SOLUTION
Write out the given parameters in the questions
![\begin{gathered} \text{standard deviation =0.4 } \\ \text{sample ,n=20} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/ol81s4bdd508gaf7fdv4axbalwtibsw2pw.png)
The critical value is the measurement used to calculate the margin of error within a set of data and is expressed as
![\begin{gathered} \text{Critical value =}1-(\alpha)/(2) \\ \text{where } \\ \alpha=0.05 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/9316r8wd32dvqr3uic1wda41j391kcd4ru.png)
Then
![\text{critical value is the z=1.960}](https://img.qammunity.org/2023/formulas/mathematics/high-school/du8j90tb1f1cmff1br7zjdzbewbtx6p5xa.png)
Therefore the critical value is1.960
Then the standard error is given by
![\begin{gathered} \sigma_{\bar{x}}=\frac{\sigma}{\sqrt[]{n}} \\ \\ \text{where } \\ n=\text{sample space=20 and }\sigma=0.4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/du0v8ehdnqfrxf0lntogx1pnjzbr7b84w5.png)
Substituting the value we have
![\sigma_{\bar{x}}=\frac{0.4}{\sqrt[]{20}}=(0.4)/(4.47)=0.089](https://img.qammunity.org/2023/formulas/mathematics/high-school/1kktw7enn8hmf4mrg4rz2kw0biflbiapx3.png)
Therefore the standard error is 0.089
The confidence interval is given by
![\begin{gathered} \text{Confidence interval=}\bar{x}\pm(1.96)(S.E) \\ \text{where S.E= standard error } \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/vbdefu42lbwjs66eie0yi6euwqsdwikgsv.png)
The mean for the sample will be
![\bar{x}=\frac{sum\text{ of data}}{n}=(3497.76)/(20)=173.988](https://img.qammunity.org/2023/formulas/mathematics/high-school/3pbwec2y9sqtauvka9xzuy3csz3sw3in8s.png)
Substitute the value to obtain the confidence interval
![\begin{gathered} \text{confidence interval=173.988}\pm1.96*0.089 \\ C.I=173.988\pm0.174 \\ C\mathrm{}I=(173.814,174.162) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/yh2ynqj8x3m88y3yz4pxmnvdxb3nyzwfr5.png)
Therefore, the confidence interval is (173.81,174.16) to 2d.p