Step by step
The question
Question: a sample of 20 has mean 15 and SD 2.5: find the 99% confidence interval for the true mean.
x̄, the sample mean, is 15: the middle of our interval.
s is the SD of the sample: 2.5.
And n is 20.
1 Why t
We do not know σ, the SD of all the numbers.
We only have s, the SD of the sample.
So we use t instead of z.
The t curve is a bit wider than the bell.
It makes the interval a bit wider, to be safe.
The t-table has a row for each df, the degrees of freedom: n minus 1.
20 minus 1 is 19.
2 Find t*
99 percent of the bell is in the middle.
That leaves 1 percent: 0.5 percent in each end.
t* marks the cut between the middle and the ends.
In the t-table we take row 19 and the 99 percent column.
Row 19, column 99%: t* = 2.861.
Careful!
df is n minus 1: 19, not 20.
3 Error bound
The error bound: t* times s over the square root of n.
2.5 over the square root of 20 is about 0.559.
Times 2.861: about 1.5993.
4 The interval
Take the error bound away from the middle: 15 minus 1.5993 is 13.4007.
And add it: 15 plus 1.5993 is 16.5993.
The 99 percent confidence interval is (13.4007, 16.5993).
5 Say it in words
In words: we are 99 percent confident that the true mean is between 13.4007 and 16.5993.
6 Check
Let's check: the middle of the interval should be 15.
13.4007 plus 16.5993, divided by 2, is 15.
Half the width should be the error bound: 16.5993 minus 13.4007, divided by 2, is 1.5993.
So it is right!
Let's look back.
First: why t.
Then: find t*.
Then: error bound.
Then: the interval.
Then: say it in words.
Last: check.