00:01
For this problem, we are told that a sample of five measurements randomly selected from a normally distributed population resulted in the following summary statistics, with x bar equals 4 .8, s equals 1 .3.
00:13
And i'll note that we have our test value or our null hypothesis value for mu would be that mu equals 6 for both parts a and b.
00:22
So we can calculate our t value, which is going to be equal to x bar 4 .8, minus mu, which is 6, divided by s, 1 .3, over the square root of n.
00:36
N is 5, so it's over the square root of 5.
00:39
So we get that t is going to be equal to negative 2 .064.
00:45
For part a, we are asked to test the null hypothesis that the mean value, or mean of the population is 6 against the alternative hypothesis, that mu is less than 6, and we are to use alpha equals 0 .05.
00:59
We'll find that the, or just by using a table, something along those lines, the critical value of t is going to be equal to 2 .7764.
01:08
If we were doing a right -tailed test, but that means that our region of rejection for our left -tailed test is going to be if t is less than negative 2 .7764, and we can see that that is not the case.
01:23
Therefore we fail to reject the null hypothesis...