# GastroenterologyWe present data relating protein concentration to pancreatic function as measured by trypsin secretion among patients with cystic fibrosis.

Modeling data distributions

Gastroenterology
We present data relating protein concentration to pancreatic function as measured by trypsin secretion among patients with cystic fibrosis.
If we do not want to assume normality for these distributions, then what statistical procedure can be used to compare the three groups?
Perform the test mentioned in Problem 12.42 and report a p-value. How do your results compare with a parametric analysis of the data?
Relationship between protein concentration $$(mg/mL)$$ of duodenal secretions to pancreatic function as measured by trypsin secretion:
$$[U/ \frac{kg}{hr}]$$
Tapsin secreton [UGA]
$$\leq\ 50$$
$$\begin{array}{|c|c|}\hline \text{Subject number} & \text{Protetion concentration} \\ \hline 1 & 1.7 \\ \hline 2 & 2.0 \\ \hline 3 & 2.0 \\ \hline 4 & 2.2 \\ \hline 5 & 4.0 \\ \hline 6 & 4.0 \\ \hline 7 & 5.0 \\ \hline 8 & 6.7 \\ \hline 9 & 7.8 \\ \hline \end{array}$$
$$51\ -\ 1000$$
$$\begin{array}{|c|c|}\hline \text{Subject number} & \text{Protetion concentration} \\ \hline 1 & 1.4 \\ \hline 2 & 2.4 \\ \hline 3 & 2.4 \\ \hline 4 & 3.3 \\ \hline 5 & 4.4 \\ \hline 6 & 4.7 \\ \hline 7 & 6.7 \\ \hline 8 & 7.9 \\ \hline 9 & 9.5 \\ \hline 10 & 11.7 \\ \hline \end{array}$$
$$>\ 1000$$
$$\begin{array}{|c|c|}\hline \text{Subject number} & \text{Protetion concentration} \\ \hline 1 & 2.9 \\ \hline 2 & 3.8 \\ \hline 3 & 4.4 \\ \hline 4 & 4.7 \\ \hline 5 & 5.5 \\ \hline 6 & 5.6 \\ \hline 7 & 7.4 \\ \hline 8 & 9.4 \\ \hline 9 & 10.3 \\ \hline \end{array}$$

2020-12-03
Step 1 Given data is:
Tapsin secreton $$[U/(kg/hr)]$$
$$\leq\ 50$$
$$\begin{array}{|c|c|}\hline \text{Subject number} & \text{Protetion concentration} \\ \hline 1 & 1.7 \\ \hline 2 & 2.0 \\ \hline 3 & 2.0 \\ \hline 4 & 2.2 \\ \hline 5 & 4.0 \\ \hline 6 & 4.0 \\ \hline 7 & 5.0 \\ \hline 8 & 6.7 \\ \hline 9 & 7.8 \\ \hline \end{array}$$
$$51\ -\ 1000$$
$$\begin{array}{|c|c|}\hline \text{Subject number} & \text{Protetion concentration} \\ \hline 1 & 1.4 \\ \hline 2 & 2.4 \\ \hline 3 & 2.4 \\ \hline 4 & 3.3 \\ \hline 5 & 4.4 \\ \hline 6 & 4.7 \\ \hline 7 & 6.7 \\ \hline 8 & 7.9 \\ \hline 9 & 9.5 \\ \hline 10 & 11.7 \\ \hline \end{array}$$
$$>\ 1000$$
$$\begin{array}{|c|c|}\hline \text{Subject number} & \text{Protetion concentration} \\ \hline 1 & 2.9 \\ \hline 2 & 3.8 \\ \hline 3 & 4.4 \\ \hline 4 & 4.7 \\ \hline 5 & 5.5 \\ \hline 6 & 5.6 \\ \hline 7 & 7.4 \\ \hline 8 & 9.4 \\ \hline 9 & 10.3 \\ \hline \end{array}$$
Step 2
1) By using Kruskal-Wallis test to compare 3 groups we get, Combining score of all the three groups, arranging them into ascending order and assigning them ra
.
$$A=group\ 1\ \leq\ 50$$
$$B=group\ 2\ =51\ -\ 1000$$
$$C=group\ 3\ =\ \Rightarrow\ 1000$$
$$\begin{array}{|c|c|}\hline \text{Observation} & \text{Rank} & \text{Groups} \\ \hline 1.4 & 1 & B \\ \hline 1.7 & 2 & A \\ \hline 2 & 3.5 & A \\ \hline 2 & 3.5 & A \\ \hline 2.2 & 5 & A \\ \hline 2.4 & 6.5 & B \\ \hline 2.4 & 6.5 & B \\ \hline 2.9 & 8 & C \\ \hline 3.3 & 9 & B \\ \hline 3.8 & 10 & C \\ \hline 4 & 11.5 & A \\ \hline 4 & 11.5 & A \\ \hline 4.4 & 13.5 & B \\ \hline 4.4 & 13.5 & C \\ \hline 4.7 & 15.5 & B \\ \hline 4.7 & 15.5 & C \\ \hline 5 & 17.5 & A \\ \hline 5 & 17.5 & C \\ \hline 5.6 & 19 & C \\ \hline 6.7 & 20.5 & A \\ \hline 6.7 & 20.5 & B \\ \hline 7.4 & 22 & C \\ \hline 7.6 & 23 & B \\ \hline 7.8 & 24 & A \\ \hline 9.4 & 25 & C \\ \hline 9.5 & 26 & B \\ \hline 10.3 & 27 & C \\ \hline 11.7 & 28 & B \\ \hline \end{array}$$
$$n_{A} = 9$$
$$n_{B} = 10$$
$$n_{C} = 9$$
$$n = n_{A}\ +\ n_{B}\ +\ n_{C} = 9\ +\ 10\ +\ 9 = 28$$
$$R_{A} = \sum\ \text{of ra for group}\ A = 2\ +\ 3.5\ +\ 3.5\ +\ 5\ +\ 11.5\ +\ 11.5\ +\ 17.5\ +\ 20.5\ +\ 24 = 99$$

