Question

Anorexia Nervosa (AN) is a psychiatric condition leading to substantial weight loss among women who are fearful of becoming fat. The article "Adipose Tissue Distribution After Weight Restoration and Weight Maintenance in Women with Anorexia Nervosa" (Amer. J. of ClinicalNutr., 2009: 1132-1137) used whole-body magnetic resonance imagery to determine various tissue characteristics for both an AN sample of individuals who had undergone acute weight restoration and maintained their weight for a year and a comparable (at the outset of the study) control sample. Here is summary data on intermuscular adipose tissue (IAT; kg).

Assume that both samples were selected from normal distributions.

a. Calculate an estimate for true average IAT under the described AN protocol, and do so in a way that conveys information about the reliability and precision of the estimation.

b. Calculate an estimate for the difference between true average AN IAT and true average control IAT, and do so in a way that conveys information about the reliability and precision of the estimation. What does your estimate suggest about true average AN IAT relative to true average control IAT?

Short answer

(a) \(\;95\% \)confidence interval: \(({\bf{0}}.{\bf{38155}},)\)

(b) \(95\% \)confidence interval: \(( - 0.0045,0.3445)\)

The confidence interval suggest that there is no difference in the true average AN IAT compared to the true average control IAT.

Step-by-step solution

a)Step 1: Find the boundaries of confident intervals

Given:

\(\begin{array}{l}n = 16\\\bar x = 0.52\\s = 0.26\end{array}\)

Let us assume that we want to calculate the confidence interval with\(95\% \)confidence (other confidence levels work similarly).

\(c = 95\% = 0.95\)Determine the t-value by looking in the row starting with degrees of freedom\(df = n - 1 = 16 - 1 = 15\)and in the column with\(1 - c/2 = \)\(0.025\)in the table of the Student's T distribution:

\({t_\alpha } = 2.131\)

The margin of error is then:

\(E = {t_{\alpha /2}} \times \frac{s}{{\sqrt n }} = 2.13 \times \frac{{0.26}}{{\sqrt {16} }} \approx 0.13845\)

The boundaries of the confidence interval then become: \(\begin{array}{l}\bar x - E = 0.52 - 0.13845 = 0.38155\\\bar x + E = 0.52 + 0.13845 = 0.65845\end{array}\)

B)Step 2: find the end point of the confidence interval

\(\begin{array}{l}{{\bar x}_1} = 0.52\\{{\bar x}_2} = 0.35\\{n_1} = 16\\{n_2} = 8\\{s_1} = 0.26\\{s_2} = 0.15\end{array}\)

Let us assume that we want to calculate the confidence interval with\(95\% \)confidence (other confidence levels work sim

\(c = 95\% = 0.95\)

Determine the degrees of freedom (rounded down to the nearest integer): \(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{{{0.26}^2}}}{{16}} + \frac{{{{0.15}^2}}}{8}} \right)}^2}}}{{\frac{{{{\left( {{{0.26}^2}/16} \right)}^2}}}{{16 - 1}} + \frac{{{{\left( {{{0.15}^2}/8} \right)}^2}}}{{8 - 1}}}} \approx 21\)

Determine the t-value by looking in the row starting with degrees of freedom \(df = 21\) and in the column with \(1 - c/2 = 0.025\) in the Student's \(t\) distribution table in the appendix:

\({t_{\alpha /2}} = 2.080\)

The margin of error is then:

\(E = {t_{\alpha /2}} \cdot \sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} = 2.080 \cdot \sqrt {\frac{{{{0.26}^2}}}{{16}} + \frac{{{{0.15}^2}}}{8}} \approx 0.1745\)

The endpoints of the confidence interval for\({\mu _1} - {\mu _2}\) are:

\(\begin{array}{l}\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E = (0.52 - 0.35) - 0.1745 = 0.17 - 0.1745 = - 0.0045\\\left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E = (0.52 - 0.35) + 0.1745 = 0.172 + 0.1745 = 0.3445\end{array}\)

The confidence interval suggest that there is no difference in the true average AN IAT compared to the true average control IAT, because the confidence interval contains \(0\)