Transcript SECTION 12.2 Tests About a Population Proportion

SECTION 12.2
Tests About a Population Proportion
NOW WHAT
In this section we are interested in the
unknown proportion, p of a population as
opposed to the unknown mean of a population.
Keep in mind, p will have an approximately normal
distribution, so it is
BACK TO THE WORLD OF z.
Our z statistic
z
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
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pˆ  p
p0 (1  p0 )
n
We don’t really know p for our standard deviation.
So, when we do a test, replace p by p0.
NOTE: When we did confidence intervals, we
used p̂ in place of p instead of p0
p̂
Assumptions
1.
2.
3.
Data are an SRS from the population
The population is at least ten times as large as the
sample (which insures independence)
A. For a significance test:
np0  10 and
n(1  p0 )  10
B. For a confidence interval:
npˆ  10 and
n(1  pˆ )  10
The Steps for a
One Proportion z-test
1.
2.
3.
State the hypothesis and name test
H0: p = p0
Ha: p ‹, ›, or ≠ p0
State and verify your assumptions
Calculate the P-value and other important values
-
4.
Done in calculator or…
Using the formulas and tables
State Conclusions (Both statistically and contextually)
- The smaller the P-value, the greater the evidence is to reject H0
Example
A coin is tossed 4040 times. There were 2048 heads. The
sample proportion of heads is
p̂ = 2048/4040 = 0.5069
That’s a bit more than one-half. Is this evidence that the
coin was not balanced?
Step 1—Parameter

The population for coin tossing contains the
results of tossing a coin forever. The parameter
p is the proportion of all tosses that lands heads
up. The null hypothesis says that the coin is
balanced. The alternative hypothesis is twosided, because we did not suspect before
seeing the data that the coin favored either
heads or tails.
H0: p = 0.5
Ha: p ≠ 0.5
Step 2—Conditions
SRS—The tosses we make can be considered
an SRS from the population of all tosses.
 Normality—Since np0=4040(.5)=2020
and n(1-p0)=4040(.5)=2020 are both at least 10,
we are safe using Normal calculations
 Independence—Since we are sampling without
replacement (?) we must have at least 40400
tosses in our population. That isn’t an issue.

Step 3—Calculations
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z
2048

0.5
pˆ  p0
 4040
 0.8810
p0 (1  p0 )
0.5(1  0.5)
4040
n
2048
pˆ 
 0.5069
4040
P-value ≈ 0.3783
SE pˆ 
Don’t forget to draw your curve.
Remember, use p0 for your
standard error calculations. Use
this standard error when drawing
the curve.
p0 (1  p0 )
.5(1  .5)

 0.0079
n
4040
Step 4—Interpretation
A proportion of heads as far from one-half (.5)
as this one would happen about 38% of the
time by chance alone, if the coin is balanced.
 For this reason, we would fail to reject the null
hypothesis.
 There is virtually no evidence that the coin is
unbalanced.
 As a reminder, this is not evidence that the null
hypothesis is true. It is still possible the coin is
unbalanced, we just don’t have strong enough
evidence to convince anyone that it is
unbalanced.

Using a Confidence Interval
For the example of the coin, it is possible that
the confidence interval would be more
meaningful than the significance test. A 95%
confidence interval is
(0.49152, 0.52235)
 We can see that 0.5 is plausible, but so are
many higher proportions, including the
proportion that we saw in our sample of 4040
tosses.
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Another Example
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Publishing scientific papers online is fast, and the
papers can be long. Publishing in a paper journal
means that the paper will live forever in libraries. The
British Medical Journal combines the two: it prints
short and readable versions, with longer versions
available online. It this OK with authors? The journal
asked a random sample of 104 of its recent authors
several questions. One question in the survey asked
whether authors would accept a stronger move toward
online publishing: “As an author, how acceptable
would it be for us to publish only the abstract of papers
in the paper journal and continue to put the full long
version on our website?” Of the 104 authors in the
sample, 65 said “Not at all acceptable.”
Do the data provide good evidence that more than half
of all authors feel that abstract-only publishing is not
acceptable?
Step 1—Parameter
The population of interest is all of the authors
for this particular journal.
 The parameter is the proportion of these
authors that disagree with the abstract-only
printing of their articles
 The null hypothesis is that there will be an
event split between those that oppose the
abstract-only printing and those in favor.
 The alternative hypothesis is that more authors
will be against the abstract-only printing.

H0: p = 0.5
Ha: p > 0.5
Step 2—Conditions
SRS—The chosen authors were a random
sample but not necessarily an SRS of all
authors from this journal.
 Normality—Since np0=n(1-p0)=52 are both at
least 10, we are safe using Normal calculations
 Independence—Since we are sampling without
replacement we must have at least 1040
authors for this magazine in the population.

Step 3—Calculations
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z
pˆ  p0

p0 (1  p0 )
n
65
pˆ 
 .625
104
P-value ≈ 0.0054
SE pˆ 
65
 0.5
104
 2.5495
0.5(1  0.5)
104
Don’t forget to draw your curve.
Remember, use p0 for your
standard error calculations. Use
this standard error when drawing
the curve.
p0 (1  p0 )
.5(1  .5)

 0.0490
n
104
Step 4—Interpretation
Because of the small P-value, there is sufficient
evidence to reject the null hypothesis.
 We can conclude that more than half of all
authors from the British Medical Journal would
be opposed to printing their articles in the
abstract-only format.
