Agricultural scientists wanted to know which of two feed types (A and B) and whi
ID: 3317895 • Letter: A
Question
Agricultural scientists wanted to know which of two feed types (A and B) and which of two anabolic steroids (1 or 2) would produce the greatest weight gain in hogs. Ultimately they wanted to know which combination of feed type and steroid would produce the most growth. To perform this experiment, they chose piglets from a highly bred line to minimize differences due to genetic variation. Each hog was randomly assigned one feed type and one steroid type. After two weeks, the researchers weighed how much mass (kg) each hog gained. For our purposes for this data set you may use a residuals plot to test for homogeneity of variance for the different factors.
What would be the best null hypothesis? The best alternative hypothesis?
A. µ1 - µ2 = 0
B. The distribution of these data are consistent with a normal distribution.
C. The group variances within these data are approximately equal.
D. Pr[1 and 2] = Pr[1] * Pr[2]
E. = 0
F. The medians () of the response for the different levels of the factor are not all equal (i.e. 1 2 3...; at least one is different
G. The means (µ) of the response for the different levels of the factor are not all equal (i.e. µ1 µ2 µ3...; at least one is different).
H. 1 - 2 > 0
I. 1 - 2 0
J. The means (µ) of the response for the different levels of the first factor are not all equal (i.e. µ1 µ2 µ3...; at least one is different).
The means (µ) of the response for the different levels of the second factor are not all equal (i.e. µ1 µ2 µ3...; at least one is different), but in some cases we may not be interested in hypothesis testing for this factor.
K. 1 - 2 0
L. µ1 - µ2 0
M. The means (µ) of the response for the different levels of the factor are all equal (i.e. µ1 = µ2 = µ3...
N. p1 > p0
O. p1 p0
P. The distribution of these data are not consistent with a normal distribution.
Q. µ1 - µ2 < 0
R. 0
S. The medians () of the response for the different levels of the factor are all equal (i.e. 1 = 2 = 3...).
T. The means (µ) of the response for the different levels of the first factor are all equal (i.e. µ1 = µ2 = µ3...).
The means (µ) of the response for the different levels of the second factor are all equal (i.e. µ1 = µ2 = µ3...), but in some cases we may not be interested in hypothesis testing for this factor.
U. Pr[1 and 2] Pr[1] * Pr[2]
V. 1 - 2 = 0
W. p1 p0
X. 1 - 2 0
Y. µ1 - µ2 0
Z. The group variances within these data are not approximately equal.
AA. 1 - 2 < 0
BB. = 0
CC. The means (µ) of the response for the different levels of the first factor are all equal (i.e. µ1 = µ2 = µ3...).
The means (µ) of the response for the different levels of the second factor are all equal (i.e. µ1 = µ2 = µ3...).
There are no interactions between the first and second factor
DD. 0
EE. p1 < p0
FF. p1 = p0
GG. p1 p0
HH. The observed data are consistent with our expectations under the null (hypothetical) distribution.
II. The observed data are not consistent with our expectations under the null (hypothetical) distribution.
JJ. µ1 - µ2 > 0
KK. The means (µ) of the response for the different levels of the first factor are not all equal (i.e. µ1 µ2 µ3...; at least one is different).
The means (µ) of the response for the different levels of the second factor are not all equal (i.e. µ1 µ2 µ3...; at least one is different).
There is an interaction between the first and second factor.
LL. µ1 - µ2 0
Explanation / Answer
Our null hypothesis will be :
CC. The means (µ) of the response for the different levels of the first factor are all equal (i.e. µ1 = µ2 = µ3...).
The means (µ) of the response for the different levels of the second factor are all equal (i.e. µ1 = µ2 = µ3...).
There are no interactions between the first and second factor.
And our alternative hypothesis will be :
KK. The means (µ) of the response for the different levels of the first factor are not all equal (i.e. µ1 µ2 µ3...; at least one is different).
The means (µ) of the response for the different levels of the second factor are not all equal (i.e. µ1 µ2 µ3...; at least one is different).
There is an interaction between the first and second factor.
Such hypotheses are chosen since we want to carry out a two-way anova on this data to check the significance of the two factors individually and the significance of the interaction between the different levels of the two factors.