In October 2013, Gallup reported that 52% of U.S. adults disapprove of the Affordable Care Act, based on random samples of 450 people, up from 45% a year earlier. Is this evidence of a real change in public opinion?

In October 2013, Gallup reported that 52% of U.S. adults disapprove of the Affordable Care Act, based on random samples of 450 people, up from 45% a year earlier. Is this evidence of a real change in public opinion?

• A. 1-Proportion Z-Test
• B. 2-Proportion Z-Test
• C. Paired T-Test
• D. Chi-Square Test

Answer: (A)

The main idea is testing whether an observed proportion differs from a baseline proportion. Here you treat the 45% from the previous year as a known baseline (p0) and test whether the October 2013 sample proportion (p̂) of 0.52 is significantly different from p0 using a one-sample z-test for proportions. Compute the standard error under the null: SE = sqrt[p0(1 − p0)/n] = sqrt(0.45 × 0.55 / 450) ≈ sqrt(0.00055) ≈ 0.0235. The z statistic is (p̂ − p0) / SE = (0.52 − 0.45) / 0.0235 ≈ 0.07 / 0.0235 ≈ 3.0. This corresponds to a p-value around 0.003 (two-tailed), well below common significance levels, so the change is statistically significant. So, the observed rise from 45% to 52% is evidence of a real change in public opinion, beyond what random sampling would typically produce. The other tests aren’t appropriate here: a two-proportion test would compare two independent samples directly, a paired test is for matched measurements on the same individuals, and a Chi-square test is for associations in contingency tables or goodness-of-fit, not this single-proportion comparison against a baseline.

The main idea is testing whether an observed proportion differs from a baseline proportion. Here you treat the 45% from the previous year as a known baseline (p0) and test whether the October 2013 sample proportion (p̂) of 0.52 is significantly different from p0 using a one-sample z-test for proportions.

Compute the standard error under the null: SE = sqrt[p0(1 − p0)/n] = sqrt(0.45 × 0.55 / 450) ≈ sqrt(0.00055) ≈ 0.0235. The z statistic is (p̂ − p0) / SE = (0.52 − 0.45) / 0.0235 ≈ 0.07 / 0.0235 ≈ 3.0. This corresponds to a p-value around 0.003 (two-tailed), well below common significance levels, so the change is statistically significant.

So, the observed rise from 45% to 52% is evidence of a real change in public opinion, beyond what random sampling would typically produce. The other tests aren’t appropriate here: a two-proportion test would compare two independent samples directly, a paired test is for matched measurements on the same individuals, and a Chi-square test is for associations in contingency tables or goodness-of-fit, not this single-proportion comparison against a baseline.