91Ó°ÊÓ

The calculation is given below:

$$\begin{gathered} \text{n = 16} \\ \stackrel{¯}{X}=\text{97.9375} \\ \text{s = 12.6463} \end{gathered}$$

$$\begin{gathered} \text{df = 16}-\text{1} \\ \text{= 15} \end{gathered}$$

$\mathrm{Î\pm }=0.20$

The critical worth of localid="1652095550050" $t=Â\pm \text{1.341}$

The confidence interval is 80%:

localid="1652095596149" $\overline{X}\frac{t×s}{\sqrt{n}}=97.9375Â\pm \frac{1.341×12.6463}{\sqrt{16}}=97.9375Â\pm 4.238=\left(93.699,102.176\right)$

$\text{The}80\%\text{confidential level}\left(93.699,102.176\right)$

The calculation is given below:

$$\begin{gathered} \text{n = 16} \\ \stackrel{¯}{X}=\text{97.9375} \\ \text{s = 12.6463} \end{gathered}$$$$\begin{gathered} \text{n = 16} \\ \stackrel{¯}{X}=\text{97.9375} \\ \text{s = 12.6463} \end{gathered}$$

$$\begin{gathered} \text{df = 16}-\text{1} \\ \text{= 15} \end{gathered}$$

$\mathrm{Î\pm }=0.20$

The critical worth of $t=Â\pm \text{1.341}$

The confidence interval is 80%:

$\overline{X}\frac{t×s}{\sqrt{n}}=97.9375Â\pm \frac{2.131×12.6463}{\sqrt{16}}=97.9375Â\pm 6.739=\left(91.199,104.676\right)$

$\text{The}95\%\text{confidential level}\left(91.199,104.676\right)$

The confidence interval 80% is (93,699,102.176) is narrower than confidence interval 95% is (91.199, 104.676)

As the level of confidence goes from 95% to 80%, the width of the confidence interval narrows. Since lowering the confidence threshold from 95% to 80% reduces the error margin, resulting in a smaller confidence interval. As the level of confidence lowers, so does our estimation of the genuine unknown population parameter to be around a specific interval.