A conservative vector field 
has curl 
. In this case,
so the vector field is not conservative.
so the vector field is not conservative.
Formulate the indicated conclusion in nontechnical terms. Be sure to address the original claim. A cereal company claims that th
1 answer:
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Answer:
1. Null hypothesis:
Alternative hypothesis: Not all the means are equal
2. D. 36
3. C. 34
4. B. 1.059
5. B. 8.02
Step-by-step explanation:
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"
Part 1
The hypothesis for this case are:
Null hypothesis:
Alternative hypothesis: Not all the means are equal
Part 2
In order to find the mean square between treatments (MSTR), we need to find first the sum of squares and the degrees of freedom.
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
We need to find the mean for each group first and the grand mean.
If we apply the before formula we can find the mean for each group
,
,
. And the grand mean
Now we can find the sum of squares between:
Each group have a sample size of 4 so then
The degrees of freedom for the variation Between is given by , Where k the number of groups k=3.
Now we can find the mean square between treatments (MSTR) we just need to use this formula:
D. 36
Part 3
For the mean square within treatments value first we need to find the sum of squares within and the degrees of freedom.
And the degrees of freedom are given by:
. N represent the total number of individuals we have 3 groups each one with a size of 4 individuals. And k the number of groups k=3.
And now we can find the mean square within treatments:
C. 34
Part 4
The test statistic F is given by this formula:
B. 1.059
Part 5
The critical value is from a F distribution with degrees of freedom in the numerator of 2 and on the denominator of 9 such that we have 0.01 of the area in the distribution on the right.
And we can use excel to find this critical value with this function:
"=F.INV(1-0.01,2,9)"
And we will see that the critical value is
B. 8.02