00:01
All right, so for this problem, the first thing that we're going to need to do is find the averages for our two samples.
00:08
So i'll call them x1 bar and x2 bar.
00:11
So we can find the averages simply by adding up all of the individual values.
00:16
So we do 65 plus 82 plus 81 and so on, dividing by the number of measurements, which would be 1, 2, 3, 4, 5.
00:23
So we'd be dividing by 10.
00:25
Or if you're using something like a graphing calculator or excel as i am here, you probably have a command for finding averages.
00:33
I'll note that i am going to be doing most of the work here by hand.
00:37
I'll just be using excel to automate things when it's going to save time, basically.
00:43
So the next thing that we're going to need to do is find the standard deviations for our two types.
00:50
So i'll call them sd1 and sd2.
00:52
To do that, we'll find the squared differences between each measurement for the two types and their mean values.
01:03
So, for instance, for type 1, we'd first have 65 minus 70 .4 to the power of 2, giving us 29 .16.
01:11
And for type 2, we'd have 64 minus 70 .2 squared, and has 38 .44, and so on.
01:22
Now our standard deviation is going to be the square root of one over our sample size minus 1, so 1 over 9, times the sum of those squared deviations from the mean.
01:36
We get 9 .264 roughly for the for type 1, and we get square root 1 over 9 .9 .66 roughly for the second.
01:54
Now, for testing the hypothesis that the two variances are equal, our test statistic is going to be an f score, which is equal to sample variance 1 squared, divided by sample variance 2 squared, and it's going to have, or it's going to be following an f distribution with n1 minus 1, so 9 numerator degrees of freedom, and 9 denominator degrees of freedom.
02:23
So our test statistic or our f score here, well, that's just going to be equal to 9 .264 squared over 9 .3666 squared.
02:34
So we get f is 0 .978, and then we can find the p value.
02:40
For instance, you can do this using a table of values.
02:43
I'm just going to be using my using excel here.
02:47
So i'll put in f .d -d -d -d -o - now let's see here.
02:54
Since we're testing simply for equality of variances, then this is going to effectively be a two -tailed test.
03:02
We do f .di or you want to do, actually, to find the p value here, we are going to, actually, pardon me, i keep on changing my mind about the approach that i want to take.
03:20
I'll instead find a rejection region.
03:25
So we would want to reject if we are less than.
03:28
F .inv 0 .02599 or more than equals f .inv .r .r .t .02599.
03:44
All right.
03:45
So we, based on the fact that our f value is within or is not less than the lower bound or more than the greater bound or the upper bound rather, we would fail to reject the null hypothesis of equality.
04:01
Therefore, we can conclude that the data does support the claim that sigma 1 squared is equal to sigma 2 squared we certainly fail to reject that hypothesis then moving on to part b we have that you use a t test or a test statistic is going to be equal to x bar 1 minus x bar 2 divided by our pooled standard deviation times the square root of 1 over n1 plus 1 over n2 where the pooled standard deviation is equal to the square root of n1 minus 1.
04:43
I'll write this out explicitly first, so it's square root of n1 minus 1 times s1 squared, plus n2 minus 1 times s2 squared, divided by n1 plus n2 minus 2.
04:58
So plugging in our values here, we have square root of 9 times 9 .264 squared plus 9.
05:08
Times 9 .366 squared divided by 10 plus 10 minus 2.
05:17
So we get that our pooled standard deviation is equal to 9 .315 roughly.
05:25
So now we have everything that we need to plug in and evaluate our t statistic.
05:30
So that's going to be equal to 70 .4 minus 70 .2 divided by 9 .315 times the of since n1 and n2 are equal to each other would be just square root of 2 over 10.
05:45
So we get that our test statistic here, t, is equal to 0 .048.
05:53
And our alternate hypothesis here, i'll note this explicitly.
05:59
Since we're testing simply for equality, we'd have that our alternate hypothesis is just that x bar 1 minus x bar 2 does not equal 0.
06:07
So this would be a two -tail test.
06:09
I'll find the critical t value for this level of significance by doing equals 2 .rt.
06:17
.inv .2t.
06:20
You want 0 .05 outside, and this is going to be with n1 plus n2 minus 2, so 18 degrees of freedom.
06:28
So the critical t value is roughly 2 .1.
06:32
The magnitude of our t score is less than 2 .1...