00:01
So for this problem, the first thing that i'm going to note is that i'll be reporting everything in units of thousands.
00:13
So if i write, for instance, 50, that actually means 50 ,000.
00:20
Just because we have everything is in multiples of 1 ,000, we might as well.
00:27
Or i'll specify that that is for the data.
00:30
The data is in units of 1 ,000.
00:34
Now we know that our level of significance is going to be 0 .05 since we are testing at a 95 % confidence level.
00:41
Our null hypothesis is going to be that there is no difference in the mean values.
00:47
The alternate hypothesis is that there is a difference in the mean values.
00:54
Mu1 minus mu2 does not equal 0.
00:57
Since we don't know the population standard deviations for either one of the values, we'll be doing this as a two -tailed t -test, independent samples.
01:15
So the first thing that we're going to have to do is compute the sample mean and sample standard deviation for the two different data sets.
01:22
So we find the sample mean by adding up all of the individual measurements, then we divide by the number of values.
01:28
Now give me a second here.
01:29
All right, so we have all of our male employees results.
01:35
Taking the sum, we have 599.
01:37
Then we divide that by the number of male employees.
01:39
So that's 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
01:43
So 599 over 10 gives a mean value of 59 .9.
01:49
And the sample standard deviation we find as follows.
01:52
We take the square root of the sum of xi minus x bar squared divided by n minus 1.
02:01
So plugging in the values, i'll begin by taking each individual measurement minus 59 .9...