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As to covariances, the variance of a difference is $\sigma^2_{x-y}=\sigma^2_x+\sigma^2_y-2\sigma_{xy}$, so covariances are somewhat involved, but they don't need to be equal.
Sphericity requires that $$\sigma^2_{i-j}=\sigma^2_i+\sigma^2_j-2\sigma_{ij}=k$$ (the same $k$) for all $i,j$. This doesn't require equal covariances. See this page for an example. This question and my answer could be useful too.

EDIT: In your example there are $N=5$ observation and $J=3$ groups, so checking for sphericity is simple, you just have to compute 15 differences (and 3 variances). If the number of observations were larger, say $N=100$, you should compute 300 differences. Using variances and covariances you just need $J=3$ variances and $J(J-1)/2=3$ covariances. In R:

> A <- c(10,15,25,35,30)
> B <- c(12,15,30,30,27)
> C <- c(8,12,20,28,20)
> # variances
> var(A)
[1] 107.5
> var(B)
[1] 74.7
> var(C)
[1] 60.8
> # covariances
> cov(A,B)
[1] 83.25
> cov(A,C)
[1] 79
> cov(B,C)
[1] 62.4

So the variance-covariance matrix is: $$\begin{bmatrix} 107.5 & 83.25 & 79 \\ 83.25 & 74.7 & 62.4 \\ 79 & 62.4 & 60.8 \end{bmatrix}$$ Now you can check for sphericity: $$\begin{split} Var(A-B) &= 107.5+74.7-2\cdot83.25=15.7 \\ Var(A-C) &= 107.5+60.8-2\cdot 79=10.3 \\ Var(B-C) &= 74.7+60.8-2\cdot 62.4=10.7 \end{split}$$

In the first method you don't need covariances. You need them in the second method. The above algebra shows that:

• if the variances are equal, then the covariances must be equal too (this is compound symmetry, which imply sphericity);
• if the variances are not equal, then the covariances must vary so that the above sums are equal.

As to covariances, the variance of a difference is $\sigma^2_{x-y}=\sigma^2_x+\sigma^2_y-2\sigma_{xy}$, so covariances are somewhat involved, but they don't need to be equal.
Sphericity requires that $$\sigma^2_{i-j}=\sigma^2_i+\sigma^2_j-2\sigma_{ij}=k$$ (the same $k$) for all $i,j$. This doesn't require equal covariances. See this page for an example. This question and my answer could be useful too.

EDIT: In your example there are $N=5$ observation and $J=3$ groups, so checking for sphericity is simple, you just have to compute 15 differences (and 3 variances). If the number of observations were larger, say $N=100$, you should compute 300 differences. Using variances and covariances you just need $J=3$ variances and $J(J-1)/2=3$ covariances. In R:

> A <- c(10,15,25,35,30)
> B <- c(12,15,30,30,27)
> C <- c(8,12,20,28,20)
> # variances
> var(A)
[1] 107.5
> var(B)
[1] 74.7
> var(C)
[1] 60.8
> # covariances
> cov(A,B)
[1] 83.25
> cov(A,C)
[1] 79
> cov(B,C)
[1] 62.4

So the variance-covariance matrix is: $$\begin{bmatrix} 107.5 & 83.25 & 79 \\ 83.25 & 74.7 & 62.4 \\ 79 & 62.4 & 60.8 \end{bmatrix}$$ Now you can check for sphericity: $$\begin{split} Var(A-B) &= 107.5+74.7-2\cdot83.25=15.7 \\ Var(A-C) &= 107.5+60.8-2\cdot 79=10.3 \\ Var(B-C) &= 74.7+60.8-2\cdot 62.4=10.7 \end{split}$$

In the first method you don't need covariances. You need them in the second method. The above algebra shows that:

• if the variances are equal, then the covariances must be equal too (this is compound symmetry, which imply sphericity);
• if the variances are not equal, then the covariances must vary so that the above sums are equal.

As to covariances, the variance of a difference is $\sigma^2_{x-y}=\sigma^2_x+\sigma^2_y-2\sigma_{xy}$, so covariances are somewhat involved, but they don't need to be equal.
Sphericity requires that $$\sigma^2_{i-j}=\sigma^2_i+\sigma^2_j-2\sigma_{ij}=k$$ (the same $k$) for all $i,j$. This doesn't require equal covariances. See this page for an example. This question and my answer could be useful too.

