1. **State the problem:** We want to test if the population proportion of households loyal to brand W cloth washing liquid is greater than 48% based on a sample of 100 households where 50 claimed loyalty. 2. **Set hypotheses:** - Null hypothesis $H_0$: $p \leq 0.48$ - Alternative hypothesis $H_a$: $p > 0.48$ 3. **Significance level:** $\alpha = 0.01$ 4. **Sample data:** - Sample size $n = 100$ - Number of successes $x = 50$ - Sample proportion $\hat{p} = \frac{50}{100} = 0.5$ 5. **Test statistic formula for proportion:** $$ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} $$ where $p_0 = 0.48$ is the hypothesized population proportion. 6. **Calculate standard error:** $$ SE = \sqrt{\frac{0.48 \times (1-0.48)}{100}} = \sqrt{\frac{0.48 \times 0.52}{100}} = \sqrt{0.002496} \approx 0.04996 $$ 7. **Calculate test statistic:** $$ z = \frac{0.5 - 0.48}{0.04996} = \frac{0.02}{0.04996} \approx 0.4003 $$ 8. **Decision rule:** - Find critical value $z_{\alpha}$ for $\alpha=0.01$ in a one-tailed test. - $z_{0.01} = 2.33$ (from standard normal table) 9. **Compare test statistic with critical value:** - Since $0.4003 < 2.33$, we fail to reject $H_0$. 10. **Conclusion:** - There is insufficient evidence at the 1% significance level to conclude that the population proportion of households loyal to brand W cloth washing liquid is greater than 48%.