Step by step
The question
Question: a sample of 18 has mean 7.2 and SD 1.6: find the 95% confidence interval for the true mean.
x̄, the sample mean, is 7.2: the middle of our interval.
s is the SD of the sample: 1.6.
And n is 18.
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.
18 minus 1 is 17.
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 17 and the 95 percent column.
Row 17, column 95%: t* = 2.110.
Careful!
df is n minus 1: 17, not 18.
3 Error bound
The error bound: t* times s over the square root of n.
1.6 over the square root of 18 is about 0.3771.
Times 2.11: about 0.7957.
4 The interval
Take the error bound away from the middle: 7.2 minus 0.7957 is 6.4043.
And add it: 7.2 plus 0.7957 is 7.9957.
The 95 percent confidence interval is (6.4043, 7.9957).
5 Say it in words
In words: we are 95 percent confident that the true mean is between 6.4043 and 7.9957.
6 Check
Let's check: the middle of the interval should be 7.2.
6.4043 plus 7.9957, divided by 2, is 7.2.
Half the width should be the error bound: 7.9957 minus 6.4043, divided by 2, is 0.7957.
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.