Step by step
The question
Question: a sample of 15 has mean 9 and SD 1.5: find the 98% confidence interval for the true mean.
x̄, the sample mean, is 9: the middle of our interval.
s is the SD of the sample: 1.5.
And n is 15.
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.
15 minus 1 is 14.
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 14 and the 98 percent column.
Row 14, column 98%: t* = 2.624.
Careful!
df is n minus 1: 14, not 15.
3 Error bound
The error bound: t* times s over the square root of n.
1.5 over the square root of 15 is about 0.3873.
Times 2.624: about 1.0163.
4 The interval
Take the error bound away from the middle: 9 minus 1.0163 is 7.9837.
And add it: 9 plus 1.0163 is 10.0163.
The 98 percent confidence interval is (7.9837, 10.0163).
5 Say it in words
In words: we are 98 percent confident that the true mean is between 7.9837 and 10.0163.
6 Check
Let's check: the middle of the interval should be 9.
7.9837 plus 10.0163, divided by 2, is 9.
Half the width should be the error bound: 10.0163 minus 7.9837, divided by 2, is 1.0163.
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.