Some researchers were interested to see if treatment with different antibiotics

Question

Some researchers were interested to see if treatment with different antibiotics limited tooth decay. 45 inbred laboratory rats were randomly selected from a larger population and assigned to one of three treatments with 15 rats each. All 45 rats were inoculated with Staphylococcus mucans, a bacterium known to contribute to the process of tooth decay. One group, the control, was given no additional treatment. A second group was given penicillin treatment, and a third was given a rifampin treatment. After six-weeks the extent of tooth decay in each group was evaluated, as a percent tooth decay. The evaluation was a subjective measurement based on the combined evaluations of three different evaluators, and it probably should be viewed as an ordinal or a categorical measurement. So one cannot assume there is a normal distribution of randomly collected measurements even though the data are presented as a percentage. Remember that factors need to be defined as factors for statistical packages to work properly.

What would be the best null hypothesis? The best alternative hypothesis?

A. Pr[1 and 2] Pr[1] * Pr[2]

B. Pr[1 and 2] = Pr[1] * Pr[2]

C.p1 p0

D.µ1 - µ2 0

E.The distribution of these data are consistent with a normal distribution.

F.µ1 - µ2 0

G.1 - 2 > 0

H.µ1 - µ2 < 0

I.p1 p0

J.The means (µ) of the three treatments are all equal (i.e. µ1 = µ2 = µ3...).

K.1 - 2 = 0

L.µ1 - µ2 0

M.p1 < p0

N.1 - 2 0

O.p1 > p0

P.The group variances within these data are approximately equal.

Q.The group variances within these data are not approximately equal.

R. µ1 - µ2 = 0

S. The observed data are consistent with our expectations under the null (hypothetical) distribution.

T. The medians () of the three treatments are all equal (i.e. 1 = 2 = 3...).

U. 0

V. 1 - 2 0

W. 1 - 2 0

X. 0

Y. The means (µ) of the three treatments are not all equal (i.e. µ1 µ2 µ3...; at least one is different).

Z. µ1 - µ2 > 0

AA. 1 - 2 < 0

BB. = 0

CC. The distribution of these data are not consistent with a normal distribution.

DD. p1 p0

EE. The medians () of the three treatments are not all equal (i.e. 1 2 3...; at least one is different).

FF. The observed data are not consistent with our expectations under the null (hypothetical) distribution.

GG. p1 = p0

HH. = 0

Explanation / Answer

The best null hypothesis will be :
T. The medians () of the three treatments are all equal (i.e. 1 = 2 = 3...).

And the best alternative hypothesis will be :
EE. The medians () of the three treatments are not all equal (i.e. 1 2 3...; at least one is different).

The reason for choosing these options are since normality assumption cannot be taken, we have to conduct non-parametric tests which requires no distribution assumption. For non-parametric test, we compare the medians for the different groups since mean is affected by skewed data but median isn't. The null hypothesis is taken to be option (T) since we want to check the medians are equal i.e all the treatments have the same effect or any of the treatments has a better effect on the tooth decay (as said in the alternative hypothesis).

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