answers:
3.5
2.875
about 3.0625
Use thing below for answers
Answer:
a) (iii) ANOVA
b) The ANOVA test is more powerful than the t test when we want to compare group of means.
Step-by-step explanation:
Previous concepts
Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".
The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"
If we assume that we have
groups and on each group from
we have
individuals on each group we can define the following formulas of variation:



And we have this property

Solution to the problem
Part a
(i) confidence interval
False since the confidence interval work just when we have just on parameter of interest, but for this case we have more than 1.
(ii) t-test
Can be a possibility but is not the best method since every time that we conduct a t-test we have a chance that we commit a Type I error.
(iii) ANOVA
This one is the best method when we want to compare more than 1 group of means.
(iv) Chi square
False for this case we don't want to analyze independence or goodness of fit, so this one is not the correct test.
Part b
The ANOVA test is more powerful than the t test when we want to compare group of means.
A.
The second and first quadrant, because the hour and temperatures are unlikely to be negative numbers. All positive numbers for both X and Y values are in the second quadrant. If the hour is before noon, then the X value will venture into quadrant one. The diagram shows the quadrants in a graph.
B. For both the time and the temperature to be negative, and in quadrant four, you would have to be located in possibly Alaska or northern Canada. In such locations colder areas would be more likely to have negative temperatures and at times before 12.
Answer:
3:1
Step-by-step explanation:
Flour : butter milk
2: 2/3
6:2
3:1
Answer:
146.41
Step-by-step explanation:
third order determinant = determinant of 3×3 matrix A
given ∣A∣=11
det (cofactor matrix of A) =set (transpare of cofactor amtrix of A) (transpare does not change the det)
=det(adjacent of A)
{det (cofactor matrix of A)} ^2 = {det (adjacent of A)}
^2
(Using for an n×n det (cofactor matrix of A)=det (A)^n−1
)
we get
det (cofactor matrix of A)^2 = {det(A) ^3−1
}^2
=(11)^2×2 = 11^4
=146.41