Step by step
The question
Question: a sample of 12 has mean 68.5 and SD 3: find the 90% confidence interval for the true mean.
x̄, the sample mean, is 68.5: the middle of our interval.
s is the SD of the sample: 3.
And n is 12.
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.
12 minus 1 is 11.
2 Find t*
90 percent of the bell is in the middle.
That leaves 10 percent: 5 percent in each end.
t* marks the cut between the middle and the ends.
In the t-table we take row 11 and the 90 percent column.
Row 11, column 90%: t* = 1.796.
Careful!
df is n minus 1: 11, not 12.
3 Error bound
The error bound: t* times s over the square root of n.
3 over the square root of 12 is about 0.866.
Times 1.796: about 1.5554.
4 The interval
Take the error bound away from the middle: 68.5 minus 1.5554 is 66.9446.
And add it: 68.5 plus 1.5554 is 70.0554.
The 90 percent confidence interval is (66.9446, 70.0554).
5 Say it in words
In words: we are 90 percent confident that the true mean is between 66.9446 and 70.0554.
6 Check
Let's check: the middle of the interval should be 68.5.
66.9446 plus 70.0554, divided by 2, is 68.5.
Half the width should be the error bound: 70.0554 minus 66.9446, divided by 2, is 1.5554.
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.