EDIT: In your example there are $N=5$ observation and $J=3$ groups, so checking for sphericity is simple, you just have to compute 15 differences (and 3 variances). If the number of observations were larger, say $N=100$, you should compute 300 differences. Using variances and covariances you just need $J=3$ variances and $J(J-1)/2=3$ covariances. In R:

> A <- c(10,15,25,35,30)
> B <- c(12,15,30,30,27)
> C <- c(8,12,20,28,20)
> # variances
> var(A)
[1] 107.5
> var(B)
[1] 74.7
> var(C)
[1] 60.8
> # covariances
> cov(A,B)
[1] 83.25
> cov(A,C)
[1] 79
> cov(B,C)
[1] 62.4

So the variance-covariance matrix is: $$\begin{bmatrix} 107.5 & 83.25 & 79 \\ 83.25 & 74.7 & 62.4 \\ 79 & 62.4 & 60.8 \end{bmatrix}$$ Now you can check for sphericity: $$\begin{split} Var(A-B) &= 107.5+74.7-2\cdot83.25=15.7 \\ Var(A-C) &= 107.5+60.8-2\cdot 79=10.3 \\ Var(B-C) &= 74.7+60.8-2\cdot 62.4=10.7 \end{split}$$

In the first method you don't need covariances. You need them in the second method.

As to covariances, the variance of a difference is $\sigma^2_{x-y}=\sigma^2_x+\sigma^2_y-2\sigma_{xy}$, so covariances are somewhat involved, but they don't need to be equal.
Sphericity requires that $$\sigma^2_{i-j}=\sigma^2_i+\sigma^2_j-2\sigma_{ij}=k$$ (the same $k$) for all $i,j$. This doesn't require equal covariances. See this page for an example. This question and my answer could be useful too.

EDIT: In your example there are $N=5$ observation and $J=3$ groups, so checking for sphericity is simple, you just have to compute 15 differences (and 3 variances). If the number of observations were larger, say $N=100$, you should compute 300 differences. Using variances and covariances you just need $J=3$ variances and $J(J-1)/2=3$ covariances. In R:

> A <- c(10,15,25,35,30)
> B <- c(12,15,30,30,27)
> C <- c(8,12,20,28,20)
> # variances
> var(A)
[1] 107.5
> var(B)
[1] 74.7
> var(C)
[1] 60.8
> # covariances
> cov(A,B)
[1] 83.25
> cov(A,C)
[1] 79
> cov(B,C)
[1] 62.4

So the variance-covariance matrix is: $$\begin{bmatrix} 107.5 & 83.25 & 79 \\ 83.25 & 74.7 & 62.4 \\ 79 & 62.4 & 60.8 \end{bmatrix}$$ Now you can check for sphericity: $$\begin{split} Var(A-B) &= 107.5+74.7-2\cdot83.25=15.7 \\ Var(A-C) &= 107.5+60.8-2\cdot 79=10.3 \\ Var(B-C) &= 74.7+60.8-2\cdot 62.4=10.7 \end{split}$$

In the first method you don't need covariances. You need them in the second method. The above algebra shows that:

• if the variances are equal, then the covariances must be equal too (this is compound symmetry, which imply sphericity);
• if the variances are not equal, then the covariances must vary so that the above sums are equal.

As to covariances, the variance of a difference is $\sigma^2_{x-y}=\sigma^2_x+\sigma^2_y-2\sigma_{xy}$, so covariances are somewhat involved, but they don't need to be equal.
Sphericity requires that $$\sigma^2_{i-j}=\sigma^2_i+\sigma^2_j-2\sigma_{ij}=k$$ (the same $k$) for all $i,j$. This doesn't require equal covariances. See this page for an example. This question and my answer could be useful too.

As to covariances, the variance of a difference is $\sigma^2_{x-y}=\sigma^2_x+\sigma^2_y-2\sigma_{xy}$, so covariances are somewhat involved, but they don't need to be equal.
Sphericity requires that $$\sigma^2_{i-j}=\sigma^2_i+\sigma^2_j-2\sigma_{ij}=k$$ (the same $k$) for all $i,j$. This doesn't require equal covariances. See this page for an example. This question and my answer could be useful too.

EDIT: In your example there are $N=5$ observation and $J=3$ groups, so checking for sphericity is simple, you just have to compute 15 differences (and 3 variances). If the number of observations were larger, say $N=100$, you should compute 300 differences. Using variances and covariances you just need $J=3$ variances and $J(J-1)/2=3$ covariances. In R:

> A <- c(10,15,25,35,30)
> B <- c(12,15,30,30,27)
> C <- c(8,12,20,28,20)
> # variances
> var(A)
[1] 107.5
> var(B)
[1] 74.7
> var(C)
[1] 60.8
> # covariances
> cov(A,B)
[1] 83.25
> cov(A,C)
[1] 79
> cov(B,C)
[1] 62.4

So the variance-covariance matrix is: $$\begin{bmatrix} 107.5 & 83.25 & 79 \\ 83.25 & 74.7 & 62.4 \\ 79 & 62.4 & 60.8 \end{bmatrix}$$ Now you can check for sphericity: $$\begin{split} Var(A-B) &= 107.5+74.7-2\cdot83.25=15.7 \\ Var(A-C) &= 107.5+60.8-2\cdot 79=10.3 \\ Var(B-C) &= 74.7+60.8-2\cdot 62.4=10.7 \end{split}$$

In the first method you don't need covariances. You need them in the second method.