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lozanna [386]
3 years ago
13

The distribution of the test statistic for analysis of variance is the F-distribution. true or false

Mathematics
1 answer:
lubasha [3.4K]3 years ago
8 0

Answer:

True. See explanation below

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 k groups and on each group from j=1,\dots,k we have k individuals on each group we can define the following formulas of variation:  

SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2  

SS_{between=Treatment}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2  

SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2  

And we have this property  

SST=SS_{between}+SS_{within}  

The degrees of freedom for the numerator on this case is given by df_{num}=k-1 where k represent the number of groups.  

The degrees of freedom for the denominator on this case is given by df_{den}=df_{between}=N-k.  

And the total degrees of freedom would be df=N-1

And the we can find the F statistic F=\frac{MSR}{MSE}

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How many 1/5 inch pieces can be cut from a sting 8 inches long
Snezhnost [94]

40 pieces

8/1 : 1/5   = 8/1*5/1   = 40/1  = 40

Another method :

1/5 = 0.2

8/0.2 = 40

5 0
2 years ago
For which values of k does the system of linear equations have zero, one, or an infinite number of solutions? [Note: not all thr
viktelen [127]

Answer:

The system has an infinite solution at k = 6, otherwise for any value of k, it has zero solution.

Step-by-step explanation:

Consider the system of linear equations:

3x_{1} -x_{2}=2\;\;\;\;\;\;(1)\\ 9x_{1} -3x_{2}=k\;\;\;\;\;(2)

The system of linear equations can have zero, one, or an infinite number of solutions:

simplify equation (1):

x_{2}=3x_{1} -2

substitute in equation (2), we get

9x_{1} -3(3x_{1}-2)=k\\9x_{1} -9x_{1}+6=k\\0+6=k

we cannot find the value of x_{1} and x_{2}.

so, there is no solution.

Multiply the equation (1) with 3 and put k is 6,

3(3x_{1} -x_{2})=3\times2\\9x_{1} -3x_{2})=6

it means both equations are overlapped. Then, the solution has infinite solutions.

Hence, the system has an infinite solutions at k is 6 otherwise for any value of k it has no solution.

4 0
3 years ago
Use the number line given to choose the best answer for the question. What value does point P represent on the number line shown
Liono4ka [1.6K]

Its -3



Mark me brainliest now :3

5 0
3 years ago
If you rotate a triangle 180 degrees with points of (1,2), (3,3), (1,4) what would the new points be?
alexgriva [62]

Answer: when you rotate the triangle 180 degrees the new points would be (-1,-2), (-1,-4), and (-3,-3).

Step-by-step explanation:

You simply flip the graph 180 degrees. Not that hard.

8 0
3 years ago
URGENT please help!!! I need a explanation please!! You own a restaurant and can reopen as long as people seated at a table are
hram777 [196]

Answer:

1) Maximum number of tables that can be kept in the dinning room is 10 tables

2)The maximum number of customers, given 6 people per table is 60 customers

Step-by-step explanation:

The parameters given are;

Distance between people seated at different tables = 6 ft

Dimension of table = 2 m by 1.5 m

Number of people at a table = 6 people

Dimension of dinning;

Width = 12 m

Length = 20 m

Dimension of entrance;

Width = 20/21*30  m

Length = 10 m

With 6 meters between tables and a table with of 1.5, we have for the arrangement in the question;

3 tables = 4.5 m

Distance between = 2 × 6 = 12

4.5 + 12 = 16 m (More room required)

With two tables, we have;

Width of tables = 2×1.5 = 3 m

Distance between = 6 m

Dining room width required = 3 + 6 = 9 m

Therefore, the maximum tables in each row = 2

Given that the dining room area extends to the entrance area, total length = 30 m.

With 4 tables we have;

Length = 4 × 2 + 6 × 3 = 26

While on the entrance side of the dinning room area which is 20 m, we have 3 tables;

Length = 3 × 2 + 6 × 2 = 18

Therefore, both arrangements are acceptable;

Just on entering by the right the length = 20/21×30 m

Therefore, with 3 tables will required number due to proximity wit the door and tables arranged along the right wall of the dinning

1) Maximum number of tables that can be kept in the dinning room = 4 + 3 + 3 = 10 tables

2)The maximum number of customers, given 6 people per table = 6×10 = 60 customers.

4 0
3 years ago
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