Given the following parameters
![\begin{gathered} \sigma\Rightarrow s\tan dard\text{ deviation=0.4} \\ n\Rightarrow\text{sample size=20} \\ \text{Significance level}\Rightarrow95\text{ \%} \\ z_{(\alpha)/(2)}=1.960 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/d9sg9p5zlmuawyclnv8r8f5hpz0m6fg0n9.png)
To find the mean of the data, we will have to use
![\begin{gathered} \bar{x}=(\Sigma x_i)/(n) \\ \Sigma x_i=3497.76 \\ n=20 \\ \bar{x}=(3497.76)/(20) \\ =174.888 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/37f424jejftv6wr50sezgyguzz3g3owqra.png)
Using the confidence interval formula of
![CI=\bar{x}\pm z_{(\alpha)/(2)}*\frac{\sigma}{\sqrt[]{n}}](https://img.qammunity.org/2023/formulas/mathematics/high-school/tekr2a84hl2mywyq3699j12pl3w64cf6me.png)
Substitute for all values to find the confidence interval.
![\begin{gathered} CI=174.888\pm1.960*\frac{0.4}{\sqrt[]{20}} \\ =174.888\pm1.960*0.0894427191 \\ =174.888\pm0.175 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/wy9pjm7bnxdtwbh3h2zptauk6d395em1bp.png)
Hence, the confidence interval is
![174.888\pm0.175](https://img.qammunity.org/2023/formulas/mathematics/high-school/l520b27i5yxnh35703d624flky24bf152u.png)
The critical value is
![1.960](https://img.qammunity.org/2023/formulas/mathematics/high-school/v2gfbr3w68no9k97kgkczp9jokr20eu05t.png)
The standard error of the mean is
![\sigma_{\bar{x}}=\frac{\sigma}{\sqrt[]{n}}=\frac{0.4}{\sqrt[]{20}}=0.089](https://img.qammunity.org/2023/formulas/mathematics/high-school/at9h4nxp5mr74520xti4kfsiqgdrqmh0nb.png)
The confidence interval is
![(174.71,175.07)](https://img.qammunity.org/2023/formulas/mathematics/high-school/o8vmx9ptr2iu8etn9szmfwlelkq6iyjnwx.png)