Question

Anorexia Treatment. Refer to Exercise 10.125 and find a $90\%$ confidence interval for the weight gain that would be obtained, on average, by using the family therapy treatment.

Short answer

$(-10.2918,-4.2282)$ is the data's $90\%$ confidence interval.

Step-by-step solution

GIven Information

The given data is,

$\begin{array}{|cccs|}\hline \text{Before}& \text{After}& \text{Before}& \text{After}\\ 83.3& 94.3& 80.5& 752\\ 86& 91.5& 81.6& 77.8\\ 82.5& 91.9& 83.8& 95.2\\ 86.7& 100.3& 82.1& 95.5\\ 79.6& 76.7& 77.6& 90.7\\ 87.3& 98& 83.5& 92.5\\ 76.9& 76.8& 89.9& 93.8\\ 94.2& 101.6& 86& 91.7\\ 73.4& 94.9& & \\ \hline\end{array}$

Explanation

Below is a table of contents.

$\begin{array}{|cccccccc|}\hline \text{Before}& \text{After}& \text{Difference=d}& d^2& \text{Before}& \text{After}& \text{Difference=d}& d^2\\ 83.3& 94.3& -11& 121& 80.5& 752& 5.3& 28.09\\ 86& 91.5& -5.5& 30.25& 81.6& 77.8& 3.8& 14.44\\ 82.5& 91.9& -9.4& 88.36& 83.8& 95.2& -11.4& 129.96\\ 86.7& 100.3& -13.6& 184.96& 82.1& 95.5& -13.4& 179.56\\ 79.6& 76.7& 2.9& 8.41& 77.6& 90.7& -13.1& 171.61\\ 87.3& 98& -10.7& 114.49& 83.5& 92.5& -9& 81\\ 76.9& 76.8& 0.1& 0.01& 89.9& 93.8& -3.9& 15.21\\ 94.2& 101.6& -7.4& 54.76& 86& 91.7& -5.7& 32.49\\ 73.4& 94.9& -21.5& 462.25& & & -123.5& 1716.85\\ \hline\end{array}$

Explanation

The sample mean is

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

$=\frac{-123.5}{17}$

$=-7.2647$

The standard deviation is,

$s_d=\sqrt{\frac{\sum d_i^2-\frac{{\left(\sum d_i\right)}^2}{n}}{n-1}}$

$=\sqrt{\frac{1716.85-\frac{(-123.5{)}^2}{17}}{17-1}}$

$=7.1574$

Explanation

The degree of freedom is,

$dof=n-1$

$=17-1$

$=16$

The level of significance's critical value is $\pm 1.746$.

The confidence level is,

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

$=-7.2647\pm (1.746)\frac{7.1574}{\sqrt{17}}$

$=(-10.2918,-4.2282)$