00:01
So this question for hotels a, b and c, we're given the sample average, sample standard deviation and also n, but n is 50 for each of these.
00:15
Their sample averages are 1 .45, 1 .80 and 1 .10 and their standard deviations are 17 .9, 22 .1 and 12 .1.
00:31
So first of all we're looking at the hotels a and b.
00:39
So looking at a and b and we want to look at the mean of a minus the mean of b.
00:50
So first of all the standard error is going to be given by the square root of sa squared over na plus sb squared over nb.
01:02
So that is the square root of 17 .9 squared over 50 plus 22 .1 squared over 50 which is 4 .0220.
01:19
Now we want a 95 % confidence interval.
01:22
So what we're going to do is we're going to note that our number of degrees of freedom in the pool is approximately na plus nb minus 2.
01:34
So 50 plus 50 minus 2 is 98 and we want to find t star such that the probability t is greater than t star is 98 degrees of freedom is 1 minus 0 .95 over 2 which is 0 .025.
01:50
So we want to look up the quantile of the t distribution that encloses 2 .5 probability to its right and we have 98 degrees of freedom and that gives us t star is 1 .9845.
02:10
So now we can find the error term which is just the standard error times t star.
02:20
So we'll do 1 .9845 times 4 .0220 and we get 7 .98 to two decimal places.
02:32
Then the point estimate x a bar minus x b bar is 145 minus 180 which is minus 35.
02:44
So then we can write that the difference in means is between the point estimate which is the difference in sample means plus the error 7 .98 to give us minus 27 .02 and on the lower end we subtract the error to get minus 42 .98.
03:03
So there's our 95 % interval...