Question

6. An article in the Journal of Applied Physiology ("Histamine Content in Sputum from Allergic and Non-Allergic Individuals", Thomas & Simmons, 1969: 535-539) reports the following data on sputum histamine level (4g/g dry weight of sputum) for a sample of 9 individuals classified as allergics and another sample of 13 individ- uals classified as nonallergics Allergics 31.0, 39.6, 64.7, 65.9,67.6, 100.0, 102.4, 1112.0, 1651.0, Nonallergics 4.7, 5.2,6.6, 18.9, 27.3, 29.1, 32.4, 34.3, 35.4, 41.7, 45.5, 48.0,48.1 (a) Use the sign test to determine if there is evidence that the median sputum histamine level of nonallergics is less than 35. Use your hypotheses. Cumulative probabilities for binomial (13,0.5 a0.05 and clearly state 0.00 1.00 2.00 3.00 4.00 5.00 6.00 700 8.00 PY vip 0.5) 0.00 0.00 0.01 0.05 0.13 0.29 0.500.71 0.87 9.00 10.00 11.00 12.00 13.00 P(Y y;p 0.5) 0.95 0.99 1.00 .00 1.00 (b) Use the two sample Wilcoxon test to determine if these data indicate there is a difference in median sputum histamine level between allergics and nonallergics. Use the normal approximation to the distribution of the test statistic and take a 0.05 and clearly state your hypotheses Recall zo.os 1.645, o.025 1.96, 20.01-2.326, o.005 2.576

Answer 1

(a)

$Here~ we~ want~ to~ test:\\ Null~hypothesis,~H_0:\theta=35~vs.~Alternative~hypothesis,~H_1:\theta<35,\\ where~\theta=population~median~of~non allergies.\\ We~replace~each~observation~greater~than~\35~by~plus~sign~and~each~observation\\ less~than~35~by~a~minus~sign.~The~sample~values~equal~to~35~is~ignored.\\ Then~r=no.~of~plus~signs=5,~s=no.~of~minus~sign=8.\\ Now,~P(R\leq 5|H_0:p=0.5)=0.29~where~p=P(observation>35).\\ Under~H_0,~p=\frac{1}{2}~and~R=random~variable~denoting~no.~of~plus~signs\\\sim Binomial(13,1/2)~under~H_0.\\ Since~P(R\leq 5|H_0:p=0.5)=0.29>0.05~so~we~fail~to~reject~H_0~at~5\%~level~of~significance\\ and~conclude~that~there~is~insufficient~evidence~that~the~median~sputum~histamine~level~of\\~non allergies~is~less~than~35.$

(b)

$Let~\theta_1~and~\theta_2~are~two~population~medians~of~Allergies~and~Non allergies~respectively.\\ Null~hypothesis,~H_0:\theta_1=\theta_2~vs.~Alternative~hypothesis,~H_1:\theta_1\neq \theta_2.$

N Sample Median
Allergies 9 67.6
Non allergies 13 32.4

We arrange the data of combined sample in increasing order. Let x=Allergies, y=Non allergies
W₁=sum of rank of x's in combined sample=151, W₂=sum of rank of y's in combined sample=102
W = max{W₁,W₂}=151.0
$Now,~Z=\frac{W-\frac{13*23}{2}}{\sqrt{9*13*23/12}}=0.1001$

$Critical ~values=\pm z_{0.025}=\pm 1.96.\\ Since~value~of~Z>1.96~hence~we~fail~to~reject~H_0~at~5\%~level~of~significance~and~\\ conclude~that~there~is~insignificant~difference~in~median~sputum~histamine~level\\ between~allergies~and~non allergies.$