Chapter 17, Problem 17.15AYK

(a)

To determine

To Explain: the confidence interval on the basis provided information.

Answer to Problem 17.15AYK

Confidence interval is 0.8176 and 1.3278

Explanation of Solution

Given:

On the basis of given output of confidence interval is 0.8176 and 1.3278 which implies that there are 95% of surety that the mean difference in tremor amplitude with ACT device off and ACT device on for the patients diagnosed with essential tremor.

(b)

To determine

To find: the mean and standard deviation.

Answer to Problem 17.15AYK

  $\begin{array}{l}\overline{x}=1.07\\ s=0.379\end{array}$

Explanation of Solution

Given:

ACT off	ACT on
1.2	0.3
1.6	0.2
2	0.6
1	0.2
1.7	0.5
1.9	0.4
0.8	0.2
1.7	0.6
1.6	0.3
0.5	0.2
1.4	0.1

Formula used:

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

  $s=\sqrt{\frac{{{\displaystyle \sum \left(d-\overline{x}\right)}}^2}{n-1}}$

Calculation:

ACT off	ACT on	difference = ACT off-ACT on	d-   $\overline{x}$	${\left(d-\overline{x}\right)}^2$
1.2	0.3	0.9	-0.17	0.03
1.6	0.2	1.4	0.33	0.11
2	0.6	1.4	0.33	0.11
1	0.2	0.8	-0.27	0.07
1.7	0.5	1.2	0.13	0.02
1.9	0.4	1.5	0.43	0.18
0.8	0.2	0.6	-0.47	0.22
1.7	0.6	1.1	0.03	0.00
1.6	0.3	1.3	0.23	0.05
0.5	0.2	0.3	-0.77	0.60
1.4	0.1	1.3	0.23	0.05
	Sum	11.8		1.44

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum d}}{n}=\frac{11.8}{11}=1.07\\ s=\sqrt{\frac{{{\displaystyle \sum \left(d-\overline{x}\right)}}^2}{n-1}}=\sqrt{\frac{1.44}{11-1}}=\text{0}\text{.379}\end{array}$

(c)

To determine

To find: the margin of error of a 95% confidence interval for the mean difference.

Answer to Problem 17.15AYK

Margin of Error = 0.2546

Explanation of Solution

Given:

  $n=11$

From the part (b)

  $\begin{array}{c}\overline{x}=1.07\\ s=0.379\end{array}$

Formula used:

  $\text{Margin}\text{of}{\text{Error = t}}_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \frac{s}{\sqrt{n}}$

Calculation:

For 0.05 level of significance and

Degree of Freedom

  $\begin{array}{c}df=n-1\\ =11-1\\ =10\end{array}$

  $t_{\raisebox{1ex}{$0.05$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,df=10}=2.228$

  $\begin{array}{c}\text{Margin}\text{of}{\text{Error = t}}_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \frac{s}{\sqrt{n}}\\ =2.228\times \frac{0.379}{\sqrt{11}}\\ =\text{0}\text{.2546}\end{array}$