00:01
So in this question, first we need to compute the confidence interval for the mean.
00:07
So here we are going to use the t interval procedure to compute this confidence interval.
00:15
So the idea here is that we should assume like use x bar plus and minus a t value times the same distribution deviation divided by the square root of n.
00:28
So the sample mean here is 23 .87.
00:33
This value here is coming from the t student distribution with degrees of freedom equals to the sample size minus 1.
00:42
So since this case, the simple size is 30, the degrees of freedom here is equals to 29.
00:49
Now, because we are computing a 95 % confidence interval, this means that this t value is the one that has an area right left to it equals to 0 .975 so with these two informations we're going to get that this value here is equal to 2 .0452 so if you plug this information here you're going to get that if you plug the value of the standard deviation and also the sample size here we are going to get the this value here is equal to 3 .42 .79.
01:36
Then we have the sample mean.
01:39
This means that the lower bound for this interval here is equals to 20 .4421...