1. **Problem statement:** We want to test if the company's claim that students increase their reading speed by 500% (5 times) is reasonable based on a sample of 100 students with an average increase of 412% and a standard deviation of 218%. We use a significance level of $\alpha = 0.025$. 2. **Set up hypotheses:** - Null hypothesis $H_0$: $\mu = 500$ (the true mean increase is 500%) - Alternative hypothesis $H_a$: $\mu < 500$ (the true mean increase is less than 500%) 3. **Test statistic formula:** $$ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} $$ where $\bar{x} = 412$, $\mu_0 = 500$, $s = 218$, and $n = 100$. 4. **Calculate the test statistic:** $$ t = \frac{412 - 500}{218 / \sqrt{100}} = \frac{\cancel{412 - 500}}{\cancel{218 / 10}} = \frac{-88}{21.8} \approx -4.04 $$ 5. **Determine critical value:** For a left-tailed test with $\alpha = 0.025$ and $df = 99$, the critical t-value is approximately $-1.984$. 6. **Decision rule:** If $t < -1.984$, reject $H_0$. 7. **Conclusion:** Since $-4.04 < -1.984$, we reject the null hypothesis. There is sufficient evidence at the 2.5% significance level to conclude that the true mean increase in reading speed is less than 500%, so the company's claim is not supported by this sample. 8. **Interpretation:** The data suggest that the average increase in reading speed is significantly less than the claimed 500%, indicating the company's claim may be overstated. --- (b) **Comparison with confidence interval:** If last week's confidence interval for the mean increase did not include 500%, it agrees with this hypothesis test result. Both methods suggest the true mean increase is less than 500%, reinforcing the conclusion that the company's claim is not reasonable based on the sample data.