First, we must find α
[tex]\alpha=\frac{1-0.95}{2}=0.025[/tex]Then, we must find Z in the Ztable,
That is z with a pvalue of
[tex]1-0.025=0.975[/tex]For this case,
[tex]Z=1.96[/tex]Then, we need to calculate the margin of error
[tex]\begin{gathered} M=Z\cdot\frac{\sigma}{\sqrt[]{n}} \\ M=1.96\cdot\frac{2.8}{\sqrt[]{32}}=0.97 \end{gathered}[/tex]Finally, the lower limit of the 95% confidence interval for the mean time for ninth-graders to play video games will be
[tex]17.1-0.97=16.1[/tex]