00:01
For this problem, we are told that the following random sample was selected from a normal distribution.
00:05
We have 4, 6, 3, 5, 9, and 3.
00:07
For part a, we are asked to construct a 90 % confidence interval for the population mean, mu.
00:13
So, first of all, we can calculate that the sample mean, x bar, is going to be equal to 5, calculating the mean as usual.
00:21
And again, using the standard formulas, we'll find that the sample standard deviation is going to be equal to 2 .28.
00:27
To construct our 90 % confidence interval then, we'll have that the confidence interval will be equal to x bar plus or minus z of 0 .05 times s, all divided by the square root of n, where we can find from a table that z of 0 .05 will be equal to, one moment here, be equal to 2 .02, which then means, oh, excuse me, we don't.
01:02
Want z, we want to be using t, t here.
01:07
So the appropriate t value will be equal to 2 .02, noting that we are using five degrees of freedom here.
01:19
So we'll find then that the confidence interval will be equal to 5 plus or minus 1 .88, or the interval from 3 .12 up to 6 .88.
01:33
Then for part b, we are asked to repeat the same process but for a 95 % confidence interval...