Question

Sleep. Refer to Exercise 10.124.

a. Determine a $90\%$ confidence interval for the additional sleep that would be obtained, on average, by using laevohysocyamine hydrobromide.

b. Repeat part (a) for a $98\%$ confidence level.

Short answer

a). $(1.1695,3.4905)$is the $90\%$confidence interval for the data.

b). $(0.5483,4.1162)$ is the data's $98\%$ confidence interval.

Step-by-step solution

Part (a) Step 1: Given Information

The data is displayed below with the values $\overline{d}=2.33,s_d=2.002$.

$1.9$	$0.8$	$1.1$	$0.1$	$-0.1$
$4.4$	$5.5$	$1.6$	$4.6$	$3.4$

Part (a) Step 2: Explanation

The degree of freedom is,

$\text{dof}=n-1$

$=10-1$

$=9$

The critical value for the level of significance is localid="1651282848943" $\pm 1.833$.

The confidence level is,

$CI=\overline{d}\pm t_{\frac{a}{2}}\frac{s_d}{\sqrt{n}}$

$=2.33\pm (1.833)\frac{2.002}{\sqrt{10}}$

$=(1.1695,3.4905)$

Part (b) Step 1: Given Information

The data is displayed below with the values $\overline{d}=2.33,s_d=2.002$.

$\begin{array}{|lllll|}\hline 1.9& 0.8& 1.1& 0.1& -0.1\\ 4.4& 5.5& 1.6& 4.6& 3.4\\ \hline\end{array}$

Part (b) Step 2: Explanation

The degree of freedom is,

$\text{dof}=n-1$

$=10-1$

$=9$

The critical value for level of significance is $\pm 2.821$.

The confidence level is,

$CI=\overline{d}\pm t_{\frac{a}{2}}\frac{s_d}{\sqrt{n}}$

$=2.33\pm (2.821)\frac{2.002}{\sqrt{10}}$

$=(0.5483,4.1162)$