Question

In: Statistics and Probability

SOLVE USING R: Twenty four rug rats were randomly assigned to four groups of six each....

SOLVE USING R: Twenty four rug rats were randomly assigned to four groups of six each. The groups were subjected to different levels of stress (high, moderate, low, and none) induced by occasional, unexpected loud noises. After one week, density of blood lym- phocytes (cells ×106/ml) were measured. Prior studies indicated that the homogeneity assumption was violated and could not be remedied.

High: 2.9, 1.8, 2.1, 1.5, 0.9, 2.8

Moderate: 3.7, 4.6, 4.2, 2.8, 3.2, 4.5

Low: 5.2, 6.2, 5.7, 4.6, 5.1, 4.9

None: 6.8, 7.9, 7.1, 6.5, 6.3, 7.6

Construct a data frame rats that has two columns: “StressLevel” and “Density.” Treatment has four values: High, Moderate, Low, and None. The second column contains the numerical values of the density of blood lymphocytes. Once you have created your data frame, you should run the Kruskal-Wallis test using the command

kruskal.test(Density ~ StressLevel, data=rats)

to test your appropriate null hypothesis. (Note: make sure you explicitly state what the null hypothesis is before running your hypothesis test and make sure that you explicitly state your conclusions with explanation!) PLEASE COPY AND PASTE R INPUT AND OUTPUT.

Solutions

Expert Solution

> #Null hypothesis H0:All the stress levels have same median
>#Alternative hypothesis Ha: At least one of the stress level have different median.

#conclusion : Since the p-value = 9.799e-05 < 0.05 level of significance so we reject the null hypothesis and conclude that At least one of the stress level have different median.


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