A light bulb consumes 1440 watt-hours per day. How long does it take to consume 6120 watt-hours?

Mathematics

1 answer:

Alex Ar [27]3 years ago

7 0

Step-by-step explanation:

Divide 24 by 4. So, 18 hours. Therefore a light bulb consumes 6840 Watt hours in 4 days and 18 hours.

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In a one-way ANOVA, what is the difference between the among-groups variance MSA and the within-groups variance MSW?

LuckyWell [14K]

Answer:

$MSR=MSA=\frac{SSR}{k-1}$

And $MSE=MSW=\frac{SSE}{N-k}$

The difference is that MSA takes incount the variation between the groups and the grand mean, and the MSW takes in count the variation within groups respect to the mean of each group .

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"

If we assume that we have $j$ groups and on each group from $j=1,\dots,j$ we have $n_j$ 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

$SST=SS_{between}+SS_{within}=6750+8000=14750$

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 $df=N-1$

We can find the $MSR=MSA=\frac{SSR}{k-1}$

And $MSE=MSW=\frac{SSE}{N-k}$

The difference is that MSA takes incount the variation between the groups and the grand mean, and the MSW takes in count the variation within groups respect to the mean of each group .

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