A random sample of 70 observations from a normally distributed population possesses a sample mean equal to 26.2 and a sample standard deviation equal to 4.1.

a. Find an approximate 95% confidence interval for

b. What do you mean when you say that a confidence coefficient is .95?

c. Find an approximate 99% confidence interval for

d. What happens to the width of a confidence interval as the value of the confidence coefficient is increased while the sample size is held fixed?

e. Would your confidence intervals of parts a and c be valid if the distribution of the original population was not normal? Explain

$\text{Samplemean}\stackrel{-}{\chi }\text{=26.2}$

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

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

The calculation is given below:

$$\begin{gathered} \text{Levelofsignificance=}\alpha =1-\text{Confidence} \\ =1-0.95 \\ =0.05 \end{gathered}$$

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

$$\begin{gathered} \text{Criticalvalue,}\pm \text{z}\alpha /2=\pm 1.96 \\ \text{Errorofmargin}\left(\text{E}\right)\text{=z}\alpha /2\times \frac{\sigma }{\sqrt{\text{n}}} \\ \text{E=1.962×}\frac{\text{4.1}}{\sqrt{\text{70}}} \\ \text{E=0.960486} \end{gathered}$$

95% confidence interval limits are shown by:

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

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

95% confidence interval is:

$=\stackrel{-}{\chi }+\text{E}$

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

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.

$\text{Samplemean}\stackrel{-}{\chi }\text{=26.2}$

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

The calculation is given below:

$$\begin{gathered} \text{Levelofsignificance=}\alpha =1-\text{Confidence} \\ =1-0.99 \\ =0.01 \end{gathered}$$

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

The closest value from the z table:

$$\begin{gathered} \text{Criticalvalue,}\pm \text{z}\alpha /2=\pm 2.58 \\ \text{Errorofmargin}\left(\text{E}\right)\text{=z}\alpha /2\times \frac{\sigma }{\sqrt{\text{n}}} \\ \text{E=2.58×}\frac{\text{4.1}}{\sqrt{\text{70}}} \\ \text{E=1.264314} \end{gathered}$$

95% confidence interval limits are shown by:

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

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

95% confidence interval is:

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

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.