from a population with
€
population proportion pˆ , large enough that both np0
(number of successes) and n(1 – p0) (number of failures)
are sufficiently large
(10 or more).
€
• State hypotheses:
Null hypothesis
Alternative hypothesis
H0 : p = p 0
Ha: p > p0, or
p < p0, or
p ≠ p0
• Calculate test statistic:
z-statistic based on H0:
z=
pˆ − p0
p0 (1− p0 )
n
• Find P-value:
Probability associated with appropriate H0:
€
P = P( Z ≥ z ), or
P = P( Z ≤ z ), or
P = 2P( Z ≥ z )
Conclusion: assess evidence against H0 in favor of Ha
depending on how small P is.
[TI-83: STAT TESTS 1-PropZTest… ]
Chapter 8
4
Choosing the sample size
The margin of error in using pˆ as an estimate for p is
m =€ z *
pˆ (1− pˆ )
n
To solve for the sample size n, we would need to know
€
the value of pˆ before we even know the size of the same
we will select to measure pˆ ! We can, however, estimate
it sufficiently
well. If we use the estimate p*, then
€
solving for n yields
€
 z* 2
n =   p*(1−
m
p*)
Using the conservative estimate p* = 0.5 produces a
€
larger value of n in most circumstances than may be
necessary (but larger samples produce even more
accurate estimates, so this is not a big problem).
Chapter 8
5
Sampling distribution of pˆ 1 – pˆ 2
An SRS of size n1 is selected
from a large population
€
€
with population proportion p1, and an independent SRS
of size n2 is selected from a large population with
population proportion p2. (Neither sample should
contain more than 10% of the population.) Compute the
sample proportions pˆ 1 and pˆ 2.
• mean of the
sampling
distribution is p1 – p2.
€
€
• standard deviation of the sampling distribution,
also called the standard error of the statistic
pˆ 1 – pˆ 2, is
€
€
p1(1− p1)
n1
+
p2 (1− p2 )
n2
• distribution becomes more nearly normal as the
€
sample size increases
Chapter 8
6
Confidence level for comparing two proportions
An SRS of size n1 is selected from a large
population with population proportion p1, and an
independent SRS of size n2 is selected from a large
population with population proportion p2.
An approximate level C confidence interval for
p1 – p2 is
pˆ (1 − pˆ1 ) pˆ 2 (1 − pˆ 2 )
( pˆ 1 – pˆ 2) ± z* 1
+
n1
n2
€
€
where z* is the upper
€
1−C
normal critical value.
2
€
[TI-83: STAT TESTS 2-PropZInt… ]
• pooled proportion
if H0: p1 = p2 is assumed true, then both samples
come from the same population; the samples can
then be combined to compute a pooled proportion
pˆ = (combined count of successes in both samples)
n1 + n 2
€
Chapter 8
7
Two-sample z test for comparing two proportions
Assumptions: Two independent SRSs are selected from
large populations, large enough that n1p1 and n2p2
(number of successes) and n1(1 – p1) and n2(1 – p2)
(number of failures) are 10 or more.
• State hypotheses:
Null hypothesis
Alternative hypothesis
H0: p1 = p2
Ha: p1 > p2, or
p1 < p2, or
p1 ≠ p2
• Calculate test statistic:
z-statistic (here, pˆ is the pooled proportion):
pˆ1 − pˆ 2
z=


€
1
1

pˆ(1− pˆ ) + 
n
n2 
 1
• Find P-value:
Probability associated with appropriate H0:
€
P = P( Z ≥ z ), or
P = P( Z ≤ z ), or
P = 2P( Z ≥ z )
Conclusion: assess evidence against H0 in favor of Ha
depending on how small P is.
[TI-83: STAT TESTS 2-PropZTest… ]