3. A survey claims that 70% of Addis Ababa residents prefer tea over coffee . A sample of 400 residents finds that 290 prefer tea. Test the claim at a 5% significance level.

Here's how to test the claim:<br /><br />1. **State the hypotheses:**<br /><br />* **Null hypothesis (H₀):** p = 0.70 (The proportion of Addis Ababa residents who prefer tea is 70%)<br />* **Alternative hypothesis (H₁):** p ≠ 0.70 (The proportion of Addis Ababa residents who prefer tea is *not* 70%) This is a two-tailed test because we are looking for a difference in either direction (more or less than 70%).<br /><br />2. **Calculate the sample proportion:**<br /><br />* p̂ = (Number who prefer tea) / (Total sample size) = 290/400 = 0.725<br /><br />3. **Calculate the test statistic (z-score):**<br /><br />* z = (p̂ - p) / sqrt[(p(1-p)) / n]<br />* z = (0.725 - 0.70) / sqrt[(0.70 * 0.30) / 400]<br />* z ≈ 1.09<br /><br />4. **Determine the critical value:**<br /><br />* Since this is a two-tailed test with a 5% significance level (α = 0.05), we look up the critical z-value for α/2 = 0.025 in a standard normal distribution table (or use a calculator). The critical z-value is approximately ±1.96.<br /><br />5. **Make a decision:**<br /><br />* Since the calculated z-score (1.09) falls within the range of -1.96 to +1.96 (i.e., it does not fall in the rejection region), we *fail to reject the null hypothesis*.<br /><br />6. **Conclusion:**<br /><br />There is not enough evidence at the 5% significance level to reject the claim that 70% of Addis Ababa residents prefer tea over coffee. The sample data does not provide statistically significant evidence to contradict the survey's claim.<br />