search for: 8x9

...the null hypothesis. However, as I am comparing the ratio of means, I do not have individual observations to randomize between the groups. The best I can think to do is create an exhaustive list of all (7x10) = 70 possible observations for m2/m1 from the data. Then create a similar list of all (8x9) = 72 possible observations for m4/m3. Pool all (70+72) = 142 observations and repeatedly randomly assign them to two groups of size 70 and 72 to represent the two ratios and compute the difference in means. This distribution could represent the distribution under the null hypothesis and I could...

...e are writing an array of size `(s0, s1) = (3, 6)` starting from `(o0, o1) = (4, 6)`. Clearly, we will exceed the width of the array, since `(s1 + o1 = 6 + 6 = 12) > (n1 = 9)`. However, now think of the array as a flattened 1D representation. In this case, the total size of the array is `n1xn2 = 8x9 = 72`, while the largest element we will access is at the largest value of `(i, j)`. That is, `i = s0 - 1 = 2`, and `j = s1 - 1 = 5`. The largest index will be `ix(i=2, j=5, n0=8, n1=9, o0=4, o1=6) = 8*2+8*4+5+6=59`. Since `59 < 72`, we are clearly at _legal_ array indices, by C semantics! The...

...gt; `(s0, s1) = (3, 6)` starting from `(o0, o1) = (4, 6)`. Clearly, we will >> exceed the width of the array, since `(s1 + o1 = 6 + 6 = 12) > (n1 = 9)`. >> However, now think of the array as a flattened 1D representation. In this >> case, the total size of the array is `n1xn2 = 8x9 = 72`, while the largest >> element we will access is at the largest value of `(i, j)`. That is, >> `i = s0 - 1 = 2`, and `j = s1 - 1 = 5`. >> >> The largest index will be `ix(i=2, j=5, n0=8, n1=9, o0=4, o1=6) = 8*2+8*4+5+6=59`. >> Since `59 < 72`, we are clearly at...

...`(s0, s1) = (3, 6)` starting from `(o0, o1) = (4, 6)`. Clearly, we will > exceed the width of the array, since `(s1 + o1 = 6 + 6 = 12) > (n1 = 9)`. > However, now think of the array as a flattened 1D representation. In this > case, the total size of the array is `n1xn2 = 8x9 = 72`, while the largest > element we will access is at the largest value of `(i, j)`. That is, > `i = s0 - 1 = 2`, and `j = s1 - 1 = 5`. > > The largest index will be `ix(i=2, j=5, n0=8, n1=9, o0=4, o1=6) = 8*2+8*4+5+6=59`. > Since `59 < 72`, we are clearly at _le...

...) = (3, 6)` starting from `(o0, o1) = (4, 6)`. Clearly, we will >>> exceed the width of the array, since `(s1 + o1 = 6 + 6 = 12) > (n1 = 9)`. >>> However, now think of the array as a flattened 1D representation. In this >>> case, the total size of the array is `n1xn2 = 8x9 = 72`, while the largest >>> element we will access is at the largest value of `(i, j)`. That is, >>> `i = s0 - 1 = 2`, and `j = s1 - 1 = 5`. >>> >>> The largest index will be `ix(i=2, j=5, n0=8, n1=9, o0=4, o1=6) = 8*2+8*4+5+6=59`. >>> Since `59 < 72...

...ng from `(o0, o1) = (4, 6)`. Clearly, we will > > > exceed the width of the array, since `(s1 + o1 = 6 + 6 = 12) > (n1 = 9)`. > > > However, now think of the array as a flattened 1D representation. In this > > > case, the total size of the array is `n1xn2 = 8x9 = 72`, while the largest > > > element we will access is at the largest value of `(i, j)`. That is, > > > `i = s0 - 1 = 2`, and `j = s1 - 1 = 5`. > > > > > > The largest index will be `ix(i=2, j=5, n0=8, n1=9, o0=4, o1=6) = 8*2+8*4+5+6=59`. > >...