1. **Problem Statement:** A 95% confidence interval for the mean reading achievement score for a population of third-grade students is given as (44.2, 54.2). We want to determine how the width of a 99% confidence interval computed from the same data compares to this 95% confidence interval. 2. **Understanding Confidence Intervals:** A confidence interval for a mean is generally calculated as: $$\text{CI} = \bar{x} \pm z^* \times \frac{s}{\sqrt{n}}$$ where $\bar{x}$ is the sample mean, $z^*$ is the critical value from the standard normal distribution corresponding to the confidence level, $s$ is the sample standard deviation, and $n$ is the sample size. 3. **Effect of Confidence Level on Interval Width:** - The critical value $z^*$ increases as the confidence level increases. - For 95% confidence, $z^* \approx 1.96$. - For 99% confidence, $z^* \approx 2.576$. 4. **Comparing Widths:** Since the sample mean $\bar{x}$, sample size $n$, and sample standard deviation $s$ remain the same, the width of the confidence interval depends directly on $z^*$. 5. **Conclusion:** Because $2.576 > 1.96$, the 99% confidence interval is wider than the 95% confidence interval. **Final answer:** The 99% confidence interval is wider.