Step by step
The question
Question: a sample of 30 has mean 31.6 and SD 5: find the 95% confidence interval for the true mean.
x̄, the sample mean, is 31.6: the middle of our interval.
s is the SD of the sample: 5.
And n is 30.
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.
30 minus 1 is 29.
2 Find t*
95 percent of the bell is in the middle.
That leaves 5 percent: 2.5 percent in each end.
t* marks the cut between the middle and the ends.
In the t-table we take row 29 and the 95 percent column.
Row 29, column 95%: t* = 2.045.
Careful!
df is n minus 1: 29, not 30.
3 Error bound
The error bound: t* times s over the square root of n.
5 over the square root of 30 is about 0.9129.
Times 2.045: about 1.8668.
4 The interval
Take the error bound away from the middle: 31.6 minus 1.8668 is 29.7332.
And add it: 31.6 plus 1.8668 is 33.4668.
The 95 percent confidence interval is (29.7332, 33.4668).
5 Say it in words
In words: we are 95 percent confident that the true mean is between 29.7332 and 33.4668.
6 Check
Let's check: the middle of the interval should be 31.6.
29.7332 plus 33.4668, divided by 2, is 31.6.
Half the width should be the error bound: 33.4668 minus 29.7332, divided by 2, is 1.8668.
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.