Question

In a completely randomized experimental design involving three assembly methods, 30 employees were randomly selected and 10 were assigned to each of the three methods. The time required to complete the task was recorded. The following information is provided: MSTR= 45.89 and MSE = 6.27. What is the critical value of F if we want to determine whether or not the means of the three populations are equal at the 5% level of significance?a. 7.32
b. 1.96
c. 2.93
d. 3.35

Answer 1

Answer:

$F = \frac{MSR}{MSE} =\frac{45.89}{6.27}=7.32$

So then the best option is:

a. 7.32

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 $3$ groups and on each group from $j=1,\dots,10$ we have $10$ 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  

$SST=SS_{between}+SS_{within}$

Where SST represent the total sum of squares.  

The degrees of freedom for the numerator on this case is given by $df_{num}=k-1=3-1=2$ where k =3 represent the number of groups.  

The degrees of freedom for the denominator on this case is given by $df_{den}=df_{between}=N-K=30-3=27$.  

And the total degrees of freedom would be $df=N-1=30 -1 =29$  

From the info given we know that $MSR=\frac{SSR}{2}=45.89$

And $MSE=\frac{SSE}{27}=6.27$

From definition the F statisitc is defined as:

$F = \frac{MSR}{MSE} =\frac{45.89}{6.27}=7.32$

So then the best option is:

a. 7.32