[tex]\begin{gathered} \bar{X}\text{ }\pm\text{ }t_{\frac{\alpha}{2}}(\frac{s}{\sqrt[]{n}})\text{ = }14.3\text{ }\pm\text{ 2.052(}\frac{2}{\sqrt[]{28}})\text{ = 1-}\alpha \\ 13.5\text{ < }\mu\text{ < }15.1 \end{gathered}[/tex]
The 1st box: 14.3
second box: 2.052
Third box: 2
fourth box: 28
fifth box: 13.5
sixth box: 15.1
Explanation:[tex]\begin{gathered} nu\text{mber in survey = 28} \\ n\text{ = 28} \\ \text{degr}ee\text{ of fr}eedom\text{ = n - 1 = 28 - 1} \\ \text{degr}ee\text{ of freedom = }27 \end{gathered}[/tex][tex]\begin{gathered} \operatorname{mean}\text{ = }\bar{X}\text{ = 14.3} \\ \text{standard deviation = s = 2} \\ 1\text{ - }\alpha\text{ = 0.95} \end{gathered}[/tex][tex]\begin{gathered} To\text{ }find_{}t_{\frac{\alpha}{2}},\text{ }wewould\text{ use the degr}ee\text{ of fr}eedom,\text{ }\frac{\alpha}{2\text{ }}\text{ and t -table} \\ \text{from the table }t_{\frac{\alpha}{2}}\text{ = 2.052} \end{gathered}[/tex][tex]\begin{gathered} \bar{X}\text{ }\pm\text{ }t_{\frac{\alpha}{2}}(\frac{s}{\sqrt[]{n}})\text{ = }14.3\text{ }\pm\text{ 2.052(}\frac{2}{\sqrt[]{28}}) \\ \end{gathered}[/tex][tex]\begin{gathered} =14.3\text{ - 2.052(}\frac{2}{\sqrt[]{28}})<\text{ }\mu<\text{ }14.3\text{ +2.052(}\frac{2}{\sqrt[]{28}}) \\ =\text{ }14.3\text{ - 0.775 }<\text{ }\mu<\text{ 14.3 +0}.775 \\ =\text{ }14.3\text{ }\pm\text{ 0.775} \\ =\text{ }14.3\text{ }\pm\text{ 0.8} \end{gathered}[/tex][tex]\begin{gathered} So,\text{ the confidence interval is:} \\ \text{ 14.3 - 0.8 < }\mu\text{ <}14.3\text{ + 0.8} \\ 13.5\text{ < }\mu\text{ < }15.1 \end{gathered}[/tex]
The 1st box: 14.3
second box: 2.052
Third box: 2
fourth box: 28
fifth box: 13.5
sixth box: 15.1