00:01
So we want to determine at a 5 % significance level, while the mean waiting time is different for these two.
00:06
And i'm just going to have to subscript them as one and two.
00:12
So find out if there's a difference.
00:15
And first of all, i found it kind of odd that they had the waiting times out to the nearest hundredth of a minute.
00:21
But in any case, i thought that was kind of interesting.
00:24
So our first sample was a sample size of 10.
00:28
The second one was of 12.
00:32
And the first sample, the mean waiting time for that first sample, was 27 .461.
00:40
And the second sample size had a mean waiting time of 25 .689 minutes.
00:46
And the two sample standard deviations, the first one was 4 .4 .0.
00:54
And the second one, in nuts and minutes, 2 .68.
00:57
And it will go 4 .5.
00:59
I don't want to do that rounding up to five.
01:02
And these are in minutes.
01:03
Now, this technique, we will be assuming that these two standard deviations are equal, and that our populations are approximately normal, and we will have to do a two sample t test.
01:15
We can't use z values.
01:18
Our sample sizes are not big enough, and we don't know that the populations are normal.
01:22
So we will do a two sample t test, and that's going to call for us to end up finding our pooled standard deviation, or our pooled variance, actually.
01:32
And we know that our degrees of freedom will end up being the 10 plus 12 less 2.
01:39
So it's going to be 20.
01:40
And why don't we just draw out before we calculate that pool variance where we will be doing our rejecting the null.
01:49
And since we're using a 5 % significance level, we'll have half of that at the upper tail, half of that at the lower tail, and we want a t value that has 20 degrees of free...