Question

Cushing's disease is characterized by muscular weakness due to adrenal or pituitary dysfunction. To provide effective treatment, it is important to detect childhood Cushing's disease as early as possible. Age at onset of symptoms and age at diagnosis (months) for 15 children suffering from the disease were given in the article "Treatment of Cushing's Disease in Childhood and Adolescence by Transphenoidal Microadenomectomy" (New Engl. J. of Med., 1984: 889). Here are the values of the differences between age at onset of symptoms and age at diagnosis:

\(\begin{array}{*{20}{l}}{ - 24}&{ - 12}&{ - 55}&{ - 15}&{ - 30}&{ - 60}&{ - 14}&{ - 21}\\{ - 48}&{ - 12}&{ - 25}&{ - 53}&{ - 61}&{ - 69}&{ - 80}&{}\end{array}\)

a. Does the accompanying normal probability plot cast strong doubt on the approximate normality of the population distribution of differences?

b. Calculate a lower\(95\% \)confidence bound for the population mean difference, and interpret the resulting bound.

c. Suppose the (age at diagnosis) - (age at onset) differences had been calculated. What would be a\(95\backslash \% \)upper confidence bound for the corresponding population mean difference?

Short answer

a. Plausible;

b. \( - 49.14\)

c. \(49.14.\)

Step-by-step solution

A)Step 1: The difference of population difference

It can be said that the data follow a linear pattern, and thus that the data (differences) is from the normal population.

B)Step 2: Find the sample standard form

The paired \(t\) confidence interval for\({\mu _D}\) is

\(\left( {\bar d - {t_{\alpha /2,n - 1}} \cdot \frac{{{s_D}}}{{\sqrt n }},\bar d + {t_{\alpha /2,n - 1}} \cdot \frac{{{s_D}}}{{\sqrt n }}} \right)\)

A one-sided confidence bound can be obtained by retaining the relevant sing \(( + \) or\( - )\), and by replacing \({t_{\alpha /2,n - 1}}\) by\({t_{\alpha ,n - 1}}\).

The \(95\% \) lower confidence bound for the population mean difference is

\(\bar d - {t_{\alpha ,n - 1}} \cdot \frac{{{s_D}}}{{\sqrt n }}\)

Step 3:

The Sample Mean \(\bar x\) of observations\({x_1},{x_2}, \ldots ,{x_n}\) is given by

\(\bar x = \frac{{{x_1} + {x_2} + \ldots + {x_n}}}{n} = \frac{1}{n}\sum\limits_{i = 1}^n {{x_i}} \)

The sample mean \(\bar d\) for the differences is

\(\bar d = \frac{1}{{15}} \cdot ( - 24 - 12 - \ldots - 80) = - 38.6\)

The Sample Variance \({s^2}\) is

\({s^2} = \frac{1}{{n - 1}} \cdot {S_{xx}}\)

where

\({S_{xx}} = \sum {{{\left( {{x_i} - \bar x} \right)}^2}} = \sum {x_i^2} - \frac{1}{n} \cdot {\left( {\sum {{x_i}} } \right)^2}\)

The Sample Standard Deviation \(s\) is

\(s = \sqrt {{s^2}} = \sqrt {\frac{1}{{n - 1}} \cdot {S_{xx}}} \)

The sample variance is

\(\begin{array}{l}s_D^2 = \frac{1}{{15 - 1}} \cdot \left( {{{( - 24 - ( - 38.6))}^2} + {{( - 12 - ( - 38.6))}^2} + \ldots + {{( - 80 - ( - 38.6))}^2}} \right)\\ = 537.3124\end{array}\)

and the sample standard deviation

\({s_D} = \sqrt {537.3124} = 23.18\)

The only missing value is

\({t_{\alpha /2,n - 1}} = {t_{0.05,14}} = 1.761\)

which was computed using a software (you can use the table in the appendix of the book).

The\(95\% \)lower confidence bound is

\(\bar d - {t_{\alpha ,n - 1}} \cdot \frac{{{s_D}}}{{\sqrt n }} = - 38.6 - 1.761 \cdot \frac{{23.18}}{{\sqrt {15} }} = \)

C)Step 4: Find the population mean difference

Similarly as in\((b)\), the \(95\% \)upper confidence bound is\(\bar d + {t_{\alpha ,n - 1}} \cdot \frac{{{s_D}}}{{\sqrt n }} = - 38.6 + 1.761 \cdot \frac{{23.18}}{{\sqrt {15} }} = 49.14\)