00:01
Hello, students, we need to write here a random sample of 5 people, the mean driving distance to the work was 22 miles.
00:07
We need to assume random variable is normally distributed.
00:11
And here, if sample size is, here sample size is less than 20, so we need to apply key distribution.
00:18
And standard aviation is given.
00:19
We need to find margin of error and 99 % confidence interval.
00:24
Let's start with our margin of error part.
00:27
The margin of error must be t alpha by 2 comma df, df is degree of freedom and this is sample standard deviation and this is n, which is sample size.
00:38
So we all know degree of freedom, df is n minus 1 which is 5, minus 1 which is 4.
00:44
And if i calculate this alpha value would be, we all know alpha value would be 1 minus 0 .99 which is point.
00:54
We need to add here this is point 0 1 so alpha y 2 is point is 0 and this is 05 so hence t of we need to write here point is 0 05 comma we need to write here 4 multiplied by s which is 5 .8 divided by under root of 5 let's erase this to make it more clear now if i write here this will come out as so t value is 4 .604 multiplied by 5 .8 divided by under root of 5 this value is calculated as 26 .7032 which must be divided by under root of 5 so 20 .7032 divided by under root of 5 so 20 07032 divided by under root of 5 this value will come out as 9 .258 759 .25875 up to two decimal places.
02:00
If i, so not not this is 9 .2585, this is, this must be 11 .94, sorry...