Transcript Slide 1
9-3
Testing a proportion
What if you wanted to test
- the change in proportion of students
satisfied with the french fry helpings at
lunch?
- the recent increase in proportion of
students who participate in the arts
program at Pingry
Instead of working with a mean, you are
working with a proportion
Underlying conditions
Assuming binomial distribution conditions
are satisfied….
np > 5 and nq > 5
Then again find a z value
Underlying conditions
Assuming binomial distribution conditions
are satisfied….
np > 5 and nq > 5
Then again find a z value
Note that p = the proportion
specified in the null hypothesis.
z
ˆ -p
p
pq
n
And the process is the same
Set your hypotheses
HO: p = k
HA: p > k
HA: p < k
HA: p ≠ k
(right tailed)
(two tailed)
(left tailed)
As before, compare the p value to α and
state the conclusion.
Proportion Example
Home Field Advantage: Is there one?
If there were no home field advantage, the home
teams would win about half of all games played. In
the 2003 major league baseball season, there were
2429 regular season games. It turns out that the
home team won 1335 of those games, or 54.96%.
Could this deviation from 50% be explained just
from natural sampling variability, or is this evidence
to suggest that there really is a home field
advantage, at least in professional sports?
The question: whether the observed rate of home
team victories is so much greater than 50% that we
cannot explain it away as just variation.
Steps
Step 1: Ho : p = 0.50 HA : p > 0.50
Step 2: Independence? Likely true.
Randomization? This is only 2003, not all years, but
likely representative.
check npo and nqo.
One proportion or two?
Step 3: find SD .01015
Step 4: Conclusion:
find z
4.89