$$R_{B} = \sum \text{of ra for group} B = 1\ +\ 6.5\ +\ 6.5\ +\ 9\ +\ 13.5\ +\ 15.5\ +\ 20.5\ +\ 23\ +\ 26\ +\ 28 = 149.5$$

$$R_{C} = \sum \text{of ra for group} C = 8\ +\ 10\ +\ 13.5\ +\ 15.5\ +\ 17.5\ +\ 19\ +\ 22\ +\ 25\ +\ 27 = 157.5$$

Hypothesis is given as:
$$H_{0}:\ \mu_{A}=\ \mu_{b}=\ \mu_{C}$$ i.e. three groups are equally effective.
$$H_{1} :\ \text{at least two of the} \mu$$ are different.
Kruskal-Wallis test statistics is given as:
$$H=\ \frac{12}{n(n\ +\ 1)}\left[\frac{R_{A}^{2}}{n_{A}}\ +\ \frac{R_{B}^{2}}{n_{B}}\ +\ \frac{R_{C}^{2}}{n_{C}}\right]-3(n\ +\ 1)$$
$$=\ \frac{12}{28\ \times\ (29)}\left[\frac{(99)^{2}}{9}\ +\ \frac{(145.5)^{2}}{10}\ +\ \frac{157.5^{2}}{9}\right]\ -\ 87$$
$$=\ \frac{12}{812}\left[\frac{9810}{9}\ +\ \frac{22350.25}{10}\ +\ \frac{24806.25}{9}\right]\ -\ 87$$
$$=\ \frac{12}{812}\left[1089\ +\ 2235.025\ +\ 2756.25\right]\ -\ 87$$
$$=\ \frac{12}{812}\left[6080.275\right]\ -\ 87$$
$$=0.01479\ \times\ [6080.275]\ -\ 87$$
$$= 89.927\ -\ 87$$
$$H = 2.927$$
$$df = k\ -\ 1 = 3\ -\ 1 = 2$$
The table of chi — square for 2 d.f. at 3% level of significance is = 5.991
The calculated value $$H = 2.921$$ is smaller than table value
Conclusion : Accept $$H_{0}$$. i.e. three groups are equally effective.
Step 3
Compare result with a parametric analysis of the data is given as:
By using excel we get anova:
Summary:
$$\begin{array}{|c|c|}\hline \text{Groups} & \text{Count} & \text{Sum} & \text{Average} & \text{Variance} \\ \hline A & 9 & 35.4 & 3.933333 & 4.9025 \\ \hline B & 10 & 54.1 & 5.41 & 11.43656 \\ \hline C & 9 & 53.5 & 5.944444 & 6.480278 \\ \hline \end{array}$$
ANOVA:
$$\begin{array}{|c|c|}\hline \text{Source of Variation} & \text{SS} & \text{df} & \text{MS} & \text{F} & \text{P-value} & \text{F-crit} \\ \hline \text{Between Groups} & 19.62735 & 2 & 9.813675 & 1.264706 & 0.29977 & 3.38519 \\ \hline \text{Within Groups} & 193.9912 & 25 & 7.59649\\ \hline \text{Total} & 23.6186 & 27 \\ \hline \end{array}$$
$$F\ -\ calculated\ value\ = 1.264706\ is\ less\ than\ F\ -\ table(critical\ value) = 3.38519.$$
Conclusion: Accept $$H_{0}.$$ i.e. three groups are equally effective.
By comparing non parametric Kruskal-Wallis test to parametric analysis of data both test have same result.