AP Stats – Chap 20
Testing Hypotheses About Proportions
We will use the Normal model and the One-Proportion
z-Test to test a hypothesis using a four-step process:
hypothesis
model
mechanics
conclusion
Hypotheses
The Null Hypothesis is a claim about an unknown
population parameter and is noted as H0 . It is called
null because it is skeptical about the world…it generally
asserts that nothing interesting or unusual is
happening or that nothing is changing. It is
proposing a model for the entire population, and we’re
going to ask if it’s wrong enough to reject.
The Alternative Hypothesis , noted as HA , will use
either a two-tailed test (are we looking for a change in
the data) or a one-tailed test (has something increased
or decreased). (Decide early in the question which of
these you are doing!)
Model
List the assumptions and check the conditions. This is
NOT optional. Use full and complete sentences and
correct vocab. It is NOT good enough to merely state
the condition np ≥ 10 , you are expected to SHOW that
you have actually checked the condition with the correct
data: np = 120 ( 0.2 ) = 24 ≥ 10 .
Name the test. State that because the conditions are
satisfied, it is okay to use a Normal model and that you’ll
perform a ____ test.
Mechanics
Write down all of your statistics (sample size, observed
number of successes, sample proportions, etc.).
Draw a curve depicting the sampling model. Locate the
P-value and shade the curve correctly. Again, this is
NOT optional.
Find the P-value.
Conclusion
Link the P-value to your decision in context.
State your decision about the null hypothesis – either you
reject it or you fail to reject it.
You enter the question assuming the null is true. Your
calculations give you a number to help you decide if it’s
likely that the results were just the result of the random
selection of your sample, or if the results are so unusual
(a P-value that’s so very low) that you need to reject the
null.
A few more things…
There is no magical P-value you’re looking for. It is a
matter of personal opinion rooted in the specific
question. A P-value of .08 as a change in teen smoking
may be very low, but the same P-value of .08 in the test
strength of the rivets that hold a plane’s wings together
may be huge.
You NEVER accept the null hypothesis. Think of the jury
and their finding of “not guilty” versus “innocent.”
When looking at the data, ask yourself:
Am I surprised? (Should I reject the null?)
How surprised am I? (What’s the P-value?)
What would not surprise me? (Find the confidence
interval.)
Smoke Detectors
A 1996 report from the US Consumer
Product Safety Commission claimed
that at least 90% of all American
homes have at least one smoke
detector. A city’s fire department
has been running a public safety campaign about
smoke detectors consisting of posters, billboards,
and ads on radio and TV and in the newspaper.
The city wonders if this concerted effort has raised
the local level above the 90% national rate.
Building inspectors visit 400 randomly selected
homes and find that 376 have smoke detectors. Is
this strong evidence that the local rate is higher
than the national average?
Hypothesis: The null hypothesis is that the city is no
different than anywhere else, that their average of
number of homes with smoke detectors is the same as
the national average. The alternative is that the safety
campaign had a positive effect, so the rate of compliance
is higher. HO: p = .90 and HA: p >.90
Model: We have a SRS of 400 homes and we assume
that’s less than 10% of all homes in the city and that
their individual use of smoke detectors is independent of
one another. We expect np = (400)(.90) = 360
successes and nq = (400)(.10) = 40 failures; both
greater than or eaual to 10, so the sample is large
enough. Therefore, we can use a Normal model and a
One-Tailed (Upper) 1-Proportion z-Test.
Mechanics: n = 400
x = 376
p = .94
SD =
pq
=
n
find SD using p and q
from null hypothesis!
(.90 )(.10 )
400
Draw Normal curve…
.94 − .90
z=
= 2.67
.015
…or…use the calculator! ☺
= .015
Conclusion: The p-value is very low (.0038), making it
unlikely that the observed value of 94% for this city’s
sample can be explained by sampling error (luck). We
reject the null hypothesis. There is strong evidence that,
in this city, more than 90% of homes have smoke
detectors.
Furthermore, we are 98% confident that between
91.2% and 96.8% of the city’s homes truly do have
smoke detectors.
Orange
Orange
According to Mars, Incorporated,
the makers of M&Ms, there are
supposed to be 20% orange in a bag.
The bag of 122 that you just opened
had only 21 orange ones. Does this contradict the
company’s 20% claim?
Hypothesis: The null hypothesis is that the proportion
of M&Ms that are orange in the bag is 20%. The
alternative hypothesis is that the bag’s proportion is
something other than 20%. HO: p = .20 and
HA: p ≠ .20
Model: It is reasonable to think that the M&M colors in
one bag are independent from every other bag, and that
the 122 candies we have are randomly representative of
all M&Ms. 122 is certainly less than 10% of all M&Ms.
np = (122)(.20) = 48.8 and nq = (122)(.80) = 97.6, both
of which are greater than or equal to 10, so the sample is
large enough. Therefore, we can use a Normal model
and a Two-Tailed 1-Proportion z-Test.
Mechanics: n = 122
x = 21
p = 21 = .172
122
Draw Normal curve…
Conclusion: The p-value is quite high, so we fail to
reject the null hypothesis. There is no evidence to
suggest that the proportion of orange M&Ms in our bag
differs from the average advertised of 20%.
Furthermore, we are 95% confident that between
10.5% and 23.9% of the M&Ms truly are orange.