Confidence Intervals for Sample Proportions: Formula, Interpretation, and Examples - Prof., Study notes of Data Analysis & Statistical Methods

How to calculate confidence intervals for sample proportions using the formula (p hat ± z * se p), where p hat is the sample proportion, z * is the critical value obtained from the standard normal distribution based on the desired confidence level, and se p is the standard error of the sample proportion. The document also discusses the concept of margin of error and provides two examples with interpretations. The first example is about a mail-order company's on-time shipping rate, and the second example is about a student organization's survey results regarding student support for a nightclub.

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Uploaded on 08/03/2009

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NOTES: Confidence intervals from sample proportions KEY
Suppose that we are estimating an unknown population proportion
p
. We do this by first finding a sample
proportion
p
ˆ
and then calculating its confidence interval. In theory, the formula for the confidence
interval would be
)
ˆ
(*
ˆpSDzp ±
or
n
pp
zp )1(
*
ˆ
±
where
*
z
depends on the Confidence Level.
But this formula involves the parameter (
p
) that we are trying to estimate! We therefore use the
Standard Error for
p
ˆ
(in other words, replace
p
with
p
ˆ
in the standard deviation formula) instead of
the Standard Deviation for
p
ˆ
. The formula for the confidence interval then becomes
)
ˆ
(*
ˆpSEzp ±
OR
n
pp
zp )
ˆ
1(
ˆ
*
ˆ
±
*
z
can be found with the following, where
C
is the given Confidence Level inputted in decimal form.
=
*
z
InvNorm
+5.0
2
C
Let L denote the left endpoint of the confidence interval and let R denote the right endpoint of the
confidence interval. Therefore, the confidence interval written in interval notation would be (L,R). The
margin of error (m) is the distance that our confidence interval branches out in either direction from
p
ˆ
.
Think of this as
)
ˆ
(RpL
mm

. Thus, the confidence interval formula can be thought of as
mp
±
ˆ
. It follows that the margin of error for these confidence intervals is
n
pp
z)
ˆ
1(
ˆ
*
. In other
words
n
pp
zm )
ˆ
1(
ˆ
*
=
. Also,
(
)
2
LR
m
=
.
INTERPRETATION: “We are ___% confident that
p
is between
L
and
.”
Suppose that the confidence level is 95%. Then there is a 95% probability that choosing any
confidence interval will give us
p
in it, because
p
is in 95% of the confidence intervals that we can
create. Once we fix
p
ˆ
and the confidence interval it generates, then
p
is either in that particular
confidence interval or it is not. Once
p
ˆ
and its confidence interval is fixed then we say that “we are 95%
confident” that
p
is in that particular confidence interval. That is, in repeated trials, we expect that 95% of
the confidence intervals in the long run, to contain
p
.
CONDITIONS/REQUIREMENTS: Before we are allowed to use the confidence interval formula certain
conditions must be met:
1. The sample proportion
p
ˆ
must be obtained from a Simple Random Sample (SRS).
2. The number of successes in the sample is at least 5, preferably at least 10, i.e.,
10
ˆ
pn
.
3. The number of failures in the sample is at least 5, preferably at least 10, i.e.,
10)
ˆ
1(
pn
.
4. The population size is at least ten times the sample size (
n
).
pf2

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NOTES: Confidence intervals from sample proportions KEY

Suppose that we are estimating an unknown population proportion p. We do this by first finding a sample

proportion p ˆ and then calculating its confidence interval. In theory , the formula for the confidence

interval would be p ˆ^ ± z * SD ( p ˆ)or n

p p p z

± where z *depends on the Confidence Level.

