1. **State the problem:** We want to test if the new exercises have a higher success rate than the traditional method, which has a success rate of 70% (or 0.7). 2. **Set up hypotheses:** - Null hypothesis $H_0$: $p = 0.7$ (new exercises are not better) - Alternative hypothesis $H_a$: $p > 0.7$ (new exercises are better) 3. **Identify the sample data:** - Sample size $n = 30$ - Number of successes $x = 26$ - Sample proportion $\hat{p} = \frac{26}{30} = 0.8667$ 4. **Use the test statistic for a proportion:** $$ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} $$ where $p_0 = 0.7$ is the hypothesized proportion. 5. **Calculate the standard error:** $$ SE = \sqrt{\frac{0.7 \times 0.3}{30}} = \sqrt{\frac{0.21}{30}} = \sqrt{0.007} \approx 0.0837 $$ 6. **Calculate the z-score:** $$ z = \frac{0.8667 - 0.7}{0.0837} = \frac{0.1667}{0.0837} \approx 1.99 $$ 7. **Find the critical value:** At a 5% significance level for a one-tailed test, the critical z-value is approximately 1.645. 8. **Make a decision:** Since $z = 1.99 > 1.645$, we reject the null hypothesis. 9. **Conclusion:** There is sufficient evidence at the 5% significance level to conclude that the new exercises have a higher success rate than the traditional method.