Width of a t confidence interval

Suppose that the weights (in kgs) of six months old babies, monitored at a healthcare facility, have $N(\mu,\sigma^2)$ distribution, where $\mu\in\mathbb{R}$ and $\sigma>0$ are unknown parameters. Let $X_1,X_2,…,X_9$ be a random sample of the weights of such babies. Let $\bar X=\frac19\sum_{i=1}^9 X_i$, $S=\sqrt{\frac18\sum_{i=1}^9(X_i-\bar X)^2}$ and let a $95\%$ confidence interval for $\mu$ based on $t$-distribution be of the form $(\bar X-h(S),\bar X+h(S))$, for an appropriate function $h$ of random variable $S$. If the observed values of $\bar X$ and $S^2$ are $9$ and $9.5$, respectively, then the width of the confidence interval is equal to __________ (round off to 2 decimal places) (You may use $t_{9,0.025}=2.262$, $t_{8,0.025}=2.306$, $t_{9,0.05}=1.833$, $t_{8,0.05}=1.86$)