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 p̂ -p

zpq

nNote that p = the proportion specified in the null hypothesis.

And the process is the same

Set your hypotheses HO: p = k

HA: p > k HA: p < k HA: p ≠ k

(right tailed) (left tailed) (two 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 find z Step 4: Conclusion:

.01015 4.89