Step by step
The question
Question: a sample of 16 has mean 22 and SD 4: find the 99% confidence interval for the true mean.
x̄, the sample mean, is 22: the middle of our interval.
s is the SD of the sample: 4.
And n is 16.
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.
16 minus 1 is 15.
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 15 and the 99 percent column.
Row 15, column 99%: t* = 2.947.
Careful!
df is n minus 1: 15, not 16.
3 Error bound
The error bound: t* times s over the square root of n.
The square root of 16 is 4.
4/4 is 1.
Times 2.947: 2.947.
4 The interval
Take the error bound away from the middle: 22 minus 2.947 is 19.053.
And add it: 22 plus 2.947 is 24.947.
The 99 percent confidence interval is (19.053, 24.947).
5 Say it in words
In words: we are 99 percent confident that the true mean is between 19.053 and 24.947.
6 Check
Let's check: the middle of the interval should be 22.
19.053 plus 24.947, divided by 2, is 22.
Half the width should be the error bound: 24.947 minus 19.053, divided by 2, is 2.947.
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.