Step by step
The question
Question: a sample of 25 has mean 75 and SD 10: find the 99% confidence interval for the true mean.
x̄, the sample mean, is 75: the middle of our interval.
s is the SD of the sample: 10.
And n is 25.
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.
25 minus 1 is 24.
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 24 and the 99 percent column.
Row 24, column 99%: t* = 2.797.
Careful!
df is n minus 1: 24, not 25.
3 Error bound
The error bound: t* times s over the square root of n.
The square root of 25 is 5.
10/5 is 2.
Times 2.797: 5.594.
4 The interval
Take the error bound away from the middle: 75 minus 5.594 is 69.406.
And add it: 75 plus 5.594 is 80.594.
The 99 percent confidence interval is (69.406, 80.594).
5 Say it in words
In words: we are 99 percent confident that the true mean is between 69.406 and 80.594.
6 Check
Let's check: the middle of the interval should be 75.
69.406 plus 80.594, divided by 2, is 75.
Half the width should be the error bound: 80.594 minus 69.406, divided by 2, is 5.594.
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.