91Ó°ÊÓ

$\text{Samplemean}\stackrel{-}{\mathrm{χ}}\text{=26.2}$

The sample size is high; the sample standard deviation can be used to approximate the overall standard deviations.

$\begin{array}{l}\mathrm{σ}=4.1\\ \text{n}=70\end{array}$

The calculation is given below:

$$\begin{gathered} \text{Levelofsignificance=}\mathrm{Î\pm }=1−\text{Confidence} \\ =1−0.95 \\ =0.05 \end{gathered}$$

$$\begin{gathered} \text{Criticalvalue=z}\mathrm{Î\pm }/2 \\ ={\text{z}}_{0.025}=1.96 \end{gathered}$$

$$\begin{gathered} \text{Criticalvalue,}Â\pm \text{z}\mathrm{Î\pm }/2=Â\pm 1.96 \\ \text{Errorofmargin}\left(\text{E}\right)\text{=z}\mathrm{Î\pm }/2×\frac{\mathrm{σ}}{\sqrt{\text{n}}} \\ \text{·¡=1.962×}\frac{\text{4.1}}{\sqrt{\text{70}}} \\ \text{E=0.960486} \end{gathered}$$

95% confidence interval limits are shown by:

$$\begin{gathered} \text{Lowerlimit=}\stackrel{-}{\mathrm{χ}}\text{-E} \\ =26.2−0.960486 \\ =25.2395 \end{gathered}$$

$$\begin{gathered} \text{Upperlimit=}\stackrel{-}{\mathrm{χ}}+\text{E} \\ =26.2−0.960486 \\ =27.1605 \end{gathered}$$

95% confidence interval is:

$=\stackrel{-}{\mathrm{χ}}+\text{E}$

localid="1651472418180" $$\begin{gathered} \text{=26.2}Â\pm \text{0.960486} \\ =(25.239514,27.160486) \end{gathered}$$

Since the approach we used performs 95% of the period, we have a confidence coefficient of 0.95that the calculated confidence interval will include the population's mean.

$\text{Samplemean}\stackrel{-}{\mathrm{χ}}\text{=26.2}$

$\begin{array}{l}\mathrm{σ}=4.1\\ \text{n}=70\end{array}$

The calculation is given below:

$$\begin{gathered} \text{Levelofsignificance=}\mathrm{Î\pm }=1−\text{Confidence} \\ =1−0.99 \\ =0.01 \end{gathered}$$

$$\begin{gathered} \text{Criticalvalue=z}\mathrm{Î\pm }/2 \\ ={\text{z}}_{0.005}=2.58 \end{gathered}$$

The closest value from the z table:

$$\begin{gathered} \text{Criticalvalue,}Â\pm \text{z}\mathrm{Î\pm }/2=Â\pm 2.58 \\ \text{Errorofmargin}\left(\text{E}\right)\text{=z}\mathrm{Î\pm }/2×\frac{\mathrm{σ}}{\sqrt{\text{n}}} \\ \text{·¡=2.58×}\frac{\text{4.1}}{\sqrt{\text{70}}} \\ \text{E=1.264314} \end{gathered}$$

95% confidence interval limits are shown by:

$$\begin{gathered} \text{Lowerlimit=}\stackrel{-}{\mathrm{χ}}\text{-E} \\ =26.2−1.264314 \\ =24.9357 \end{gathered}$$

$$\begin{gathered} \text{Upperlimit=}\stackrel{-}{\mathrm{χ}}+\text{E} \\ =26.2−1.264314 \\ =27.4643 \end{gathered}$$

95% confidence interval is:

localid="1651473127163" $$\begin{gathered} \stackrel{-}{\mathrm{χ}}+\text{E} \\ \text{=26.2}Â\pm 1.2\text{64314} \\ =(24.935686,27.464314) \end{gathered}$$

If the confidence coefficient is raised, the width increases.

Yes, it will be validbecause of the central limit theorem, sample means are about distributed normally when the sample size is larger than 30.