1. **State the problem:** We want to test if more than 45% of the bank customers have saving accounts based on the sample data. 2. **Identify the sample data:** Total customers $n=835$, customers with saving accounts $x=401$. 3. **Calculate the sample proportion:** $$\hat{p} = \frac{x}{n} = \frac{401}{835} \approx 0.4802$$ 4. **Set up hypotheses:** - Null hypothesis $H_0$: $p = 0.45$ (45% have saving accounts) - Alternative hypothesis $H_a$: $p > 0.45$ (more than 45% have saving accounts) 5. **Use the test statistic for proportion:** $$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$$ where $p_0=0.45$. 6. **Calculate the standard error:** $$SE = \sqrt{\frac{0.45 \times 0.55}{835}} = \sqrt{\frac{0.2475}{835}} \approx \sqrt{0.000296} \approx 0.0172$$ 7. **Calculate the z-score:** $$z = \frac{0.4802 - 0.45}{0.0172} = \frac{0.0302}{0.0172} \approx 1.756$$ 8. **Interpret the z-score:** For a significance level of 0.05, the critical z-value for a one-tailed test is approximately 1.645. 9. **Conclusion:** Since $z=1.756 > 1.645$, we reject the null hypothesis and conclude that there is sufficient evidence to support the claim that more than 45% of the bank customers have saving accounts.