Chapter 10, Problem 10.66RE

(a)

To determine

To Calculate: a 95% confidence interval for the population mean.

Answer to Problem 10.66RE

For 95% Confidence intervals are

(5.8473, 7.4119)

Explanation of Solution

Given:

  

  $\begin{array}{c}c=95\%=0.95\\ n=6\end{array}$

Formula used:

  $\overline{x}=\frac{{\displaystyle \sum x}}{n}$

Calculation:

The mean is

  $\begin{array}{c}\overline{x}=\frac{9+6+8+\mathrm{...}+9+7+7}{28}\\ =\frac{179}{28}\\ =6.6296\end{array}$

The variance is

  $\begin{array}{c}s=\sqrt{\frac{{\left(9-6.6296\right)}^2+\mathrm{....}+{\left(7-6.6296\right)}^2}{28-1}}\\ =2.0172\end{array}$

The t-value by seeing in the row starting with degrees of freedom $df=n-1=28-1=27$ and in the column with $p=\left(1-c\right)/2=0.025$ in the table of the $T$ distribution:

  $t_{\alpha /2}=2.052$

The margin of error is

  $E=t_{\alpha /2}\times \frac{s}{\sqrt{n}}=2.052\times \frac{2.0172}{\sqrt{28}}=0.7823$

Confidence intervals are

  $\begin{array}{c}\overline{x}-E=6.6296-0.7823=5.8473\\ \overline{x}+E=6.6296+0.7823=7.4119\end{array}$

(b)

To determine

To Explain: that the need to know about these students in order to interpret result in part (a).

Answer to Problem 10.66RE

It is need to know that the students would be answered truthfully and that all students who were asked the question also answered the question.

Explanation of Solution

It is need to know that the students would be answered truthfully and that all students who were asked the question also answered the question because else the result. Might not be representative of the population