A researcher was interested in knowing if preterm infants with late metabolic ac

Question

A researcher was interested in knowing if preterm infants with late metabolic acidosis and preterm infants without the condition differ with respect to urine levels of a certain chemical. The mean levels, standard deviations, and sample sizes for the two samples studied were as follows:

.

c) If we don?t want to assume the urine levels of this chemical are normally distributed but want to test if the medians of the urine level of this chemical for the two populations are equal, what test shall we use? The urine levels of this chemical in the 2 samples are shown below. Test this hypothesis at ?=0.10 level (state H0, HA, and p-value).

With condition:9.7, 1.5, 7.9, 4.9, 3.4
Without condition: 11.1, 9.6, 7.4, 8.5, 9.8, 6.5, 8.1, 7.3, 10.2

Explanation / Answer

HO:there is no difference in median between with condition and without condition

HA: there is difference in median between with condition and without condition

there will be t test for difference of two sample median

median of sample of with condition is 4.9

median of  sample of without condition is 8.5

n1=5 , n2 =9

t=(4.9-8.5)/square root of( (1/n1+1/n2)S2)

S2=n1s12+n2s22/(n1+n2-2)=(5*3.3*3.3+9*1.5*1.5)/12=6.225

now t=-9.679

p value for t at 12 df is given by 2.54x10-7

p value is less than the alpha=0.1 , so we rejet null hypothesis and accepty HA

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