Respuesta :
Explanation:
The given values in the question are
[tex]\begin{gathered} CL=0.90 \\ \end{gathered}[/tex]We will forst of all calculate the margin of error using the formula below
[tex]\begin{gathered} E=z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}} \\ where, \\ \sigma=0.7 \\ n=43 \\ z_{\frac{\alpha}{2}}=1.645(using\text{ z-score tbles for 90\% confidence interval\rparen} \end{gathered}[/tex]Hence,
The margin of erro will be
[tex]\begin{gathered} E=z_{\frac{\alpha}{2}}\frac{\sigma}{n} \\ E=1.645\times\frac{0.7}{\sqrt{43}} \\ E=0.1756 \end{gathered}[/tex]Hence,
The lower limit and upper limit will be calculated using the formula below
[tex]\begin{gathered} lowerlimit=\bar{x}-E \\ Upperl\imaginaryI m\imaginaryI t=\bar{x}+E \\ where,\bar{x}=6.8 \end{gathered}[/tex]By substituitng the values, we will have
[tex]\begin{gathered} lowerl\imaginaryI m\imaginaryI t=\bar{x}-E=6.8-0.1756=6.6244 \\ Upperl\imaginaryI m\imaginaryI t=\bar{x}+E=6.8+0.1756=6.9756 \end{gathered}[/tex]Hence,
The final answers to to decimal places is given below as
[tex](6.62,6.98)[/tex]