00:01
Let me take notes for this question given values.
00:03
So the sample size denoted by n, this is 12 year, and the sample mean given here which is 79 .3, and the sample standard deviation which is given as 7 .8, and the confidence level which is given as 95 % as a decimal number, that would be 0 .95.
00:19
So we're going to find the confidence interval for the number of words per minute, that means the population mean.
00:25
So we have to find the confidence interval for the population mean which is denoted by mu here.
00:30
Let me write the formula first.
00:31
This is the sample mean, plus or minus.
00:33
Because we know the sample standard deviation, so we have to use the t distribution.
00:38
This is t alpha over 2 and sample standard deviation divided by root n.
00:42
Let's get this t value first.
00:44
So the alpha is 1 minus confidence level, but we need alpha over 2.
00:48
This is 1 minus 0 .95 and divide by 2 which would be 0 .025...