00:01
For this problem, in part a, the best estimate of the common standard deviation, which i'll be designating using the symbol sigma, is going to be found by taking the square root of the mean squares within, msw, which we aren't given explicitly on the table, but we can calculate that by taking the square root of the sum of squares within, the ssw, divided by the degrees of freedom within groups, dfw.
00:38
Looking at our given values on the table, we have that the sum of squares within groups is 4 .71, the degrees of freedom within groups is 30, so our estimate would be root 4 .71 over 30 for a result of 0 .396.
00:57
For part b, the value of the test statistic that is going to be f, we can find by taking the sum of squares between groups, ssb, divided by the degrees of freedom between groups, dfb, over the sum of squares within groups, divided by the degrees of freedom within groups, noting that this would be the same thing as calculating out msb divided by msw.
01:31
We're just doing it in one step as opposed to calculating out the msb and msw separately first.
01:37
So ssb, the sum of squares between groups is 1 .4, divide that by 3.
01:42
The sum of squares within groups is 4 .71 with 30 degrees of freedom, so 1 .4 over 3 divided by 4 .71 over 30 gives a result of 2 .972.
01:57
For part c, regarding the valid conclusions of the hypothesis test at the alpha equals 0 .025 level of significance, we need to note that we can see from our table that the p -value is equal to 0 .0474, which is greater than our level of significance.
02:32
So that means that we would fail to reject ftr, we fail to reject the null hypothesis...