7. A common symptom of otitis media in young children is the prolonged presence

Question

7. A common symptom of otitis media in young children is the prolonged presence of fluid in the middle ear, known as middle-ear effusion. The presence of fluid may result in temporary hearing loss and interfere with normal learning skills in the first 2 years of life. One hypothesis is that babies who are breast-fed for at least 1 month build up some immunity against the effects of the disease and have less prolonged effusion than do bottle-fed babies. A small study of 24 pairs of babies is set up, where the babies are matched on a one-to-one basis, according to age, sex, socioeconomic status, and type of medications taken. One member of the matched pair is a breast-fed baby whereas the other member is a bottle-fed baby. The outcome variable is the duration of middle-ear effusion after the first episode of otitis media. The results are given in table below.

Pair Duration of Duration of Pair Duration of Duration of effusion in numbereffusion in effusion innumbereffusion in breast-fed bottle-fed baby (days) baby (days) 20 breast-fedbottle-fed baby (days) baby (days) 52 39 15 21 28 35 15 12 30 182 17 28 15 65 10 27 58 223 19 20 21 12 39 17 17 12 57 76 186 29 10 23 24 19 34 25 28 20 12 1) Why might a nonparametric test be useful in testing the hypothesis. A nonparametric test would be useful because the distribution of duration of effusion is very skewed and the assumption about normality of the underlying distribution are unlikely to hold. 2) Which nonparametric test should be used here?

Explanation / Answer

1) The R -code is;

x=c(20,11,3,24,7,28,58,7,39,17,17,12,52,14,12,30,7,15,65,10,7,19,34,25)
y=c(18,35,7,182,6,33,223,7,57,76,186,29,39,15,21,28,8,27,77,12,8,16,28,20)
shapiro.test(x) # Test for normality
shapiro.test(y) # Test for normality

And the output is ,

> shapiro.test(x)

Shapiro-Wilk normality test

data: x
W = 0.86009, p-value = 0.003384

> shapiro.test(y)

Shapiro-Wilk normality test

data: y
W = 0.66424, p-value = 3.513e-06

Here both p-values are less than 0.05 thus the data is not follow normality.

Hence we go to non parametric test.

2) Now here, we have to test

H0 : average duration of effusion in breast-fed babys is same as average duration of effusion in bottel-fed babys.

Thus we use wilcoxon test ;

The R-code is given as;

x=c(20,11,3,24,7,28,58,7,39,17,17,12,52,14,12,30,7,15,65,10,7,19,34,25)
y=c(18,35,7,182,6,33,223,7,57,76,186,29,39,15,21,28,8,27,77,12,8,16,28,20)
wilcox.test(x,y)

And the output is as;

> wilcox.test(x,y)

Wilcoxon rank sum test with continuity correction

data: x and y
W = 215.5, p-value = 0.1371
alternative hypothesis: true location shift is not equal to 0

Here p-value 0.1371 which is greater than 0.05 thus we accept H0 .

And conclude that,

babies who are breast-fed for at least 1 month build up some immunity against the effects of the disease and have equal prolonged effusion than do bottle-fed babies.

Related Questions

Navigate

• Browse (All)
• Browse 7
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.