Chapter 10.2, Problem 10.47E

To determine

To find: a 95% confidence interval for the population mean and additional information does the confidence interval provide.

Answer to Problem 10.47E

95% for Confidence interval

(2.7167, 12.0113)

Explanation of Solution

Given:

  $\begin{array}{c}n=11\\ \overline{x}=7.364\\ s=6.918\\ c=95\%=0.95\end{array}$

Formula used:

  $E=z_{a/2}\times \frac{s}{\sqrt{n}}$

For Confidence interval

  $\begin{array}{c}\overline{x}-E\\ \overline{x}+E\end{array}$

Calculation:

The t-value by looking in the row starting with degrees of freedom $df=n-1=11-1=10$ and in the column with $p=\left(1-c\right)/2=0.025$ in the table of the Student $T$ distribution:

  $t^{\ast }=2.228$

The margin of error is

  $E=t_{a/2}\times \frac{s}{\sqrt{n}}=2.228\times \frac{6.918}{\sqrt{11}}=4.6473$

Confidence intervals

  $\begin{array}{c}\overline{x}-E=7.364-4.6473=2.7167\\ \overline{x}+E=7.364+4.6473=12.0113\end{array}$

The confidence interval is giving a range of possible values for the population mean $\mu$ , while the significance tests only tests one possible value for the population mean $\mu$ .