1. **Problem:** A publication reports 42% of adults believe radio stations should wait until after Thanksgiving to play holiday music. A researcher believes the true percentage is less than 42% and surveys 400 adults; 156 agree with the statement. Which statement is false? 2. **Hypotheses:** - Null hypothesis: $H_0: p = 0.42$ - Alternative hypothesis: $H_a: p < 0.42$ (since the researcher believes the true proportion is smaller) 3. **Sample proportion:** $$\hat{p} = \frac{156}{400} = 0.39$$ 4. **Test statistic formula:** $$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$$ where $p_0=0.42$, $n=400$ 5. **Calculate standard error:** $$SE = \sqrt{\frac{0.42 \times 0.58}{400}} = \sqrt{\frac{0.2436}{400}} = \sqrt{0.000609} \approx 0.0247$$ 6. **Calculate z:** $$z = \frac{0.39 - 0.42}{0.0247} = \frac{-0.03}{0.0247} \approx -1.215$$ 7. **P-value:** For $z = -1.215$, the P-value (left tail) is approximately 0.112 (close to 0.12). 8. **Interpretation:** - At significance level $\alpha=0.10$, since P-value $\approx 0.112 > 0.10$, we fail to reject $H_0$. 9. **Practical vs statistical significance:** The difference between 0.42 and 0.39 is small; statistical significance depends on P-value, but practical significance depends on real-world impact. 10. **One-sided vs two-sided test:** Since the researcher believes $p < 0.42$, a one-sided test is appropriate and provides stronger evidence against $H_0$ if true. 11. **Check statements:** - A is true. - B is true. - C is true. - D is false because a small difference with a non-significant P-value means results are neither statistically nor practically significant. - E is true. **Final answer:** Statement D is false.