But this formula involves the parameter ( p ) that we are trying to estimate! We therefore use the

Standard Error for p ˆ^ (in other words, replace p with p ˆ^ in the standard deviation formula) instead of

the Standard Deviation for (^) p ˆ. The formula for the confidence interval then becomes

p ˆ ± z * SE ( p ˆ )

OR

n

p p p z

z *can be found with the following, where C is the given Confidence Level inputted in decimal form.

z * = InvNorm^  

C

Let L denote the left endpoint of the confidence interval and let R denote the right endpoint of the

confidence interval. Therefore, the confidence interval written in interval notation would be ( L , R ). The

margin of error ( m ) is the distance that our confidence interval branches out in either direction from p ˆ^.

Think of this as ( L p ˆ R )

m m ← →. Thus, the confidence interval formula can be thought of as

p ˆ^ ± m. It follows that the margin of error for these confidence intervals is^ n

p p z

. In other

words n

p p m z

= (^). Also,

R L

m

INTERPRETATION: “We are ___% confident that p is between L and R .”

Suppose that the confidence level is 95%. Then there is a 95% probability that choosing any

confidence interval will give us p in it, because p is in 95% of the confidence intervals that we can

create. Once we fix p ˆ^ and the confidence interval it generates, then p is either in that particular

confidence interval or it is not. Once p ˆ and its confidence interval is fixed then we say that “we are 95%

confident ” that p is in that particular confidence interval. That is, in repeated trials, we expect that 95% of

the confidence intervals in the long run, to contain p.

CONDITIONS/REQUIREMENTS : Before we are allowed to use the confidence interval formula certain

conditions must be met:

  1. The sample proportion (^) p ˆ must be obtained from a Simple Random Sample (SRS).
  2. The number of successes in the sample is at least 5, preferably at least 10, i.e., np ˆ ≥ 10.
  3. The number of failures in the sample is at least 5, preferably at least 10, i.e., n ( 1 − p ˆ)≥ 10.
  4. The population size is at least ten times the sample size ( n ).
  1. As part of a quality improvement program, your mail-order company is studying the

process of filling customer orders. According to company standards, an order is shipped on time if it is sent within 3 working days of the time it is received. You select an SRS of 100 of the 5000 orders received in the past month for an audit. The audit reveals that 86

of these orders were shipped on time. Find a 95% confidence interval for the true proportion of the month’s orders that were shipped on time. Also, interpret the confidence interval.

ANSWER:

ᡦ 㐄 100 0. 86 ㎙ 1. 960 㒕

⡨.⡶⡴䙦⡩⡹⡨.⡶⡴䙧 ⡩⡨⡨

ᡨ̂ 㐄

⡶⡴ ⡩⡨⡨

㐄 0.86 (0.7920 , 0.9280)

INTERPRETATION #

I am 95% confident that the proportion of all orders last month that were shipped on time is between 0.792 and 0.928.

INTERPRETATION # I am 95% confident that between 79.2% and 92.8% of all last month’s orders were shipped on time.

  1. A student organization wants to start a nightclub for students under the age of 21. To

assess support for this proposal, they will select an SRS of students and ask each respondent if he or she would patronize this type of establishment. In the time available,

they were able to obtain 356 responses of which 70% said “yes.” Calculate and interpret a 90% confidence interval for this situation. What is the margin of error?

ANSWER:

ᡦ 㐄 356 0. 70 ㎙ 1. 645 㒕

⡨.⡵⡨䙦⡩⡹⡨.⡵⡨䙧 ⡱⡳⡴

ᡨ̂ 㐄 0.70 (0.6600 , 0.7400)

INTERPRETATION #

I am 90% confident that the proportion of all students under the age of 21 that would patronize this type of nightclub is between 0.66 and 0.74.

INTERPRETATION # I am 90% confident that between 66% and 74% of all students under the age of 21 would patronize this type of nightclub.

MARGIN OF ERROR WAY #1 WAY #

ᡥ 㐄 1.645㒕

⡨.⡵⡨䙦⡩⡹⡨.⡵⡨䙧 ⡱⡳⡴

ᡥ 㐄

䙦⡨.⡵⡲⡹⡨.⡴⡴䙧 ⡰

㐄 0.03995 = 0. 㐆 4 % = 4%