00:01
For this problem, we are told that the following sample of six measurements were randomly selected from a normally distributed population.
00:08
The first thing that i'll note is that we'll find that the sample standard deviation is going to be equal to...
00:14
One moment here.
00:15
All right, just wanted to double check a value i found.
00:19
All right, so this is going to be s equals 3 .098 for the sample standard deviation.
00:26
And we'll find that the sample mean, x bar, is actually going to be equal to 3.
00:31
In part a, we are asked to test the null hypothesis that the mean of the population is 2 against the alternative hypothesis, mu, less than 2, where we are to use alpha equals 0 .05.
00:45
So our first step here, since we have a very small sample size, will be to calculate the t value, which will be equal to x bar minus mu, divided by s over the square root of n, which will then become clearly 3.
01:01
098, or excuse me, it would be 3 minus 2 rather, 3 minus 2 over 3 .098, divided by the square root of 6, and we'll get that t is going to be equal to 0 .79.
01:21
I'll note that we'll use the same t value for both of these tests.
01:28
Additionally, we'll want to, using this level of significance, and noting that we have a one -tailed test, term the critical t value, in which case we'd have that with five degrees of freedom, the critical t value is going to be negative 2 .015.
01:43
We can see that our calculated t value, let's see here, t equals 0 .79, is greater than our critical t value, but we have to keep in mind this is a one -tailed lesser or to the left sort of test, which means that this, that we are not, not in the rejection region so we fail to reject fail to reject in part b we are asked to to test the null hypothesis that the mean of the population is two against the alternative hypothesis that mu does not equal to where we are to use the same alpha level so for referencing the appropriate tables we'd want to be using the alpha over 2 value of 0 .025 and we'll find that the critical t value here would be equal to let's see 2 .5706.
02:39
2 .5706.
02:41
So our region of rejection is going to be where the absolute value of t is greater than 2 .5706, but we can see that 0 .79 is less than 2 .5706.
02:56
So we fail.
03:00
One moment here...