First, we must find α

Then, we must find Z in the Ztable,
That is z with a pvalue of

For this case,

Then, we need to calculate the margin of error
![\begin{gathered} M=Z\cdot\frac{\sigma}{\sqrt[]{n}} \\ M=1.96\cdot\frac{2.8}{\sqrt[]{32}}=0.97 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/t7www3xztw13s645u7g7zxu3zkm5ym5t5j.png)
Finally, the lower limit of the 95% confidence interval for the mean time for ninth-graders to play video games will be
