A manufacturer is running an experiment using the standard (control) infant formula, and two new formulas, A and B. The goal is to boost the immune system in infants. 120 infants in the study are randomly assigned to each of three groups: Group A, Group B, and a Control Group. The study is run for 12 weeks. At the end of the study the variable measured is Total IGA (in mg per dl) score. A one-way ANOVA on these data is employed. The hypothesis tested by the one-way ANOVA F test are:

A.

H0: the mean IGA score is the same for all three formulas.


Ha: the mean IGA score is not the same for all three formulas.

The Group of answer choices are as below:

A.

H0: the mean IGA score is the same for all three formulas.


Ha: the mean IGA score is not the same for all three formulas.

B.

H0: the mean IGA score is the same for all three formulas.


Ha: the mean IGA score is higher for at least one of the two treatment groups than the control group

C.

H0: the mean IGA score is the same for all three formulas.


Ha: the mean IGA score is higher for both treatment groups A and B than the control group.

D. No answer text provided.

Explanation:

Given that a manufacturer is running an experiment using the standard (control) infant formula, and two new formulas, A and B to boost the immune system in infants.

The variable that is measured is Total IGA (in mg per dl) score.

We have to state the hypotheses for one-way ANOVA F test to test if the mean IGA score is the same for all three formulas.

In null hypothesis H0, the mean of all groups are equal and in alternative hypothesis, Ha, the mean of the groups are not the same.

Therefore, the hypothesis tested by the one-way ANOVA F test are:

H0: the mean IGA score is the same for all three formulas.

Ha: the mean IGA score is not the same for all three formulas.

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d

step-by-step explanation:

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.