9.2a
Tests about a Population
Proportion
Target Goal:
I can check the conditions for carrying out a
test about a population proportion.
I can perform a significance test for a
sample proportion.
h.w: pg 548: 27 – 30, pg 562: 41, 43, 45
Carrying Out a Significance Test
Suppose a basketball player who claimed to be an
80% free-throw shooter. In an SRS of 50 free-throws,
he made 32. His sample proportion of made shots,
32/50 = 0.64, is much lower than what he claimed.
Does it provide convincing evidence against his
claim?
To find out, we must perform a significance
test of
H0: p = 0.80
Ha: p < 0.80
where p = the actual proportion of free throws the
shooter makes in the long run.
Tests About a Population Proportion

 For
a hypothesis test where
Ho: p = po, use po to estimate p.
 For a confidence interval:
we used p̂ as an estimate of p .
Assumptions for Inference about
a Proportion:
To perform a significance test about
a proportion, check the same conditions.
1. Random: SRS
2. Independent: Population  10n
3. Normal:
np0  10 and nq0  10 for a hypothesis test.
npˆ  10 and nqˆ  10 for a confidence interval.
 To
test hypothesis Ho: p = po,
pˆ  po
z
poqo
n

The One-Sample z Test for a Proportion
One-Sample z Test for a Proportion
Choose an SRS of size n from a large population that contains an unknown
Use this
test
when H : p = p , compute the
proportion p of successes.
To test
theonly
hypothesis
0
0
the
expected
numbers
of
successes
z statistic
ˆ
pand
 p 0n(1 - p ) are
and failures
np
0
z 0
p0 (1 the
p0 )population
both at least 10 and
is at least 10 times nas large as the
sample.
Find the P-value by calculating the probability of getting a z statistic this large
or larger in the direction specified by the alternative hypothesis Ha:

Tests About a Population Proportion
The z statistic has approximately the standard Normal distribution when H0
is true. P-values therefore come from the standard Normal distribution.
Here is a summary of the details for a one-sample z test for a
proportion.
Ex. Binge Drinking in College
 In
a representative of 140 colleges and
17592 students, 7741 students identify
themselves as binge drinkers. Considering
this SRS, does this constitute strong
evidence that more than 40% of all college
students engage in binge drinking?
Step 1: State - What hypotheses do you want to test, and
at what significance level? State the hypothesis in words
and symbols.
 We
want to test a claim about the
proportion of all U.S. college students
who have engaged in binge drinking at the
α = .05 level. Our hypotheses are
 Ho: p = .40 40% of all college students
are binge drinkers
 Ha: p > .40 more than 40% of all U.S.
college students have engaged in binge
drinking.
7741
pˆ 
17,592
 0.44 (need to find z)
Step 2: Plan - Choose the appropriate inference
procedure. Verify the conditions for using the
selected procedure.
To test claim, we will use the one prop z test .
Check conditions.
 Random: SRS? We are given SRS.
 Independent:
Total population > 10 n:
10(17592) = there are more than 175,920 college
students in the country. ( to use sample σ) yes
independent.
 Normal: npo and nqo ≥ 10?
17592(.40) = 7036.8 ≥ 10
17592(.60) = 10,555.2 ≥ 10
Yes, we can use normal approximation.
Step 3: Do - If the conditions are met, carry
out the inference procedure.
 Calculate
pˆ  po
z
poqo
n
 With
z statistic
0.44  0.40

 10.83
0.40(0.60)
17,592
a z score this large, the P-value is
approx. 0.
Step 4. Conclude - Interpret the results
in the context of the problem
p-value (= 0) this small, < α = .05, tells
us that we have no chance of obtaining a
sample proportion
 The
as far away from .40 as pˆ  .44.

We reject Ho and conclude that more than
40% of U.S. college students have
engaged in binge drinking.
Ex. Is that Coin Fair?
 The
French naturalist Count Buffon tossed
a coin 4040 times and counted 2048
heads. The sample proportion of heads is
p̂ = 0.5069

 Is
this evidence that Buffon’s coin was not
balanced?
Step 1: State
 The
parameter p is the probability of
tossing a head. The population contains
the results of tossing the coin forever.
 Our hypotheses are:

Ho: p = .50 The coin is balanced.

Ha: p 0.5
Buffon’s coin is not balanced.
The null hypothesis gives p the value po = .50.
Step 2: Plan
We will use the one prop z test.
Check conditions.
 SRS? Yes
 Total population > 10 n?
The population of tosses is infinite.
and nqo ≥ 10?
npo = 4040(0.5) = 2020 ≥ 10
nqo = 4040(0.5) = 2020 ≥ 10
 npo
Step 3: Do - If the conditions are met, carry
out the inference procedure.
 Calculate
z statistic
pˆ  po
0.5069  .5
z
, z
 .88
poqo
.5(.5)
n
4040
 P-value
= (two sided so)
 Normcdf(.88,E99)
2(.1894) = .3788
Step 4. Conclude - Interpret the results in
the context of the problem.
 The
proportion of heads as far away from
1/2 as Buffon’s would happen 38% of the
time. This provides little evidence against
Ho. Thus, Buffon’s result doesn’t show that
his coin is unbalanced.
 Note:
calculator much quicker.
 STAT: TESTS:1-Prop Z Test
 Read
pg. 549 - 555