Chemically it is established that the percentage of iron in five samples of ore extracted in a certain mine has an average of $22.3\%$ and a standard deviation of $1.8\%$. In six samples of ore extracted from another mine the average content was $20\%$ and the standard deviation of $1.3\%$. It is estimated that the variances of the percentages are the same.
Establish the limits of a $95\%$ confidence interval, for the difference in mean iron content between the minerals extracted from the two different mines. Interpreting the results.
My attempt
We know the variance of the percentages are the same, and
\begin{align} n_x &= 5 & \bar{x} &= 22.3\% & s_x &= 1.8\% \\ n_y &= 6 & \bar{y} &= 20\% & s_y &= 1.3\% \end{align}
As we want a $95\%$ confidence interval, then
$$\text{let}\quad 1-\alpha=0.95 \quad\to\quad \alpha=0.05 \quad\implies\quad \frac{\alpha}{2}=0.025$$
The interval of confidence have the form:
$$I=(p_1,p_2),$$
where
\begin{align} p_1&=\bar{x}-\bar{y}-t_{\frac{\alpha}{2}}(n_x+n_y-2)\sqrt{\frac{n_x+n_y}{n_x\times n_y}\frac{n_xs^2_x+n_ys^2_y}{n_x+n_y-2}} \\ p_2&=\bar{x}-\bar{y}+t_{\frac{\alpha}{2}}(n_x+n_y-2)\sqrt{\frac{n_x+n_y}{n_x\times n_y}\frac{n_xs^2_x+n_ys^2_y}{n_x+n_y-2}} \end{align}
I know
$$t_{\frac{\alpha}{2}}(n_x+n_y-2)=t_{0.025}(9)=2.2622,$$
but I'm stuck when I'm trying to calculate $\bar{x},\bar{y},\ldots$ etc. Can someone help me?