**Exercise: Confidence Interval Estimation** A random sample of 11 items is drawn from a population whose standard deviation is unknown. The sample mean is \( \bar{x} = 810 \) and the sample standard deviation is \( s = 20 \). Use Appendix D to find the values of Student’s t. **(a)** Construct an interval estimate of \( \mu \) with 98% confidence. *(Round your answers to 3 decimal places.)* The 98% confidence interval is from _______ to _______. **(b)** Construct an interval estimate of \( \mu \) with 98% confidence, assuming that \( s = 40 \). *(Round your answers to 3 decimal places.)* The 98% confidence interval is from _______ to _______. **(c)** Construct an interval estimate of \( \mu \) with 98% confidence, assuming that \( s = 80 \). *(Round your answers to 3 decimal places.)* The 98% confidence interval is from _______ to _______. **(d)** Describe how the confidence interval changes as \( s \) increases. - The interval stays the same as \( s \) increases. - The interval gets wider as \( s \) increases. (Correct) *Note: Please refer to the table in Appendix D for the appropriate t values for constructing these confidence intervals.*

From the given information,

Sample size n =11

sample mean, x̅=810

sample standard deviation, s=20

Confidence level = 98%

Degrees of freedom: 24(=25-1).

Using Excel function, “=T.INV.2T (0.02,10)”, the critical value for two-tailed test at 98% confidence level is 2.7638.

The standard error is:

$\begin{array}{rcl}SE& =& \frac{s}{\sqrt{n}}\\ & =& \frac{20}{\sqrt{11}}\\ & =& 6.0302\end{array}$

Margin of error:

$$\begin{gathered} E=t_{(ꭤ/2)}SE \\ =2.7638\times 6.0302 \\ =16.666 \end{gathered}$$

Confidence interval:

$$\begin{gathered} CI=\mathrm{x̅}\pm E \\ =810\pm 16.666 \\ =(793.334,826.666). \end{gathered}$$

The 98% confidence interval is from 793.334 to 826.666.