00:01
So in this question, we've been given observation 7, 12, 5, 34, 14, 8, 0, 9, and 36.
00:13
So this implies that a sample size n is 9.
00:18
So for our part a, we are asked to find for the mean.
00:22
And you know the mean is equal to the summation xi for n.
00:25
So you are going to sum everything over here.
00:27
This is equal to 7 plus 12 plus 5 plus 4 24 plus 14 plus 8 plus 0 plus 9 plus 36 divided by 9 so this gives us 135 for 9 and this is 15 for part b you have to find the confidence interval so you know the confidence interval it gives by x bar plus or minus t alpha on 2 comma n minus 1 times s over the root of n why this is the margin of error so let's find for t alpha on 2 comma n minus 1 so alpha is 0 .01 so divided by 2 so you have t 0 .01 over 2 you have 9 minus 1 so when we read it from our t t 3 .35 so now we can find for our confidence interval.
01:43
So we have 15 plus or minus 3 .355 times our s is 14 .8577 divided by the rate of 9.
02:04
So how do we even find for our standard deviation, which is s over here? so the formula is the square root of the summation xi minus.
02:14
X bar squared over n minus 1 so when we substitute everything over here you are going to get a 14 .85 there so this implies getting for the lower limit you're going to get negative 1 .6159 so lower limit so the upper limit will give us 31 .6159 so this is for the upper limit so our confidence interval is negative 1 .6159 and comma 31 .6 .6 6159...