Construct the lower bound of a confidence interval using the following information: $n=18$; mean = 51.056; standard deviation $=32.288$; alpha $=0.05$. Remember that alpha $=0.05$ is for a $95 \%$ confidence interval and alpha $=0.01$ is for a $99 \%$ confidence interval.

Step 1 ：Given the following information: sample size \(n=18\), mean \(=51.056\), standard deviation \(=32.288\), and alpha \(=0.05\). Remember that alpha \(=0.05\) is for a \(95\%\) confidence interval.

Step 2 ：To construct the lower bound of a confidence interval, we need to use the formula: mean - Z * (standard deviation / sqrt(n)), where Z is the Z-score corresponding to the desired confidence level.

Step 3 ：For a \(95\%\) confidence interval, the Z-score is approximately \(1.96\).

Step 4 ：Substitute the given values into the formula to find the lower bound of the confidence interval: \(51.056 - 1.96 * (32.288 / \sqrt{18})\).

Step 5 ：Calculate the above expression to get the lower bound of the confidence interval, which is approximately \(36.14\).

Step 6 ：Final Answer: The lower bound of the \(95\%\) confidence interval is \(\boxed{36.14}\).