Step by step
The question
Question: a sample of 9 has mean 100 and SD 12: find the 98% confidence interval for the true mean.
x̄, the sample mean, is 100: the middle of our interval.
s is the SD of the sample: 12.
And n is 9.
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.
9 minus 1 is 8.
2 Find t*
98 percent of the bell is in the middle.
That leaves 2 percent: 1 percent in each end.
t* marks the cut between the middle and the ends.
In the t-table we take row 8 and the 98 percent column.
Row 8, column 98%: t* = 2.896.
Careful!
df is n minus 1: 8, not 9.
3 Error bound
The error bound: t* times s over the square root of n.
The square root of 9 is 3.
12/3 is 4.
Times 2.896: 11.584.
4 The interval
Take the error bound away from the middle: 100 minus 11.584 is 88.416.
And add it: 100 plus 11.584 is 111.584.
The 98 percent confidence interval is (88.416, 111.584).
5 Say it in words
In words: we are 98 percent confident that the true mean is between 88.416 and 111.584.
6 Check
Let's check: the middle of the interval should be 100.
88.416 plus 111.584, divided by 2, is 100.
Half the width should be the error bound: 111.584 minus 88.416, divided by 2, is 11.584.
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.