AP Statistics 6.9 Justifying a Claim Based on a Confidence Interval for a Difference of Population Proportions- FRQs - Exam Style Questions

State:
Let \( p_{07} \) be the population proportion of U.S. adults who would answer “yes” in December \(2007\).
Let \( p_{08} \) be the population proportion of U.S. adults who would answer “yes” in December \(2008\).
\( H_0: p_{07} = p_{08} \)
\( H_a: p_{07} \neq p_{08} \)

Plan:
We will use a two-sample \( z \)-test for proportions.
Conditions:
1. Random: Both samples were randomly selected.
2. Normal: All counts are at least \(10\):
• \(2007\): \(622\) yes, \(1,020 – 622 = 398\) no
• \(2008\): \(676\) yes, \(1,009 – 676 = 333\) no
3. Independent: Population of U.S. adults is much larger than \(10 \times 1,020\) and \(10 \times 1,009\).

Do:
\( \hat{p}_{07} = \frac{622}{1020} \approx 0.6098, \quad \hat{p}_{08} = \frac{676}{1009} \approx 0.6700 \)
\( \hat{p}_c = \frac{622 + 676}{1020 + 1009} = \frac{1298}{2029} \approx 0.6397 \)
\( z = \frac{0.6098 – 0.6700}{\sqrt{0.6397(1-0.6397)\left(\frac{1}{1020} + \frac{1}{1009}\right)}} \approx \frac{-0.0602}{\sqrt{0.2304 \times 0.00197}} \approx \frac{-0.0602}{0.0213} \approx -2.82 \)
\( p \)-value \( = 2 \times P(Z < -2.82) \approx 2 \times 0.0024 = 0.0048 \)

Conclude:
Since the \( p \)-value (\(0.0048\)) is less than \(\alpha = 0.05\), we reject \( H_0 \). There is convincing evidence that the proportion of U.S. adults who would answer “yes” changed from December \(2007\) to December \(2008\).